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How many ways there are to conclude that Eo=mc^2 ?

  1. Jan 2, 2016 #1
    One way to conclude that E0=mc2 is to look at the work done by constant force:

    W = integral[ F dx ] = mc2(1/√1-(v/c)2)-1)

    and from that you get that E_kin = Etot - E0 , where E0 = mc2

    How many other ways there are to conclude that E0 = mc2 ?
     
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  3. Jan 2, 2016 #2

    Fredrik

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    You can simply define the energy of a particle as the 0 component of the four-momentum p. If p is defined as the four-vector that, in the particle's rest frame, is equal to the mass m times the four-velocity v (which in the rest frame has componennts (1,0,0,0)), then (if we use a -+++ metric and units such that c=1) we have ##p^2=m^2v^2=m^2(-1)=-m^2## and ##p^2=-E^2+\vec p^2##. So ##-E^2+\vec p^2=-m^2##. This implies ##E^2=m^2+\vec p^2##. If we restore factors of c, this turns into ##E^2=m^2c^4+\vec p^2c^2##. If ##E_0## denotes the energy when velocity (and momentum) is zero, this implies that ##E_0=mc^2##.

    If four-momentum is defined as the four-vector p formed by the generators of translations in the 0,1,2,3 directions, then we can define the mass as the non-negative m such that ##-m^2=p^2##. Then we can proceed as above.
     
  4. Jan 3, 2016 #3

    bcrowell

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    Welcome to PF! By the way, the method you're using to write up your math is going to be very slow and painstaking for you, will not produce very readable results, and will not work for more complex math. A better way to do it is using LaTeX: https://www.physicsforums.com/help/latexhelp/

    I don't think your reasoning in #1 is correct. Your integral is a definite integral, so any constant term cancels out. There is also a more subtle issue, which is that one needs to provide some justification for the assumption that ##W=\int F dx## holds without modification in relativity. In Einstein's first 1905 paper on relativity, he does this in section 10, using an intricate argument that starts from the fact that for a charged particle that is instantaneously at rest, ##ma=qE##.

    Einstein's original derivation of ##E=mc^2## is not in his first paper on SR. It's the topic of his second paper, and it uses an argument involving an object that emits light simultaneously in two opposing directions. For a critique of this argument, see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 . My SR book has a presentation of Einstein's argument in section 4.2, along with follow-up discussions in section 4.4 and ch. 9 that take care of the issues that Ohanian complains about.

    The most common approach today is to search for a four-vector that is conserved in collisions. An example along these lines by @DrGreg is here: http://www.physicsforums.com/showthread.php?p=2416765 .

    This may also be of interest: Sonego and Pin, Deriving relativistic momentum and energy, http://arxiv.org/abs/physics/0402024
     
  5. Jan 3, 2016 #4

    Fredrik

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    We have ##W=\int F\mathrm dx=\int \dot p\dot x\mathrm dt##, where the integration on the right is over a time interval [a,b] such that the velocity is zero at time a. The integrand ##\dot p\dot x## can be rewritten as ##\frac{d}{dt}(m\gamma)## (in units such that c=1). So we end up with
    $$W=\int_a^b \frac{d}{dt}(m\gamma)\mathrm dt =\gamma_bm-\gamma_am.$$ If we drop the b from the notation and restore factors of c, we get
    $$W=\gamma mc^2-mc^2=E-E_0.$$
     
  6. Jan 3, 2016 #5

    bcrowell

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    But that doesn't imply that ##E_0## has any special interpretation. If this argument were correct, then there would have been no reason for Einstein to publish his second 1905 paper on SR, with a separate argument. He would simply have noted it as a consequence of his argument about mechanical work in his first paper.
     
  7. Jan 3, 2016 #6

    Fredrik

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    I think I see what you meant when you mentioned a constant before. If we define the total energy E by ##E=\gamma m## (times ##c^2##), then ##E_0## is the total energy of an object at rest. But we could add a constant to that definition. We would have ##E=\gamma m+A##, but we would still have ##W=E-E_0##. So what is the correct value of A?

