Yes, (it is not what I am saying per se--it is what Einstein said 100 years ago and what has been proven over and over again experimentally).

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1 Date: Tue, 20 Dec :41: From: Brian Greene Reply-To: Organization: Columbia University X-Accept-Language: en-us, en To: "Dr. Rainer Glaser" Subject: Re: [Fwd: Re: Einstein's Equation] X-No-Spam-Score: Local 1 Dr. Rainer, Dr. Rainer Glaser wrote: Re: [Fwd: Re: Einstein's Equation] Dr Greene: I have no problem with the equation in the sense that it tells us how to convert E into m or m into E. We can take energy and make the energy into particles (amazing, but proven, people can do it). We can take particles and turn them into energy (amazing, also proven, people cam do it). The problem that separates you and me right now seems to be that you say ANY energy content is equal to some mass and if energy is liberated, well, then mass was lost. Yes, (it is not what I am saying per se--it is what Einstein said 100 years ago and what has been proven over and over again experimentally). But just because energy has a mass >>equivalent<<, by way Einstein's equation, the equation >>does not mean that all energy changes are associated with mass changes<<. Yes it does. They are one and the same. To make your argument understandable (if you are right), you need to address the electrostatic part of Schroedinger's equation. Opposite charges attract each other, the energy drops. This term is independent of the masses of the carriers of the charges (e.g. proton & electron). But once they are in attraction, the energy is lower, and according to you, the mass has been reduced. That is what you are saying (or at least what I think I am reading) and that is where the controversy is. To me the masses of the carriers of the charges have NOT changed. It is the mass of the combined system, proton plus electron plus their interaction. When particles are interacting, as in your example, you can't properly speak of the masses of the individual constituents. (When we talk about the "mass of the proton" for example, we generally mean an isolated proton that is at rest. If the proton is not at rest or not in isolation, its mass differs from its isolated rest mass. Literally, if you weighed a non-resting/non-isolated proton it would register a different value.) The electrostatic energy has changed, charges are interacting with electric fields. -- The flashlight example I have not throught 1

2 through. 2 You should think it through. You will conclude that the mass decreases, as I've said. I do not think that "a pot of water is getting heavier as it gets hotter," and, no offense, that notion seems rather odd to me. Well, put such a pot on a scale and heat it up. If your scale is (fantastically accurate) you will see that the reading will go up. It seems odd to you (no offense) because you have not fully absorbed relativity. Higher rotational and some higher vibrational levels are populated, translational energy of the molecules is increased. The energy content went up, yes. We could compute a >>mass equivalent<< of that energy increase, yes. But the water in that pot after the heating either will have the same mass and be hot (motion!, observed), or have a higher mass and be cold (ever observed?, I don't think so.). As above--the reading on the scale would increase. If I understand you correctly, you want it both ways; heating the water and getting it both hot and heavy. -- Keep in mind that there is the kinetic term in Schroedinger's equation. On heating, the kinetic energy goes up and there will be some relativistic mass increase. In that part of the equation, Schroedinger's equation, some of the energy goes into mass and some stays in kinetic energy (amazing, proven, we agree on that too). But the Coulomb term of the equation, that is where we need to look to clear this up. Do you see what I am trying to say? -- I am not trying to give you a hard time. I do indeed. But you are not right. Let me give you yet another example. You mentioned the "mass of a proton" above. Do you know where the mass of a proton comes from? Of course, you know that a proton is composed of three quarks. But if you look up the masses of the proton's constituent quarks, you'll find that their sum is far less than the mass of a proton. How could that be? The answer is that the quarks interact through the strong force which is mediated by gluons. Now gluons, like photons have zero rest mass. However, because the gluons are zipping around inside the proton, they have energy. And it is the ENERGY of these moving gluons--their kinetic energy if you will--that makes up the bulk of the MASS of the proton. The energy in the gluons is not merely FORMALLY equivalent to a corresponding amount of mass. The energy in the gluons IS mass, and this 2

