I've seen force defined in special relativity as the rate of change of 4-momentum

$$ {\bf{F}} = \frac{d {\bf{p}}}{dt} $$

Can anyone comment on the following derivation of that relation?

Take one dimension of space. If I'm moving with 4-velocity ${\bf U}(t) = \frac{d{\bf x}}{d \tau}$, then I'd experience an acceleration of $\frac{d{\bf U}}{d \tau}$. (Quick clarification: since ${\bf U}$ has constant norm, it will be orthogonal to its derivative, so ${\bf U} \cdot \frac{d{\bf U}}{d \tau} = 0$. And since in my momentarily co-moving reference frame (MCRF), ${\bf U}$ is entirely in the time direction, my acceleration, $\frac{d{\bf U}}{d \tau}$, will be entirely in the space direction.) So, in my MCRF,

$$ \frac{d{\bf U}}{d \tau} = a \left(\begin{array}{c}0\\1\\\end{array}\right) = a \cdot \vec{{\bf e}}_x $$

Here's the step I'm unsure about: would it be correct to equate the acceleration I feel, $a$, with the force my rocket engine applies on me, divided by my mass, $F/m$? That would give us

$$ F \vec{{\bf e}}_x = m \frac{d{\bf U}}{d \tau} $$

Generalizing to three spacial dimensions, you'd get

$$ {\bf F} := F_x \vec{{\bf e}}_x + F_y \vec{{\bf e}}_y + F_z \vec{{\bf e}}_z = m \frac{d{\bf U}}{d \tau} $$

Finally, in my MCRF, $d \tau = dt$, so you'd get the original force law. Is this a correct way to derive the force law in special relativity?


1 Answer 1


No, isn't correct, because saying that the acceleration $a$ you feel is $F \over m$ implies that you're using the law of motion: $$ F = ma $$ Which is valid in classical mechanics, not in special relativity. The correct relation con be derived formally by a least action principle

In special relativity, the law we deduce has to be invariant under Lorentz transformation. In other words, the law must have the same form in every inertial reference frame who agrees that light speed is c. The problem rise by the fact, that the free particle Lagrangian in Classical Mechanics: $$ L = \frac{1}{2} m v^2 $$ Is not invariant under Lorentz transformation, so, if you use the Euler-Lagrange equations for determine the law of motion (which is F=ma), this would be incorrect. Instead you should ''build'' a Lagrangian which is Lorent-invariant, for instance:

$$ L = -mc^2\sqrt{1 - \frac{v^2}{c^2}} $$

This is Lorentz invariant, so, if you now use the Euler-Lagrange equations (which I recall are derived by a least action principle) you will obtain the correct equation:

$$ \vec{F} = \frac{d}{dt} (m \gamma_u \vec{v}) $$

  • $\begingroup$ It would be helpful if you could elaborate on the correct relation. For instance, what is the definition of force you are now (implicitly) using when you say they are not equal? $\endgroup$
    – Abhinav
    Apr 20, 2016 at 3:37
  • $\begingroup$ Sorry, I've edited the answer $\endgroup$ Apr 20, 2016 at 7:17

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