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Does E = mc^2?

  1. Nov 7, 2006 #1
    In the past I have argued that E does not always equal mc^2 under all conditions. I will have to put a hold on that assertion for now until I've fully studied this phenomena. It appears that there are two definitions regarding this situation and each of two well known relativists have used these two, one of each. So until I fully comprehend why there are two definitions and how this expert fould be wrong then I'm putting a hold on my assertions and assumptions of this matter. The two relativists are Wolfgang Rindler and Fritz Rohrlich. Each published a paper on similar subjects, presumably with different definitions. Oy!

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  3. Nov 7, 2006 #2

    James R

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    [itex]E=mc^2[/itex] is always true, provided that m is taken to be the relativistic mass.

    These days, physicists generally don't like to use the concept of relativistic mass, so they quote the following formula instead:

    [tex]E^2 = (pc)^2 + (mc^2)^2[/tex]

    where it is understood that m is the REST mass (not the relativistic mass), and p is the relativistic momentum. E is the total energy (kinetic + rest energy).
  4. Nov 7, 2006 #3
    You do realize that I wouldn't have created this thread if I didn't have a very good reason for suspecting that E does not always equal mc^2. That is subject to debate. I am retrracting my comment until I have researched the problem along the lines of another definition I came across. Please read


    The following is an article from the American Journal of Physics on this topic by Wolfgang Rindler and Jack Denur

    (yawn) I believe I know that fact more than anyone on this forum. Its repeated in nearly every other thread. :biggrin:

    But don't jump to conclusions. Physicists use whatever term is useful to them in their work. A particle physicist is always interested in the the proper mass of particle just as a cosmologist is awlays equal to the gravitational mass = inertial mass = relativistic mass.

    Best wishes

  5. Nov 8, 2006 #4
    Well instead of wasting forum space, our time and being pompus you could have simply posted all your ideas in one go instead of 'reserving' this spot.

  6. Nov 8, 2006 #5

    The old and new concepts of physics-An open-dialogue journsl is on the way to publish the paper

    Sharma A.

    Comment by Kant S.
  7. Nov 9, 2006 #6
    Sorry, pmb_phy, but, since in the document: http://www.geocities.com/physics_world/sr/mass_momentum_density.htm;
    is said that the force is Lorentz invariant, I would like to take the opportunity to ask you what does it mean exactly. I understand why rest mass and relativistic interval (or proper time) are Lorentz invariant, but I would say force is not. How does it happens it is?
    Thank you.
  8. Nov 9, 2006 #7
    as far as I know from Robert Resnick "Introduction to Special Relativity" Wiley 1968 p.147 as long as the particle on which the force acts has only OX component the OX component of the force has the dame magnitude in all inertial reference frames in relative motion
    sine ira et studio
  9. Nov 9, 2006 #8
    In that particular case, under a Lorentz transformation from S to S' the x-component of the force in S has the same numerical value as that in S'. That is what is meant by Lorentz invariant in this particular instance.

    However I'm beginning to think that this is an incorrect usage of the term "invariant." I had in mind a paper by Rindler and Denur when I created that web page and that was the term they used except that they didn't use the term "Lorentz" in "Lorentz invariant." I'll send you a PM on this. Please read it. Thanks.

    Best wishes

  10. Nov 9, 2006 #9
    E^2 = (pc)^2 + (mc^2)^2
    Can be reduced to;
    only when the momentum “p” is zero.
    Thus if “p” is zero the movement of the object in question has a “speed” of zero,
    Giving the Energy as seen from the one frame of reference where there is no kinetic Energy therefore NO MOVEMENT.
    There is no “relativistic” anything involved – that was the whole point Einstein was making; reducing the whole formula to the short E = mc^2; to solve for the Potential Energy held within a mass.
    There is no relativistic use of mc^2, except to define the Potential Energy intrinsic within the mass. Any motion, or measurement from a frame that requires motion to exist, requires the use of the whole formula
    not just E^2 = (mc^2)^2
    Last edited: Nov 10, 2006
  11. Nov 9, 2006 #10
    You're entitled to your opinion.
    I've been through the derivations more times than I can count.
    Wrong. I had previously had gone through derivations which showed that E = mc^2 is not always true. E.g. see the example given here


    That is an example of a object with finite dimension having stress exerted on it, i.e. a non-closed systems.
    It depends onn what you mean by "m". If you mean the inertial mass of a point object subject to external forces there are several ways. Here are 4 that I wrote up

    If and only if m is proper mass. This thread was not about m being proper mass. It has to do with the claim that relativistic mass is the same as inertial energy.

    Recall Rindler's intro to sr text where he writes on page 150 under the section "Relativistic Mechanics of Continua"
    As far as what Einstein meant by E = mc^2, that changed with time. I do know, however, that he used it in his last update to The Meaning of Relativity to mean that m = relativistic mass. If you need other references to this fact then please read both of Schut's books, i.e. "A first course in general relativity" and "Gravitation from the ground up" where he demonstrates that inertial mass is a function of pressure in the most general case. He gets the same result I do in this page

    Best wishes

  12. Nov 9, 2006 #11
    With comments like this please do not expect me to read anymore of your posts. Welcome to my ignore list.

    Best wishes in your future acts of insulting people

  13. Nov 9, 2006 #12
    The point is; what is the meaning of the formula E = mc^2.
    Sure even Einstein could revise the original meaning by replacing m (mo) with a relativistic m (mr) as a short hand way of saying:
    E^2 = (pc)^2 + (moc^2)^2
    That doesn’t mean he was right in doing so – I don’t consider the man infallible.
    It is a new and different formula, and not addressing the same issue as
    E = moc^2 the energy in mass.
    Instead E = mrc^2 attempts to address TOTAL ENERGY.
    That short hand assumes momentum p= mv by applying a relativistic mr of:
    mo/sqrt(1-V^2) times v
    to get the momentum p.

    IMO it is as valid and more consistent to instead define and apply a relativist velocity vr of:
    v/sqrt(1-V^2) times mo
    to get the momentum p.

    Then the infinity that comes as v nears c (V nears 1) is not loaded onto the “mass” in the problem. But onto to the “apparent” or “relativistic” velocity.
    This is a least consistent with the apparent distance to be covered in the “other” reference frame approaching infinity; as the Lorentz Conversion tells use the unit measure of length in that frame becomes infinitely short (in our view) as V approaches 1.

    Leaving just one proper mass that makes much more sense.
    I see no reason to oppose the current view of invariant mass.
    Last edited: Nov 10, 2006
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