This question pertains specifically to the case of a constant pure 4-force in special relativity, of the form $\textbf{F} = (f_0, f_x, 0, 0)$, where $f_0 = \frac{γ}c \frac{dE}{dt}$. I understand how to solve the case of a constant 3-force in special relativity, but as far as I can tell one cannot use the usual trick of $p(t) = p(0) + f_xt$ since the 3-force is time dependent.

To find a solution, would you need to use the pure force result $\textbf(u \cdot f) = \frac{dE}{dt}$? Any guidance would be helpful!

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    $\begingroup$ Constant four force means constant proper acceleration i.e. the solution is the Rindler coordinates. $\endgroup$ – John Rennie Jun 15 at 11:30
  • $\begingroup$ Forgive my ignorance - how is this different from the constant 3 force case? I thought a constant 3 force led to a hyperbolic solution? $\endgroup$ – alex1stef2 Jun 15 at 11:39
  • $\begingroup$ Take a look at this en.wikipedia.org/wiki/Hyperbolic_motion_(relativity) $\endgroup$ – Run like hell Jun 15 at 12:24
  • $\begingroup$ @JohnRennie your reply is sort of partly right, but more wrong than right IMO; see my answer below. $\endgroup$ – Andrew Steane Jun 15 at 14:00
  • $\begingroup$ @AndrewSteane yes, I agree. I made exactly the mistake you mention in your answer. $\endgroup$ – John Rennie Jun 15 at 16:47

The confusion here is that many very able physicists have picked up the idea that motion at constant proper acceleration is somehow motion at "constant 4-acceleration" and thus "constant 4-force", but this is simply not true.

For rectilinear motion at constant proper acceleration $a_0 = A^\mu A_\mu$, the invariant size of the 4-acceleration is fixed and so is the direction in space, but the direction in spacetime is not. The direction in spacetime is orthogonal to the 4-velocity and the 4-velocity is certainly changing, and so is the 4-acceleration.

So the problem here is really about clarity and precision of expression. Many people say "constant 4-force" or "constant 4-acceleration" when what they mean is "constant size of 4-force" or "constant size of 4-acceleration".

Compare it to another case: motion in a circle at constant speed in Newtonian physics. Would we say of such motion that the acceleration is constant? We probably would, but we are familiar enough with this case that we don't confuse ourselves: we know the acceleration vector is changing, but its size is constant. Would we say the velocity is constant? No we wouldn't. Would we say the force is constant? We might, or we might not. Strictly, as a vector quantity, it is not constant.

So now let's come back to special relativity. If someone says "calculate the case of motion under a constant 4-force" then really the careful student has little option but to take the statement at face value and try to solve for that case. The solution is difficult and such a force certainly is not pure, and has very little relevance to physics. If the question was "calculate the case of motion under a 4-force of fixed spatial direction and constant invariant size" then the student (whether careful or not) can breath a quick sigh of relief and get on and tackle that standard problem. But if the question said the first while intending to mean the second, then it was an ill-posed question.

(An example of this came up recently in an Oxford physics exam; we need to improve our procedures to catch such things.)


The four-force is the rate of change of four-momentum in respect to the proper time, that is $$ {\bf F} = \frac{d{\rm P}}{d\tau}$$ If it was constant that means that $$ {\bf P}(\tau) = {\bf P}_0 + {\bf F}\tau$$ However that would mean that the mass $$m^2(\tau) = {\bf P}(\tau)\cdot {\bf P}(\tau) = m_0^2 + {\bf P}_0\cdot{\bf F}\tau + {\bf F}\cdot{\bf F}\tau^2 $$ is in general time-dependent. A particle with constant mass can have constant four-acceleration only if ${\bf P}_0\cdot{\bf F}= {\bf F}\cdot{\bf F}$. Assuming $m_0>0$ you can however choose a frame in which ${\bf P}_0=(m_0,0,0,0)$, ${\bf F}=(F_0,\vec{F})$ you have $$ F_0 m_0 = 0 = F_0^2 - \vec{F}\cdot\vec{F}$$ and that means that $F_0=0$, $\vec{F}=\vec 0$, ${\bf F}=0$.

In conclusion, you cannot act with a constant four-force on a massive particle with constant mass.

If you allow for mass to change, you have four-velocity $$ {\bf u}(\tau) = \frac{1}{m(\tau)}{\bf P}(\tau) = \frac{{\bf P}_0 + {\bf F}\tau}{\sqrt{m_0^2 + {\bf P}_0\cdot{\bf F}\tau + {\bf F}\cdot{\bf F}\tau^2}}$$ and the four-position $$ (t(\tau),\vec x(\tau)) = {\bf x}(\tau) = \int \frac{{\bf P}_0 + {\bf F}\tau}{\sqrt{m_0^2 + {\bf P}_0\cdot{\bf F}\tau + {\bf F}\cdot{\bf F}\tau^2}} d\tau$$Knowing $t(\tau)$ and $\vec x(\tau)$ you may be able to recover $\vec x(t)$.

  • $\begingroup$ The question itself specifies that it is a constant pure force, which would require that the mass is a constant. $\endgroup$ – alex1stef2 Jun 15 at 12:47
  • $\begingroup$ @alex1stef2 Witch 5-force being a constant four-vector, it's impossible. I suspect that it may be constant in amplitude, but change in direction so that at any time ${\bf P}\cdot{\bf F}=0$; then the mass could be constant. Do you think this is what they mean? I'm not familiar with term "pure force". $\endgroup$ – Adam Latosiński Jun 15 at 14:08
  • $\begingroup$ A 4-force is said to be "pure" when it does not change the rest mass of the entity on which it acts. Your answer is correct and helpful. $\endgroup$ – Andrew Steane Jun 15 at 16:31

Under John Rennie's guidance; let me know if there's something wrong.

$\textbf{F} = d \textbf{P}/d\tau = m \cdot d \textbf{U}/d\tau$ in the case of a constant force.

$\textbf{F} = m \cdot \textbf{A}$

$\textbf{A} \cdot \textbf{A} = a_0^2$ - > $\textbf{F} \cdot \textbf{F} = \frac{a_0^2}{m^2}$

So we have $\frac{a_0^2}{m^2} = f_x^2 - f_0^2$ for this particular 4-force.

Using the result $sinh(\eta) = \frac{1}{c}\int \:a_0\left(t\right)dt$ yields

$$\frac{\beta}{\sqrt{1-\beta ^2}} = \frac{a_0 t}{c} = \frac{ m \sqrt{f_x^2 - f_0^2} t }{c} $$

which can then be inverted to find the components of the 4-velocity of the particle as a function of time!

  • $\begingroup$ With the constant four-force the mass won't be constant. $\endgroup$ – Adam Latosiński Jun 15 at 12:33

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