# B My inquiries about light

1. Jun 23, 2017

How does gravity affect light if its massless? And how does light apply force if its massless?

2. Jun 23, 2017

### Staff: Mentor

Gravity affects everything. It bends spacetime. Light simply follows straight lines in this curved spacetime, which looks like a curved path seen "from outside".
It does not have mass, but it has energy and momentum. If you change the momentum of light, you get an opposite momentum change in the object that changes the light.

3. Jun 23, 2017

### Orodruin

Staff Emeritus
In general relativity gravity is not described as a force but as the geometry of space-time and light, like other things, moves in this geometry and is affected by it. Also, the source of gravitation in GR is not mass but energy, momentum and stress. In the limit where small masses are moving slowly, the mass energy dominates all other contributions and you essentially recover mass as the source of gravity as in Newtonian gravitation.

4. Aug 10, 2017

### Creative name456

This is one of the problems with Isaac Newton's laws of gravity. According to Newton, in order to have a gravitational pull between two objects, both objects must have mass. However, light bends when under the effects of gravity because of something called the principle of equivalence, which was conceived of by Einstein. Let's say you put a person in a box(let's call it box A), and place them on the Earth. Box A is being pulled down at 9.8 m/s^2. Then you put another person in another box(let's call it box B). Box B has a rocket strapped to the bottom of it, and is blasting off into space accelerating at 9.8 m/s^2. Now, no amount of measurements or calculations you could take would be able to tell you whether you're in box A or box B. Thus, the principle of equivalence. Now, let's say you flipped on a flashlight and shined it through the universe. The light would appear to bend because of the fact that you have travelled upwards compared to where you flipped on the light. So, light bends under the effects of gravity.

5. Aug 10, 2017

### Creative name456

Light does not have energy. This is what allows it to travel as fast as it does. According to the formula E=Mc^2, the energy of light is zero because it has no mass. 0=(0)c^2

6. Aug 10, 2017

### Staff: Mentor

Wrong; it does.

Which does not apply to light. This formula is actually a special case of the correct general formula, which is $E = \sqrt{m^2 c^4 + p^2 c^2}$, where $p$ is the momentum. The famous $E = mc^2$ is what you get from this general formula when $p = 0$, but that is only possible for an object with nonzero rest mass $m$. Light can never have zero momentum (because it always moves at $c$), so it always has energy even though it has zero rest mass.

7. Aug 10, 2017

### Ibix

Someone should tell solar cell manufacturers!

8. Aug 10, 2017

### Orodruin

Staff Emeritus
Light bends because of the geometry of space-time. The principle of equivalence gives you a heuristic argument for a local observer with proper acceleration. There is nothing local about gravitational lensing.

When given alone, I usually consider it as an expression for the rest energy relation to the low-energy inertial mass, which is a fundamental insight. More appropriate might be to write it $E_0 = mc^2$. Also worth mentioning is that the full formula reduces to $E = pc$ in the massless case.

9. Aug 10, 2017

### Mister T

It's true that in the newtonian approximation, things have to have mass to be affected by gravity. That is an approximation that works so well physicists thought it was universally valid. But in the last 100 years all that has changed. Theory predicts, and observation confirms, that you don't have to have mass to participate in the gravitational interaction.

10. Aug 10, 2017

### Mister T

That's a perfectly good reason for rejecting the relation $E=mc^2$ and replacing it with $E_o=mc^2$, as Einstein himself did. For a single ray of light, $E=pc$ because $m=0$. You can see this by looking at the more general case where $m$ may or may not be zero and $E=\sqrt{(pc)^2 + E_o^2}$.

Another special case is when $m \neq 0$ and $p=0$. In that case $E=E_o$. A frame where $p=0$ is called a rest frame, and this is where we get the name rest energy for $E_o$. This leads to some confusion because it's not possible for the momentum $p$ of a single ray of light to be zero, but we still say that its rest energy is zero, an unfortunate linguistic circumstance.

