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Proof of E = mc^2

  1. May 30, 2014 #1
    What is the proof of Einsteins famous equation E = mc2 ?
     
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  3. May 30, 2014 #2
  4. May 30, 2014 #3

    Matterwave

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    I believe this ancillary paper is a more direct proof of what you want: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

    Einstein does use results from his previous paper though, so you might want to read both.
     
  5. May 30, 2014 #4

    russ_watters

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    In science, "proof" generally refers to evidence. Nuclear energy and particle accelerators are two good pieces of evidence.
     
  6. May 30, 2014 #5

    bcrowell

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    There are many different possible ways of proving this. For a critique of some of the technical issues in Einstein's original derivation, see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 . In my SR book http://www.lightandmatter.com/sr/ I give a recap of the Einstein argument in section 4.2, and resolutions of the issues Ohanian complains about in sections 4.4 and 9.2. It's hard to give a really rigorous treatment without using the stress-energy tensor, which hadn't been invented when Einstein wrote his 1905 proof.
     
  7. Jun 1, 2014 #6

    stevendaryl

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    There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this:
    1. Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: [itex]E = g(v) m[/itex] (By "mass", I mean rest mass).
    2. Assume that the momentum of a particle is proportional to the mass and is in the same direction as the velocity: [itex]\vec{p} = f(v) m \vec{v}[/itex]. I'm also assuming that the unknown function [itex]f(v)[/itex] depends only on the magnitude of the velocity, and not it's direction.
    3. Assume that the expression for momentum approaches the Newtonian result, [itex]\vec{p} = m \vec{v}[/itex] for low speeds. That implies that [itex]f(0) = 1[/itex]
    4. Assume that the expression for kinetic energy (total energy minus rest energy) approaches the Newtonian result: [itex]m g(v) - m g(0) \Rightarrow 1/2 m v^2[/itex]
    5. Assume that in a collision of particles, the energy and momentum are conserved.

    With all those assumptions, we can prove that
    [itex]f(v) = \gamma = 1/\sqrt{1-v^2/c^2}[/itex]
    [itex]g(v) = \gamma m c^2[/itex]

    To see this, consider an experiment where we collide two identical pieces of clay, each of mass [itex]m[/itex], so that they stick to form a single, larger chunk of clay of mass [itex]M[/itex]. We look at the collision in three different rest frames:
    1. Frame 1: Pick a frame in which one mass is traveling at velocity [itex]-v[/itex] in the y-direction, and the other at velocity [itex]+v[/itex] in the y-direction. Neither has any velocity in the x-direction.
    2. Frame 2: Pick another frame whose spatial origin is moving at velocity [itex]u[/itex] in the x-direction, relative to the first frame. Assume that [itex]u \ll v[/itex].
    3. Pick a third frame whose spatial origin is moving at velocity [itex]u[/itex] in the y direction. Again, choose [itex]u \ll v[/itex].

    In Frame 1: By symmetry, afterward the composite mass [itex]M[/itex] is stationary. Conservation of energy tells us that [itex]2 g(v) m = g(0) M[/itex].

    In Frame 2: The total momentum in the x-direction before the collision is approximately (ignoring higher-order terms in [itex]u[/itex]) [itex]-2 f(v) m u[/itex]. The total momentum afterward is approximately (using the Newtonian approximation, since [itex]u[/itex] is small): [itex]-M u = -2 g(v) m u/g(0)[/itex]. Conservation of momentum in the x-direction gives: [itex]f(v) = g(v)/g(0)[/itex].

    In Frame 3: One mass is moving at velocity [itex]v_1 = \dfrac{v-u}{1-uv/c^2} \approx v - u (1-v^2/c^2)[/itex] in the y-direction. (In the approximation, I did a Taylor expansion, and only kept the lowest order terms in [itex]u[/itex]). The other mass is moving at velocity [itex]v_2 = -\dfrac{v+u}{1+uv/c^2} \approx -v - u (1-v^2/c^2)[/itex]. After the collision, mass [itex]M[/itex] is moving with velocity [itex]-u[/itex] in the y-direction.

    Since we are assuming that momentum has the form [itex]p = f(v) m \vec{v}[/itex], if we change frames so that the velocity changes slightly by [itex]\delta v[/itex], the momentum will change by an amoung [itex]\delta p = m (\frac{d}{dv}(v f(v))) \delta v[/itex] (ignoring higher-order terms in [itex]\delta v[/itex])

    So the first particle will have momentum in the y-direction: [itex]f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)[/itex]

    The second particle will have momentum in the y-direction: [itex]-f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)[/itex]

    The total momentum before the collision in the y-direction is:

    [itex]-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2)[/itex]

    The momentum after the collision in the y-direction is:
    [itex]- Mu = -2mg(v)/g(0) u = -2m f(v) u[/itex]
    (where I used the results from the other two frames).

