Force in Special Relativity

Question

I have a question regarding how force on a body works in the framework of special relativity. As far as I am aware, the equation for force in special relativity is:
$$F=m\alpha$$ Where $\alpha$ is the proper acceleration. Would this imply that, if the proper acceleration is constant, the force on the body is always constant? For example, if the proper acceleration was a constant $\mathrm{1\ ms^{-2}}$ and the rest mass was 10 kg, the force would be a constant 10N, regardless of the velocity?

Answer 1

You're right, acceleration has to be redefined in special relativity. In Newtonian mechanics, the definitions $F=m\ddot{x}$ and $F=\dot{p}$ are equivalent, where $p$ is momentum. However, in special relativity, the quantity $m\dot{x}$ is not all that special, and instead what becomes important is $p=m\gamma\dot{x}$ (where here, $m$ is always taken to be the actual mass. The rest mass.)

So, once that redefinition of $p$ occurs, the above two definitions of force are no longer equivalent! The more important one turns out to be $F=\dot{p}$.

Now, with $x$ being a vector, you might ask, "wait a second, expanding $F=\dot{p}$ I get $F=m(\dot{\gamma}\dot{x}+\gamma \ddot{x})$, isn't that cumbersome? Especially, in certain problems, with the added complication of transforming $(ct,x)$ vectors between different reference frames?" You'd be absolutely right! My relatively uninformed opinion on this is that that is what tensors, proper time, and the development of intuition [by working lots of practice problems] is for. But, it's worthwhile; Most notably it leads to profound insights into magnetism.

Answer 2

Force is not really a very useful concept in relativity, but anyway this WP article has a brief treatment. We have the force four-vector

$$ \textbf{F}=\frac{d\textbf{p}}{d\tau} \qquad ,$$

where the derivative is with respect to proper time. This is the same as the OP's $F=m\alpha$. The reason it has to be with respect to proper time is that if we want the result to be a four-vector, we have to divide the four-vector $d\textbf{p}$ by a Lorentz scalar.

Because the derivative is with respect to proper time, the spacelike part of this force four-vector is the force that would be measured by an inertial observer who was instantaneously comoving with the object being acted on. It's not the three-force $\textbf{f}=d\textbf{p}/dt$ that would be measured by an observer in whatever frame we're using to describe all these vectors. The spacelike part of $\textbf{F}$ differs from $\textbf{f}$ by a factor of $\gamma$,

$$\textbf{F}=(\ldots,\gamma \textbf{f}) \qquad .$$

So say passenger Amy is in a spaceship that has a constant proper acceleration of $g$. She's standing on the deck. A second passenger Bob jumps off a table or something so he's inertial but instantaneously comoving with Amy. Because Bob is instantaneously comoving, his proper time is the same as Amy's, he sees no relativistic corrections, and he judges the force of the deckplates acting on Amy's feet to be just $F=mg$. It's constant for as long as the ship maintains constant proper acceleration.

However, observer Carl back on earth sees the ship moving at relativistic speeds. In his coordinates, $f=F/\gamma=mg/\gamma$ varies

Would this imply that, if the proper acceleration is constant, the force on the body is always constant?

So the answer depends on the frame of reference. In the comoving frame, the force is constant. In a fixed inertial frame, it's changing.