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E^2=(mc^2)^2.(pc)^2 proof?

  1. Dec 21, 2012 #1
    Hello everyone,

    I understood the energy mass equation with the common box and photon example. But how do you prove E^2=(mc^2)^2.(pc)^2 ? Thanks for help.
     
  2. jcsd
  3. Dec 21, 2012 #2

    bcrowell

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    Welcome to PF!

    You'll get better responses to your posts if you mark up the math properly using LaTeX. I've done that here for your equation: [itex]e^2=(mc^2)^2+(pc)^2[/itex]. To see how I did that, click on the
     
    Last edited: Dec 21, 2012
  4. Dec 21, 2012 #3

    PeterDonis

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    Did you mean this formula?

    [tex]E^2 = \left( m c^2 \right)^2 + \left( p c \right)^2[/tex]

    Note the plus sign. For a simple proof, consider what this looks like in the object's rest frame, where [itex]p = 0[/itex]:

    [tex]E^2 = \left( m c^2 \right)^2[/tex]

    which of course is the "energy mass equation" you refer to. Now the [itex]m[/itex] in this equation is an invariant; it's the object's rest mass. So if we transform to any other frame, we must still have the same value of [itex]m[/itex], even though [itex]E[/itex] changes. In a frame where the object is moving at velocity [itex]v[/itex], we have

    [tex]E = \gamma m c^2[/tex]

    where [itex]\gamma = 1 / \sqrt{1 - v^2 / c^2}[/itex]. If we note that the momentum [itex]p[/itex] is given by

    [tex]p = \gamma m v[/tex]

    we can see that

    [tex]E^2 - \left( p c \right)^2 = \gamma^2 \left( m c^2 \right)^2 - \gamma^2 \left( m v c \right)^2 = \gamma^2 \left( m c^2 \right)^2 \left( 1 - \frac{v^2}{c^2} \right) = \left( m c^2 \right)^2[/tex]

    which rearranges to the general formula I gave at the start of this post. In other words, the formula just expresses how the invariance of rest mass is maintained, by the energy and momentum changing in concert as you change frames.
     
  5. Dec 21, 2012 #4
    You've got your signs mixed up!
     
  6. Dec 21, 2012 #5

    bcrowell

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    Thanks!

    Re PeterDonis's #3, this works if you already believe in [itex]E=m\gamma[/itex] and [itex]p=m\gamma c[/itex], but that does lead to the question of how to prove those things, and it doesn't cover m=0 unless you explicitly appeal to taking the limit. Really what's going on here is that [itex]m^2=E^2-p^2[/itex] is more fundamental than [itex]E=m\gamma[/itex] and [itex]p=m\gamma c[/itex].
     
  7. Dec 21, 2012 #6

    PeterDonis

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    Yes, that's true. As you say, it depends on what you think is more "fundamental".

    Yes, although if you accept [itex]E = pc[/itex] for light, then that already gives you the m = 0 version. But again, that depends on what you think is fundamental.

    I agree, this is a better way to look at it.
     
  8. Dec 21, 2012 #7
    Found these two videos explained things very well (and easy to understand):



     
    Last edited by a moderator: Sep 25, 2014
  9. Dec 25, 2012 #8
    I know what is [itex]E=mc^2[/itex] is. I was asking how to prove [itex] E^2=(mc^2)^2+(pc)^2 [/itex].
     
    Last edited by a moderator: Sep 25, 2014
  10. Dec 25, 2012 #9

    PeterDonis

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    Well, have you read the posts we've made giving possible proofs? Do you have questions about them? Issues with them?

    Also, as bcrowell pointed out, what counts as a "proof" depends on what you accept as already proven. If you accept [itex]E^2 - \left( p c \right)^2 = \left( m c^2 \right)^2[/itex] as already proven, then the proof of [itex]E^2=(mc^2)^2+(pc)^2 [/itex] is a one-liner. If you accept the formulas for [itex]E[/itex] and [itex]p[/itex] in terms of [itex]\gamma[/itex], [itex]m[/itex], and [itex]v[/itex] that I gave, the proof isn't much longer. Where do you want the proof to start?
     
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