I´d like some help with E=mc^2!

[SwoleMan]

Hello infidels. Very glad to have discovered this oasis of reason.

I am debating this extremely annoying catholic fundamentalist (he does not accept the present pope, and is a creationist and a geo(ego)centrist as well).

When I tried to explain why the theory of special relativity is something of a fundamental law, he came up with this: “E=mc^2 does not necessarily imply atomic power. Post hoc ergo propter hoc is a fallacy, if you did not know it.�

I replied: “The equation E=mc^2 preceded the practical experiments.� (Of course I also gave a longer answer, but it is not needed here).

His response was as follows: “That does not prove that the results of them (the experiments) depends on the equation. That the equation is older than the experiments is no proof that it is their cause (my remark: thank you for repeating yourself – I did not understand what you meant with the first sentence). The physicist (my remark again: a physicist that I mailed) has himself confessed that something in the equation could be not fully constant (celerity) but without changing the outcome of the experiments. That means that he can not deny the possibility that the law of energy conservation could be false. Or more precise: kinetic energy does not exist before the reaction – it appears with it. With other words, the law of energy conservation, if it is being restricted to a limited, well defined energy, is refuted."

Well, I think I have a little clue what to answer, but I am really not sure. Because I am not a physicist, I would very much appreciate some good suggestions from someone here more cunning than me.

I am sorry if my English is not that great, and it is probably my (or my antagonists) fault if you do not understand the problem (it is translated from Swedish).

[Rymmie1981]

All math is just language that we use to describe the universe. E=mc^2 no more causes nuclear reactions than purple causes rain. E=mc^2 is just a way to describe the relationship between matter and energy; i.e., they are equivelant. All theories are just explanations of what we observe. When we observe something outside of the boundaries of the theory, the theory must change or be discarded. Special relativity works so well that have not had need to discard it.

I'm sorry, but your friend is mostly right. Until he gets into his rant about energy conservation, that is. There he goes waaaay into left field with some weird logic. I'm not sure how to approach his energy conservation agruement without knowing what experiments he was talking about.

If they were related to E=mc^2, then he needs a refresher course in nuclear energy. If he means c is not constant, then he is confused. c is the speed of light in a vacuum: 186,000 miles per second. However, the speed of light changes depending on the medium that it travels through. The bent straw illusion you get when you put a straw in water is because light travels slower through water than through air. But, c in the equation is always the same. The energy released in a nuclear explosion is not dependant on where it explodes. That's why they can do accurate tests of nuclear weapon payloads by detonating them underground.

I hope that helps some, but more information is needed before I can give more help.

[atheist]

Originally posted by Andreas83
When I tried to explain why the theory of special relativity is something of a fundamental law, he came up with this: “E=mc^2 does not necessarily imply atomic power. Post hoc ergo propter hoc is a fallacy, if you did not know it.�

I replied: “The equation E=mc^2 preceded the practical experiments.� (Of course I also gave a longer answer, but it is not needed here).

First off, you might ask him to (re)learn what the post hoc, ergo propter hoc fallacy actually means. Scientific theories can make predictions. If the predictions hold true, then the theory remains valid. This is not a fallacy!

post hoc, ergo propter hoc: "It happened after, therefore it was caused by."

scientific prediction: "If we control X variables, we predict Y results."


His response was as follows: “That does not prove that the results of them (the experiments) depends on the equation. That the equation is older than the experiments is no proof that it is their cause


It is not so much a cause than it is a prediction, or something (an equation) that describes the behaviour. You poster doesn't seem to understand the difference.


(my remark: thank you for repeating yourself – I did not understand what you meant with the first sentence). The physicist (my remark again: a physicist that I mailed) has himself confessed that something in the equation could be not fully constant (celerity) but without changing the outcome of the experiments.

How can something be 'not fully constant'? Either it is or it isn't. Perhaps he misunderstood the physicist. Celerity (or 'c' in the equation) refers to the speed of light in a vacuum, which is indeed constant. However, the speed of light is less when travelling through some medium (i.e. air, transparent glass).

I'd suggest that the poster pick up a copy of Einstein's "Relativity: The Special and General Theory". It's quite inexpensive (~$9 USD).


That means that he can not deny the possibility that the law of energy conservation could be false.

Non sequitur fallacy, don't you know? :p


Or more precise: kinetic energy does not exist before the reaction – it appears with it. With other words, the law of energy conservation, if it is being restricted to a limited, well defined energy, is refuted."

Perhaps your poster should brush up on Newtonian physics, also.

If the energy doesn't 'exist' as kinetic energy, then it exists as a potential energy somewhere; The Law of the Conservation of Energy is not refuted. If your poster can demonstrate otherwise, I can think of many real physists (probably including his own contact) that would just love to see it.


Well, I think I have a little clue what to answer, but I am really not sure. Because I am not a physicist, I would very much appreciate some good suggestions from someone here more cunning than me.

I'm not a physicist (yet!) either. However, I think it's very obvious your poster isn't either!

Hope I could help - or at least give this thread a bump.

[Tetlepanquetzatzin]

Originally posted by Andreas83
Hello infidels. Very glad to have discovered this oasis of reason.

I am debating this extremely annoying catholic fundamentalist (he does not accept the present pope, and is a creationist and a geo(ego)centrist as well).
Greetings. Why debate someone you find annoying?
His response was as follows: “That does not prove that the results of them (the experiments) depends on the equation. That the equation is older than the experiments is no proof that it is their cause (my remark: thank you for repeating yourself – I did not understand what you meant with the first sentence). The physicist (my remark again: a physicist that I mailed) has himself confessed that something in the equation could be not fully constant (celerity) but without changing the outcome of the experiments. That means that he can not deny the possibility that the law of energy conservation could be false. Or more precise: kinetic energy does not exist before the reaction – it appears with it. With other words, the law of energy conservation, if it is being restricted to a limited, well defined energy, is refuted."
It appears that your annoying friend thinks that conservation of energy is claimed to apply to kinetic energy separately and that this claim is wrong. Your annoying friend is wrong on the first account and right on the second: Kinetic energy is not conserved nor does any physicist claim that it is. Total energy, i.e. the sum of the kinetic and potential energies of all components, is conserved in isolated systems, though.

[aychamo]

Great 1st reply.

I have something for you to ask/tell your friend. "Dinosaurs."

[SwoleMan]

Why debate someone you find annoying?

Hehe, I hope he finds me annoying as well.

All math is just language that we use to describe the universe. E=mc^2 no more causes nuclear reactions than purple causes rain. [B]

Of course. Sorry, maybe I should have been more precise. What I meant was that it is very unlikely that someone would have discovered nuclear power just by using the "trial and error" method (the experiments preceding the theory).

If he means c is not constant, then he is confused. c is the speed of light in a vacuum: 186,000 miles per second. However, the speed of light changes depending on the medium that it travels through.

Well, I think he probably knows that. What the physicist I mailed told me, was that some physicists discuss, is if c maybe have slowed down a bit since big bang.

Perhaps your poster should brush up on Newtonian physics, also.

Here is the most annoying (or funny?) part; he does not accept Newtons theories of gravity. I have told him about experiments with cesium watches being speeded up in abscense of earth gravity and that the mass of the sun represents approximately 98% of the mass of the whole solar system, the later wich he denies by saying that "it is a (sneeze) heliocentric statement based on Newtons equations" (how do you measure the mass of the sun without using them, anyway?).

If the energy doesn't 'exist' as kinetic energy, then it exists as a potential energy somewhere

I am going to point that out to him, but I think he already knows. I do not know exactly how good he is at physics, but I can assure you that he is not as stupid as I may have made him look.

Non sequitur fallacy, don't you know?

A bit embarrasing...what does that mean? :)

I have something for you to ask/tell your friend. "Dinosaurs."

Haha, we have not discussed that. He probably has made up a good lie for himself even here (the usual creationist arguments I believe).

Thanks for the quick answers!

[atheist]

non sequitur: "It doesn't follow."

Your poster said:
The physicist (my remark again: a physicist that I mailed) has himself confessed that something in the equation could be not fully constant (celerity) but without changing the outcome of the experiments. That means that he can not deny the possibility that the law of energy conservation could be false.

I don't see how his physist theorizing about the speed of light slowing down a little since the Big Bang means that energy is not conserved. Does he mean to say that energy from faster-speed-of-light era of the universe would exist as an excess amount when the speed of light supposedly slowed down? I don't see how this follows either: If 'c' changes, then the potential energy 'left over' from a faster-speed-of-light era would change too, no? E=mc^2 would still be E=mc^2. Does his physist friend think that Einstein's equation was incorrect in this faster-speed-of-light era?

Perhaps your poster needs to describe/defend his position better.

edited to add:

Isn't this what "without changing the outcome of the experiments" means?

[SwoleMan]

I don't see how this follows either: If 'c' changes, then the potential energy 'left over' from a faster-speed-of-light era would change too, no? E=mc^2 would still be E=mc^2.

I have given this a thought as well, and I agree with you. Should be interesting to see what defence he will set up for his science-hating claims this time.

Does his physist friend think that Einstein's equation was incorrect in this faster-speed-of-light era?

No no, he his doing physics the catholic fundamentalist way himself; the physicist is on my side (I would not have mailed him else) ;)

Isn't this what "without changing the outcome of the experiments" means?

Japp. Or so I think (it is not always fully clear what he means, but he think he is rather clever himself)

[Godless Wonder]

Originally posted by Andreas83
[ . . . ] Well, I think he probably knows that. What the physicist I mailed told me, was that some physicists discuss, is if c maybe have slowed down a bit since big bang.

Google "supernova 1987a "speed of light"" or just click this link
http://www.evolutionpages.com/SN1987a.htm

As a result of that event, we were able to make a direct trigonometric measurement of the distance to that supernova of nearly 167,000 light years. This is now the most distant object ever measured by direct triangulation. Also, other measurements have been made that confirm the speed of light has not changed appreciably since it left SN1987A. Doesn't go all the way back to the big bang, but it does show 167000 years worth of constant c.

[Oregon Slim]

E = mc^2 was not really very closely related to the development of nuclear energy. In fact, it also applies to other forms of energy, eg chemical. If you burn something and release heat energy, you have actually converted a small amount of mass into that heat. It just happens that when you split some atomic nuclei, like uranium or plutonium, the nuclei that you get after the fission don't weigh quite as much as the nuclei that you started with. The difference is mostly converted to energy, but you can explain energy from fission without knowing anything about relativity.

The protons in a nucleus are repelled from each other by an electric force, but attracted to each other by a strong nuclear force. It happens that at very short distances, such as those within a nucleus, the attractive force is much stronger than the repulsive force, so that most of the time, nothing happens, and the nucleus can stay together.

