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I've heard said by many physicists things along the lines of "Since light has no mass, we need general relativity to explain why light is affected by gravity."

But why is this necessary?

We know that light has energy $E=h\nu=\frac{hc}{\lambda}$, and we know that gravity is really based on total energy and not just rest mass, so therefore light has gravitational mass equivalent to $m=\frac{h\nu}{c^2}=\frac{h}{\lambda c}$.

So why can't we just say that it makes perfect sense that light is affected by gravity?

Matt
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    Does this answer your question? What would be the photon's effective mass in Newton's Law of Gravitation? – Eletie Mar 04 '21 at 17:15
  • This might help. Why can't I do this to get infinite energy? – mmesser314 Mar 04 '21 at 17:35
  • Eletie, thanks, that's very helpful. It explains why GR relativity is necessary to explain the details of light's orbits. But it technically doesn't answer why people are explaining it as having anything to do with light's lack of mass. – Matt Mar 04 '21 at 17:43
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    @Matt I've actually not seen people make the statement you mention; only that GR is needed to correctly explain how light interacts with gravity. The point is that you're trying to use Special Relativistic formula with Newtonian gravity, which we know results in contradictions, but I think I see your argument. The counter-argument is that it just doesn't work when trying to interpret the mass in Newtonian gravity as the relativistic mass (which is bad practice anyway); so that's reason enough to only use the invariant mass in Newtonian gravity... – Eletie Mar 04 '21 at 18:41
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    @Matt ... which of course leads to the conclusion that light being massless means no gravitational interaction. This is how I'd interpret the statement anyways. Either way, we need to resort to GR to fit observation.

    The answers here (and links therein) regarding why relativistic mass in Newtonian gravity doesn't work may be useful too https://physics.stackexchange.com/questions/521369/why-cant-relativistic-mass-equation-be-put-in-newtonian-gravitation-equation?noredirect=1&lq=1

     – Eletie Mar 04 '21 at 18:43
  • @Eletie, thanks! – Matt Mar 04 '21 at 21:13

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we know that gravity is really based on total energy and not just rest mass

Before GR, it was thought that gravity acted on rest mass only. So even just to say gravity acts on total energy shows the need for a new theory.

Furthermore, the whole concept of rest mass doesn't exist without SR at the very least.

Señor O
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  • This is more of a historical answer. Meanwhile, people are still saying this. – Matt Mar 04 '21 at 17:41
  • Also, worth noting that Einstein came out with $E=mc^2$ the same year as special relativity (in one of his infamous 1905 papers), so a good bit before he came up with general relativity. – Matt Mar 04 '21 at 18:26
  • I don't really get where you're seeing a legitimate issue here. I guess you're proposing we could say gravity affects everything based on its total energy, but that's flawed because that's identical to saying gravity affects everything the same because literally everything has some total energy – Señor O Mar 04 '21 at 18:37
  • Well gravity does affect everything that has some total energy. I never said "the same" though, obviously it depends on the amount of energy. – Matt Mar 04 '21 at 21:17
  • @Matt thats a nonsensical statement. EVERYTHING has some total energy. So it has nothing to do with it having energy, it has to d with it existing. – Señor O Mar 04 '21 at 23:59
  • I mean, I didn't make that statement, you did. I was only confirming that it's true. – Matt Mar 05 '21 at 01:44
  • Wait what? It does affect everything the same. – Señor O Mar 05 '21 at 02:12
  • Well you can also say that any law of physics affects everything the same, but even so they depend on the properties of these things. In this case, the total virtual force exerted on both objects depends on the product of both object's energies. When one object is much more massive than the other, than you can ignore the force exerted on it, and thus ignore the mass the of the smaller object. But theoretically it still matters. – Matt Mar 05 '21 at 13:25