    I don't think this is really an issue. Why not just say that we choose A=0 because this is the choice that gives us the nicest interpretation of E?

    The approach to SR that I like best goes like this:

    1. Define mathematical terms. (Minkowski spacetime, world line, proper time, momentum, work, etc.)
    2. Choose a set of correspondence rules. (Assumptions that specify how to interpret the mathematics as predictions about results of experiments).
    3. Use 1-2 to make predictions about results of experiments.
    4. Do experiments to determine the accuracy of those experiments.

    In this approach, the definitions in step 1 don't need to be justified. Definitions never do. We may be able to show e.g. that two concepts defined in step 1 are related in a way that's very similar to how they're related in pre-relativistic classical mechanics. If such a demonstration makes it easier for us to develop our intuition about SR, then I would consider it a worthwhile thing to do. But I wouldn't consider it essential. The only thing that can really justify what we did in step 1 is step 4.

    I would say that the calculation of the work required to accelerate a mass m from speed 0 to speed v is such a demonstration. It tells us that if we define work, force, momentum and energy in certain ways (ways that are similar to how these terms are defined in pre-relativistic classical mechanics), the relationship between work and kinetic energy is the same as in pre-relativistic classical mechanics.
     
    Last edited: Jan 3, 2016
  8. Jan 4, 2016 #7
    I don't get it. Can you explain that in detail?
     
  9. Jan 4, 2016 #8

    Fredrik

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    It's explained in some detail (for the more general case of a force that isn't necessarily constant) in post #4. What I left out is the proof of the claim ##\dot p \dot x=\frac{d}{dt}(\gamma m)##. It goes like this (still using units such that c=1):
    \begin{align*}
    & \gamma=\frac{1}{\sqrt{1-\dot x^2}}=(1-\dot x^2)^{-\frac 1 2}\\
    & \dot\gamma =-\frac 1 2(1-\dot x^2)^{-\frac 3 2}(-2 \dot x\ddot x) = \gamma ^3\dot x\ddot x\\
    & 1+\gamma^2\dot x^2=1+\frac{\dot x^2}{1-\dot x^2}=\frac{1-\dot x^2+\dot x^2}{1-\dot x^2}=\gamma ^2\\
    & \dot p=\frac{d}{dt}(\gamma m \dot x)=\dot\gamma m\dot x+\gamma m \ddot x =\gamma ^3\dot x\ddot x m\dot x+\gamma m \ddot x =\gamma m\ddot x(\gamma^2\dot x^2+1) =\gamma ^3 m\ddot x\\
    & \dot p \dot x =\gamma^3 m\ddot x \dot x = m\dot\gamma =\frac{d}{dt}(\gamma m).
    \end{align*}
     
  10. Jan 4, 2016 #9
    How does #4 explains Eo = m·c²?

    Edit: See bcrowell's post for details (and no, #6 doesn't explain it either).
     
    Last edited: Jan 4, 2016
  11. Jan 4, 2016 #10

    Fredrik

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    I explained how to do the calculation described in the text you quoted, and I had previously explained what that calculation accomplishes. If that's not what you wanted explained, then you shouldn't have asked for an explanation of "that". You should have told us what you wanted explained.

    It is impossible to explain why ##mc^2## is the energy of a particle at rest, without a definition of "energy" (and "at rest"). What definition would you like to use?
     
  12. Jan 4, 2016 #11
    Just read the title of this thread.

    The same as Einstein used for his derivation of Eo=mc².
     
  13. Jan 4, 2016 #12

    Fredrik

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    If you quote a post and then ask for an explanation of "that", then you're asking for an explanation of what was said in that quote, not for an explanation of what was said in the thread title. That's how quotes work. The whole point of them is to show the reader what you're responding to.

    If you want to discuss that definition here, you're going to have to post it here along with reference (preferably a link to a page where you can download the article). You may have to provide other definitions as well, but this is the minimum requirement to get a discussion started.

    And @bcrowell, if you have something to say, you should say it. Why are you "liking" these posts?
     
  14. Jan 4, 2016 #13
    And I'm still waiting for the explanation how

    results in

    PS: You do not need to answer unless you are going to provide this explanation. I will ignore everything else.
     