If, however, you can shake things loose just a little bit, the strong attractive force diminshes much more quickly with distance than the repulsive electric force. The thing flies apart from the repulsion and all hell breaks loose.

Because so much energy is released, the mass goes down, but that's just another aspect of the results of the operation of the basic forces.

[Sven]

First a remark: IMHO discussing with someone who still believes in geocentrity is entirely futile. We have sent probes to the other planets and their moons with astonishing high precision - using heliocentricity. If he doesn't accept this argument and rather wishes to believe in epicycles (do a google search if you don't know about this), he is beyond hope.


What I meant was that it is very unlikely that someone would have discovered nuclear power just by using the "trial and error" method (the experiments preceding the theory).

I'm actually not sure about this. As far as I know, the Curies for example were not influenced by Einstein's theory. But I don't know if Hahn and Meitner were influenced by it. You may have a point here.


[...] some physicists discuss, is if c maybe have slowed down a bit since big bang.

As far as I know, these "physicists" who discuss this are almost all creationists grasping for straws.


Here is the most annoying (or funny?) part; he does not accept Newtons theories of gravity.

Then he should perhaps explain why we were able to sent probes to other planets using this theory. Sure, it's no proof that the theory is correct, but proof that it works. Does he give an alternative? Lots of tiny angles pushing the probes around?


that the mass of the sun represents approximately 98% of the mass of the whole solar system

Nitpick: I think it's about 99.8%.


how do you measure the mass of the sun without using them, anyway?.

How does he measure his own weight without using weighing machines constructed using Newton's theory of gravity?

[Sven]

Originally posted by Oregon Slim
E = mc^2 was not really very closely related to the development of nuclear energy. In fact, it also applies to other forms of energy, eg chemical. If you burn something and release heat energy, you have actually converted a small amount of mass into that heat.

I take issue with this statement. As a theoretical chemist I can assure you that it's entirely false. The heat released in chemical reactions all stems from re-organisation of the electrons around the atomic nuclei. No change of mass ever occurs in a chemical reaction.

A good example for E=mc^2 is the annihilation of an electron and a positron to exactly the energy obtained by putting their masses in this equation.

[SwoleMan]

First a remark: IMHO discussing with someone who still believes in geocentrity is entirely futile. We have sent probes to the other planets and their moons with astonishing high precision - using heliocentricity. If he doesn't accept this argument and rather wishes to believe in epicycles (do a google search if you don't know about this), he is beyond hope.

I have read a lot about heliocentricity vs geocentricity since the debate started. I think no defender of the heliocentric system has used the arguments of high precision planet landing and orbiting with very good results, but I do not remember exactly why. Please correct me if I am out on slippery ice (If one can say so in Enhlish ;)).

Nitpick: I think it's about 99.8%.

Damn! I have to buy a new, enhanced memory. I looked this number up right now in my national encyclopedia, and it actually says it is about 99,85% :)

Lots of tiny angles pushing the probes around?

Maybe not the probes, but the planets and stars! And I can not even find a really good argument to crush him with (of course I can laugh, but no, he is too annoying) :mad:

By the way, does anyone here know if a probe put in orbit "counterclockwise" around the earth, is being slowed down much faster (of course there must be some affection by the moon?) than a probe put in orbit "clockwise". If not so, I may with this show him that there is no mystic "force" in outer space that can grab a rocket or a probe to make it take a full turn around the earth once/24 hours (I have already pointed this out to him, but have been unable to give any facts). Sigh...I know, this is quite a strange question...

[sophie]

Oregon Slim : If, however, you can shake things loose just a little bit, Quite literally. The shaking rate varies for different compounds. I predict the Einsteinian equation moves towards: E*(Tr) = m, where Tr is the related shake effort needed to revert the mass m back to its energy form E. This is an easy experiment to set up. I'll quote the preliminary example of shaking glass hard enough to cause it to rupture.

[Sven]

Originally posted by Andreas83
Damn! I have to buy a new, enhanced memory. I looked this number up right now in my national encyclopedia, and it actually says it is about 99,85% :)

You only need to know that Jupiter is (by far) the largest planet and has about 1/10 of the diameter of the sun. If we assume that Jupiter and the sun have the same density, we easily arrive at 99.9%. I subtracted an additional 0.1% for the other planets - voila! (how do I make an accent on the a?) :)

Sophie: I suppose your post above was a joke.

[atheist]

Originally posted by Andreas83

By the way, does anyone here know if a probe put in orbit "counterclockwise" around the earth, is being slowed down much faster (of course there must be some affection by the moon?) than a probe put in orbit "clockwise". If not so, I may with this show him that there is no mystic "force" in outer space that can grab a rocket or a probe to make it take a full turn around the earth once/24 hours (I have already pointed this out to him, but have been unable to give any facts). Sigh...I know, this is quite a strange question...

There are significant ways that gravitational forces act on orbiting/rotating celestial bodies. For example, the rotation of the moon is apparently 'backwards' for what a stable orbit would be. The reason why the moon doesn't fly off into space, or crash into the earth is because of the 'speed' in which the moon orbits the earth, along with the strength of earth's gravity well. I suspect (again, I'm not yet a physicist!) that the gravity exerted by the sun (and even other plantets) may also have significant effects on the orbits of planetary satellites/moons.

I recall visiting a website that had loads of Java applets that demonstrated stable/unstable orbits. I can't remember the address, though. :(

[Abacus]

Originally posted by atheist
I recall visiting a website that had loads of Java applets that demonstrated stable/unstable orbits. I can't remember the address, though. :(

Is this the one? (http://burtleburtle.net/bob/physics/index.html)

[SwoleMan]

Interesting site, but unfortunately, it does not seem to answer my question. I meant satellites put in orbit counterclockwise the earths daily rotation around its own axis - an "axis of evil" if you ask my debate opponent.

I have learned though, that it is obviously a fact that the launch of a satellite in the "wrong" direction requires a greater amount of energy.

[Oregon Slim]

Originally posted by Sven

I take issue with this statement. As a theoretical chemist I can assure you that it's entirely false. [/B]

OK. You say it's false and I say that it's true. Let us decide this scientifically, by looking at experimental evidence. Do you have any apparatus that can measure the tiny change in mass resulting from such a small change in energy? Does anyone?

Based on hydrogen fusion reducing mass by only 1 part in a couple thousand, my swag guess is that the change in mass from combustion is gonna be down around one part in 10^16 or so. But the theory says that it is going to happen. You've got the same pieces arranged with different energy, so the mass must change, just as it does when you put protons and neutrons together to make helium.

[SwoleMan]

I have learned though, that it is obviously a fact that the launch of a satellite in the "wrong" direction requires a greater amount of energy.

Hmm...this may of course seem a little strange seen from a geocentric perspective. If the earth is fixed, it is the rest of the universe that rotates, and does not that mean that simple logic would say that the launch of a satellite would require less energy if pointed clockwise its direction (the universe)? But according to astronomers, it is the other way round. Because of that, the earth must simply rotate, or am I wrong?

[atheist]

Perhaps Newton is a little too modern for your poster. Maybe he should start by reading Copernicus' On the Revolution of Heavenly Spheres. It describes, by reason and mathematics, why the Earth should be considered to rotate, and not the rest of the Universe, like your poster seems to believe. It's a classic example of Occam's Razor.


Abacus:

Yeah! That's the site!

[Shadowy Man]

Originally posted by Andreas83
But according to astronomers, it is the other way round.

You'd be better off asking the military. They launch far more satellites than astronomers do.

[SwoleMan]

It's a classic example of Occam's Razor.

Is the math of a heliocentric system simpler than that of a geocentric one? It looks to me that retrograde motion of the planets messes things up for a geocentrist who wants to refer to Occam´s razor, which my poster actually did (is not good old Occam on my side? :mad: ).

[Loren Pechtel]

Originally posted by Sven

I take issue with this statement. As a theoretical chemist I can assure you that it's entirely false. The heat released in chemical reactions all stems from re-organisation of the electrons around the atomic nuclei. No change of mass ever occurs in a chemical reaction.

A good example for E=mc^2 is the annihilation of an electron and a positron to exactly the energy obtained by putting their masses in this equation. [/B]

No--the mass change does occur. It's just so small that you can't measure it and do not consider it in figuring out what happens. It's the same as our ignoring relatavistic effects in dealing with the real world. (Note, however, that if you're designing a GPS receiver you do need to consider relativity. The satellites are way up there, moving very fast and you need a lot of digits of precision.)

[Jesse]

Speaking of geocentrism, this is kind of cool:

A new geocentric map of the universe (http://www.aleph.se/andart/archives/2004/01/mappa_mundi.html) (click here (http://www.astro.princeton.edu/~mjuric/universe/all100.gif) to see a large image of just the map itself, without comments)

(of course, 'geocentric' here doesn't mean the authors think the earth is the center of the universe, just that are showing the entire cosmos in terms of distances from the earth, with distances on a logarithmic scale so they can show both nearby objects like planets along with very distant ones like clusters of galaxies)

[Friar Bellows]

Originally posted by Sven
I take issue with this statement. As a theoretical chemist I can assure you that it's entirely false. The heat released in chemical reactions all stems from re-organisation of the electrons around the atomic nuclei. No change of mass ever occurs in a chemical reaction.

That's simply untrue, and I can't understand how a theoretical chemist could get that wrong. Surely you have a chemical handbook that lists the weights of common molecules (grams/mole or whatever crazy units they use in chemistry) to a high degree of accuracy (I think CRC would do). Consider the famous reaction:

2 H2 + 02 ---> 2 H20 + energy

You will find that the mass of 2 hydrogen molecules + 1 oxygen molecule is slightly more than the mass of 2 water molecules. And the mass difference is the just energy released divided by c^2. But we're talking about a mass difference that is probably 1 part in billions and I don't think most chemists are too concerned about that.

[Sven]

Originally posted by Friar Bellows
That's simply untrue, and I can't understand how a theoretical chemist could get that wrong. Surely you have a chemical handbook that lists the weights of common molecules (grams/mole or whatever crazy units they use in chemistry) to a high degree of accuracy (I think CRC would do). Consider the famous reaction:
2 H2 + 02 ---> 2 H20 + energy
You will find that the mass of 2 hydrogen molecules + 1 oxygen molecule is slightly more than the mass of 2 water molecules. And the mass difference is the just energy released divided by c^2. But we're talking about a mass difference that is probably 1 part in billions and I don't think most chemists are too concerned about that.
Umm, sorry, you're possibly right. I'm coming from chemistry, not from physics as many other theoritical chemists. If it's really only about 1 ppb, OK, I could accept this.

