The problem 3.20 in Rindler's book Relativity asks to calculate the proper acceleration of a particle in circular uniform motion with radius r; here is the exact text:

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As you can see, the solution says that the acceleration is $\dfrac{\gamma^2\, u^2}{r}$.

In an answer to another post on this site it is stated that the force is proportional to $\gamma^3$; if I undestood correctly, the origin of that $\gamma^3$ factor is the mass of the particle, so now I am confused: where does that $\gamma^2$ of Rindler's solution come from?

Is the given solution wrong?, or am I just too confused?


1 Answer 1


So there is no error: if a particle moves at uniform speed $u$ around a circular path of radius $r$, as seen from an inertial observer, then it describes (in that observer's coordinates) a position vector $\vec r(t) = r [\cos(\omega~t), \sin(\omega~t)].$ We can directly turn this into a 4-position that everyone can agree on, by pairing it with the time coordinate to get the 4-position $(ct,~\vec r(t)).$ Strictly speaking this is a path (a set of points) and the $t$ coordinate does not mean anything special in relativity; the objective thing is the helix of points in spacetime, and not this particular parameterization of them or these particular coordinates for those points.

However we do know that the Lorentz transforms preserve the metric, so if you have any two 4-vectors $(a, b, c, d)$ and $(w, x, y, z)$ then there is a product between them $wa - bx - cy - dz$ that everyone agrees on. This can get a little bit subtle because you usually also want to include translations in your group of coordinate transforms, the so-called Poincaré group, and then you need to say that the 4-position is some sort of "proto-4-vector" because it doesn't preserve this product with translations. But the difference between any two proto-4-vectors is a proper 4-vector.

So if $\big(a_{0,1},~~b_{0,1},~~c_{0,1},~~d_{0,1}\big)$ are two (proto-)4-vectors and we invent the symbol $\Delta Q = Q_1 - Q_0$ to describe differences between these two vectors, then every inertial frame will agree on $\Delta a^2 - \Delta b^2 - \Delta c^2 - \Delta d^2.$ Shrinking this to a differential and specializing on our coordinates, when we say a time $dt$ passes, we can put this into the path and say that other people might disagree on the actual time but between these two nearby points on the helix, an "interval" $ds^2 = dt^2 (c^2 - \vec v\cdot \vec v)$ passes. We can then describe the whole helix in terms of this coordinate $s$ instead, which everyone will agree upon: $s = ct/\gamma.$ We are lucky that $\gamma$ is basically just a constant for us for this helix. And of course we define the "proper time" $\tau = s/c$ to get this in more familiar units.

What this means is that the 4-position can also be described as $$\big(c~\gamma~\tau,~~ r \cos(\omega~\gamma~\tau),~~ r \sin(\omega~\gamma~\tau),~~0\big)$$ for constant $\gamma, \omega, r.$ Taking two derivatives with respect to $\tau$ gives, $$-r~\omega^2\gamma^2 ~\big(0,~~ \cos(\omega~\gamma~\tau),~~ \sin(\omega~\gamma~\tau),~~0\big),$$ where we're explicitly using the fact that $\gamma$ is not changing with $t$ and hence not changing with $\tau$ either. (Keep remembering that $\tau$ is just a parameterization of the path which everyone can objectively derive.)

The magnitude of this is of course just $i~r \omega^2 \gamma^2 = i~\gamma^2~(u^2/r),$ very simply. It is imaginary (spacelike), as it must be: the 4-velocity of any particle is a timelike vector of magnitude $c,$ and that magnitude must be preserved, so of course you can only accelerate and change it with perpendicular vectors that don't affect that magnitude -- and all of the vectors perpendicular to a timelike vector are spacelike.

It happens to contain a $\gamma^2$ factor because the acceleration is perpendicular to the velocity, which is a somewhat subtle point. The reason that you're seeing a $\gamma^3$ term in part comes down to a mnemonic which many people have memorized, that proper acceleration is $\gamma^3$ times coordinate acceleration. This is a dangerous mnemonic, but you can see why it looks very useful and memorable. The danger is that this assumes a particle is accelerating in the same direction that it is travelling, which obviously doesn't apply in this case. It can instead be derived that the proper expression is $\gamma^3/\gamma_\perp.$

Please also note that at no point did we have to invoke the mass of the particle.


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