  15. Jan 4, 2016 #14

    stevendaryl

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    I'm not sure if it's more or less intuitive than the mathematical proof that assumes energy and momentum are part of a Lorentz 4-vector, but you can give a hand-wavy argument for the relativistic form of momentum and energy by considering collisions, conservation in multiple frames, and the nonrelativistic limit (where the forms for momentum and kinetic energy should reduce to the Newtonian expressions).

    You can first establish that spatial momentum is given by [itex]\vec{p} = m \gamma \vec{v}[/itex] (where [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]) It is basically the only choice that reduces to Newtonian momentum in the nonrelativistic limit and which, if it is conserved in one frame, then it is conserved in every frame.

    Then, you can try the guess that the energy of a free particle has the form [itex]E(v) = g(v) m[/itex] for some unknown function [itex]g(v)[/itex], which depends on the magnitude of velocity (but not the direction). Now, consider a completely inelastic collision in which two wads of "gum" of mass [itex]m[/itex] that collide and form a larger wad of mass [itex]M[/itex].

    Let [itex]F[/itex] be the center-of-mass frame. In this frame, the initial velocity of one wad is [itex]+v[/itex] (let it be in the x-direction), and the initial velocity of the other wad is [itex]-v[/itex]. Afterward, the larger mass [itex]M[/itex] that results is stationary. Conservation of energy gives:

    [itex]2 g(v) m = M g(0)[/itex]

    Let [itex]F'[/itex] be a frame that is moving at speed -u in the x-direction, relative to [itex]F[/itex]. Assume that [itex]u \ll v[/itex]. In this new frame, one wad is traveling at speed [itex]v' \gt v[/itex], and has momentum [itex]p' = + m \gamma' v'[/itex]. The other wad is traveling at speed [itex]v'' \lt v[/itex], and has momentum [itex]p'' = - m \gamma'' v''[/itex], where [itex]\gamma' = \frac{1}{\sqrt{1 - \frac{v'^2}{c^2}}}[/itex] and [itex]\gamma'' = \frac{1}{\sqrt{1 - \frac{v''^2}{c^2}}}[/itex]

    Now, using the relativistic velocity addition formula and some slightly tedious algebra, you can show, to first order in [itex]u[/itex], that:
    [itex]p' \approx \gamma m v + \gamma m u[/itex]
    [itex]p'' \approx - (\gamma m v - \gamma m u)[/itex]

    So the total momentum in this frame, before the collision is:
    [itex]p_{total} = 2 \gamma m u[/itex]

    After the collision, the big wad of mass [itex]M[/itex] is just moving at speed [itex]u[/itex], which we assumed was very small, so we can use a nonrelativistic approximation: [itex]p = M u[/itex].

    So conservation of momentum in this frame gives:
    [itex]2 \gamma m u = M u \Rightarrow M = 2 \gamma m[/itex]

    Now, plugging this back into our equation for conservation of energy in the original frame gives:

    [itex]2 m g(v) = M g(0) \Rightarrow 2 m g(v) = 2 \gamma m g(0)[/itex]

    So we conclude that [itex]g(v) = \gamma g(0)[/itex]

    At this point, we compute the form of kinetic energy for low velocity:

    [itex]K(v) = E(v) - E(0) = m g(v) - mg(0) = m \gamma g(0) - m g(0)[/itex]

    We can use a Taylor series to write [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \approx 1 + \frac{1}{2} \frac{v^2}{c^2} +[/itex] terms of order [itex]v^4/c^4[/itex]. So to lowest order,

    [itex]K(v) \approx m g(0) \frac{1}{2} \frac{v^2}{c^2}[/itex]

    Comparing with the Newtonian limit, [itex]K(v) = \frac{1}{2} m v^2[/itex], tells us that [itex]g(0) = c^2[/itex]

    So that tells us the complete form for relativistic energy is [itex]E(v) = m g(v) = m \gamma g(0) = m \gamma c^2[/itex]

    Note that the assumption that energy depends only on mass and velocity is kind of bold. In classical physics, that's only true if you ignore "internal energy" such as heat.
     