But after all, we are able to predict the reaction energies of chemical reactions with high precision using quantum chemistry programs which don't incorporate relativistic effects. All these programs work under the assumption that the masses of the particles don't change - and we get the very good answers with this assumption. Edited to add: please see below before answering this.
(And don't forget one of Dalton's laws: " The Law of Conservation of Mass can be stated as follows: The total mass of the reactants in any chemical reaction is exactly equal to the total mass of the products." from http://www.ucdsb.on.ca/tiss/stretton/chem1/atomic1.html.
(granted, this is from a time when differences of ppb could not be measured) )

Edited to add: OK, after thinking for a while about this, I came to the following conclusion. Maybe it's only a different point of view. I said that the different energies are caused by a different organisation of the electrons (and the nuclei of course). This different total energy of course corresponds to a different total mass (according to E=mc^2). Thus, one could attribute the released energy to this change in energy - and both of us are right in a sense. Can we agree on this or am I totally confused now?
If the above is right, I think the different organisation of the electrons is at least an explanation which is more reasonable to work with in (theoretical) chemistry. The changes in the mass are so tiny that it doesn't really makes sense to "explain" the energy release in chemical reactions by these, that is this view does not help to understand "what is going on".

[Loren Pechtel]

Originally posted by Sven

If the above is right, I think the different organisation of the electrons is at least an explanation which is more reasonable to work with in (theoretical) chemistry. The changes in the mass are so tiny that it doesn't really makes sense to "explain" the energy release in chemical reactions by these, that is this view does not help to understand "what is going on".

Right. While the mass does change it's not signifigant to the chemist. There's no reason to model it because it won't affect the outcome of what you do. It's just like engineers normally use Newtonian physics rather than Einstienian.

[atheist]

Originally posted by Andreas83
Is the math of a heliocentric system simpler than that of a geocentric one? It looks to me that retrograde motion of the planets messes things up for a geocentrist who wants to refer to Occam´s razor, which my poster actually did (is not good old Occam on my side? :mad: ).

The Ptolemaic World requires not only orbits, but other "adjustments" like epicycles, and is generally less acurate (and more difficult) in making predictions. Indeed, Occam is on your side. :cool:

http://en.wikipedia.org/wiki/Ptolemaic_cosmology

[Jesse]

Originally posted by atheist
The Ptolemaic World requires not only orbits, but other "adjustments" like epicycles, and is generally less acurate (and more difficult) in making predictions. Indeed, Occam is on your side. :cool:

http://en.wikipedia.org/wiki/Ptolemaic_cosmology True, but although Copernicus had some good general arguments for why a heliocentric model was conceptually simpler (explaining retrograde motion of planets, along with other facts like why mercury and venus always stay near the sun in the sky), his actual detailed model for predicting the position of planets in the sky required about as many epicycles as the Ptolemaic one to give accurate predictions. Only after Kepler made his discoveries about elliptical orbits, and planet's changing speeds as they orbited, could the planets' position in the skies be accurately predicted without the need for epicycles.

[sophie]

That equation should be E = M*k such that kRM (k is related to M) and there is a solution for some atomic matter M. where k = c-squared. Unfortunately there seems to be specific k's for specific classes of matter M, a grave problem for physicists.

For example if you burn logs in your fireplace the 'k' is related to the type of wood burned, and the energy released (heat) is given by E = M1*K1 such thet K1 is related to M1. One can notice some types of wood give more heat (energy) than others.

The relation of E = M*c-squared to chemical equations is again E = M*k but rewritten as E = (M+deltaM)*K. Two chemicals combined are E1 = (M1+deltaM1)*K1 and E2 = (M2+deltaM2)*K2. The potential energy is E = (M1*K1 + M2*K2) and the kinetic energy is (deltaM1*K1 + deltaM2*K2). Finding what K1 and K2 for each chemical is, is quite a task if one wishes to use this form of the equation to yield practical results.

However E is equal to M*c-squared for atomic mass.

[sophie]

In case we forget E = M*c-squared is valid over infinitesimal distances so the question would be why would there be an E = M*K which is valid over alll distances such that M is related to K.

Good question. We already know with a fair degree of certainity that E=M*c-squared is valid, but only over tiny distances. Through observation we propose that E is proportional to M for all types of mass. This we rewrite as: for all mass M there seems to be an amount of energy embedded within the mass which is represented as E. Why is E proportional to M such that there would exist a constant K for each type of mass M? We note generally that if we increase the mass M to 2M we generally obtain more energy from 2M. To increase E we have to increase M. Therefore if we use a constant of proportionality K and rewrite the equation to be E = M*K what can we say about K. Firstly for atomic matter K = c-squared. We observe for other types of mass M, the energy yields vary from mass to mass. A pound of paper does not yield the same energy as a pound of chicken. Therefore because these two items both yield energy and not of the same quantities the K must be different for both. So E(chicken) = M*K(chicken) and E(paper) = M*K(paper). So to write this a little simnpler we say E=M*K such that M is related to K, meaning the constant of proportionality differs for each genre of mass M.

Another point to note which is beyond the scopê of this post is the fact that the time (t) over which the energy is available also varies. I mean there is a time (t) over which the energy becomes available.

*Now how's that for psuedo-science?

[Tetlepanquetzatzin]

Originally posted by Sven
But after all, we are able to predict the reaction energies of chemical reactions with high precision using quantum chemistry programs which don't incorporate relativistic effects. All these programs work under the assumption that the masses of the particles don't change - and we get the very good answers with this assumption. Edited to add: please see below before answering this.
On the other hand, some quantum chemistry calculations do take relativistic effects into account. This is done by basing the calculations on the Dirac equation, rather than the Schrödinger equation. The relativistic correction to the ground state energy of H2Po is 7%.
Edited to add: OK, after thinking for a while about this, I came to the following conclusion. Maybe it's only a different point of view. I said that the different energies are caused by a different organisation of the electrons (and the nuclei of course). This different total energy of course corresponds to a different total mass (according to E=mc^2). Thus, one could attribute the released energy to this change in energy - and both of us are right in a sense. Can we agree on this or am I totally confused now?
We can agree on that.

[Sven]

Originally posted by sophie
That equation should be E = M*k such that kRM (k is related to M) and there is a solution for some atomic matter M. where k = c-squared. Unfortunately there seems to be specific k's for specific classes of matter M, a grave problem for physicists.
For example if you burn logs in your fireplace the 'k' is related to the type of wood burned, and the energy released (heat) is given by E = M1*K1 such thet K1 is related to M1. One can notice some types of wood give more heat (energy) than others.

So what? You seem to have problems applying this equation. The M is not the mass of the logs you burn, but the difference of the masses of the logs and all the end products (ash, water vapor and CO2 gas). Or did you mean this, and still think that there is "a grave problem for physicists"? If yes, could you please provide references for the claim "'k' is related to the type of wood burned"?


The relation of E = M*c-squared to chemical equations is again E = M*k but rewritten as E = (M+deltaM)*K.

And what is deltaM, please?


Two chemicals combined are E1 = (M1+deltaM1)*K1 and E2 = (M2+deltaM2)*K2. The potential energy is E = (M1*K1 + M2*K2)

Why is this a potential energy?


and the kinetic energy is (deltaM1*K1 + deltaM2*K2).

Why is this a kinetic energy?


Finding what K1 and K2 for each chemical is, is quite a task if one wishes to use this form of the equation to yield practical results.

Since nobody needs this equation to get practical results, we can safely ignore the problem of finding the K's.


In case we forget E = M*c-squared is valid over infinitesimal distances so the question would be why would there be an E = M*K which is valid over alll distances such that M is related to K.

Why do you think that the validity of E=mc^2 has anything to do with distance? What do you mean by this after all?


Through observation we propose that E is proportional to M for all types of mass.

(1) E=mc^2 is a result of the theory of relativity, thus the proportionality is not only a result of observations.
(2) What do you mean by "types of mass"?


Firstly for atomic matter K = c-squared.

Since all matter is composed out of atoms, why there should be a difference if we look at larger units?


We observe for other types of mass M, the energy yields vary from mass to mass. A pound of paper does not yield the same energy as a pound of chicken.

When we do what exactly with these two pounds?


Another point to note which is beyond the scope of this post is the fact that the time (t) over which the energy is available also varies. I mean there is a time (t) over which the energy becomes available.

Could you give references for this weird claim?


*Now how's that for psuedo-science?

Perhaps I have again severe problems understanding what you are trying to say - but the above strongly looked like pseudo-science to me.

[Sven]

Originally posted by Tetlepanquetzatzin
On the other hand, some quantum chemistry calculations do take relativistic effects into account. This is done by basing the calculations on the Dirac equation, rather than the Schrödinger equation. The relativistic correction to the ground state energy of H2Po is 7%.

Yes, I know that there are programs based on the Dirac equations. I'm actually one of the authors of a paper on Pt-complexes, including contributions from someone else wo did some relativistic calculations on these complexes to investigate if our conclusions still hold if these effects are included. I was surprised how large the differences, for example on ground state energies as you noted, really are.

The point I was (falsely) trying to make was that we get the right results without including relativistic effects (at least for molecules composed out of "light" atoms). I realized later at which point my reasoning was wrong.

[sophie]

Sven : You seem to have problems applying this equation. The M is not the mass of the logs you burn If E is proportional to M then M is the mass of the log.

Sven : Or did you mean this? No that was not what I meant.

And what is deltaM, please? The decrease in M.

Since nobody needs this equation to get practical results, we can safely ignore the problem of finding the K's.Something one can debate.

[Jesse]

Sven:
You seem to have problems applying this equation. The M is not the mass of the logs you burn

sophie:
If E is proportional to M then M is the mass of the log.

But the energy released is proportional to the mass that was actually "destroyed", not just mass that was taken away from the log. For example, if I chop off a section of the log weighing one pound, the log will decrease in mass by one pound, but that doesn't mean a pound of wood was converted to energy! Likewise, if I burn a log and a pound of it is converted to smoke and ash, that also doesn't mean a pound of wood was converted to energy. Only the mass that actually disappears entirely, rather than just moving somewhere else, can be used in the equation E=mc^2 to determine how much energy was released.

[sophie]

Jesse : Only the mass that actually disappears entirely, rather than just moving somewhere else, can be used in the equation E=mc^2 to determine how much energy was released. So why cannot we write E = (m1+m2)*c-squared. What remains is m1 and what disappears is m2, is this not also correct? Is there not atomic waste associated with nuclear reactions?

[Jesse]

sophie:
So why cannot we write E = (m1+m2)*c-squared. What remains is m1 and what disappears is m2, is this not also correct?

No. The way the equation works is that to find the energy released, you use dM, the difference in total mass before and after the energy-releasing process happens. So if the initial mass was M=m1+m2, and then after the process m2 had disappeared with m1 left over, then the difference dM would be M-m1=m2, so the energy released would be equal to E=m2*c^2.