Note though that for a collection of light rays it is possible to find a rest frame where their total momentum is zero. And in these cases the collection does indeed have a nonzero mass, even though each ray separately has no mass. Thus we see that the mass of a collection is not in general equal to the sum of the masses of its constituents.

All of this is part of the meaning of one of Einstein's greatest discoveries, the equivalence of mass $m$ and rest energy $E_o$.

Last edited: Aug 10, 2017
11. Aug 10, 2017

### Staff: Mentor

In Newtonian gravity, you could interpret light as particle with a tiny mass, then it would be bent by gravity just like regular matter would.
In General Relativity, you can calculate that the real bending is twice the value you get from Newtonian gravity - and this factor 2 (confirmed during a solar eclipse 1919) was very important to establish the theory.

12. Aug 11, 2017

### vanhees71

Newtonian physics and light fits not well together since light is as relativistic as something can get, because it's nothing else than electromagnetic radiation, which is adequately described only by the Maxwell equations, which are relativistic, because they are massless fields.

You can mathematically prove that massless fields do not make any sense within Newtonian physics by analyzing the unitary ray representations of the Lorentz group. As proven by Enönü and Wigner in a famous paper, proper unitary representations of the classical Galilei group do not lead to wave equations that are interpretible as describing some physically sensible dynamics. You have to introduce a non-zero mass as a central charge of the Galilei Lie algebra to get usual non-relativistic quantum mechanics right, and the massless limit doesn't make sense.

That's why you have to treat electromagnetism as a relativistic field theory. To treat gravity relativistically you end up almost inevitably with Einstein's GR or very similar theories. Within Einstein's GR the sources of the gravitational field is the energy-momentum-stress tensor of "matter" (i.e., all fields there are in nature except the gravitational field). As soon as you consider everything within a consistent relativistic theory, all apparent paradoxes concerning the gravitational influences on (and of) light are resolved immediately.

13. Aug 11, 2017

### sweet springs

Hi.
Newtonian gravity is a mutual interaction, that product $m_1 m_2$ matter. Einstein gravity is individual. $T^{\mu\nu}_1$ changes geometry of its around and $T^{\mu\nu}_2$ changes its around independently. In this sense amount of $T^{\mu\nu}_2$ does not matter with gravity effect of $T^{\mu\nu}_1$ on body 2 and vice versa. I write here $T^{\mu\nu}$ not m with the same reason that the above posts. Anyway massless or I would rather say even energyless is OK as recipient of gravity originated from another body or energy. Best.

Last edited: Aug 11, 2017
14. Aug 11, 2017

### Orodruin

Staff Emeritus
This is not really true, the Einstein field equations are non-linear. However, the Newtonian theory of gravity is linear and so the exact opposite of your assertion here is true.

15. Aug 11, 2017

### sweet springs

Not force but acceleration more exactly geometry I take care in my #13. Is massless or energyless of recipient OK?

Thanks for your teaching of non linearity. Non linearity matters when we have more than two gravity sources in the sense that linear superposition of their contributions does not work.
In case one gravity source ,i.e. there exist only two bodies that each one get influence of the other I am not sure whether non linearity matters? You count self interaction also for non linearity? Best.

Last edited: Aug 11, 2017
16. Aug 11, 2017

### Orodruin

Staff Emeritus
If there are two bodies, then there are two sources. The general two-body problem in GR does not have a closed form solution. You can only find some approximations, such as the one where you assume one of the bodies to act as a test particle.

17. Aug 11, 2017

### sweet springs

Thanks. Schwartzshild metric is an example of your test particle case I suppose where only one mass m appears in the metric. In approximation no mass or energy of second particle appears in geometry. In your strict sense manner of falling into a black hole varies according to masses of falling bodies for an example.
In OP's case second "particle" is light so tiny energy compared to the first particle of gravity source usually, your test particle case may apply.

Last edited: Aug 11, 2017