    So conservation of momentum in this frame gives:

    [itex]-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2) = -2m f(v) u[/itex]

    So we have a first-order differential equation for [itex]f(v)[/itex]:

    [itex](\frac{d}{dv}(v f(v))) (1-v^2/c^2) - f(v) = 0[/itex]

    The unique solution satisfying [itex]f(0) = 1[/itex] is

    [itex]f(v) = 1/\sqrt{1-v^2/c^2}[/itex]

    Now going back to [itex]g(v)[/itex]. We showed that [itex]g(v)/g(0) = f(v) = 1/\sqrt{1-v^2/c^2}[/itex]. Expanding for low-velocities gives: [itex]g(v) = g(0) (1 + 1/2 v^2/c^2 + ...)[/itex]. So the kinetic energy at low velocities is [itex]m (g(v) - g(0)) = m g(0) 1/2 v^2/c^2 + ...[/itex]. For this to agree with the newtonian result, [itex]g(0) = c^2[/itex]

    So we have the two results:
    [itex]\vec{p} = 1/\sqrt{1-v^2/c^2} m \vec{v} = \gamma m \vec{v}[/itex]
    [itex]E = g(v) m = m g(0) f(v) = m c^2 1/\sqrt{1-v^2/c^2} = \gamma m c^2[/itex]
     
  8. Jun 1, 2014 #7

    bcrowell

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    Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc2.
     
  9. Jun 1, 2014 #8

    stevendaryl

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    Yeah, as I said, it's a heuristic derivation, not a proof.
     
  10. Jun 1, 2014 #9
    Is there a justification for this assumption? Why not [itex]E = E_0 + g(v) m[/itex]?
     
  11. Jun 1, 2014 #10

    stevendaryl

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    Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.
     
  12. Jun 1, 2014 #11

    PeterDonis

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    If energy and momentum form a 4-vector, then rest energy has to be proportional (equal, if you choose units properly) to the invariant length of the 4-vector, i.e., to invariant (rest) mass.
     
  13. Jun 1, 2014 #12

    PeterDonis

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    ##E_0## is just ##m## (in units where ##c = 1##, which you can always choose), so this equation reduces to stevendaryl's equation (you just redefine ##g(v)## to include the constant ##1##).
     
  14. Jun 1, 2014 #13
    Me too.

    Here is a derivation of the relativistic momentum without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):

    https://www.physicsforums.com/showpost.php?p=4737048&postcount=4

    Starting with the result of this derivation I get

    [tex]dE = F \cdot ds = v \cdot dp = \frac{{m \cdot v \cdot dv}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3 }}[/tex]

    The integration results in

    [tex]E - E_0 = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} - m \cdot c^2[/tex]

    With your assumption #1 you simply set the constant of integration to zero resulting in Eo=m·c². That turns your derivation into a circular argumentation.
     
  15. Jun 1, 2014 #14

    Fredrik

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    This is a bit of an overstatement. You're using that ##F=\frac{dp}{dt}## and that ##p=\gamma m v##. These are non-trivial assumptions. They can be considered definitions of the left-hand sides, but you can also take ##E^2=p^2c^2+m^2c^4## to be the definition of E, and then you don't have to do any calculations at all. This definition needs no justification other than the fact that the theory makes very accurate predictions (as already mentioned by Russ).
     
  16. Jun 1, 2014 #15
    Remember the topic! Eo=m·c² has to be proofed. Thus you can't take it as a given.
     
  17. Jun 1, 2014 #16
    ##F=\frac{dp}{dt}## is Newton's second law and ##p=\gamma m v## the result of the derivation.
     
  18. Jun 1, 2014 #17

    Fredrik

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    Right, but so is ##F=m\frac{d^2x}{dt^2}##, and this formula doesn't lead to the desired result. If you want the calculation to be considered a proof, or just a fairly solid argument, you have to explain why you're using that specific statement of Newton's 2nd. If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.

    Can you justify the last equality in the first line without this?
     
  19. Jun 1, 2014 #18
    No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.

    The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.

    Stick to facts!
     
  20. Jun 1, 2014 #19

    Matterwave

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    If the argument is regarding whether Newton's second law states ##F=ma## or ##F=\frac{dp}{dt}##, the answer is actually that his language is explicitly given as that of the second formula, although he explicitly assumes that the momentum in the second formula is given by ##p=mv##.

    Reading the Principia you find that Newton stated his second law thus (taken from the translation by Andrew Motte):

    "The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed".

    Here "motion" is roughly Newton's word for "momentum". He also only ever used the word "proportional" and not equal to, so there could be some as yet defined constant. So his language technically says ##\frac{dp}{dt}\propto F##. However, earlier in the text, Newton defined "motion" (momentum) thus:

    "The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly".

    He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes ##p \propto mv## and NOT ##p=\gamma mv## (obviously he would have no idea about the second case as that arose 300 years after him).

    So, although his language technically might agree with DrStupid, I think the spirit of his language agrees with Frederik.
     
  21. Jun 1, 2014 #20

    Fredrik

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    Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR? If it's the former, you certainly do have to explain why you favor dp/dt over ma. If it's the latter, then you have to explain why you consider your definitions (of momentum, work, etc) more fundamental than the end result, which can also be taken as a definition.
     
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