Is there not atomic waste associated with nuclear reactions?

In nuclear fission, yes, but the total mass of the atomic waste would be less than the total mass of the nuclear material (uranium rods, say) before the nuclear reaction. So again, you'd use dM = (initial mass) - (leftover mass) and plug that into E=dM*c^2 to find the energy released.

[sophie]

Jesse : So again, you'd use dM = (initial mass) - (leftover mass) and plug that into E=dM*c^2 to find the energy released .Note the emphasised word for effect.

Jesse, think you can write a paper on energy available vs. energy released?

[Jesse]

Originally posted by sophie
Note the emphasised word for effect.

Jesse, think you can write a paper on energy available vs. energy released? In the context of this discussion about E=mc^2, it seems most natural to define "energy available" as the amount of energy that would be released if 100% of the mass were converted to energy. In other contexts it might mean something a little different--in thermodynamic terms, not all the energy released by a process can be used to do useful work, a certain fraction will be "waste heat" which cannot be used at all, so a thermodynamicist might use the phrase "energy available" to mean "energy available to do work". But in this context it seems simpler to define "energy available" as the maximum amount of energy that could possibly be released by a system, unless you have some reason for wanting to define it in a different way.

[Sven]

Sophie:
Thanks for some answers. You said that deltaM is the decrease in M. Do you mean the decrease in the mass of the log, or do you mean the decrease in the mass of (the log plus the other products)? I put this is brackets to make it more clear.

And I would say, if M1+deltaM1/M2+deltaM2 is the total mass of the two chemicals, then surely the energy available from both of them can be calculated by setting K1 = K2 = c^2. The problem, how much energy will be released in a chemical reaction is an entirely different one.
But
(1) we can measure this energy directly by using calorimeters.
(2) we can calculate this energy with quantum chemical programs based on the Schroedinger or Dirac equation - with quite good precision for small molecules.

Thus I see no need at all to determine the constants K1/K2 for chemical reactions, especially because the change in mass is so tiny that it's much easier to measure the released energy directly.

Do you intend to answer my questions
(1) Why is this a potential energy?
(2) Why is this a kinetic energy?
(3) Why do you think that the validity of E=mc^2 has anything to do with distance? What do you mean by this after all?
(4) Since all matter is composed out of atoms, why there should be a difference if we look at larger units?
(5) Could you give references for this weird claim? [regarding: "Another point to note which is beyond the scope of this post is the fact that the time (t) over which the energy is available also varies. I mean there is a time (t) over which the energy becomes available."]

[Sven]

OK, I thought about posting two actual examples so that we can better talk about what is actually meant by M1, M2 etc.

You (sophie) gave the example of two different types of wood giving different amount of energies when burnt. To keep this simple, I'll reduce this to the burning of pure carbon (C) versus burning of methane (CH_4). In the following, a _ means that a subscript (index) follows.

First case:

C + O_2 ----> CO_2 + E_I

Second case:

CH_4 + 2 O_2 ----> CO_2 + 2 H_2O + E_II

E=mc^2 now tells us, that E_I is equivalent to a mass m_I and E_II equivalent to a mass m_II (note that it's generally agreed on to use a small "m" for mass). E_I and E_II (and thus m_I and m_II) of course are different.

We thus have the following equations for the masses:

m(C) + m(O_2) = m(CO_2) + m_I

and

m(CH_4) + 2 m(O_2) = m(CO_2) + 2 m(H_2O) + m_II

Or to state it another way: the sum of the masses of the products is smaller than the sum of the masses of the reactants, provided that E_I/II is positive. If E_I/II is negative, the mass of the products will be greater. If you multiply the equations above with c^2, you get the equations for the energy (neglecting of course any kinetic energy of the particles and all potential energies).

That's the way physicists understand the relationship of E=mc^2 to chemical reactions. Now it's your turn to apply your M1, deltaM1, K1, etc. to the above and to explain what they mean in this context.

Note that the total energy available from C and CH_4 is of course E=m(C)*c^2 and E=m(CH_4)*c^2. But this is entirely irrelevant for chemical reactions.

[sophie]

Jesse : But in this context it seems simpler to define "energy available" as the maximum amount of energy that could possibly be released by a system, Excellent, with new terchniques available to make the appropriate conversion, the goal posts wiil change, won't it?

[Sven]

Originally posted by sophie
Excellent, with new terchniques available to make the appropriate conversion, the goal posts wiil change, won't it?
You possibly misunderstood Jesse. The "goalposts" are not defined by our technique, but the laws of thermodynamics. Or do you think that with new technique we can circumvent these laws?

As an aside: This will be my last post till monday. Sorry that you'll have to wait until then for my answers if you intend to answer one of my other posts.

[sophie]

Why is this a potential energy? I misused the term, perhaps it is latent energy or more or less available energy.

Why is this a kinetic energy? Perhaps it is energy available to transform itself from one form into another.

Why do you think that the validity of E=mc^2 has anything to do with distance? What do you mean by this after all This is irrelevant, so we can forget it.

Since all matter is composed out of atoms, why there should be a difference if we look at larger units? My mistake.

Could you give references for this weird claim? Sorry I can't so we shall have to pass on this one.

Now it's your turn to apply your M1, deltaM1, K1, etc. to the above and to explain what they mean in this context Sorry I'll pass on that turn.

[sophie]

Sven : Since all matter is composed out of atoms, why there should be a difference if we look at larger units? I just had a brain wave. I am not sure if we can make a rational stab at this, but I'll certainly try.

Say you are chewing ona piece of bacon, you will say that bacon is completely made up of atoms and I agree. Now I like three to four slices of bacon per serving which is roughly (yummy) seventeen grams. Now just imagine consuming seventeen grams of bacon - what an immense amount of energy at m*c-squared. I'm sorry it is not evident because I am not jet powered after eating bacon sometimes I even become sick.

What does this mean? It could be that any rational person may realise the energy of E=m*c-squared is not released (providing it exists) so there must be some other thing (hmm - yes) some other thing which relates the mass M of the bacon (seventeen grams in case it slips our mind) which gives us the energy. Ahh, it may be the fat of the bacon, but that is made up of atoms, therefore the only rational explaination must be the atoms are carrying energy which is not the m*c-squared kind. Rational - yes.

Therefore there must be some sort of relationship between the mass M and the energy E seeing when i literally pig out (have two portions instead of one) I can run around all night (well almost).

Sven does this shift the paradyne in your mind at all?

[Jesse]

Originally posted by sophie
I just had a brain wave. I am not sure if we can make a rational stab at this, but I'll certainly try.

Say you are chewing ona piece of bacon, you will say that bacon is completely made up of atoms and I agree. Now I like three to four slices of bacon per serving which is roughly (yummy) seventeen grams. Now just imagine consuming seventeen grams of bacon - what an immense amount of energy at m*c-squared. I'm sorry it is not evident because I am not jet powered after eating bacon sometimes I even become sick.
But most of the mass of that bacon is not destroyed, it just goes through your digestive system and comes out the other end. If you and the bacon were placed on a very precise scale, the scale would not measure a loss of seventeen grams when the bacon disappeared into your stomach. Using m=17 grams is just like ignoring the mass of the ashes and smoke when you figure out the energy released when burning a log--you're not using E=mc^2 correctly if you do that.

[sophie]

Jesse : But most of the mass of that bacon is not destroyed, it just goes through your digestive system and comes out the other end. If you and the bacon were placed on a very precise scale, the scale would not measure a loss of seventeen grams when the bacon disappeared into your stomach. Using m=17 grams is just like ignoring the mass of the ashes and smoke when you figure out the energy released when burning a log--you're not using E=mc^2 correctly if you do that You have somehow managed to miss the gist of my post.

[Jesse]

Originally posted by sophie
You have somehow managed to miss the gist of my post. OK, so were you just asking where the energy comes from, if it doesn't come from mass being lost? The answer would be that a miniscule amount of mass is lost in the chemical reactions that happen in the digestive process--you might have some reaction like "molecule A + molecule B = molecule C + molecule D + energy", where the energy is equal to the difference in mass dM = (molecule A + molecule B) - (molecule C + molecule D), multiplied by c^2.

[sophie]

Jesse : where the energy is equal to the difference in mass dM = (molecule A + molecule B) - (molecule C + molecule D), multiplied by c^2. In this case are you sure that c-squared is not better off as K such that M is-related-to K. How do you know for sure that the atom was split? How do you know that the atoms were not acting as carriers of energy something emergent at larger distances from the nucleus? How? How? How?

[Jesse]

Originally posted by sophie
In this case are you sure that c-squared is not better off as K such that M is-related-to K. How do you know for sure that the atom was split? How do you know that the atoms were not acting as carriers of energy something emergent at larger distances from the nucleus? How? How? How? Well, a chemist can calculate the difference in mass between reactants and products, and perhaps confirm this experimentally as well--you'd have to ask someone more knowledgeable about chemistry for details. But anyway, it is possible to confirm that the energy released is proportional to dM*c^2, this isn't just an assumption.

[sophie]

Jesse : But anyway, it is possible to confirm that the energy released is proportional to dM*c^2, this isn't just an assumption. You have avoided my line of reasoning. Do yourealise that c-squared is such an incredibly large number that the mass which is lost would have to be so fractionally tiny in order to balance the equation. My seventeen grams of bacon would be so difficult to chew if we needed such large amounts of energy to break the bonds, or split the atoms. My stomach would literally hurt.

Think about the difference between chewing on uranium in contrast to chewing on bacon. I will have to think before I reply again. Thanks.

Oh, try to reflect on this: carriers of energy something emergent at larger distances from the nucleus? Weaker bonding - less energy. Closer to the nucleus - stronger bonding - more energy.

[Shadowy Man]

Originally posted by sophie
You have avoided my line of reasoning. Do yourealise that c-squared is such an incredibly large number that the mass which is lost would have to be so fractionally tiny in order to balance the equation. My seventeen grams of bacon would be so difficult to chew if we needed such large amounts of energy to break the bonds, or split the atoms. My stomach would literally hurt.


I think you are completely missing what Jesse is saying. Yes, the energy of an atom, if the mass is converted 100% into energy is enormous. Luckily that's not what you are doing when you eat bacon. The gamma radiation alone from the conversion would probably kill you.

The chemical bonds between atoms have significantly less energy (and hence mass) than the rest-mass energy of the atoms themselves. What you are doing when chewing the bacon, or burning wood, is breaking those chemical bonds, not splitting atoms. In fact, the energy between the atoms is so small, relative to their rest-mass energy, that you can pretty much ignore the negligible (yet still non-zero) mass change.

