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Most textbooks define the geodesic equation for a particle with unit mass, such that it looks like:

$$ \ddot{x}^{\mu} + \Gamma^{\mu}_{\alpha \beta} \dot{x}^\alpha\dot{x}^\beta = 0$$ Where "dot" denotes derivative w.r.t. proper time (say). My question is, if we are working in a system of units where $c = G = M$ , where M is the mass of the black hole in a Schwarzschild metric, does the above formulation of the geodesic equation make sense? It would seem that it describes the path of a particle with mass equal to that of the black hole, which for sure wouldn't follow a simple geodesic trajectory, since its own gravitational effect is obviously not negligible.

Urb
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Johnny
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2 Answers2

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if we are working in a system of units where ==

This isn't really possible. There is a lot of freedom in choosing units, but you have to actually go through the process. For example, the speed of light naturally has dimensions of distance divided by time. We can choose units where distance and time are measured by the same unit, in which case $c=1$, and is considered dimensionless because the distance and time cancel out. We can also choose to measure mass in those same units, so that the units in $G$ will similarly cancel out.

When talking about black holes, we frequently choose mass units fixed to the mass of the black hole. But since there is nothing to cancel with that mass, the result cannot be dimensionless. Maybe you've seen this written as $M=1$, but that would be abuse of notation; it should be something more like $M=1M$, because that last $M$ is the unit. So you can choose units where $c=G=1$ and $M=1M$, but there's still a unit in such a system.

Now, forget about units for a second. There seems to be a second and entirely separate point of confusion here about what the geodesic equation is actually measuring.

it describes the path of a particle with mass equal to that of the black hole

The geodesic equation describes a curve in a fixed metric — that's it. I emphasize the fixed part of this, which implies that if there is a "thing" moving along this geodesic, we assume that it is not affecting the metric in any way. A geodesic does not describe the exact path that a massive particle would actually take. Instead, we talk about "test particles", which are assumed to have such small mass (or more precisely, stress energy) that we can ignore the mass and pretend that the metric is not altered by their presence. The OP was correct in physically intuiting that

its own gravitational effect is obviously not negligible.

To deal with an object that has non-negligible "gravitational effect", you need more than just the geodesic equation. ("Self-force", "radiation reaction", and "backreaction" are the relevant keywords here.) Which is to say that the geodesic equation really has nothing at all to say about mass in any units whatsoever; it assumes that the mass is zero — which doesn't change with units.

(See related discussion here.)

Mike
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The geodesic equation is, as pointed out in a comment, geometrical. An easy way to see that it doesn't depend on units is to recognize that it has the same form in any set of coordinates, and a change of units is a change of coordinates. If you want to work in units where $x$ is in feet, $y$ in furlongs, $z$ in cubits, and $t$ in centuries, you can do that.

So as long as the mass of the test particle is insignificant compared to the source, it follows one distinct trajectory regardless of it's mass?

Basically yes. This is what the equivalence principle expresses. The particle also needs to have small dimensions, and you need to assume an energy condition. There are formal theorems that state the conditions rigorously.

user295271
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