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Is this a legal derivation?

  1. Nov 16, 2012 #1
    Hey guys,

    I was trying to reverse engineer Einstein's formula for energy, E=γmc^2 by re-engineering Newton's Law of motion, F=ma. I was talking with my physics prof about deriving energy from this because I got two different answers but it gets weird because the incorrectly derived formula works.

    F = ma = dp/dt -> F dx = mv dv -> E = ∫ F dx = ∫ mv dv = .5mv^2 + C

    Then I did this

    F = dp/dt = v dp (dx/dx) -> F dx = v dp -> E = ∫ v dp = vp + C

    My prof told me that my last integral, ∫ v dp, is an illegal operation and that v must be converted into p/m which makes sense because it then follows that E = .5mv^2 = p^2/2m.

    I did some fiddling around though because I was curious and I was able to derive E = γmc^2 and the formula always works. What I derived from the above was:

    E = vp + C = vp + mc^2/γ, p=γmv

    I'm just curious if anyone can point out why it works.

    Also, I know that energy for a photon is equal to |p|c. When m=0 then v=c and I find it interesting that the rest mass, m/γ, is introduced above given energy equivalence. So, bad math or is there something to this?
     
  2. jcsd
  3. Nov 17, 2012 #2

    robphy

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    Your professor is correct in that "∫ v dp = vp + C " is incorrect because v is not a constant... it is a function of p... namely (p/m) as you were told.

    There are some unclear/inconsistent variable uses here:

    " F dx = v dp -> E = ∫ v dp "
    implies that E is the [relativistic] kinetic energy (via the work-energy theorem)

    " E = γmc^2 "
    implies that E is the relativistic energy and m is the rest-mass

    but then you say
    "find it interesting that the rest mass, m/γ,"
    then m in this sentence must be the so-called "relativistic mass"

    Now, when you say "E = vp + C = vp + mc^2/γ, p=γmv"
    then, making the substitution for p=γmv [where m must be the rest mass],
    one gets
    E = vp + C = v(γmv) + mc^2/γ = m (γv^2+ c^2/γ)
    where the m is factored out to unravel the expression.... which is not recognizable as anything meaningful.


    So, I think you have to go back and fiddle around some more... but be consistent in the meaning of your variables and don't do any illegal mathematical operations.
     
  4. Nov 17, 2012 #3
    Thanks for the reply.

    I still can't figure out why it works. At first thought that since dv = 0 the first integral is legal but that makes F = 0... So I still have no idea.

    I also messed up in my original response, M/γ is actually the rest mass divided by gamma, not the formulation for rest mass.

    I went a little further with it and also found a formula for kinetic energy using this:

    Ek = PV(γ + 1) + Mc^2 (γ^-1 - γ^2) = γmc^2 - mc^2, P=γmc^2

    Again, the math works but is it significant to anything or just a silly way of saying (gamma)mc^2 - mc^2?
     
    Last edited: Nov 17, 2012
  5. Nov 17, 2012 #4
    After reading that over that is exactly what I got but by my calculations:

    E = vp + C = v(γmv) + mc^2/γ = m (γv^2+ c^2/γ) = γmc^2 when I tried plugging in a few values assuming a=0.
     
  6. Nov 17, 2012 #5
    E.g.:

    Particle with a mass of 1000 MeV/c^2 is traveling at 0.6c -> γ = 1.25

    E = 1.25 * 1000 MeV/c^2 * c^2 = 1250 MeV

    Alternatively:

    E = 1000 MeV/c^2 (1.25 * (.6c)^2 + c^2 / (1.25)) = 1000 MeV/c^2 (.45 c^2 + 0.8 c^2) = 1000 MeV/c^2 (1.25 c^2) = 450 MeV + 800 MeV = 1250 MeV
     
  7. Nov 17, 2012 #6
    Just figured it out, I inadvertently made an expansion.

    (γc^2) = γv^2 + c^2/γ

    E = m(γc^2) = m(γv^2 + c^2/γ) = PV + mc^2/γ
     
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