"null geodesic" - Dirac

In Dirac's book on relativity, he begins and ends his section on proving the stationary property of geodesics with references to "null geodesics". His last sentence is: "Thus we may use the stationary condition as the definition of a geodesic, except in the case of a null geodesic."

What is a null geodesic? Does he mean a null interval, like for a photon, with ds^2 = 0?

George Jones
Staff Emeritus
Gold Member
Yes.

So what path does a photon take then, if not those geodesic equations?

George Jones
Staff Emeritus
Gold Member
Dirac doesn't say that light does not follow a geodesic; Dirac writes "Thus we may use the stationary condition as the definition of a geodesic, except in the case of a null geodesic."

Dirac's writing has to be unpacked very carefully. Just my personal opinion, but I think that Dirac's book is not a good book to use to teach oneself general relativity.

Last edited:
PeterDonis
Mentor
Dirac doesn't say that light does follow a geodesic

I think you mean "doesn't" here, correct?

George Jones
Staff Emeritus
Gold Member
I think you mean "doesn't" here, correct?

Yes. I have edited my post to reflect this.

OK, are you saying that a photon DOES follow those geodesic equations? Please be explicit.

I see in the derivation - Hamilton's principle, stationary property, etc. - that I divided through by ds. Since ds equals zero for a photon, it seems to me that thus a photon would NOT follow the geodesic equations that result. I can't make sense of them for a photon, which is why I asked the question in the first place.

If I minimize the component time (provided there are no off-diagonal terms for the time degree of freedom in the metric, static, etc.) then I get some different geodesic equations. Haven't managed to integrate them yet to check against the empirical data, but they are different.

BTW, what is the latest and greatest / most accurate / data for the photon grazing the sun problem?

PeterDonis
Mentor
are you saying that a photon DOES follow those geodesic equations?

Not in the form you're using them. See below.

I see in the derivation - Hamilton's principle, stationary property, etc. - that I divided through by ds. Since ds equals zero for a photon, it seems to me that thus a photon would NOT follow the geodesic equations that result.

That's not because photons don't follow geodesics; it's because "path length" ##ds## is not a good parameter along null geodesics, because it doesn't uniquely label each point on the geodesic with a different parameter value (it can't, since ##ds^2 = 0## everywhere on a null geodesic). You have to choose some other parameter that does uniquely label each event on the geodesic with a different parameter value. (Coordinate time in a suitable coordinate chart will be such a parameter.)