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  16. Jan 5, 2016 #15

    Fredrik

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    If you're not going to post the definitions you want to use (this is the absolute minimum that would be required from any poster), you're not going to get a better reply than you already got from me.

    Don't put something that someone else said in a quote with my name on it.
     
    Last edited: Jan 5, 2016
  17. Jan 5, 2016 #16
    Let [itex]F[/itex] be the center-of-mass frame. In this frame, the initial velocity of one wad is [itex]+v[/itex] (let it be in the x-direction), and the initial velocity of the other wad is [itex]-v[/itex]. Afterward, the larger mass [itex]M[/itex] that results is stationary. Conservation of energy gives:

    [itex]2 g(v) m = M g(0)[/itex]


    Wait a moment, what happens if i insist that g(0) = 0 ?
    Then this equation does not make any sense, unless g(v) = 0 which is trivial solution.
    Should there be then some additional energy after the collision in the forms of photons, some other particles, potential energy etc such that:
    [itex] E_{something else} = 2 g(v) m [/itex]
    and total momentum:
    [itex] p_{something else} = 0 [/itex]
    to make the larger mass stationary? or is it just impossible possibility that g(0) = 0 ?

    I don't think your reasoning in #1 is correct. Your integral is a definite integral, so any constant term cancels out. There is also a more subtle issue, which is that one needs to provide some justification for the assumption that W=∫Fdx" style="position: relative;" tabindex="0" id="MathJax-Element-9-Frame" class="MathJax">W=∫FdxW=∫Fdx holds without modification in relativity. In Einstein's first 1905 paper on relativity, he does this in section 10, using an intricate argument that starts from the fact that for a charged particle that is instantaneously at rest, ma=qE" style="position: relative;" tabindex="0" id="MathJax-Element-10-Frame" class="MathJax">ma=qEma=qE.

    Ok thanks:) Yes it is definite integral so there may be a constant term .

    There is one other way to calculate the work integral, which is definite integral. But it is just a mathematical trick, nothing else:
    [itex] W = \int F dx = \int p dv = p_{end}v_{end} - \int v dp = p_{end}v_{end} - (1 - sqrt{ 1 - (v_{end}/c)^{2}}) [/itex]
    which suggest that:
    [itex] g(v) = pv/m + A = \gamma v^{2} + A [/itex]
     
    Last edited: Jan 5, 2016
  18. Jan 5, 2016 #17

    stevendaryl

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    That doesn't sound consistent to me. The left-hand side is nonzero.

    The three equations that I have for this collision are:
    1. [itex]2 g(v) m = M g(0)[/itex] (from energy conservation in the COM frame)
    2. [itex]2 \gamma m = M[/itex] (from momentum conservation in another frame)
    3. [itex]m (g(v) - g(0)) = \frac{1}{2} v^2 + [/itex] higher-order terms (from nonrelativistic limit)
    That's why I conclude [itex]g(0) \neq 0[/itex].

    As I said in the post, the argument assumes that energy is proportional to mass, and works out the proportionality. That's actually a pretty strong assumption, because in nonrelativistic physics, it's only true for particles with no internal energy.

    On the other hand, it's possible it's just a matter of definition. If you want to allow internal energy, then that internal energy will be associated with a "pseudo-mass", that will also have momentum.
     
  19. Jan 5, 2016 #18

    stevendaryl

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    I understand that you have to define energy in order to prove anything about it. But you can look at it from the point of view of collisions: If momentum is conserved in every frame, then so is the quantity [itex]E_{total} = \sum_j m_j \gamma_j c^2[/itex]. Whether that's the energy or not is a matter of definition, but it's certainly an interest quantity by the fact that it's conserved.
     
  20. Jan 5, 2016 #19
    This is an interesting derivation, but the basic assumption is problematic. In classical mechanics it results in an velocity-independent energy which is obviously wrong. Could you show that E:=g(v)·m is allowed in special relativity if we wouldn't know that the result is correct?
     
  21. Jan 5, 2016 #20
    And the question is, how do you get this term? I still do not understand how your calculation results in Eo=m·c².
     
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