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Derivation of E = mc^2

  1. May 28, 2008 #1

    ehj

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    The kinetic energy of an object can be shown to be:
    Ekin = m*c^2 - m0*c^2

    Where m is the relativistic mass, m0 is the rest mass and c is the speed of light.

    Is it acceptable to derive E = mc^2 from this equation by saying the kinetic energy of a photon is mc^2 because it has a rest mass of zero, and thereby making a relation between photon/light energy and mass. Or is this argumentation too weak? Does the "zero rest mass" photon actually come from the derived equation?
    If this derivation isn't that strong I would like to ask for another derivation which uses the relativistic momentum, relativistic kinetic energy and/or Lorentz-transformations etc. I would like to avoid: derivations that use the photon momentum from theories concerning electromagnetism, and derivations that use Taylor expansion since I havn't learned that yet. Is this possible :)?
     
  2. jcsd
  3. Jun 2, 2008 #2
    E=mc^2 is used in the formula you gave, i.e. its not sufficient as a proof.
    Most derivations i'm familiar with are fairly complex, but i recall being shown a surprisingly simple "derivation" a while ago... let me see if i can find it ...

    http://www.sciam.com/article.cfm?id=significance-e-mc-2-means
    you should be able to follow that, good luck
     
  4. Jun 3, 2008 #3

    Gib Z

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  5. Jun 3, 2008 #4
    The actual formula is

    E^2 = m^2.c^4 + p^2.c^2

    where p is the momentum....
    now if the object has rest mass. we use E = mc^2 as at rest, momentum will be 0.

    But for particles like electrons and photons, rest mass is 0 but they have momentum.
    So the equation for them reduces to E = pc
     
  6. Jun 3, 2008 #5

    Gib Z

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    Electrons have zero rest mass, what a profound statement...

    Even ignoring that you don't know what an electron is, the energy of a photo in given by E=hf which is not equal to the product of its momentum and the speed of light, since pc = hv, and v is not equal to f.

    Overall, even if you were correct, you helped the OP in no way at all.
     
    Last edited: Jun 3, 2008
  7. Jun 3, 2008 #6
    I made a big error... it's not electron.. it's photon only... don't laugh mister.

    E = pc
    pc gives mc^2... dimensionally same.. and this equation is correct... I'm sure you've never come across it ;)
     
  8. Jun 3, 2008 #7

    Gib Z

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    Ahh even more godly revelations. E = pc because the units are equal. Why bother even proving E=mc^2, just check the units guys! I have never come across it true, but that this equation is correct is false.
     
  9. Jun 3, 2008 #8
    I wonder what happens if we click this link, aye?

    h**p://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

    write http
    ;)
     
  10. Jun 3, 2008 #9

    Gib Z

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    And nowhere does it say pc = mc^2.
     
  11. Jun 3, 2008 #10
    I didn't equate them parametrically, I was trying to explain to you that the equation is true, and not false as you believed.
    Anyway go through it.
    have to go now bye... and by the way can we add our signatures?
     
  12. Jun 3, 2008 #11

    Gib Z

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    What kind of wish washy logic do people go through when they reason:

    I define p = mv. We have E^2 = (mc^2)^2 + (pc)^2. Now if we let m=0, we are only left with E= pc. Lets ignore the fact p = mv = 0 here, in fact lets change the definition to something else right about now.

    I see now that i was incorrect: E= pc = hf, if we take p = h/lambda. But that is still far besides the point - your first post gives no useful info on how to prove E=mc^2, as you assume a deeper result.

    EDIT: P.S - There you go : https://www.physicsforums.com/profile.php?do=editsignature
     
    Last edited: Jun 3, 2008
  13. Jun 3, 2008 #12
    Do you need to be sarcastic to prove your superior intellect or is it just in your nature?
     
  14. Jun 3, 2008 #13

    ehj

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    Gib Z the derivation you linked to uses taylor expansion, which was one of the things I would like to avoid :).
     
  15. Jun 3, 2008 #14
    As best I can tell, E = mc^2 comes from the assumption that the formula for kinetic energy (which is derived in the wikipedia article using pretty simple integral calculus):

    Ek = ymc^2 - mc^2

    represents the difference between _total_ particle energy (ymc^2) and _rest_ energy (mc^2). In other words, saying a particle's rest energy is E=mc^2 is an interpretation of the kinetic energy formula.
     