[Jesse]

sophie:
You have avoided my line of reasoning. Do yourealise that c-squared is such an incredibly large number that the mass which is lost would have to be so fractionally tiny in order to balance the equation. My seventeen grams of bacon would be so difficult to chew if we needed such large amounts of energy to break the bonds, or split the atoms. My stomach would literally hurt.

Have you forgotten that dM is not seventeen grams in this case? Once again, the actual change in mass that takes place in the chemical reactions that take place in your digestive system is tiny--the difference in mass between the reactant molecules and the product molecules is very, very small. The vast majority of the mass of the bacon is not destroyed, it comes out of your body as waste. And the tiny amount of mass dM that is destroyed is directly proportional to the energy released, according to E=dM*c^2. This can be confirmed by considering any one of the chemical reactions that take place in digestion in isolation--say, "bacon fat molecule A + digestive molecules B and C = molecules D and G + energy" and figuring out the exact mass of the molecules on the left hand side compared with the mass of the molecules on the right hand side.

[Loren Pechtel]

Originally posted by Shadowy Man
I think you are completely missing what Jesse is saying. Yes, the energy of an atom, if the mass is converted 100% into energy is enormous. Luckily that's not what you are doing when you eat bacon. The gamma radiation alone from the conversion would probably kill you.

The chemical bonds between atoms have significantly less energy (and hence mass) than the rest-mass energy of the atoms themselves. What you are doing when chewing the bacon, or burning wood, is breaking those chemical bonds, not splitting atoms. In fact, the energy between the atoms is so small, relative to their rest-mass energy, that you can pretty much ignore the negligible (yet still non-zero) mass change.

How about a WAG on the energy levels involved:

Normal electron binding energy is measured in a few electron volts (as exhibited by the wavelengths emitted by such reactions). The annihlation energy of a single proton is nearly a billion electron volts.

[Oregon Slim]

Protons aren't annihilated in nuclear reactions such as occur in weapons or reactors. The change in mass for the atoms participating in the reactions is way down around a part per 1000. Much bigger than in everyday reactions that involve the electrical force, but not enough to change the neutrons or protons into pure energy.

[Loren Pechtel]

Originally posted by Oregon Slim
Protons aren't annihilated in nuclear reactions such as occur in weapons or reactors. The change in mass for the atoms participating in the reactions is way down around a part per 1000. Much bigger than in everyday reactions that involve the electrical force, but not enough to change the neutrons or protons into pure energy.

The point of chemical bonding energies vs total annihlation energy is that that's also the % of mass change.

If making H20 liberates 1EV (I have no idea of what it actually liberates) then the mass change is about 1 part in 14 billion. That's why chemists don't worry about it.

P.S.
I went looking with Google to see if I could find the energy above. Google's results hint that it's 540EV but there was no cached page and the link pointed to something else by now. However, I did find this from what looks like some physics problems:

Calculate the mass of the 4He nucleus in atomic mass units by subtracting the mass of two electrons from that of a neutral 4 He atom. (See Appendix D and give four decimal places.) (b) The procedure suggested for part (a) is theoretically incorrect because we have neglected the binding energy of the electrons. Given that the energy needed to pull both electrons far away from the nucleus is 80 eV, is the correct answer larger or smaller than implied by part (a), and by how much (in u)? In stating the mass of the nucleus, about how many significant figures can you give before this effect would show up?

This is specifically discussing the mass change due to binding energy of the electrons.

[Nil Desperandum]

Originally posted by Andreas83
By the way, does anyone here know if a probe put in orbit "counterclockwise" around the earth, is being slowed down much faster (of course there must be some affection by the moon?) than a probe put in orbit "clockwise". If not so, I may with this show him that there is no mystic "force" in outer space that can grab a rocket or a probe to make it take a full turn around the earth once/24 hours (I have already pointed this out to him, but have been unable to give any facts). Sigh...I know, this is quite a strange question... [/B]

Well, without digging into specifics, what can be said about retrograde launches is that they obviously require more fuel; it is really a comparison to a prograde orbit, which are launched from Cape Canaveral Air Station in Florida. Because of the proximity to the equator, and additional 1000+ mph is added to the launch velocity due to the natural rotation of the Earth. Because of this, we can launch heavier satellites, or the same types of satellites with less fuel required to launch them into space on a polar/retrograde orbit, because of the extra boost.

I guess this is somewhat like a comparison to running against the wind or with it. I'm not quite sure what a retrograde orbit is going to do in terms of atmospheric drag, but you have to take into consideration that the lowest orbiting satellites are 300 miles out, so there really is more of a concern with solar wind, solar pressure, solar radiation, as opposed to Earth-based effects. Obviously, gravity plays a part in all of this...

I do know that 17,500 is the escape velocity of a satellite, and it's eccentricity = 0 at that point, so it's orbit is essentially a circle around the earth, given minor perturbations due to solar weather effects... we use this eccentricity to maintain geosynchronous orbits @ 10,998 miles out. At that distance and speed, the GPS satellites orbit the earth twice every day; it is beyond me to do the math to figure out how slow it would have to go to stay in one spot over the earth, and hence, rotate with the earth once every 24 hours, but your retrograde orbits [with an inclination >90 degree to 180 degrees] are slowed down in the fact that they are not gaining speed from the rotation of the earth, but once you get beyond atmospheric drag, the solar effects take over, and those effect satellites regardless of which way they are going.

Chris

[sophie]

Sophie : Quite literally. The shaking rate varies for different compounds. I predict the Einsteinian equation moves towards: E*(Tr) = m, where Tr is the related shake effort needed to revert the mass m back to its energy form E. This is an easy experiment to set up. I'll quote the preliminary example of shaking glass hard enough to cause it to rupture.
Sven : Sophie: I suppose your post above was a joke.
Here is quite a modern approach:

Spheres covered with two different compounds, for example, will release from the shaking platforms at different times.
(http://www.nature.com/nsu/000608/000608-7.html)

SO, what now?

[sophie]

As modfern science expands the need to posture differently becomes greater. Here is an example from the world of Chemistry where the inadequacy of the simpler methods are currently being demonstrated.

Introduction (http://vergil.chemistry.gatech.edu/research/bondbreaking/intro.html)

The Performance of Single-Reference Approximations (http://vergil.chemistry.gatech.edu/research/bondbreaking/single-ref.html)

The Performance of Multi-Reference Methods (http://vergil.chemistry.gatech.edu/research/bondbreaking/multi-ref.html)

E=m*k - anyone?

[Sven]

Here is quite a modern approach:

Spheres covered with two different compounds, for example, will release from the shaking platforms at different times.
(http://www.nature.com/nsu/000608/000608-7.html)

SO, what now?
Yes, so what now?
The article above talks about breaking chemical bonds. This has nothing to do (as explained before) with E=m*k or whatever a priori. You originally claimed: "E*(Tr) = m, where Tr is the related shake effort needed to revert the mass m back to its energy form E." Since only chemical bonds are broken here, where is this mass which is reverted to energy? Hint: mass and energy are not converted into one another, they are equivalent.


As modfern science expands the need to posture differently becomes greater. Here is an example from the world of Chemistry where the inadequacy of the simpler methods are currently being demonstrated.

Surprise, surprise. Anyone who learned a little bit about quantum chemistry knows that single-reference methods often have problems and multi-reference methods are needed instead. This has been known for decades. I learned this in my first year of study of quantum chemistry. The point you apparently missed in these articles is that QC can calculate the bond breaking energies using multireference methods. The only problem with QC methods is that they are no "black box". You have to know which method is reliable in which case and for which reasons. That's one of the reasons why you have to learn many things before you can proudly claim anything about energies in chemical reactions.

But your E=m*k simply isn't very helpful - since there are literally billions of chemical compounds which can react with one another, we would have to determine the "k" for each of these reactions - but what do we do with this "k" when we have it? And how do you think we should determine this "k"? With one of the two methods I proposed above (calorimeter or quantum chemistry) or do you have another idea?


Sven does this shift the paradyne in your mind at all?

If I'd find the word "paradyne" in any dictionary, I could perhaps answer this question, too.

[sophie]

Sven : But your E=m*k simply isn't very helpful - since there are literally billions of chemical compounds which can react with one another,

Here is another modern way of looking at things :

The value of the new constant depends on the type of gas as opposed to the universal gas constant, (http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html)

[sophie]

Sven : Hint: mass and energy are not converted into one another, they are equivalent If mass and energy were equivalent Einstein's equation would not be E=m*c-squared it would be E=m, this is what equivalence means the last time I researched it.

Suppose there are two partitions with a trap door and a small passage seperating them and an evil monkey was attending to the door and the passage. Each partition is said to be a system closed upon itself, but the monkey is so quick that no one notices the door opening and closing and heat cannot escape into the passage. Further in partition A there are two hydrogen atoms while in partition B there is a single H2. The monkey shakes partition B until the H2 becomes 2 H atoms, then the monkey deftly switches one H from partition A with an H from partition B. Is there now more energy in partition A and less energy in partition B?

[sophie]

Sven you wrote: You have to know which method is reliable in which case and for which reasons. then you went on to write:
. . .since there are literally billions of chemical compounds which can react with one another, we would have to determine the "k" for each of these reactions You are currently using a quantative method!

Anyway this is not what I am saying about E=m*k where k is related to m. What I am saying is the further away from the nucleus, the bonds which constitute the mass m are weaker in relation to the bonds which are closer to the nucleus. Why do you think they devised the atom bomb the way they did?

[Sven]

If mass and energy were equivalent Einstein's equation would not be E=m*c-squared it would be E=m, this is what equivalence means the last time I researched it.

Are you serious? Do you also think (for example) that 1 J = 4.184 J/cal * 0.239 cal are not equivalent? :confused:
The c^2 is simply a factor to convert one energy unit (kg) into another (Joule). You perhaps confuse the mathematical meaning of "equivalent" with the physical meaning. You can attribute a mass to every energy. And an energy to every mass. This is meant by "equivalent".


Suppose there are two partitions with a trap door and a small passage seperating them and an evil monkey was attending to the door and the passage. Each partition is said to be a system closed upon itself, but the monkey is so quick that no one notices the door opening and closing and heat cannot escape into the passage. Further in partition A there are two hydrogen atoms while in partition B there is a single H2. The monkey shakes partition B until the H2 becomes 2 H atoms, then the monkey deftly switches one H from partition A with an H from partition B. Is there now more energy in partition A and less energy in partition B?
This reminds me of Maxwell's demon. To split H2 into 2 H, the monkey (and everyone else) needs exactly the binding energy. Thus, after shaking partition B (and thus adding energy), all four hydrogen atoms have exactly the same energy and are indistinguishable - if we ignore the kinetic energy of the particles and possible potential energies for the moment. What's the point of your question?
Perhaps you don't realize that H2 has less mass/energy than 2 H - and this "missing" mass/energy is equivalent to the (negative) binding energy.