3 mass IS the bulk of the proton's mass. 3 Best, BG It might look that way, but it's really not personal. Best, RG Original Message Subject: Re: Einstein's Equation Date: Tue, 20 Dec :58: From: Brian Greene <greene@physics.columbia.edu> Reply-To: greene@physics.columbia.edu Organization: Columbia University To: Dr. Rainer Glaser <glaserr@missouri.edu> References: <p bfcd0e1e431b@[ ]> <43A7748D @physics.columbia.edu> <p bfcde3bf5546@[ ]> <43A84D9A @physics.columbia.edu> <p bfcdffd4ea6d@[ ]> Dear Dr. Glaser, I presume there must be hundreds of places you can find this written down--take a look at some good books on relativity (maybe John Wheeler's book, as an example). But I do wonder why seeing things written in a book is somehow more convincing than just thinking through the physics. The lesson from E = m c^2 is that mass and energy are (up to a conversion factor, c^2) the SAME thing. In units where c=1, the equation is E = m. Energy and mass are equal. Thus, if energy is released in a chemical reaction, so that the remaining E has decreased, you immediately learn that m has decreased (they are equal afterall). It is as simple as that. This has NOTHING to do with nuclear reactions. Einstein's equation is a fundamental feature of the physical universe. It is not tied to one particular kind of physical process. Here's another way of saying the same thing: Einstein showed that gravity couples to total mass-energy (not just rest mass as Newton would have said). This, of course, is the essential content of general relativity. 3

4 So when you weigh something on a scale, the scale's reading reflects the total energy of the object. Some of this energy comes from the object's rest mass, but other contributions come from the object's internal energy (kinetic motion, energy of chemical bonds, etc.) Gravity does not care about the source of energy-- it couples to total energy. Thus, if any contribution to the total energy changes (such as releasing energy stored in chemical bonds, or changing the potential energy between charged constituents as in your example), the scale's reading will change. It really is as straightforward as that. 4 Do you agree with my example of a flashlight getting lighter as it emits light? Do you agree with my example of a pot of water getting heavier as you heat it up? If you do agree, then you agree completely with my article. On the other hand, if you don't agree with these examples, you are quite mistaken about this aspect of nature. Let me know if it is now clear. With best wishes, BG Dr. Rainer Glaser wrote: Re: Einstein's Equation Dear Dr. Greene: Please convince me. Do give me a journal reference that says that the combined mass of a proton and an electron changes with the distance between them. That is what you are saying, right? Also, do give me one good chemistry textbook (intro or advanced) that makes this point. As to the "nonsense". I am sorry, I was bold, and I will apologize if I have to. I hope you are right. It would be much worse for you to be wrong on this, than it is for the chemists (including myself). I will start conversations with some colleagues who are more into pure theory than I am. Best, RG. 4

5 5 Thanks for your reply. Please note that the piece clearly states that E = mc^2--while applying in all interactions-- has its largest, most manifest appearance in nuclear interactions. So, you are right in saying that non relativistic quantum mechanics is very accurate in its assumption that mass is conserved. But that is beside the point. However small the effect, mass DOES change if the energy stored in chemical bonds changes--with the change given by Einstein's formula. If energy is released in a chemical interaction then the energy which remains, clearly, has decreased. And a decrease in energy is equivalant, a la Einstein, to a decrease in mass. Period. Frankly, it is amazing to me that professional scientists, such as yourself are not aware of this. (As mentioned in my first , you are not alone in this confusion. A number of your colleagues, some of whom are in the National Academy, were similarly confused and wrote me (although most were rather more polite--your use of the word "nonsense" in refering to my article is beyond the pale). After a few s, all have agreed that indeed the article is perfectly correct; some thanked me for bringing to their attention a point they had previously failed to grasp--an unexpected but welcome admission. With best wishes, BG Dr. Rainer Glaser wrote: Re: Einstein's Equation Hi again: I hope you enjoyed my Feynman adaptation and I am delighted about you quick reponse. 5