  16. Jun 3, 2008 #15

    Fredrik

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    My favorite "derivation" is the one from Weinberg's QFT book. It goes something like this: A Hilbert space of one-particle states has the property (by definition) that every vector is an eigenvector of P2, and they all have the same eigenvalue, so we'll just write this eigenvalue as -m2 and call m the mass of the particle. Done-diddly-done. (If the eigenvalue is positive, we'll write the eigenvalue as n2 instead and call n the mass. In this case the particle is called a tachyon).

    But this calculation is pretty cool too (even though the notation is kind of sloppy):

    Choose units such that c=1. The work W performed accelerating a mass m from speed 0 to v is

    [tex]W = \int F dx = \int \dot p dx=\int\frac{d}{dt}(\gamma m\dot x) \dot x dt = \int(\frac{d}{dt}(\gamma m \dot x^2)-\gamma m \dot x\ddot x)dt[/tex]

    Since [itex]\dot\gamma=\gamma^3\dot x\ddot x[/itex], we have

    [tex]\gamma m\dot x\ddot x =\frac{m}{\gamma^2}\gamma^3\dot x\ddot x=\frac{m}{\gamma^2}\dot\gamma=-\frac{d}{dt}(\frac{m}{\gamma})[/tex]

    so

    [tex]W=\int\frac{d}{dt}(\gamma m \dot x^2+\frac{m}{\gamma})dt =\gamma m v^2+\frac{m}{\gamma}-0-m =\frac{m}{\gamma}(\gamma^2 v^2+1)-m =\frac{m}{\gamma}\gamma^2-m =\gamma m -m[/tex]

    We can restore factors of c if we want:

    [tex]W=\gamma m c^2-mc^2[/tex]

    The mass m is often written as m0 by people who prefer to call it the "rest mass". They define [itex]m=\gamma m_0[/itex] and call that the "relativistic mass". This notation turns the result into

    [tex]W=mc^2-m_0c^2[/itex]
     
  17. Jun 3, 2008 #16

    Fredrik

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    We can always write [itex]p^2=-m^2[/itex], where p2 is the Minkowski space square of the four-momentum, so the equation really says that [itex]-E^2+\vec p^2=-m^2[/itex], or equivalently [itex]E^2=\vec p^2+m^2[/itex]. With c included explicitly, it's [itex]E^2=\vec p^2c^2+m^2c^4[/itex]. The m on the right-hand side is the rest mass, so if you prefer the m0 notation, you should write [itex]E^2=\vec p^2c^2+m_0^2c^4[/itex].

    Note that if we simply choose to write the left hand side as mc2, this m must have dimensions of mass, so we can do this if we take this to be the definition of "relativistic mass" m. This is one of the reasons why I find "relativistic mass" to be such a lame and useless concept.
     
  18. Jun 4, 2008 #17

    ehj

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    Hello Fredrik. I have used a similar method (although without having to do the "c=1" as you did) of deriving the relativistic kinetic energy formula - by using the force's work. The problem is not deriving the formula for kinetic energy as you did, because I've already done that. But how do you get from that, the relativistic kinetic energy, to E = mc^2 ?
    I'm not that familiar with minkowski space, four-momentum yet as I'm only in high school. :|
     
    Last edited: Jun 4, 2008
  19. Jun 4, 2008 #18

    Gib Z

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    It only uses Taylor expansion to show that the relativistic equation for kinetic energy is approximately the Newtonian expression for v << c. To get to E=mc^2 it only requires some basic integration.

    EDIT:

    Both?
     
  20. Jun 4, 2008 #19
    Here you go:
    dE=Fdx
    dE=d(mv)/dt.dx = d(mv)/dt.dx/dt.dt
    = v d(mv)
    = v^2 dm + mv.dv
    m=m0/(rt (1-v^2/c^2)) [m0=rest mass]
    =>m.(rt(1-v2/c2)) = m0
    On differentiating the above equation, we get
    c^2.dm=v^2.dm + mvdv

    Thus dE = v^2.dm + mvdv = c^2.dm
    Thus dE= c^2.dm
    On integrating,
    E=mc^2
     
  21. Jun 4, 2008 #20

    ehj

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    that derivation doesn't work spideyunlimit.
    You forget the arbitrary constant upon integration, and as you're using the work done by the force, the kinetic energy, you can't just throw the constant away.
     
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