# How does gravity effects both time and light if they have no mass [duplicate]

I've been reading about how black holes can effect both time and light with gravity. So I was wondering, doesn't something have to have mass to be effected by gravity? And if so, does this mean both light and time have mass? And if not, how can gravity effect something that has no mass?

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## marked as duplicate by Waffle's Crazy Peanut, Manishearth♦May 21 '13 at 13:57

Hi Tom. Welcome to Physics.SE. Before asking questions, it's always good to search for them or atleast have a look at the Related list. For now, your question has already been solved by many answers. A possible duplicate: physics.stackexchange.com/q/22876/11062 –  Waffle's Crazy Peanut May 21 '13 at 13:39
Gravity is the curvature of space-time and it affects energy, not only mass. There must be some post explaining this. –  jinawee May 21 '13 at 13:41
@ϚѓăʑɏβµԂԃϔ: IMO the question also mentions time, so it's not a dupe. –  Manishearth May 21 '13 at 13:52
@Manishearth: Hi Manish. I told that there are many questions that can cover that. OP has asked 2 in 1. For example, here's the question covering time..! Actually, I sped up and so, I was unable to insert more posts. In fact, the tags have almost all related posts I hope so ;-) –  Waffle's Crazy Peanut May 21 '13 at 13:55
@ϚѓăʑɏβµԂԃϔ: Good point, closed :) –  Manishearth May 21 '13 at 13:57

Firstly, the way gravity affects time is completely different from the way gravity affects mass. Do a search on "gravitational time dilation". Secondly, gravity affects light because it has energy. This is governed by the Einstein Field Equation. Thirdly, the black-holes tag is irrelevant since all objects do have gravitational pull. Fourthly, the visible-light tag doesn't apply here. It applies to optics only, that too in the visible spectrum. Even if you want to talk about light, there is the electromagnetic-radiation tag. It is clearly stated in the description.

Edit:

Time dilation is very different from objects getting attracted to gravity. Instead, it is because of the curvature of spacetime.

Gravitational time dilation is given by

$$\frac{c_0^2\mbox{d}t^2}{\mbox{d}s^2}$$

For a Schwarzschild metric, assuming the velocity of the observer is 0 all the time, for example, it is

$$t=\tau\sqrt{1-\frac{r_s}{r}}$$

Where t is coordinate time and tau is proper time.

This is obtained from the Einstein Field Equation (EFE).

It can be obtained by applying Hamilton's principle/Principle of stationary action to the Einstein Hilbert Lagrangian Density:

$$\mathcal L=\lambda R$$

Then, it can be found that

$$G_{\mu\nu}=\frac{1}{2\lambda}T_{\mu\nu}$$

Where $T_{\mu\nu}$ is the Stress-Energy-Momentum Tensor and we defined the Einstein tensor $G_{\mu\nu}$ as:

$$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$$

By ensuring that GR goes to Newtonian Gravity at the classical limit,

$$\frac{1}{2\lambda}=\frac{8\pi G}{c_0^4}$$

So clearly this involves the stress energy momentum tensor which has momentum in its componnents, obviously by definition.

So, in conclusion, it makes NO sense to say time has mass, its only a dimension and light has no (rest) mass but has momentum, which can be given by:

$$p=\frac{\hbar\omega}{c_0}$$

The momentum also contributes to its stress-energy momentum tensor and thus to the gravitational pull.

Finally, one may calculate an SR approximation for the gravitational effect between a massive body and light or light and light using Newtonian gravitation and (special) relativistic mass.

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This really doesn't answer the question, and answers pointing the user to look for an answer elsewhere are discouraged. This is probably why it's getting downvoted, you haven't really explained anything here :/ –  Manishearth May 21 '13 at 14:04
I edited your first line "gravity affecting time is completely different from gravity affecting mass" as I did a double take on reading it: it might have been read as saying there were two different "kinds of gravity", which is clearly not what you mean. Hope you agree. –  WetSavannaAnimal aka Rod Vance Sep 1 '13 at 5:46
@WetSavannaAnimalakaRodVance: Ok, thanks, that's probably better. –  Dimensio1n0 Sep 1 '13 at 5:52