[Sven]

Sven you wrote: You have to know which method is reliable in which case and for which reasons. then you went on to write:. . .since there are literally billions of chemical compounds which can react with one another, we would have to determine the "k" for each of these reactions
You are currently using a quantative method!

Sorry, I (once again) have no idea what you are trying to say here.


Anyway this is not what I am saying about E=m*k where k is related to m. What I am saying is the further away from the nucleus, the bonds which constitute the mass m are weaker in relation to the bonds which are closer to the nucleus. Why do you think they devised the atom bomb the way they did?
What kinds of bonds do you think are there in matter? There are bonds between the nucleons (very strong, but short-ranged), and there are bonds between atoms, almost entirely due to the valence electrons. So yes, for these two types of bonds you're right. Do you think there are other types of bonds? If yes, please explain which ones. If no, what's your problem here? And why do you return to the nucleus at all, I thought we were talking about (quantum) chemistry?

[Sven]

Here is another modern way of looking at things :
The value of the new constant depends on the type of gas as opposed to the universal gas constant, (http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html)
So what?? This has nothing to do with energy! And I also don't understand why you think this (what is discussed at the link) is anything special or modern. If we divide a universal constant (R) by the mass per mole of the gas (M), the latter being of course dependend on the type of gas, it is no wonder that we get a new constant (R', I introduced this new symbol to avoid confusion), which also depends on the type of gas. But this is only a great example why your E=k*m is equally useless.

R is known. M can be calculated by a short look in the periodic table. But what's the use of the new constants R'? Nobody needs them because in every specific case, the equation for the ideal gas can be solved without calculating these first. It's the same with you "k" - there is no need to calculate and tabulate them separately, because the energy/mass can be determined in every single case the usual way. And I repeat: How do you think these "k's" should be determined if not following one of the ways I presented?

[Friar Bellows]

I think there's too much talk in the popular literature about the equivalence of mass and energy, and not enough talk about the distinction. As far as special relativity is concerned, energy is the time-component of the four-momentum, while mass is the magnitude of the four-momentum. Only when the spatial components of the four-momentum are zero does the mass have the same value as the energy. This is most easily expressed in the following equation:

E^2 = p^2 * c^2 + m^2 * c^4

(E is energy, p is momentum, m is rest mass, c is speed of light)

So when p = 0, then:

E = m * c^2

and we can say that energy has the same value as mass. Of course, the factor of c^2 is irrelevant, just an annoying constant; you can think of it as merely a conversion factor between joules and kilograms. In fact, when you do general relativity, this factor is usually ignored altogether (i.e. c is set equal to 1).

(Disclaimer: this is my meagre understanding of the situation, and is therefore subject to being revised when someone smarter and more knowledgable than me points out my misconceptions and errors.)

[Sven]

I think there's too much talk in the popular literature about the equivalence of mass and energy, and not enough talk about the distinction. As far as special relativity is concerned, energy is the time-component of the four-momentum, while mass is the magnitude of the four-momentum. Only when the spatial components of the four-momentum are zero does the mass have the same value as the energy. This is most easily expressed in the following equation:

E^2 = p^2 * c^2 + m^2 * c^4

(E is energy, p is momentum, m is rest mass, c is speed of light)

So when p = 0, then:

E = m * c^2

This is one of the reasons why I wrote "if we ignore the kinetic energy of the particles", that is simply assume that v=0 and thus p=0 holds. OK, originally I didn't think about your point when I wrote this ;) :p
But by making the additional assumption that v=0 holds, my post still should be correct (We of course ignore that v=0 is a ridiculous assumption after shaking the partition B; chemical bonds simply don't break this way. But it's only a gedankenexperiment after all.)


(Disclaimer: this is my meagre understanding of the situation, and is therefore subject to being revised when someone smarter and more knowledgable than me points out my misconceptions and errors.)
I have to second this :)

[sophie]

Sven, are you always on the same page? On the 29th of March you wrote: Yes, so what now?
The article above talks about breaking chemical bonds. This has nothing to do . . . which was in response to this link (http://www.nature.com/nsu/000608/000608-7.html) which I posted. Soon after, on the 30th of March you subsequently wrote:(We of course ignore that v=0 is a ridiculous assumption after shaking the partition B; chemical bonds simply don't break this way. But it's only a gedankenexperiment after all.) Now tell me which is it?

Sven: Can chemical bonds be broken by shaking or not? A simple yes or no answer will suffice.

[Jesse]

E^2 = p^2 * c^2 + m^2 * c^4

(E is energy, p is momentum, m is rest mass, c is speed of light)

So when p = 0, then:

E = m * c^2
But also, if you use p = gamma*m*v and substitute it into that equation (with gamma = 1/squareroot(1 - v^2/c^2) ), it turns out to be equivalent to E = gamma*m*c^2, or E = m'*c^2 where m' is the "relativistic mass" instead of the rest mass.

[Sven]

(We of course ignore that v=0 is a ridiculous assumption after shaking the partition B; chemical bonds simply don't break this way. But it's only a gedankenexperiment after all.)
Can chemical bonds be broken by shaking or not? A simple yes or no answer will suffice.
Yes, I think they can. You misunderstood my statement above. I simply stated that the particles can not have zero velocity/momentum after breaking the bond in reality, that's all. Note that the "ridiculous assumption" was that "v=0", nothing more.

[sophie]

Sven: Anyway think of shaking as external to a closed system, it won't add energy to the system, if the system heats up it is conserving energy from within. Shaking is a not very well understood mechanism. Note things like shaking babies (GOD bless their poor souls (or whatever we are supposed to say on this site)) cause them to die.

In the demonstration I scored from Maxwell's (via instructions from a modern reliable elder male source) when the H2 bond is broken (let's say if the bond is broken) the state of each H will be unknown due to some form of uncertainity (the particular states of both H atoms at the moment they rupture). This rupturing by shaking would not add energy to the system unless it is coming from the container itself, that is the container starts to leak, so let us add the assumption that the container will not leak energy from its mass. When the evil monkey exchanges atoms between the two containers, a probability function for each container would have to be proposed since the H atoms may fuse together in both containers. This is not the same answer as you have given.

[sophie]

Sven : March 17, 2004, 06:43 AM in the Mental Meandering on Quantum Physics. (http://www.iidb.org/vbb/showthread.php?t=79266&page=6&pp=10)
My point was that energy is an abstract concept and not something which "does" anything by itself.Have you since changed your position seeing you have claimed mass and energy are equivalent? This would sort of imply mass is also an abstract concept, which seems sort of far out to me.

Anyway energy density can be viewed as mass. At the fundamental levels of our universe mass and energy are used interchangeably but they are not equivalent because there is some form of conservation which transforms one to the other and the other to one. If you want to misuse the terms you should give credit to its aliasing - not before, but if you are incorrect then so be it.

[Friar Bellows]

But also, if you use p = gamma*m*v and substitute it into that equation (with gamma = 1/squareroot(1 - v^2/c^2) ), it turns out to be equivalent to E = gamma*m*c^2, or E = m'*c^2 where m' is the "relativistic mass" instead of the rest mass.

Yes, as unfashionable as it is today, there are some good reasons to keep using the concept of relativistic mass. Thinking of mass in different ways helps one to understand the overall concept better.

[Sven]

Anyway think of shaking as external to a closed system, it won't add energy to the system

First I don't know how this should work: in reality, shaking certainly adds energy to the system. Second, when you don't add energy by shaking, why shake the system at all? If you don't influence the system, nothing will change.


if the system heats up it is conserving energy from within.

Huh? :confused: Heat is simply kinetic energy of particles. Where do you think this energy comes from if not from the outside?


Shaking is a not very well understood mechanism.

If you say so. I have no idea if you're right here. Perhaps you can substantiate this by some references.


Note things like shaking babies (GOD bless their poor souls (or whatever we are supposed to say on this site)) cause them to die.

So what? Do you think that this isn't well understood? Perhaps you should ask a doctor.


In the demonstration I scored from Maxwell's (via instructions from a modern reliable elder male source)

Is the "modern [...] elder male" important in any way? I'm much more interested why you think this source is reliable.


when the H2 bond is broken (let's say if the bond is broken) the state of each H will be unknown due to some form of uncertainity (the particular states of both H atoms at the moment they rupture)

Hmm. Let me give my point of view here. If we add the energy to break the bond in a predetermined way, it should be clear in which state the two atoms end. Only if the two H atoms are separated by an infinite distance the two possible states (singlett and triplett, at least I think this is what your reliable source meant) have exactly the same energy. As long as the distance is finite, one of those states should be lower in energy and one higher - thus adding the "right" amount of energy, we get only one of these state.


This rupturing by shaking would not add energy to the system

You get no rupture without adding energy. Did you sleep through all you basic chemistry and physics classes?

This is not the same answer as you have given.
Yes. So what? See above.


Have you since changed your position seeing you have claimed mass and energy are equivalent? This would sort of imply mass is also an abstract concept, which seems sort of far out to me.

You again confuse the meaning of equivalent in mathematics with the meaning in physics. But yes, mass is also in some sense an abstract concept - the same as with charge etc. We only say that something has mass because we need a force to accelerate it and because it attracts other masses. But what is mass? What is charge? etc.
Can you give a concrete definition without resorting to phenomenons?


but they are not equivalent because there is some form of conservation which transforms one to the other and the other to one.

This is simply wrong. As I said, you can assign an energy to every mass and a mass to every energy. There's no transformation between mass and energies, only between different kinds of energy (for example mass and kinetic energy). If you think otherwise, please provide references or go and ask some guys at a physics forum. We also could conduct a poll here, but I think we agree that this would be absurd for such a minor point. You can also listen to Einstein himself at http://www.aip.org/history/einstein/voice1.htm
Excerpt: "The mass and energy were in fact equivalent, according to the formula mentioned before."

Perhaps you could conceed that you were wrong just once.

[sophie]

ven: First I don't know how this should work: in reality, shaking certainly adds energy to the system. Second, when you don't add energy by shaking, why shake the system at all?Ok, here we shall delve into different forms of energy namely energy provided by mechanical means and energy provided by electro-chemical means. One can note one of the greatest obstacles surmounted by our ancestors was the ability to change one form of energy into another. We now have gas powered motors turning wheels and traversing distances and it is taken for granted.

Shaking a chemical system is sort of the same principle. The energy needed to shake and the resulting vibrations are not directly available to heat the system as say applying a bunsen burner or a packet pf photons. When these two different types of systems meet (high speed shake and chemical bondings)we have a basis stress applied on the bonding (this will become more apparent when the new pulse lasers are readily put to use) of the pairs of atoms or the combinations of atoms.

