# Photons and gravity

1. Sep 7, 2007

### JPC

was wondering :

photons are said to have a m0 = 0 (a mass of 0 at rest), but since they are constantly moving , they must have a mass, because of the energy of their movement ? because isnt energy mass ?

and if they have mass, it must mean that they dont move in straight lines, but in slighly curved lines, because of the gravity of nearby objects ?

am i right ? half right ? or completely wrong ?

2. Sep 7, 2007

### pervect

Staff Emeritus
There are a couple of issues here. One of them is that the term "mass" has at least a couple of different definitions. (If you look even further, you can find more than two!). But the ones most relevant to your problem are discussed in a couple of the sci.physics.faqs:

http://www.math.ucr.edu/home/baez/physics/ParticleAndNuclear/photonMass.html [Broken]

Does mass change with speed?

I'll quote from the second in part, you should probably read both of the above links for a fuller explanation.

There is a third issue here. You are assuming, based on Newtonian theory, that only "mass" is relevant to gravity. This is correct in Newtonian theory, but not necessarily true in general. I could get into more detail, but I don't want to digress too far from your original question. And the answer to your original question is that photons have energy, and they have momentum, they are even (occasionally) said to have "relativistic mass", but they have a zero "invariant mass".

Because of the later fact, and the widespread use of "invariant mass" in physics, aying that a photon has "mass", without any qualifications, is generally regarded as *wrong*.

Last edited by a moderator: May 3, 2017
3. Sep 7, 2007

### DaveC426913

And yes, photons are affected by massive objects. That's how we get gravitational lensing.

4. Sep 8, 2007

### JPC

ok, so photons do not move in exact straight lines

-----------------------------------------------

lets say we have a universe, with nothing in it except an object and a photon

a) if the object is in front of the photon :
> does the photon stay at speed c because it cannot go faster than c ?

b) if the object is behind the photon :
> does the photon decrease in speed, going slower than c ? meaning that its speed will constantly decrease with time (this decrease would decrease as it goes further away) ? or does it gets closer to c as it moves away from the object (the photon getting back its normal speed when less disrupted by gravitational forces) ?

c) if the object is sideways of the photon
> will it constantly accelerate sideways (this acceleration will decrease as the photon goes further away) ?

Last edited: Sep 9, 2007
5. Sep 13, 2007

### Shooting Star

You are trying to model the situation with the photon as a test particle moving in the gravitational field of much more massive body using Newtonian gravitation. Let me try to answer you qualitatively, as a first approximation to the real phenomenon.

Let's take the massive body as the frame of reference. The speed of the photon does not increase as it falls toward the body, but instead its energy increases, and so its frequency increases. Similarly, the energy decreases as it recedes from the body, along with its frequency. When, as you put it, it’s moving sideways, its energy and frequency is maximum when it’s closest to the body.

So, the speed is always c, but it's the energy that increases or decreases, which is also true of an object moving in a Newtonian gravitational scenario. But the path of the photon here, using General Relativity, is slightly different from that of a test object using Newtonian gravitation.

Last edited: Sep 13, 2007
6. Sep 14, 2007

### JPC

Ok , so meaning that masses around us slightly change the colors we see (changes too small for it to make a difference to our eyes) ?

but now, about frequency, if the photon has a frequency , it means that it repeats a small movement (oscillates) as it advances ?
Because if like the photon moves forward in a straight lines, but that during that it moves right and left, it would mean that its true movement (frame of reference : earth) would look like a sinx or cosx equation ? meaning that photons dont really move in straight lines ?

if this is true :
- what distance left and right they oscilate ?

7. Sep 14, 2007

### Staff: Mentor

No, it does not. The wavelike nature of light causes diffraction and interference, but it does not mean that photons actually wiggle from side to side as they travel. Electrons and other particles also display diffraction and interference effects, and they don't wiggle from side to side, either. Welcome to quantum mechanics!

8. Sep 14, 2007

### meopemuk

I cannot agree with your idea that the frequency of light changes when light passes through gravity field. Suppose that the light source is at the height H above the ground and you are measuring its frequency by a device on the ground. Suppose also that the light source emits radiation of frequency f. This means that there are 1/f wave crests generated per second in the reference frame associated with the source (at elevation H). I think that observer on the ground should also see exactly 1/f wave crests per second. Otherwise, where the additional crests would come from?

This is similar to the situation when somebody at elevation H drops stones at the rate of 1 stone per second. The stones must hit the ground at the same rate 1 stone per second. Otherwise we would have a trouble with the law of conservation of stones.

Eugene.

9. Sep 14, 2007

### Shooting Star

You've got that part right, even though the change in colour would be impossible to visually detect by us under normal conditions. But modern instruments can detect the change in frequency of light coming from the bottom of a high building when it reaches the top.

As for the frequency of a photon, when a large number of them is present, the phenomenon can be treated like a classical wave, and it is the frequency of this wave we are talking about. You must look up a highschool level text book to learn about the basic concepts.

10. Sep 14, 2007

### Shooting Star

Not my idea really. Started with Einstein, and now about 99% of the scientific community of the world subscribe to it. The rest 1% do believe in the result but have got different theories.

You have not taken into account that time flows at a different rate at the bottom of the tower than at the top because of the different gravity. The two are exactly related such as to conserve your number of crests or number of stones. In fact, counting the light crests was the argument given by Einstein in his warm-up to his Theory of General Relativity.

11. Sep 14, 2007

### meopemuk

There is an alternative explanation of gravitational red shift experiments (e.g., Pound & Rebka), which does not assume the change of photon's frequency as it travels through the gravity field. This explanation seems more logical to me and to other authors, e.g.,

L. B. Okun, K. G. Selivanov, V. L. Telegdi, "On the interpretation of the redshift in a static gravitational field" Am. J. Phys. 68 (2000), 115; http://www.arxiv.org/abs/physics/9907017

In this explanation, it is acknowledged that the rate of any periodic process and the frequency of photons emitted by a physical system depends on the separation of energy levels in the system. In the gravitational field, the energy of levels is lowered, so the energy separation is decreased too. Therefore, all clocks go slower on the Earth surface than in the free space. Atoms on the Earth surface emit lower energy photons than identical atoms in space. So, the red shift is explained by the effect of gravity on the sources of photons rather than by its effect on the photon's energy and frequency.

Eugene.

Last edited: Sep 14, 2007
12. Sep 15, 2007

### JPC

what's the 'separation of energy levels in the system' ?

13. Sep 15, 2007

### meopemuk

Take, for example, a hydrogen atom. We know that this bound system has discrete energy levels $E_0, E_1, E_2, \ldots$. It emits radiation in transitions between such energy levels. The energy of emitted photons is equal to the separation of corresponding atomic levels $h \nu = E_i - E_f$.

In an attractive gravitational field all atomic energies are reduced

$$E_i' = E_i (1 - \frac{GM}{rc^2})$$

Correspondingly, all energy differences $E'_i - E'_f$ and energies of emitted photons are reduced too. This is called "the gravitational red shift".

Eugene.