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I read that the photon doesn't have mass, but my teacher says that the photon has mass because the sun can attract it (like in the experiments to prove the theory of relativity).

I think that there is another reason to explain that. How can I explain that the photon doesn't have mass and the sun attracts photons?

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Did your teacher mention the experiments that prove the theory of relativity? If so, it's a bit confusing that he/she thinks that the experiments also prove that photons have mass. – JiK Aug 29 '14 at 14:09

To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to show photons are attracted by gravity even in Newtonian gravity.

If you have a large mass $M$ attracting a small mass $m$ and the distance between the two masses is $d$ then the force between them is given by Newton's equation:

$$ F = \frac{GMm}{d^2} $$

To get the acceleration $a_m$ of the small mass $m$ we use Newton's second law $F = ma$ so:

$$ a_m = \frac{F}{m} = \frac{GMm}{md^2} = \frac{GM}{d^2} $$

Note that the mass of the small object has cancelled out, so the acceleration doesn't depend on $m$ at all. That means a massless object like a photon experiences exactly the same acceleration as a massive object. So even in Newtonian gravity we expect the path of a light ray to be deflected by gravity. In fact with some head scratching the equation for the expected deflection can be derived, and it is:

$$ \theta_{Newton} = \frac{2GM}{c^2d} $$

where $d$ is the distance of closest approach and $\theta_{Newton}$ is the angle that the light ray is bent.

As I mentioned at the start, to properly describe the light ray you need general relativity and using this we find that the deflection is actually twice as big as Newtonian gravity predicts:

$$ \theta_{GR} = \frac{4GM}{c^2d} $$

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did Newton predict that deflection? – Py-ser Aug 29 '14 at 6:46
@Py-ser: Newton didn't derive the equation, but in Optiks he did say Do not Bodies act upon Light at a distance, and by their action bend its Rays, and is not this action strongest at the least distance?. However your point is taken and I've edited my answer accordingly. – John Rennie Aug 29 '14 at 6:52
ok thanks for the clarification! – Py-ser Aug 29 '14 at 7:06

What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force.

Let me briefly state what the theory of relativity has to say about gravity without going into the complicated mathematics. Imagine you try to measure the distance between two points on the surface of the earth. generally, the shortest distance between two points is a straight line. But because earth is a sphere (more or less) and hence the surface is "curved", the shortest distance between two points is not really a straight line. In fact, you can never draw a straight line on a curved surface.

Mr. Einstein says that just like the two dimensional earth surface has a geometry, the four dimensional "spacetime" also has a geometry. the geometry of spacetime is generally flat but in the vicinity of an object with substantial mass, the geometry is curved. It's the curvature of spacetime that we call gravity.

We know from classical mechanics that a moving object tends to keep moving in a straight line with a constant velocity. This is true in the theory of relativity as well but because a massive object curves spacetime, a straight line through spacetime is not always straight. When a moving object comes near a massive body, it falls into the curved region and therefore bends towards it and appears to be "attracted" by it.

Hope this answers your question.

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@John Rennie, ma = GMm/d^2 generally you can cancel the small m as it appears in both sides of the equation but I am not exactly sure if you can do that in this particular case as the mass of a photon is zero. In other words, 3x = 3y implies x = y but 0x = 0y doesn't always mean x = y. is there something I am missing? – Rahat Aug 29 '14 at 7:41
I'd rather consider it to be the mass obtained due to the energy. Note that photons only have a zero STATIONARY mass, but if they are in motion, especially in a curved manifold, you cannot expect the mass to be zero thanks to e=mc^2.... – GRrocks Aug 29 '14 at 8:27
Yes you can, and indeed that's how the Newtonian prediction for the deflection is calculated. You're quite correct that you can't simply divide zero by zero, but you can take the limit as $m \rightarrow 0$. – John Rennie Aug 29 '14 at 8:28
Thanks for the clarification. – Rahat Aug 29 '14 at 8:44

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