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I tried to derive a relativistic formula for kinetic energy in a different way but my results didn't match to the actual equation and I got negative kinetic energy.Here is the method: $$d(K.E)=F.ds$$ where $$F = \frac{(mv.dv)}{ds}$$ and $$m = \frac{m_{rest}}{\sqrt{1-\frac{v^2}{c^2}}}$$ Now,$$d(K.E)=\frac{m_{rest}v}{\sqrt{1-\frac{v^2}{c^2}}}.\frac{dv}{ds}.ds$$ $$d(K.E)=\frac{m_{rest}v}{\sqrt{1-\frac{v^2}{c^2}}}{dv}$$ On integrating both sides, $$\int d(K.E) = \int_0^v\frac{m_{rest}v}{\sqrt{1-\frac{v^2}{c^2}}}{dv}$$ on itegatgrating and doing some algebra
$$K.E={m_{rest}}c^2{\sqrt{1-\frac{v^2}{c^2}}}-m_{rest}c^2$$ The above expression shows the K.E energy to be negative But this can't be possible. Please tell me where I am wrong

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Your expression for force is incorrect. It is more properly written as $$F = \frac{dp}{dt} = \frac{d(\gamma m v)}{dt} = m\left(v\frac{d\gamma}{dt} + \gamma\frac{dv}{dt}\right)$$ where $p$ is momentum, $m$ is the rest mass[1] of the object, and $\gamma$ is the expression $1/\sqrt{1-v^2/c^2}$.

$$dKE = Fds = \frac{dp}{dt}ds = dp\frac{ds}{dt} = m\left(vd\gamma + \gamma dv\right)v = m\left(v^2d\gamma + \gamma vdv\right)$$

Integrating yields $$KE = m\left(\int v^2d\gamma + \int \gamma vdv\right)$$ which will result in the correct expression for kinetic energy. $$KE = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$$ Your derivation is missing the $d\gamma$ integral.


[1] Very few physicists use $m = \gamma m_{rest}$ anymore because of mistakes like this. They just say that $m$ is the invariant mass, which is the same as the rest mass.

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  • $\begingroup$ Thanks for the answer.But I am still unable to understand the logical reason behind why can't we use $ m=\Gamma m_{rest}$ for the above derivation. $\endgroup$ Commented May 12, 2017 at 9:47
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    $\begingroup$ @Nancy Relativistic mass is not illogical. It can be used consistently with difficulty. Physicists (including Einstein) just do not find the concept useful. Momentum takes on the Newtonian form when you use relativistic mass, but that doesn't work for any other formula. Relativistic kinetic energy is not $K = (1/2)\gamma mv^2$. $\endgroup$
    – Mark H
    Commented May 12, 2017 at 11:15

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