When I say if the system heats up it would be because the atoms are reacting to the stress applied to their bondings. Therefore we are not really adding energy to break the bond it is more as if we are punching the bonds until they collapse. As I said pulse lasers will demonstrate this effect clearly except with a twist or two.

No I did not sleep through my basic chemistry and physics classes, I skipped them to provide food for my then young family yet I managed to deduce the answers on the exams.

Finally we are once again back with the evil monkey switching the two H atoms from the two partitions and they might just fuse in both cases in each partition with a little bit of additional warmth generated.

[sophie]

Sven: This is simply wrong. As I said, you can assign an energy to every mass and a mass to every energy. Only when things are taken out of context or if the universe of discourse is restricted to fundamental physics. If what you say were actually universally true then the heavier atoms of the periodic table which are not naturally occuring would remain stable over a far longer period of time when their equivalent GeV are sustained in the labs.

Actually the Alchemists would have manufactured gold a long long time ago, which shows things don't work simply because one says it works that way or because there is a percieved equivalence or because in some cases xGeV is shown as equivalent to some mass m. The concept of universal equivalence must be reigned in in order for the concept to work. I simply can't take the energy given off by an atom bomb and create the bomb itself. This is a ridiculous extrapolation.

As an aside: because Einstein said something does not automatically make it true. Each appeal must be taken on its individual merit.

[Sven]

Shaking a chemical system is sort of the same principle. The energy needed to shake and the resulting vibrations are not directly available to heat the system as say applying a bunsen burner or a packet pf photons.

OK, let's look at your example again. We have a H2 molecule and shake the container wherein it rests. Thus energy can only be transmitted to the molecule when hitting the walls (neglecting black box radiation for the moment). If the molecule hits the wall, it of course gets kinetic energy and vibrational energy.


When these two different types of systems meet (high speed shake and chemical bondings)we have a basis stress applied on the bonding (this will become more apparent when the new pulse lasers are readily put to use) of the pairs of atoms or the combinations of atoms.

Yes. When the vibrations get "high" enough, the bond will break. For this to happen, the vibrational energy has to be at least as high as the bond breaking energy. What's your point here?


When I say if the system heats up it would be because the atoms are reacting to the stress applied to their bondings.

See up. I see no way to only get vibrations and no motion of the whole molecule by shaking the container. Thus heat is contained in internal as well as in external degrees of freedom.


Therefore we are not really adding energy to break the bond it is more as if we are punching the bonds until they collapse.

Huh? :confused: Do you really think that no energy is contained in the vibration of the mulecule? Perhaps you should look into a basic physics book.


No I did not sleep through my basic chemistry and physics classes, I skipped them to provide food for my then young family yet I managed to deduce the answers on the exams.

But you apparently missed that vibrations contain energy. Your family is apparently older now, so I suggest to use the time to catch up on the basic education before discussing these things here.


Finally we are once again back with the evil monkey switching the two H atoms from the two partitions and they might just fuse in both cases in each partition with a little bit of additional warmth generated.
Two particles of zero kinetic energy can only combine if you find a way to dissipate the energy released by the combination - in this case this only would be possible by a collision with the container before the molecule splits again into two atoms. If the particles additionally have kinetic energy, the situation is of course worse. A good starting point to understand this kind of "kinetics" would be "P.W.Atkins: Physical Chemistry".

BTW, I still don't see the point of your Gedankenexperiment.

This is simply wrong. As I said, you can assign an energy to every mass and a mass to every energy.
Only when things are taken out of context or if the universe of discourse is restricted to fundamental physics.
Where exactly did I take something out of context? The above is universally true, and Einstein agrees with me. Did you look up the link I provided? (also see below)

If what you say were actually universally true then the heavier atoms of the periodic table which are not naturally occuring would remain stable over a far longer period of time when their equivalent GeV are sustained in the labs.
What do you mean by "sustain" here? I once again have no idea what you're trying to say, sorry. And it's not some cases, it's every case (so far). Did you ever hear about that the orbit of mercury could be explained by relativity because of exactly the fact that mass and energy are equivalent?


Actually the Alchemists would have manufactured gold a long long time ago, which shows things don't work simply because one says it works that way or because there is a percieved equivalence or because in some cases xGeV is shown as equivalent to some mass m.

You perhaps missed my comment: "There's no transformation between mass and energies, only between different kinds of energy". This also means that different masses (different elements in this case) are different energies, which have to be transformed into one another. I nowhere said that this transformation is always practicable. You don't seem to be able to grasp the meaning of equivalent here: "you can assign an energy to every mass and a mass to every energy". Nothing more, nothing less.


As an aside: because Einstein said something does not automatically make it true. Each appeal must be taken on its individual merit.

You're joking, right? :eek: :confused: :banghead:
Einstein derived the formula - don't you think that he could perhaps know better what it means than a layman? You again and again show that you have no idea how science uses terms and in some cases also that you don't know the science. But nevertheless you think you know better what this formula is supposed to mean than its "inventor".

[Paul of Eugene OR]

[/b]
I take issue with this statement. As a theoretical chemist I can assure you that it's entirely false. The heat released in chemical reactions all stems from re-organisation of the electrons around the atomic nuclei. No change of mass ever occurs in a chemical reaction.

A good example for E=mc^2 is the annihilation of an electron and a positron to exactly the energy obtained by putting their masses in this equation.

Uh - Sven - actually, the mass does get lost, but the amount is so utterly insignifigant that normal mass measurement techniques can't measure it.

Oops, I see that point was made earlier! Forgive me, Sven.

Trivial question of the day:

Can somebody calculate from e=mc^2 the amount of foot pounds of energy in one pound of matter?

[V-Bird]

Hi.

This internal link with its additional references may be of some assistance.

http://www.iidb.org/vbb/showpost.php?p=1517732&postcount=19

Regards,

V.

[sophie]

Sven:

We have a H2 molecule and shake the container wherein it rests. Thus energy can only be transmitted to the molecule when hitting the walls (neglecting black box radiation for the moment). If the molecule hits the wall, it of course gets kinetic energy and vibrational energy.If this is how you wish to interprest the experiment I have no quarrel with you, however this is not the way I would set up the experiment.


But you apparently missed that vibrations contain energy.Yes they do but to be useful energy within a system it must be converted (put to use). Remember energy can be used in a variety of ways.

What do you mean by "sustain" here? I once again have no idea what you're trying to say, sorry. And it's not some cases, it's every case (so far). Did you ever hear about that the orbit of mercury could be explained by relativity because of exactly the fact that mass and energy are equivalent?I beg your pardon, but in the physical world, to obtain energy from mass a process of transformation must be undergone. Yes E=m*c-squared, but to obtain E it implies processing m. Obtain(E) -> Process(m). This is not equivalent in any way. Also Obtain(m) -> transform(E).

You don't seem to be able to grasp the meaning of equivalent here: "you can assign an energy to every mass and a mass to every energy". Nothing more, nothing less.On paper, in limited experiments one can use the mass-energy equivalent, but never in reality are mass and energy equivalent. It's not said that mass can be viewed as an energy density for simply no reason, the term density implies a process associated with the energy. What's so difficult for you to grasp here? I noted that the heavier atoms only exist momentarily in the lab when its equivalent mass-energy is applied, does this not trigger alarm bells in your scientific mind, that something is missing?

Your family is apparently older now, so I suggest to use the time to catch up on the basic education before discussing these things here.And what is this supposed to mean? Do you know anything about my family?

[Sven]

We have a H2 molecule and shake the container wherein it rests. Thus energy can only be transmitted to the molecule when hitting the walls (neglecting black box radiation for the moment). If the molecule hits the wall, it of course gets kinetic energy and vibrational energy.
If this is how you wish to interprest the experiment I have no quarrel with you, however this is not the way I would set up the experiment.

How else do you get the molecule vibrating? Please explain what use is shaking the container if it apparently does not interact with the molecule. Or do you want to suggest that it is possible to only excite vibrations by the collisions without speeding up the whole molecule? Please explain what exactly do you have in mind here.


Yes they do but to be useful energy within a system it must be converted (put to use). Remember energy can be used in a variety of ways.

If the molecule vibrates "high" enough, the bond breaks because the two atoms travel away from one another. Where exactly is this conversion of energy you talk about?


I beg your pardon, but in the physical world, to obtain energy from mass a process of transformation must be undergone. Yes E=m*c-squared, but to obtain E it implies processing m. Obtain(E) -> Process(m). This is not equivalent in any way. Also Obtain(m) -> transform(E).

You still don't understand. You say "to obtain energy from mass". The point is that mass is equivalent to energy but you need a process of transformation to get other kinds of energies out of mass. That's the crux. The process transforms different kinds of energy into one another (for example mass and kinetic energy), not mass as something different to energy into kinetic energy. Perhaps we can agree on that this is only a semantic difference? (meaning of equivalent)

On paper, in limited experiments one can use the mass-energy equivalent, but never in reality are mass and energy equivalent.

Your problem is that you simply don't grasp the meaning of "equivalent" here. See above. And you have not answered my point about mercury where we observe exactly this equivalence. Another point: Why do photons have a mass? Which can be observed and matches the calculated mass obtained by h nu = m c^2? (h: Planck's constant, nu: frequency of the light)


It's not said that mass can be viewed as an energy density for simply no reason

Who says this?


I noted that the heavier atoms only exist momentarily in the lab when its equivalent mass-energy is applied, does this not trigger alarm bells in your scientific mind, that something is missing?

Sorry, I still don't understand your problem here. Please explain this further - in particular what you mean by "when its equivalent mass-energy is applied".


Your family is apparently older now
And what is this supposed to mean? Do you know anything about my family?
Sorry if I got to personal. But you wrote "my then young family". I thought the "then" implied that this is no longer the case. That's all.

[sophie]

Sven:

Please explain this further - in particular what you mean by "when its equivalent mass-energy is applied".The perodic table made the prediction of the heavier atoms possible. Using calculations based on the expected mass of the predicted heavier atoms their proposed energy content was derived. The energy of the predicted atoms their 'mass-energy' equivalent was found using the equation E=m*c-squared. However when faced with the real proposition in the laboratory and subsequently generating the equivalent energy, the mass which were the proposed atoms, did not materalise except momentarily(?). This demonstrates (at least to me) that this equivalence (the mass-energy equivalence) is only numerical but not existential in the sense that it is meaningful in and by itself when trying to construct mass out of energy. However it is meaningful when we speak about obtaining Energy from the destruction of Mass and this is apparent because of atomic energy and atomic bombs.

My point is the equivalence does not seem to be symmetrical and there is not enough experimental evidence to claim the equivalence as symmetrical. Further to this please do not confuse the use in experiments of a 'mass-energy' equivalence where the energy supplied to a laboratory process is based on a calculation obtained from some mass m, using the E=m*c-squared equation.

