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If a constant force $F$ acts on a particle of rest-mass $m_0$, starting from rest at $t=0$, then what is its total momentum $p$ as a function of time? What is the corresponding energy $E$ as a function of time?

So I know $p=\gamma mu$ and $E=\gamma mc^2$

I know that $t'=\gamma (t-(v/c^2)x)$

I rearranged to get $\gamma$ by itself and setting $t=0$ I get $\gamma = t'/(-(v/c^2)x)$

My new equations are $p=t'mu/((-u/c^2)x)$ and $E=t'mc^2/(-1/c)x$

Are these new equations correct? I'm hesitant about this as no part of this equation mentions point of reference, but I couldn't find any other way to relate momentum and energy to time.

coconut
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ChillingK
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2 Answers2

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It depends on what you mean by a constant force.

If this means that the accelerating observer feels a constant force, i.e. a constant acceleration $a=F/m$, then this is the relativistic rocket problem. As discussed in this question the velocity measured by a non-accelerating observer is given by:

$$ v = \frac{at}{\sqrt{1 + (at/c)^2}} $$

The momentum is then simply given by:

$$ p = \gamma m v $$

Alternatively if you mean that the force is constant in the non-accelerating observer's frame then this means that $dp/dt$ is constant for the non-accelerating observer so we end up with the boring result that the momentum is just proportional to time. Of more interest in this situation would be the velocity as a function of time, which is obtained by solving:

$$ \frac{d}{dt}\left(\gamma m v\right) = F $$

for the constant force $F$. Offhand I don't know if this has a closed form solution.

Community
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John Rennie
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    The second equation does actually have a closed-form solution. You obtain $v/\sqrt{1 - v^2/c^2} = Ft/m + C$; this can then be rearranged to get $v$ as a function of $t$, which is a separable ODE that can be integrated to obtain $x(t).$ – Michael Seifert Dec 08 '16 at 14:17
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Your confusion is understandable. Most courses (the ones I've seen anyway) dealing with special relativity spend so much time on the Lorentz Transformation that they don't really get to much else. It was only when I took the second semester of the Electricity and Magnetism grad course that I started to feel like I had the beginning of a grasp of special relativity.

That being said, if we take the equation from the previous answer

$v=\frac{at}{\sqrt{1+(\frac{at}{c})^2}}$

It's not too hard to find gamma.

$\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$

$\frac{1}{\gamma^2}=1-(\frac{v}{c})^2$

From the first equation we get

($\frac{v}{c})^2=\frac{(\frac{at}{c})^2}{1+(\frac{at}{c})^2}$

Finish the work and you'll have $\gamma$.

For energy, we have $E=\gamma m c^2$

David Elm
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