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Why E^2=(mc^2)^2+(hv)^2?

  1. Nov 13, 2015 #1
    I understand that E=mc^2 and E=hv can't be used to set mc^2 equal to hv, but why would the total equation be E=(mc^2)^2+(hv)^2 instead of E=mc^2+hv? I'm sorry if this question is stupid.
     
  2. jcsd
  3. Nov 13, 2015 #2
    Why is the Pythagorean theorem ##c^2 = a^2 + b^2## instead of ##c = a + b##? That's just how space works.

    ##E^2 = (mc^2)^2 + (pc)^2## is just an application of the Pythagorean theorem in spacetime.
     
  4. Nov 13, 2015 #3
    Ok, thanks. By the way, what is the difference between pc and hv? I see them used interchangeably.
     
  5. Nov 13, 2015 #4
    Ok, I gave a misleading answer and I think I need to address it. Space-time has a different metric than space, and Pythagorean theorem works differently. In familiar 3D, the metric is ##dr^2 = dx^2 + dy^2 + dz^2## so Pythagorean theorem is as above.
    But in 4D space time, time has an opposite sign to space, so the corresponding metric is ##ds^2 = -dt^2 + dx^2 + dy^2 + dz^2## or ##ds^2 = dt^2 - dx^2 - dy^2 - dz^2## depending on the convention you adopt.

    So you can rewrite ##E^2 = (mc^2)^2 + (pc)^2## as
    ##(mc^2)^2 = E^2 - (pc)^2##
    ##(mc^2)^2 = E^2 - (p_xc)^2 - (p_yc)^2 - (p_zc)^2##

    The invariant mass "m" is the 4D length of the energy and momentum 4-vector. But length is calculated using something like the Pythagorean theorem but with negative sign in front of the momentum. This is a result of the negative sign in the metric. This is clearer if you drop all the c constants from the equation.
    ##m^2 = E^2 - p_x^2 - p_y^2 - p_z^2##
    You can drop c if you measure length and time in compatible units. If you measure time in seconds, you should measure length in light-seconds, and then you can set c = 1 light-second/second and then make it go away.
     
  6. Nov 13, 2015 #5
    You shouldn't be using ##h\nu## in this equation. It isn't correct to interchange ##h\nu## with ##pc## except for massless particles.
     
  7. Nov 13, 2015 #6
    Thank you.
     
  8. Nov 13, 2015 #7

    jtbell

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    Staff: Mentor

    Who says it looks like that?
     
  9. Nov 13, 2015 #8
    My bad, I meant E^2
     
  10. Nov 13, 2015 #9
    E=hv also for a massive particle.
     
  11. Nov 13, 2015 #10
    What is "v" here?
     
  12. Nov 13, 2015 #11

    jtbell

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    Staff: Mentor

    I was referring to the (hv)^2 part, not the E on the left which should indeed be E^2.

    I'm pretty sure it's supposed to mean "nu" (##\nu##) for frequency, not "v" for velocity.
     
  13. Nov 14, 2015 #12
    ##\nu##
     
  14. Nov 16, 2015 #13
    What do you mean?
     
  15. Nov 16, 2015 #14

    jtbell

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    Staff: Mentor

    A particle's wave function has a time-dependent part whose frequency is given by ##\nu = E/h##.

    For example, a free particle has ##\Psi(x,t) = Ae^{i(px-Et)/\hbar} = Ae^{2 \pi i(px-Et)/h}## so the time dependent part is ##e^{-2 \pi i Et / h}##. One form of an oscillator in complex-number space is ##e^{-i \omega t} = e^{-2 \pi i \nu t}##. Compare the two, and you get ##\nu = E/h##.
     
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