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I'm so confused due to how many different equations for energy are there. For example:

E = m$c^2$,

E$^2$ = (m$c^2$)$^2$ + (pc)$^2$,

E = mγc$^2$ - mc$^2$

What does each equation represent? What is the relativistic kinetic energy? Why is m and $m_0$ sometimes used interchangeably?

BarCouSeH
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    Rest energy. Total energy. Kinetic energy. (assuming $m$ is rest mass in the first equation). For more info see https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation – Kibble Sep 03 '16 at 00:15
  • What about E = mγc$^2$ ?? Is that rest energy too? If so then what's the point of γ if v=0 ?? – BarCouSeH Sep 03 '16 at 04:35
  • @Kibble What else could $m$ be? There is no other mass other than rest mass. – garyp Sep 03 '16 at 12:24

1 Answers1

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The equation for the total energy of a relativistic object is:

$$ E^2 = p^2c^2 + m^2c^4 \tag{1} $$

If we consider only massive particles, so this excludes photons, the relativistic momentum is $p=\gamma mv$ we can rewrite equation (1) as:

$$ E = \sqrt{\gamma^2m^2v^2c^2 + m^2c^4} $$

Then take out a factor of $mc^2$ to get:

$$\begin{align} E &= mc^2\sqrt{\gamma^2\frac{v^2}{c^2} + 1} \\ &= mc^2\sqrt{\frac{v^2/c^2}{1-v^2/c^2} + 1} \\ &= mc^2\sqrt{\frac{v^2/c^2 + 1-v^2/c^2}{1-v^2/c^2}} \\ &= mc^2\sqrt{\frac{1}{1-v^2/c^2}} \end{align}$$

Which is just:

$$ E = \gamma mc^2 \tag{2} $$

So for massive particles the equations (1) and (2) are equivalent. However for massless particles equation (2) is no use while equation (1) still applies. That's why most physicists will use equation (1) though you will still see equation (2) from time to time.

Finally, if the total energy is $E$ you can subtract off the rest energy $mc^2$ to give the equation for kinetic energy you have as your third equation:

$$ KE = \gamma mc^2 -mc^2 \tag{3} $$

Your first equation, the famous $E=mc^2$, is the total energy of the object in its rest frame i.e. the frame in which its velocity is zero.

John Rennie
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  • So if I was observing (as a stationary observer) an electron that is travelling at 99.997% the speed of light, how would I go about calculating it's total energy, rest energy, and kinetic energy, based on my frame of reference? Also, how would I calculate these energies from the electron's frame of reference? – BarCouSeH Sep 03 '16 at 12:34
  • @BarCouSeH: the rest energy is always $m_ec^2$, regardless of the observer. That's because the rest energy is defined as the total energy in the rest frame of the electron. In the rest frame of the electron its kinetic energy is of course zero, since the electron rest frame is the frame in which the electron isn't moving. – John Rennie Sep 03 '16 at 12:34
  • In any frame in which the electron is moving use equations (1) or (2) for the total energy (both equations give the same answer) and equation (3) for the kinetic energy. – John Rennie Sep 03 '16 at 12:35
  • You said earlier that we could use equation (1) to calculate the energy of massless particles. How would you go about doing so? – BarCouSeH Sep 03 '16 at 12:42
  • @BarCouSeH: for a photon $m=0$ and $p=h/\lambda =h\nu/c$. Jut substitute these values in equation (1). See If photons have no mass, how can they have momentum? for more on this. – John Rennie Sep 03 '16 at 13:45
  • For calculating the kinetic energy of any object in classical physics we always use E$_K$=$\frac{1}{2}$mv$^2$. So if we were to calculate the kinetic energy of a fast moving object while taking into account relativistic effects we should be able to use the same equation but with the $\gamma$ in there too, i.e. E$_K$=$\frac{1}{2}$$\gamma$mv$^2$. But this equation is not consistent with E$_K$ = γmc$^2$ - mc$^2$, so why is it incorrect? – BarCouSeH Sep 04 '16 at 09:55
  • @BarCouSeH: you say we should be able to use the same equation but with the $\gamma$ in there too but that's wrong. I'm guessing you're assuming that mass can always be replaced by $\gamma m$, but it can't. You can, with a bit of effort, show that $\tfrac{1}{2}mv^2$ is the low energy approximation for the full relativistic expression. – John Rennie Sep 04 '16 at 10:00
  • when I try to use the relativistic kinetic energy equation for small speeds the energy is zero. What is wrong here? Shouldn't the relativistic equation apply for all situations? – BarCouSeH Sep 04 '16 at 15:23
  • @BarCouSeH: the comments aren't intended to be used for discussion. You could ask a new question, or I'm in the chat room at the moment if you want to ask there. – John Rennie Sep 04 '16 at 15:34