I didn’t want to be the first one to reply… because I’m just now taking GR and don’t understand it all completely myself. There really is no simple explanation I can point to to why this ends up happening this way other than to say that the solutions to the Field Equations work out that way. There are definitely books you can pick up on the subject. For a lay intro to GR, I recommend my professor’s book Time Travel in Einstein’s Universe (R. Gott, Houghton Mifflin, 2001). Otherwise, some serious tensor and metric learning is in order.
Maybe you’ve done that, in which case we can go through the equations and see how a photon ends up deflected by twice the amount predicted by Newton when travelling in a vacuum near a mass. (by the by, one of the interesting results of the field tensor is that gravity is not only produced by mass, but also by pressure!)
A single photon will also be deflected by twice the Newtonian amount in passing by a massive object-- You don’t need a pair of photons. Does this provide an easy way to distinguish between a massless particle and a massive particle? Not really… A very lightweight particle, like a neutrino (which is hence moving at very close to the speed of light, at a given energy) will be deflected by an amount almost equal to the amount for light. There is no experimental evidence, nor is there likely ever to be, that the photon is truely massless; all we can say experimentally is that if the photon does have mass, it’s an insanely small one.
Chronos, my reading of the OP, and the quote from Ring’s subsequent post, is that the question isn’t whether the photon is deflected by twice the Newtonian amount. It’s whether a photon, under GR, is deflected by twice the amount of a massive particle, also under GR.
Are you saying the answer to “Does a single photon passing a gravitating body experience double the deflection a massive particle would?” is “no”? And, of course, do you have a cite or derivation? Certainly “yes” would be a counter-intuitive answer (to me, anyway), but Ring’s quote seems to say that “yes” is correct.
I’m not sure Chronos’ point was that unclear: a nearly massless particle will have a deflection in passing a massive body that will be almost indistinguishable from that for a truely massless one. The latter happens to be twice the Newtonian result, but this is no more than a convenient comparison - it is, after all, wrong.
More formally, Chronos is claiming that the deflection has a well-behaved limit in the limit of the mass tending to zero and that this is equal to the massless case. I don’t have an explicit cite - and I doubt Chronos can dig one up either - but that’s largely because physicists take such statements as obvious, unless there’s some trap. As you say, ZenBeam, it would be counter-intuitive. And rather than give a derivation, I’m going to point in the direction of section 11.1 in Schutz A First Course in General Relativity (Cambridge, 1985), where he lays out the orbit equations in the massive and massless cases alongside each other. The key comparison is between (11.8) and (11.9), which superficially differ by one extra term in the massive case. However, if you put the mass of the particle in explicitly, then it’s clear that that term vanishes in the m tends to zero limit. Not entirely rigorous, but you’re always free to slog through the full derivations of the deflections in both cases.
Frankly, I don’t understand the quote Ring passed on and, unfortunately, I don’t have immediate access to Tolman’s book.
Incidently, I don’t know of any intuitive argument for why the massless answer comes out exactly twice the Newtonian answer. I seem to remember that Weinberg gives a slick derivation of Rutherford scattering in an appendix to The Discovery of Subatomic Particles (?), up to an undetermined overall constant. I suspect you could similarly argue that the form of the deflection formula is fixed by such an argument and so the only thing that can differ is a simple numerical factor. But why exactly 2?
I think I have part of it. First, some equations from Pauli’s Theory of Relativity (slightly rearranged):
For a point mass:
1 / r[sup]4[/sup] * (dr / d[sym]f[/sym])[sup]2[/sup] + 1/r[sup]2[/sup] - 2 m c[sup]2[/sup] / (B[sup]2[/sup]r) - 2m/r[sup]3[/sup] = 2 E / B[sup]2[/sup] (Eq 427 a)
For a photon:
1 / r[sup]4[/sup] * (dr / d[sym]f[/sym])[sup]2[/sup] + 1/r[sup]2[/sup] - 2m/r[sup]3[/sup] = 1 / [sym]D[/sym][sup]2[/sup] (Eq 427 b)
where (he writes) “the law of areas
r[sup]2[/sup] * d[sym]f[/sym] / d[sym]t[/sym] = const. = B (Eq 425)
is seen to be valid”
Note that m is not the particle mass, but rather is proportional to the mass generating the field (e.g. the Sun, not the proton zipping by). Also not that the variable in Eq. (425) is proper time [sym]t[/sym], not observer time t. [sym]f[/sym] and r are normal polar coordinates.
He writes “These equations completely determine the required paths”. “If the last term on the left-hand side [of equation (427 b)] were not present, the light ray would be a straight line, at a distance [sym]D[/sym] from the origin.”
At any rate, as the speed of the massive particle increases towards C, d[sym]f[/sym] / d[sym]t[/sym] at the minimum distance from the Sun increases without bound. This means the term 2 m c[sup]2[/sup] / (B[sup]2[/sup]r) in (427 a) approaches zero. He doesn’t say what E is, but energy seems a good guess. In that case, E will also increase without bound, presumably keeping E / B[sup]2[/sup] finite.
All that’s left is to show 1 / [sym]D[/sym][sup]2[/sup] = 2 E / B[sup]2[/sup] in the limit as the massive particle speed approaches C (or not, as the case may be).