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

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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

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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.

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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I think you mean "doesn't" here, correct?Dirac doesn't say that light does follow a geodesic

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George Jones

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Yes. I have edited my post to reflect this.I think you mean "doesn't" here, correct?

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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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Not in the form you're using them. See below.are you saying that a photon DOES follow those geodesic equations?

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.)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.