"null geodesic" - Dirac

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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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  • #2
George Jones
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Yes.
 
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So what path does a photon take then, if not those geodesic equations?
 
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George Jones
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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.
 
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George Jones
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I think you mean "doesn't" here, correct?
Yes. I have edited my post to reflect this.
 
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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?
 
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PeterDonis
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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.)
 

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