# Freedom in physically interpreting the Schwarzschild coordinates leads to trouble

For the Schwarzschild metric

$ds^2 = -(1-\frac{M}{r}) dt^2 + (1-\frac{M}{r})^{-1} dr^2 + r^2(d\theta^2 + \sin^2{\theta}~ d\phi^2)$

most references (Landau-Lifshitz, Schutz or Misner-Thorne-Wheeler) say that the physical meaning of the coordinate $r$ is $C/2 \pi$ where $C$ is the circumference around the central massive object. They also give a definition of coordinate time $t$ (using the concept of an observer at infinity).

Now, if I try to operationally define the coordinates in a different way, I run into a trouble. I label the newly defined coordinates with primes ($t',r',\theta',\phi'$).

New definitions: I choose the central point of the mass to be at $r'=0$. Next, I define $dr'=ds$, where $ds$ is the length measured along the radially outward direction using a ruler. While doing so (measuring the length radially outward), I also define $d\theta' = d\phi'=0$. Hence with these ad-hoc definitions, the Schwarzschild length element becomes (for two events which represent the ends of the rod when rod is used to make a radial length measurement)

$dr'^2 = -(1-\frac{M}{r'}) dt'^2 + (1-\frac{M}{r'})^{-1} dr'^2 + r^2\times(0^2 + \sin^2{\theta'}~ 0^2)\\ \implies dr'^2 \left( (1-\frac{M}{r'})^{-1} -1 \right) = (1-M/r')dt'^2$

and so on. Note that in terms of old coordinates ($t,r,\theta,\phi$), $dt=0$, since the ends of the rod are read simultaneously. But this is no longer true in terms of $t'$, since $ds \neq 0 \implies dr' \neq 0 \implies dt' \neq 0$. So clearly the physical interpretations of the two sets ($t',r',\theta',\phi'$) and ($t,r,\theta,\phi$) are very different.

The problem: Now, the two sets of coordinates obey the same math equations; e.g. both their length elements are of the Schwarzschild form (the very first equation above). Hence, any physical problem (an apple falling towards the earth) solved in either of the two sets of coordinates will yield a solution which is mathematically of the same form. But, there arises an ambiguity in physically interpreting the results. For example the result in unprimed coordinates may be $r=t^2$, and in primed coordinates it will be $r'=t'^2$. Now, since ($t',r',\theta',\phi'$) and ($t,r,\theta,\phi$) have different physical meanings, the two mathematical results imply two different physical interpretations. Which one is correct?

That's wrong. The mathematical expression of the solution won't look the same since the math expressions of initial conditions will look different; e.g. initial conditions may be $t=r=10, \theta= \phi=0$ in unprimed ones and $t'=21, r';=19, \theta'= \phi'=0$ in primed ones, because same physical situation have different mathematical representations in the two coordinates. This will lead to math solutions whose expressions will look different (in the two coordinate systems). But they actually will have the same physical interpretation, since the operational definitions of the two coordinates (primed and unprimed) are not the same.