# Gravity affecting space

1. May 11, 2014

### apurvmj

It is said that light particle being mass-less charged particle are not affected by gravity but space, which is mass-less and charge-less get warped by gravity. How come?

2. May 11, 2014

You need to learn General relativity to understand that.

3. May 11, 2014

### ChrisVer

Well, there is not a distinction between gravity and spacetime. That's why some people refer to General Relativity as geometrodynamics (as stated in wiki, it's a loose use).
The concept is that the photon (light's particle) is not really affected by gravity- it's massless. So it follows "straight lines"... However those straight lines are determined by the geometry of spacetime, which is affected by matter and energy through Einstein's equation.
So the photon follows straight lines, but straight lines on a given geometry... For example, if you take a ball (sphere) and you want to sketch a line from a point A to point a B, is that going to be "straight"? I'd call straight the line which has the less physical length (something someone can understand by classical mechanics through Hamilton's least principle)... In the case of the sphere, these lines can be circles...

4. May 11, 2014

### ChrisVer

Of course it's difficult to get the concept without General Relativity or differential geometry background... The only things someone could understand is the general "ideas"/concept behind everything.

5. May 11, 2014

### Bill_K

There is a big distinction. Gravity is the curvature of spacetime. Spacetime exists even in Special Relativity, where there is no gravity.

Geometrodynamics is not the same thing as General Relativity. Geometrodynamics was Wheeler's attempt to replace all of elementary particle physics with curved spacetime.

Photons possess energy and momentum and are affected by gravity, just as particles with mass are affected by gravity.

6. May 11, 2014

### ChrisVer

The distinction does not exist in General Relativity, because spacetime is affected by gravity. You cannot speak of spacetime, without being able to work on it with a metric, and metric's being determined by Einstein's equation... If you try to speak of Special Relativity, you just have to take the curvature to be zero (flat metric).

For geometrodynamics:
http://en.wikipedia.org/wiki/Geometrodynamics
(Einstein's geometrodynamics) and I made it clear as said in the paragraph that it's a loose term. But what I wanted to emphasize with it, is that studying general relativity is equivalent to studying the geometry of spacetime.

Photons however follow geodesics (no matter what energy they have)... that means they follow straight lines on the specific geometry, as they do as "massless" on a flat spacetime. The thing is that those straight lines, are not what someone would at first call "straight"... It depends on the spacetime/gravity. The connection, which appears in the geodesics equation, is something like an extra force a moving particle will "feel", and it's not determined by its energy or mass (even massless particles will feel it), but by the metric of spacetime, which is determined by the differential equations existing in Einstein's equation.

Last edited: May 11, 2014
7. May 11, 2014

### bcrowell

Staff Emeritus
8. May 11, 2014

### Bill_K

The idea that photons and other particles follow geodesics is a weak field approximation, neglecting the gravitational field produced by the particle itself.

Consider a photon passing by a central body. Both objects possess energy, and together they orbit about their common center of energy. The gravitational field carried by the photon affects the central body, moving it back and forth. And the path followed by the photon is determined by the combined fields of both objects, and is not the same as a geodesic in the original spacetime.

9. May 11, 2014

### ChrisVer

You are just making it more complicated (smile) by inserting self-interactions... especially for a photon on itself...If we want a complete theory, we would have to say that the photon can produce a pair of particle/antiparticle, and so on... (I also said that that energy/matter is playing a role in Einstein's eq, so would the energy of photons)
However I think that this approximation (weak field as you named it) can explain quite well the bending of light while passing next to a gravitational source (like the sun or a black hole)... which I think reflects the OP question...

Last edited: May 11, 2014
10. May 11, 2014

### WannabeNewton

Yes it is. For example, deduction of the concept of the redshift of electromagnetic radiation (or of a semi-classical photon) in a gravitational field, as usual in the geometric optics approximation, does not require space-time curvature, but rather just require special relativity and good old fashioned Newtonian gravity. There are countless other examples of this of course, but also examples of electromagnetic phenomena in the presence of gravitational fields that cannot be explained e.g. by conformally flat metric theories of gravity.

11. May 12, 2014

### apurvmj

Sapce time curvature is gravity right. But without mass curvature in space doesn't exist. It is dificult me to understand mass can affect absolute nothing ie space, but can't affect energy particle like photon.

12. May 12, 2014

### Staff: Mentor

It does. See posts #5, #7, and #10.

13. May 12, 2014

### Mordred

the volume of space is never empty, there is always some form of energy/mass-density residing in every region of space. How space time geometry is described is essentially a distribution of the energy-density relations.

This article will help, its designed to explain geometry in terms of the FLRW metric, which is on a far larger scale than a localized gravitational body, however the same principles apply.

https://www.physicsforums.com/showpost.php?p=4720016&postcount=86

think of a fluid, though that fluid is various forms of energy/mass, dust, radiation and gases. mass contributes to the positive pressure of that fluid , vacuum energy or the cosmological constant is the negative pressure influence. the energy-density to pressure relations are covered on this page.
http://en.wikipedia.org/wiki/Equation_of_state_(cosmology)

Localized gravity essentially affects the pressure distributions in a given region.

Last edited: May 12, 2014
14. May 12, 2014

### Staff: Mentor

This is (to the best of our knowledge) true of our universe, but it is important to note, first, that for much of the universe's volume, while the energy density is not, strictly speaking, zero, it is very, very, very, very small, much too small for us to measure directly, and there are other places, such as the interiors of neutron stars, where the energy density is very, very, very large, thirteen or more orders of magnitude larger than the density of water (for a total variation in energy density of something like 40 orders of magnitude or more). The FLRW model, with a constant energy density everywhere at a given instant of time, is an approximation, only meant to be valid on a very large distance scale, something like a billion light years or larger.

Second, we use other approximate models in which the energy density is exactly zero except for an isolated region; the most commonly used such model is Schwarzschild spacetime, describing the vacuum region around an isolated spherically symmetric body. So as far as models go, it is not always true that there is energy density everywhere. That means that this...

...is also not really correct, or at least, it's highly misleading, because the whole point of models like Schwarzschild spacetime is that spacetime curvature can vary even in a vacuum region; it is not necessarily the same everywhere even in a region of spacetime where the energy density is exactly zero everywhere. (In fact, the maximally extended Schwarzschild spacetime, which describes an eternal black hole, has zero energy density *everywhere*, and yet spacetime is curved and the curvature varies.)

And the energy density, plus all the other caveats above.

15. May 12, 2014