My final point is energy is obtained from mass but so far mass has not been experimentally derived from energy. And it is this point which may be confusing when the terms energy and mass are said to be equivalent, there is simply no proof that this equivalence is symmetrical.

[Sven]

Sven:
The perodic table made the prediction of the heavier atoms possible. Using calculations based on the expected mass of the predicted heavier atoms their proposed energy content was derived. The energy of the predicted atoms their 'mass-energy' equivalent was found using the equation E=m*c-squared.

Who did these calculations? I never heard about them, sounds quite interesting. Do you have a reference?

However when faced with the real proposition in the laboratory and subsequently generating the equivalent energy, the mass which were the proposed atoms, did not materalise except momentarily(?).

Are you talking about collision experiments of heavy nuclei here or how do you suppose those "very heavy" nuclei were generated? References would be welcomed here also.

This demonstrates (at least to me) that this equivalence (the mass-energy equivalence) is only numerical but not existential in the sense that it is meaningful in and by itself when trying to construct mass out of energy.

Or that you misunderstood something. Please provide the references, let me study this and then we can discuss this further.

My point is the equivalence does not seem to be symmetrical

Do you mean that mass is never created out of other kinds of energy or what? Counterexample: virtual particles, appearing all the time. Research the Casimir effect for example. And see below.


Further to this please do not confuse the use in experiments of a 'mass-energy' equivalence where the energy supplied to a laboratory process is based on a calculation obtained from some mass m, using the E=m*c-squared equation.

I don't know which kind of laboratory processes do you mean here. So there is no problem of confusion.


My final point is energy is obtained from mass but so far mass has not been experimentally derived from energy.

Look at a particle accelerator. In every collision, loads of particles with mass are generated simply out of the kinetic energy of the colliding particles. The equivalence is symmetrical, face it.

Bye for the week-end - but I hope I find some time to already answer you on saturday or sunday.

I hope that you also answer the other questions of my previous post.

[sophie]

Look at a particle accelerator. In every collision, loads of particles with mass are generated simply out of the kinetic energy of the colliding particles. The equivalence is symmetrical, face it.Interpreted this way surely. Are the particles to which you refer sub-structures? Never heard of the creation of an atom, more likely the destruction of one. I think you have mixed two or three different types of colors and claim the white seen is indicative of a single unique process.

Element 110 (http://pearl1.lanl.gov/periodic/elements/110.html)

Element 118 (http://pearl1.lanl.gov/periodic/elements/118.html)

[sophie]

Sven: Do you mean that mass is never created out of other kinds of energy or what? Counterexample: virtual particles, appearing all the time. Research the Casimir effect for exampleI will agree that mass is created from energy when something naturally occuring can be reproduced in the lab and exhibited to the world in the normal way in which we see and appreciate material things in life.

All the time we use combinations of mass to create composites which can be viewed as new M = m1 + m2. Never has there been an M emergent from energy. Things simply cannot be made from heat or EMR with our current knowledge. You seem to not want to concede this basic point. This is the reason why I say that the mass-energy equivalence is not symmetrical.

If you claim this symmetry between mass and energy, the onus is now upon you to produce mass made from energy to support your point. If you can do so at this stage in our evolution I will concede the point.

[Paul of Eugene OR]

Sven:


My final point is energy is obtained from mass but so far mass has not been experimentally derived from energy. And it is this point which may be confusing when the terms energy and mass are said to be equivalent, there is simply no proof that this equivalence is symmetrical.

But mass HAS been derived from energy. I read about that when I was a youngster - gamma rays, a form of electromagnetic radiation and hence qualifying as energy in anybody's book, passing near a postive charged nucleous of an atom, turned into a pair of particles with mass - an electron and a positron.

[Sven]

Interpreted this way surely. Are the particles to which you refer sub-structures? Never heard of the creation of an atom, more likely the destruction of one. I think you have mixed two or three different types of colors and claim the white seen is indicative of a single unique process.
In a collision in a particle accelerator all kinds of particles are formed: electrons, positrons, muons, anto-muons, pions, protons, neutrons etc. Thus, all compounds of an atom are there (electrons, protons, neutrons).

But this wasn't your original claim. You said that mass is never created from energy - you were not talking about atoms specifically. Every collision in an accelerator contradicts your claim.

See, for example:
"every such collision between lead nuclei will produce about a thousand particles" from an experiment at CERN http://www.lbl.gov/Science-Articles/Archive/NA49-experiment.html
"Since an antiproton (or proton) is almost 2000 times heavier than an antielectron (or electron), it takes a lot more energy to create them." and "CERN's Super Proton Synchrotron (SPS) became a 300 GeV proton - antiproton collider, and in 1983 the UA1 experimental team, led by Carlo Rubbia, saw two new particles, the W boson and Z boson, being produced in the SPS collisions." from http://livefromcern.web.cern.ch/livefromcern/antimatter/history/AM-history02-b.html

Or perhaps you simple study the following article Transforming Energy into Mass: Particle Creation (http://www.phys.virginia.edu/classes/252/particle_creation.html) by Michael Fowler, a professor at physics at the University of Virginia.

If the above is not enough to convince you that you are wrong, then I see no point in continuing this discussion.

Thanks for the links for elements 110 and 118. But they unfortunately don't clarify what you were talking about above. For example "the mass which were the proposed atoms, did not materalise except momentarily(?)". Yes, the new nuclei are unstable - exactly as predicted beforehand. What's your point here? Why do you think this contradicts the equivalence of mass and energy?

[Sven]

All the time we use combinations of mass to create composites which can be viewed as new M = m1 + m2. Never has there been an M emergent from energy. Things simply cannot be made from heat or EMR with our current knowledge. You seem to not want to concede this basic point.
I would be cautious with such bold claims - they simply demonstrate that you have not done your "homework". I won't concede this basic point since you are simply wrong. The creation of an electron-positron pair out of EMR (to be specifically, gamma rays with energies > 1022 keV) is known for over 50 years. See, for example http://www.prestoncoll.ac.uk/cosmic/cascade/cascades.htm, especially Figure 1. This happens all the time in our atmosphere, has been reproduced in the laboratory and is over all almost completely understood.

Also see http://www.soem.ecu.edu.au/units/scp2211/Lecture03-2p.pdf, pages 32-35 (numeration of the slides)
and " When a high energy photon goes through a strong transverse field, it can create a real electron-positron pair." from http://www-acc-theory.kek.jp/members/cain/cain21b.manual/node61.html (looks very technical).

Case closed.

[sophie]

But mass HAS been derived from energy. I read about that when I was a youngster - gamma rays, a form of electromagnetic radiation and hence qualifying as energy in anybody's book, passing near a postive charged nucleous of an atom, turned into a pair of particles with mass - an electron and a positron. I cannot argue with this. However look at the experimental setup, did it end up saying: mass derived solely from gamma rays? (Raises hand to answer question) No sir, it was a combination experiment, they also used the nucleus of an atom (perhaps not just any old atom, but an atom nevertheless).

Therefore I say yes, and I say no.

[sophie]

(1) Sophie: On paper, in limited experiments one can use the mass-energy equivalent, but never in reality are mass and energy equivalent. I am clarifying reality to mean reality of the present day condition.

(2) Sophie: If you claim this symmetry between mass and energy, the onus is now upon you to produce mass made from energy to support your point. If you can do so at this stage in our evolution I will concede the point.You have done so in a very limiting sense which is not enough to satisfy my understanding.

(3) Sophie: Things simply cannot be made from heat or EMR with our current knowledge. You seem to not want to concede this basic point.Note I never said place each next to a boiling pot of soup.


(A) Sven: Thus, all compounds of an atom are there (electrons, protons, neutrons). Sub-structures.



(B) Sven: But this wasn't your original claim. You said that mass is never created from energy - you were not talking about atoms specifically. Every collision in an accelerator contradicts your claim. Possibly but not completely. If you wished to say that in the early universe mass and energy were equivalent and there were constant fluctuations between the two I would agree with you wholeheartedly. However my view of the universe as it exists today points to an asymmetrical relationship betwen mass and energy.

Let me explain why. Take fissionable material and subject it to some process P, which yields useful energy and a lot of waste. Surely E=m*c-squared? Yes sir it does indeed. Now take Johnny de Mal and place him in a situation with some device P' which has the ability to make mass from Energy. Now Johnny's arch enemy Sadie has just completed a bust from a large block of wood. Johnny seeing this immediately applies his device P' and turns the bust back into a block of wood with a tremendous burst of energy, thereby rendering Sadie's efforts useless. This just will not happen, and why I feel this way is because the conditions to create mass from energy does not exist any more. Therefore the relationship is now asymmetrical. I agree particles in a limited sense are coaxed out of the system but never from pure energy.


(B) Sven: If the above is not enough to convince you that you are wrong, then I see no point in continuing this discussion. A rational decision. Thanks. I am not convinced, and thus until we meet again.

[sophie]

Sven: I'm also not very interested in your definition of "energy".

Sven: I hope that you also answer the other questions of my previous post.I do not think this to be a rational idea on my part. On another thread I gave a definition of Energy which you did not think was suitable to your model of mind. Using this outcome from the last encounter, seeing fully well that any answers I would have given would be based on the same form of idealogy, it may as well be considered absurd on my part to attempt such an effort. Therefore I humbly and honorably decline your most gracious invitation. Thanking you in advance.

[V-Bird]

Energy... Movement.

Force... the energy that keeps that movement to a path or from it.

[sophie]

Mr. V-Bird: Energy... Movement.Then does this not beg the question of a whole new field for physics to explore - The physics of Energy?

[V-Bird]

Been done... tho you could try to start your own...

[Dragar]

This just will not happen, and why I feel this way is because the conditions to create mass from energy does not exist any more. Therefore the relationship is now asymmetrical. I agree particles in a limited sense are coaxed out of the system but never from pure energy.

But haven't we seen electron-positron pairs produced from high energy electromagnetic waves (gamma rays)? Well, seen the paths they leave in cloud chambers, at any rate. Is this not mass from pure energy (light)?

Dragar

[sophie]

Quoted from another source : The strong force binds with energy derived from EMR [the purest form of energy] The weak force is the decay, the 'radiation' is also energy it is derived from EMR. Gravity is mass, mass is formed by the strong force keeping pure energy 'in check', this forms matter, matter is then itself bound by the remaining pure energy that produces gravity. It seems like the gravity is the interaction of the strong force 'in check' with the "loose" energy 'out of check'. Anyway it sure looks like an "impending mate". Reminds me of the old saying :They who know and know they know - they are wise, follow them... they who know not they know not - they are fools so ignore them, the rest in between really don't matter. :banghead: :boohoo: :notworthy

[Sven]

Note I never said place each next to a boil