1. Jul 6, 2011

### Trevormbarker

I have two main questions about light, one does gravity affect it? Two, does it have mass?
I have read lots of posts and articles and many say that light is directly affected by gravity while others say that gravity "bends" the spacetime arround it so the light appears to be curved but is in reality travelling straight through the curved spacetime. Also for the question about mass I have read that light has no mass but it has momemntum? My understanding of momentum is that is it mass x velocity so I do not understand how it can have no mass but have momentum.

2. Jul 6, 2011

### WannabeNewton

Photons are affected by gravity in the sense that they trie to follow the 'straightest possible' lines or geodesics of the curved space- time so there is a bending of the light that the photons comprise. Since photons have no rest frame they also have no rest mass so Einstein's energy - mass relation reduces to E = pc for a photon so it does indeed have a momentum based on its energy.

3. Jul 6, 2011

### Trevormbarker

so would it be correct to say that photons themselves are not directly affected by the gravity but they are travelling "straight" through the spacetime which is curved by gravity?

4. Jul 6, 2011

### khemist

Light travels on geodisks that are created from the presence of matter. In 3d space, it is generally a curved line. In 4d, it is straight.

I believe light is able to curve space time as well, but you need a lot of energy to make a serious "dent."

5. Jul 6, 2011

### Trevormbarker

is that because the light has momentum and therefor non-rest mass? so enough of it could curve the space time like normal matter?

6. Jul 6, 2011

### WannabeNewton

It has a certain energy and in GR the source of the curvature is the stress - energy tensor so even energy density contributes, if negligible, to the curved geometry.

7. Jul 7, 2011

### ZealScience

Yes, it does bend under gravitational field, and is because of mass!

But it is not the inertial mass or rest mass. But according to E=mc2, if it has energy, it must have mass.

8. Jul 7, 2011

### Drakkith

Staff Emeritus
Isn't that exactly how gravity affects everything?

9. Jul 7, 2011

### ZealScience

But that is only confined to the interpretation of general relativity...

10. Jul 7, 2011

### WannabeNewton

Except that we are talking about photons and that equation does not apply to photons.

11. Jul 7, 2011

### ZealScience

Why?

12. Jul 7, 2011

### WannabeNewton

A photon has no rest frame so it has no rest mass. E = m$c^2$ only applies to particles in their rest frames.

13. Jul 7, 2011

### Drakkith

Staff Emeritus
Is there another theory that explains it better than GR?

14. Jul 7, 2011

### ZealScience

No, for rest frame m is the rest mass or inertial mass. But for photon m is exactly the mass. Because the mass increase is inertial mass plus mass of kinetic energy, so using E=mc2 gotta be the mass of photon.

15. Jul 7, 2011

### WannabeNewton

For a particle not in its rest frame the mass - energy equation reads $E = ((m_{0}c^{2})^{2} + (pc)^{2})^{1/2}$ where $m_0$ is the rest mass of the particle meaning its mass in its rest frame. A photon has no such frame so the equation reduces to E = pc. E = m$c^2$ does not apply to photons.

16. Jul 7, 2011

### ZealScience

In your equation, m0 for photon is 0, so it is E=pc=mc2

Why it's not correct? I think E=mc2 applies to all energies, otherwise objects emitting photons would not have conservation of energy/mass

17. Jul 7, 2011

### Drakkith

Staff Emeritus
Photons don't have rest mass, which is what the M stands for in the equation. But this really isn't an issue, as the FULL equation, as WannabeNewton posted above can easily be used in its full form for a photon. The mc^2 part just becomes 0. Since pc^2 + 0 equals pc^2, we simply shorten the equation up to E=pc

18. Jul 7, 2011

### ZealScience

So whats the problem with using E=mc2

19. Jul 7, 2011

### Drakkith

Staff Emeritus
For photons? The problem is that E=0 for photons if you use that, which isn't true. E = 0 x c^2 since m = 0. That results in E=0.

20. Jul 7, 2011

Staff Emeritus
ZealScience, what you write is fundamentally incorrect, as people have been trying to tell you. You are misusing formulas.

In addition, it's experimentally excluded - if you use this to predict the deflection of a beam of light by a gravitational source, you will be off by a factor of 2.