6 6 I know quantum chemistry and read some relativistic quantum chemistry. For the lighter elements, nonrelativistic quantum chemistry is very accurate. Neither of these treatments consider mass changes as a function of electrostatic interactions. That is the point that needs explanation (or provision of a reference). Best, RG. Dear Dr. Glaser, Many thanks for your . Regarding the point you raise, indeed there is no error in my piece. Any (good) textbook on chemistry Which ones are you referring to? notes that mass is not strictly conserved in chemical reactions, a fact the world learned in 1905 with the advent of special relativity. Allow me to cut and paste a response to similar queries I received (almost exclusively from chemists who, it seems, have yet to fully digest Einstein's discoveries), in order to avoid typing the explanation anew I thank the reader for his ; I am gratified that he has given my piece such a close read. However, the reader's criticism is wrong. The reader recounts the conventional description, which people believed prior to 1905, in which mass is conserved in chemical reactions. After 1905, this view was realized to be wrong. The error is that the energy in the chemical bonds 6

7 to which the reader refers, contributes to the mass of the substance (gasoline, say). The standard description ignores this. That is a wonderfully accurate approximation since the mass contribution of the bonds is E/c^2 (E = bond energy), which is typically very small. But it is an approximation nonetheless. 7 The lesson that has yet to be widely appreciated is that ALL forms of energy (thermal, potential, rest mass, etc)--all of them-- contribute to the total mass of a system. Chemical bonds are one manifestation of energy; they contribute to a system's mass. If you change the energy in chemical bonds, you change the mass of the system. Thus, the reader is right in saying that the energy released in chemical reactions arises from the difference in bond energies. But the reader is failing to realize that this difference in bond energies results in a difference in masses, with the difference given by Einstein's formula. As this is a subtle idea, a couple of other, simpler examples, may help understanding. When you turn on a flashlight, the flashlight weighs less than it did before you turned it on. The energy released from its batteries (in the form of the beam of emitted light as well as heat) causes the mass of the flashlight to decrease in accord with Einstein's formula. Put the turned-on flashlight on a fantastically accurate scale, and you'd watch the reading on the scale go down, precisely in accord with Einstein's formula. Even in the most mild of "chemical" reactions in which you heat a pot of water, the energy you put into the water from your stove's flame causes the mass of the pot of water to increase. A hot pot of water weighs more than an otherwise identical pot of water that is cold. And the difference in their masses is given by the difference in their (thermal) energies divided by c^2, in accord with Einstein's formula. My piece was meant, in part, to disabuse people of the incorrect notion that E = mc^2 operates solely in nuclear reactions. It operates in all interactions. From the reader's , it seems he may be one of the people I had in mind. I hope the piece, and this , helps 7

8 to clarify the confusion Indeed, I hope this helps you as well.let me know if you have further questions. With best wishes, <>BG The error in that conventional description is that the energy in those chemical bonds you refer to actually contributes to the mass of the substance (gasoline, say). The standard description ignores this. That is not a bad approximation since the mass contribution of the bonds is E/c^2 (E = bond energy), which is typically very small. But it is an approximation nonetheless. So, you are right that the energy released, say, is the difference in bond energies. But this difference in bond energies results in a difference in masses, with the difference given by Einstein's formula. Best, BG Thanks for your . Your comments are wrong. As in my piece, E = mc^2 applies to ALL interactions, NOT just nuclear reactions. 8

