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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?

Qmechanic
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Anders Gustafson
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3 Answers3

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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.

Jan Bos
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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.

Peter Diehr
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The equation for KE does NOT come from integrating force through some distance/time/velocity. You don't even need calculus. Of course that does give you the correct equation, it's more just a fun quick way to get the equation.

It's all from the true energy-mass equivalence. This works regardless of the speed you have. The approximation of it, is the standard KE used for non-relativistic stuff (as expected, since newtonian mechanics is an approximation)

Namely,
$E^{2} = (mc^{2})^{2} + (pc)^{2}$

That is the energy-mass equivalence, notice that if momentum is zero, you get the famous $E=mc^2$ (rest energy)

Then to get KE, you just need to subtract resting energy.
So,

$KE = \sqrt{(mc^{2})^{2} + (pc)^{2}} - mc^2 = mc^2(\sqrt{1+(\frac{p}{mc})^2}-1)$

Like that. If you taylor expand that square root to approximate it, you get $KE=\frac{1}{2}mv^2$. And if you sub in $p=\gamma mv$ for speeds close to c, you'll get $KE=(\gamma -1)mc^2$ after some simplification.

Vynneve H
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