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In Thomas A Moore's General Relativity Workbook in Chapter 8 titled Geodesics, it is argued that the geodesic equation does not fix the scale of any worldline's proper-time $\tau$. The argument for this goes something as follows. If $x^\mu(\tau)$ is some curve that satisfies the geodesic equation:

$$\frac{d}{d\tau}\Big(g_{\alpha \beta}\frac{dx^\beta}{d\tau}\Big)-\frac{1}{2}\partial_\alpha g_{\mu \nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0$$

then so does the curve $x^\mu(b\tau)$ where $b$ is some constant. The only way to fix the scale of $\tau$ would be to use the fact that the dot product of the four-velocity with itself is $-1$ in any reference frame. In other words $x^\mu(\tau)$ satisfies the equation,

$$g_{\mu \nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$

whereas $x^\mu(b\tau)$ does not. My questions on this are as follows:

  1. Can someone help we with the math by showing exactly how $x^\mu(b\tau)$ satisfies the geodesic equation but does not satisfy the second equation given that $x^\mu(\tau)$ satisfies both?

  2. What exactly is meant by fixing the 'scale' of the proper time? What are we trying to prove with this line of reasoning?

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3 Answers 3

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  1. By the chain rule $$\frac{d}{d\tau}(x(b\tau)) = b \cdot\left(\frac{dx}{d\tau}\right)(b\tau)$$ (indices suppressed for brevity). The rest is simple algebra.

  2. Proper time is a special case of a world line parameter. It serves the purpose of giving you the velocity seen from the locally comoving observer at any point of the worldline, by just deriving w.r.t. proper time. You could choose other parameterizations of the wordline, without losing the physics, but then you would lose that side benefit.

Proper time is to relativity what the arc length parameterization is for ordnary spatial curves. For example, if you parameterize a circle of radius $R$ by its arc length, you could write $$\vec r(s) = R\cdot\left( \cos (\frac{s}{R})\atop \sin (\frac{s}{R}) \right)$$ Without calculating, you would already know that $$\int_0^s \sqrt {r^2(\sigma)} d\sigma = s$$ and (hence) that the derivative of the curve parameterization is the (unit) tangent vector $$\left|\frac{d\vec r}{d s}\right|=1$$ You could again choose a different parameterization, but then you would have to perform some gymnastics to get the line integral and/or the tangent vector.

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  • $\begingroup$ Shouldn't the chain rule be $\frac{dx(b\tau)}{d\tau}=\frac{dx(b\tau)}{d(b\tau)}\frac{d(b\tau)}{d\tau}=b\frac{dx(b\tau)}{d(b\tau)}$?? $\endgroup$
    – Ethan
    Jun 4, 2021 at 18:17
  • $\begingroup$ @Aryamann: sure, I just abbreviated the result a little bit. Now I have added the argument of the derivative for greater clarity. $\endgroup$
    – oliver
    Jun 4, 2021 at 18:39
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Based on your comment to oliver's nice answer, I thought I'd mention that this is one of those times in which Leibniz notation (e.g. $\frac{dx}{d\tau}$) is misleading.

Consider the function $f$ which eats a number and spits out its square. We normally express this as $f(x)=x^2$ or $f(t)=t^2$, but in these two expressions the quantities $x$ and $t$ don't mean anything - they're just placeholders which make it easy to see explicitly what happens when we plug some quantity into $f$. We could equally well write $f(\bullet) = \bullet ^2$.

The derivative of $f$ is the function which eats a number and multiplies it by two. So what should we call this? Often times we call it $\frac{df}{dx}$, but this doesn't really make sense because, in the absence of further information, I have no idea what $x$ is. The typical thing to do is to tacitly assume that $x$ is the argument of $f$, but this can become slightly confusing and ambiguous (as you have found).

In contrast, the Newtonian notation $f'$ is unambiguous. $f'$ is the derivative of $f$, making no reference to the argument of the function.


Shouldn't the chain rule be $\frac{dx(b\tau)}{d\tau}=\frac{dx(b\tau)}{d(b\tau)}\frac{d(b\tau)}{d\tau}=b\frac{dx(b\tau)}{d(b\tau)}$?

Using a mix of Leibniz and prime notation, the chain rule can be expressed as

$$\frac{d}{d\tau} x(b\tau) = x'(b\tau) \cdot \frac{d}{d\tau}(b\tau) = x'(b\tau) \cdot b$$

Your expression is correct, but you should understand $\frac{dx(b\tau)}{d(b\tau)}$ to be the derivative of the function $x$, evaluated at the value $b\tau$ - written in prime notation, $x'(b\tau)$.

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You received several good answers explaining the mathematical point of view of the invariance of the geodesics eq. with respect to time scaling. I shall therefore concentrate on the physical aspect.

The geodesics eq.is actually invariant with respect to an affine rescaling of the curve parameter. This means that if $x^\alpha(\tau)$ is a solution, then also $x^\alpha(a \tau+b)$ is a solution, where $a$ and $b$ are constants.

What does this physically mean? It simply means that it is not important if you measure time in seconds or minutes or ...martian days (the $a$ scaling) and you can start counting now, or.....now, or....whenever you feel like (the $b$ shift).

Is this surprising? It shouldn't. After all the geodesics eq. describes curves of extremal parameter "length" and this obviously does not depend on whether you use metric, imperial or any other unit of measurement.

However the geodesics eq. is a 2nd order ODE , so it needs initial conditions for $x^\alpha(\tau)$ and $\frac{dx^\alpha}{d\tau}(\tau)$ if you want unique solutions. The first initial condition will fix the $b$ shift and the second will fix the $a$ scaling.

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