Discontinuity in the zero mass limit of geodesics in general relativity

Question

In general relativity, a test mass is by definition an pointlike particle whose mass $m$ tends to zero, so that it does not affect the nearby spacetime. As a result, the orbit of a test mass is given by a timelike geodesic.

While the mass tends to zero, a timelike geodesic is quite different from a null geodesic, which is the worldline of a massless particle.

Therefore, it seems that the limit $m\to 0$ of the orbit is not continuous. My question is what is the fundamental reason behind this discontinuity?

I am aware of the difference between representations of massless and massive particles in group theory, but I don't see a clear relation to this classical problem.

Answer 1

Ali Seraj wrote: "I am aware of the difference between representations of massless and massive particles in group theory, but I don't see a clear relation to this classical problem."

I don't know anything about group theory, but in the relativistic energy equation for time independend metrics E=-pₜ=constant you replace the particle's rest mass m with the photon's hf/c², where f is the frequency with which the photon started or arrives at infinity.

Ali Seraj asked: "what is the fundamental reason behind this discontinuity?"

There is no discontinuity, if you take an orbit with v/c=0.999999... you'll barely notice any difference to an orbit with v=c. The more 9s you add after the decimal point, the closer you get.

With v=c (and therefore ds=0) the τ in the equations of motion is no longer the proper time though, but the photon's affine parameter. In static coordinates this is equivalent to the integrated local shell times.

The proper time would go to zero if v was really c, so you can only approach, but never reach c with ds≠0, but plotting by coordinate time the ultrarelativistic particle's and the photon's trajectory look almost the same.

For example, if you plot an orbit around the photon sphere, the photon with v=c will make infinite revolutions, and a particle with v/c=999999... will need more 9s after the decimal point to make more revolutions:

At the end of the simulation the particles that started with v<c also show v=c in the numeric display when they approach r=rₛ, but that is just a rounding artifact since there's only so much space for the 9s after the decimal point.