3
$\begingroup$

I am assuming that zero (rest) mass particles don't interact gravitationally with each other and other particles. Does that mean they experience a "flat" spacetime instead of a curved one? I find it a little hard to imagine that. Also, does this mean massless particles can pass through an event horizon (not the singularity itself, I assume there would be some other way of interaction that would prohibit that).

$\endgroup$
  • $\begingroup$ As @M.Herzkamp edited into his answer (though not explicitly) and netom states in his (again not explicitly), mass and energy are two sides of the same coin. They are equivalent. And as John Rennie points out in his answer, all physical objects in GR obey the geodesic equation. $\endgroup$ – Wouter Jul 21 '14 at 12:32
3
$\begingroup$

No, a massive body is able to bend light around it, which is called gravitational lensing. This has been observed multiple times.

EDIT

Photons are massless. Otherwise, they would not travel at the maximum speed, which is called speed of light.

Keep in mind, that gravitational lensing is not a part of Newtonian mechanics. You need general relativity for that. And in the context of general relativity, it is not mass, which exerts gravitation, but energy. And photons clearly carry energy. So, you could produce a gravitational well (even a black hole) entirely without massive particles.

$\endgroup$
  • $\begingroup$ Ok, so you are saying: 1) photons are massless (are they?) and 2) photons are affected by gravity. But doesn't this make gravity non-symmetric in that a mass-ful particle interacts with a mass-less particle through gravity, but not the other way around? All other forces seem to be more symmetric (on the other hand, we havn't observed gravitational repulsion, have we ;-) ) $\endgroup$ – kutschkem Jul 21 '14 at 12:10
  • $\begingroup$ updated answer to answer your questions. $\endgroup$ – M.Herzkamp Jul 21 '14 at 12:19
3
$\begingroup$

The trajectory of an object, whether massless or not, in a curved spacetime is given by the geodesic equation.

$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$

Solving this is a formidable problem unless there is some helpful symmetry that simplifies the working, but in the particular case of light we have a short cut because any massless particle will follow a null geodesic, and these are usually a lot easier to calculate. For example in my answer to the question Why is a black hole black? I calculate the null geodesic for light trying to escape from a black hole and show that light at or below the event horizon can't escape.

So the answer to your question is that massless particles do indeed experience the curvature of spacetime in the sense that they are deflected by it. As M. Herzkamp says in his answer, this is why gravitational lensing of light happens.

$\endgroup$
3
$\begingroup$

I am assuming that zero (rest) mass particles don't interact gravitationally with each other and other particles.

That's not a valid assumption in general relativity. Particles with zero invariant mass have energy and momentum and, thus, gravitate. Essentially, in general relativity, the density and flux of energy and momentum are the sources for spacetime curvature.

For example, see "Do two beams of light attract each other in general theory of relativity".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.