In galilean relativity $$p=mv$$ and $$KE=\frac{1}{2}mv^2$$

If I understand it in special relativity the equation for momentum is $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$ In galilean relativity there is the equation $$\frac{p}{m}=v$$ while in special relativity the equation is $$\frac{p}{m}=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}$$ In galilean relativity $$KE=\frac{1}{2}m\frac{{p}^2}{{m}^2}$$ and if this equation holds in special relativity it produces the equation $$KE=\frac{1}{2}m\frac{v^2}{1-\frac{v^2}{c^2}}$$ Is this a correct way to express kinetic energy in special relativity?

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    $\begingroup$ No, because like all the previous equations, $E=p^2/2m$ is not true in special relativity. $\endgroup$ – Javier Mar 4 '16 at 22:28
  • $\begingroup$ The correct relationship between energy and momentum is $p = Ev/c^2$. You then have to subtract off $mc^2$ and Taylor expand to get kinetic energy. $\endgroup$ – knzhou Mar 4 '16 at 23:17

It is useful to understand how the expression for kinetic energy is arrived at in non relativistic mechanics and apply that for the relativistic case. Let's say we have a body at stand still and apply a force $F = \frac{d(mv)}{dt}$ to get it moving. The energy gained by the body is then

$ E_k = \int_0^t Fdx = \int_0^t \frac{d(mv)}{dt} dx = \int_0^tmv dv = \frac{1}{2} m v^2 $

Doing the same in relativistic mechanics you need to take into account the body gets heavier (or more "inertial") by a factor $\gamma = (1-\frac{v^2}{c^2})^{-1/2}$ with $m$ now its rest mass and the integral works out differently. Calculating the work done we will find:

$ E_k = \int_0^t Fdx =\int_0^t \frac{d(\gamma mv)}{dt} dx = \int_0^tv d(m\gamma v) $

This integral is slightly more involved but will result in $E_k = (\gamma -1 ) mc^2$. Physically a lot more work has to be done to accelerate the body as it gets heavier.

Note that in the classical case you can write $E_k = \int_0^t \frac{p}{m} dp = \frac{p^2}{2m}$ but not anymore in relativistic mechanics as you can see from the integral expression in the above which becomes

$\int_0^tv d(m\gamma v) = \int_0^t\frac{p}{\gamma m} dp$.

So your expression for $E_k$ in terms of momentum is incorrect.

  • $\begingroup$ Thank's! I was also wondering what would be the equation for kinetic energy if $\gamma = (1-\frac{v^2}{c^2})^{-1/2}$ was replaced with $\gamma = \frac{v^2}{c^2}+1$? $\endgroup$ – Anders Gustafson Mar 5 '16 at 5:43
  • $\begingroup$ I found $E_k = \frac{1}{2} m v^2 (1 + \frac{3v^2}{4c^2})$ in that case but why are you interested in that? Did you mean $\gamma = 1 + \frac{v^2}{2c^2}$? That would be the correct $\gamma$ expanded to first order in $\frac{v^2}{c^2}$. $\endgroup$ – Jan Bos Mar 6 '16 at 12:22

Kinetic energy in Special Relativity is $KE=(\gamma -1)mc^2$, where $\gamma $ is the Lorentz factor. This is derived from the relativistic equation for total energy, $E^2 = (pc)^2 + ((mc)^2)^2$; the negative term removes the rest energy from the total energy.

The Galilean kinetic energy is the limit of this expression for small velocities.

  • $\begingroup$ This should be a comment, an answer should explain why that is correct and why OP's expression is not. $\endgroup$ – Ryan Unger Mar 4 '16 at 23:11

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