9 In nuclear reactions--such as standard fission and fusion--the energy produced via E = mc^2 arises by releasing that stored in nuclear bonds. 9 In chemical reactions, the energy produced via E = mc^2 arises by releasing that stored in chemical bonds. The only difference is that the greater strength of nuclear reactions means more energy can be extracted in this manner. When you turn on a flashlight, the flashlight weighs less than it did before you turned it on. The energy released from its batteries (in the form of the beam of emitted light as well as heat) causes the mass of the flashlight to decrease in accord with Einstein's formula. Even in the most mild of "chemical" reactions in which you heat a pot of water, the energy you put into the water from your stove's flame causes the mass of the pot of water to increase. A hot pot of water weighs more than an otherwise identical pot of water that is cold. And the difference in their masses is given by the difference in their (thermal) energies divided by c^2, in accord with Einstein's formula. My piece was meant, in part, to disabuse people of the wrongheaded notion that E = mc^2 operates solely in nuclear reactions. From your , you are one of the people I had in mind. I hope the piece, and this , helps to clarify your confusion. All the best, BG Thanks for your . The missing point is that the energy in the chemical bonds (that you imagine releasing) contributes to the mass of the initial vat of gasoline. That is, the mass of the gasoline is not just the sum of the masses of the individual atoms that constitute it. Instead, the mass of the gasoline is also partly due to the energy in the bonds between these 9

10 10 constituents. When you release this energy, the mass of the remaining products is therefore reduced. The overall point is that E = mc^2 tells us that mass and energy are really, truly the same thing. Thus if "energy is released" regardless of how that happens (chemical reaction, nuclear reaction, etc) then that can equally well be thought of as "mass is released". And, if mass is released, there is less than there was before it was released. Similarly, if energy is put into a system, that increases the system's mass. When you compress a spring it has more mass than when it was uncompressed. It's molecular make-up has not changed at all, but the energy stored in the compressed spring shows up as additional mass. The compressed spring weighs more than the uncompressed one. Let me know if that helps. Best, Dr. Rainer Glaser wrote: Dear Dr. Greene: I enjoyed reading your article in the New York Times a while back about Einstein's Famous Equation. I read with interest the sections in which you explain that the combustion engine is based on Einstein's famous equation. And so are batteries, and physiological functions... Reading that, I thought "Surely, you are joking Dr. Greene!" For a professor in physics and in mathematics, a professor from Columbia University, to write such nonsense in the New York Times, in an homage to Einstein, is nothing short of surreal. Yet, this story is getting even better. When asked for a correction, the New York Times was not interested (and considering your status and affiliation why should they be interested.) In this situation, I think it is your responsibility to correct those errors in your article in some way in public. On the other hand, feel free to provide me with any references from the peer-reviewed literature that supports what you wrote about combustion and I will gladly throw out all the many textbooks that explain all these processes perfectly well with chemical thermodynamics. Best, 10

11 Rainer Glaser "We know accurately only when we know little; with knowledge doubt increases." J. W. von Goethe Visit us on the web at Dr. Rainer Glaser MM MM UU UU Professor in Chemistry Tel: (573) MMM MMM UU UU Department of Chemistry Fax: (573) MMMM MMMM UU UU University of Missouri Home: (573) MM MMMM MM UU UU Columbia, MO MM MM MM UUUUUUU ========================================================================= -- "We know accurately only when we know little; with knowledge doubt increases." J. W. von Goethe Visit us on the web at Dr. Rainer Glaser MM MM UU UU Professor in Chemistry Tel: (573) MMM MMM UU UU Department of Chemistry Fax: (573) MMMM MMMM UU UU University of Missouri Home: (573) MM MMMM MM UU UU Columbia, MO MM MM MM UUUUUUU ========================================================================= -- "We know accurately only when we know little; with knowledge doubt increases." J. W. von Goethe Visit us on the web at Dr. Rainer Glaser MM MM UU UU Professor in Chemistry Tel: (573) MMM MMM UU UU Department of Chemistry Fax: (573) MMMM MMMM UU UU University of Missouri Home: (573) MM MMMM MM UU UU Columbia, MO MM MM MM UUUUUUU ========================================================================= 11