# "null geodesic" - Dirac

## Main Question or Discussion Point

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?

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George Jones
Staff Emeritus
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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
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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
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2019 Award
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.)