# Energy shift in a photon passing by a gravitational well

1. May 5, 2013

### elevin

Gravity effects the energy of a photon per the equation GM/c^2r. I understand that a photon traveling on a crash course with a black hole will be blue shifted according to the hole's mass, however I'm a little confused as to what happens when a photon simply passes nearby a massive object. For example, if photon A passes by black hole B, it will be affected in proportion to its distance to the object. Is it fair to say that the photon will be blue shifted as it's distance lessens and then redshifted in equal amount when it passes by and the distance increases? Or would the sum total energy of the photon be lessened by having this interaction? How about if there is two black holes and the photon passes directly in between them?

2. May 5, 2013

### Simon Bridge

A photon falling through a gravity well gains/loses energy the same as anything does, though it's world-line is somewhat special. Since no observer measures a change in speed, the energy changes manifest as changes in color.

Think of a thought-experiment with a beam of photons which gets sampled by detectors at different positions about the massive body(s).

3. May 5, 2013

### Popper

That is incorrect. Choose any observer at rest with respect to the black hole. That observer wil measure no change in frequency of a photon. What changes is what is observed by different observers located at different distances from the black hole.

All that will happen is that it will change direction. The frequency will remain unchanged. Think of it like this: any particle moving in a static gravitational field will have a constant energy due to energy conservation. Therefore the energy will remain unchanged and so will thus so will the frequency. It's only different observers who measure different values of energy/frequency. Observers lower in the field will see photons coming at the with higher frequency than observers who are further out. Observers who emit light at positions closer to the black hole will be seen as red shifted by observers further away.

As the poster above said - Think in terms of observers.

4. May 6, 2013

5. May 6, 2013

### Staff: Mentor

The delay isn't due to the light slowing down, it's due to it having to travel farther.

6. May 6, 2013

### Bill_K

Not if it's aimed directly in between them, but I do believe in the case of two black holes orbiting each other a well-directed light ray [why do we always call it a "photon"? This is purely classical!] could be blue-shifted, ala the "slingshot effect" or gravity assist that's used to accelerate spacecraft on planetary missions.

7. May 6, 2013

### pervect

Staff Emeritus
Pretty much. In the Schwarzschld metric $E = \sqrt{ \left| E^0 E_0 \right|} = \sqrt{ \left| g^{00} E_0 E_0 \right| } = \frac{1}{\sqrt{1-2GM/rc^2}} E_0 \approx E_0 \left(1 + GM/rc^2 \right)$

Where E is the energy of the photon as measured in the local frame, and $E_0$ is the so-called "energy at infinity", which is a constant of motion. At infinity the two are equal (hence the name).

If you have one stationary black hole, the energy at infinity is a constant of motion, and the locally measured energy E will approach the energy-at-infinity as r approaches infinity. Thus the locally measured energy E will be the same before and after passing near the massive object.

In more general cases, one needs to do the calculations.

8. May 6, 2013

### Simon Bridge

@Bill_K: isn't a "corpuscle of light" a classical concept?
But I digress - it is not very helpful to discuss what happens to individual photons here.

@elevin: you appear reluctant to do conduct your though experiments as per post #2. Is there a special reason for this? Do give it a go - it should help with your understanding.

Note: the speed of light is measured the same for all observers ... it follows that EMR cannot "slow down" as suggested in post #4. Therefore the observation must be incorrect. But you knew all this already - there is nothing here you cannot figure out yourself.

Everything follows from the invariance of the speed of light, conservation of energy, and the mass-energy relation, all of which you know about. Just make sure you include the observer in your thought experiments. Different observers will agree and disagree about some of the things they observe depending on their frames of reference. That's what relativity is.

9. May 6, 2013

### elevin

Hey Simon - I agree with you (also I know EMR doesn't slow, was a slip of the tongue). The last math class I took was high school algebra so basically everyone on here has me beat. I apologize for my lack of experimentation.

I'm in a cosmology for dummies class at Harvard Extension (yes that's supposed to be funny), working on a paper and have literally been sitting at my computer for the probably 100 hours over the past week typing this thing. The original mechanics I was basing my paper on had to do with GM/c^2r where a light wave would lose energy near a massive object (as an explanation for tired light). I then learned it's not necessarily the case, that light entering a gravitational well gains energy and then leaving the well loses it with a net zero effect (hence why Pound-Rebka indicates a blue shift).

I can't argue with physics, but this is just so dang counter-intuitive to me. Per the principle of equivalence it would seem that I could find an equivalent gravitational red shift substitution for acceleration. All hypothetical, but if my body somehow exerted a tremendous gravitational force, enough to warp space-time and pull a star toward me, unless that star followed the curved shape of space and started accelerating toward me, that it would intuitively be accelerating away by simply holding its position. If it did hold it's position despite my curving space toward me, then it would be red shifted, no?

Is there a difference between what I described and the pound rebka? I know this has to do with the object emitting the light and not the light wave itself, but it's analogous to my standing on the surface of the earth looking outward. Anything not falling into the earth's gravitational well seems it would be redshifted (by a minor minor minor amount), but that's not the case. The light gains energy. Likewise if I'm unmoving on a hypothetical black hole, all light would be blue shifted. Seems weird.

Any answers to this are appreciated, however I don't speak math so...

10. May 6, 2013

### Simon Bridge

"Counter-intuitive" is normal - which is why you need the math.

In relativity you have to get really pedantic about reference frames and observers and who is observing what - especially when you are starting out.

I'll try give you an idea about reference frames.
Stand up and hold a pen out in front of you at arms length.
Look at the pen.
You, the pen, and the chair you were sitting on, are all stationary in the same reference frame.
This will be the rest-frame of the chair.

Now turn on the spot, keep turning, keeping the pen in front of you.
The pen is still stationary in your reference frame, but it is now accelerating towards you in the rest-frame of the chair.
In your reference frame you notice some odd stuff - for instance, the pen keeps trying to pull away from you. Since it is accelerating towards you, you'd think it would push on your hand but it pulls instead. That is because it is stationary with respect to you but would rather not be: you have to hold it there. This is the centrifugal effect and you see it because you are in a non-inertial reference frame. [*]

The equivalence principle is like this - it basically says that when you experience a gravitational field, you are experiencing an effect of being in a non-inertial reference frame. Objects can move about in that reference frame just fine - and be doing something different in another reference frame.

With the pen-experiment, you would intuitively prefer to call the rest-frame of the chair the really "real" frame, and everything else is somehow less real, producing illusions etc. The big difference with Relativity is that there is no preferred reference frame. This is the big counter-intuitive leap you need to make.

I think you should get used to special relativity before tackling general. Have a read through:
http://www.physicsguy.com/ftl/
... should be accessible and clue you in on some of the math.

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

[*] there is a limit to this analogy - it is pretty clear that the frame in which you are spinning and the one in which you are still and everything else (except the pen) is spinning about you are not equivalent - after all, you are the one getting dizzy. You can stop now BTW.

11. May 7, 2013

### Agerhell

Really? I would say that the delay mainly is because light is slowing down. Of course, locally, clocks are slowing down at the same rate so using local clocks you will not be able to notice...

Sure there are some bending of the path of the light which will cause the light to travel further, however, this will actually work to minimize the delay as light will choose the path that minimises the amount of coordinate time it takes to get from A to B...

12. May 7, 2013

### Simon Bridge

You would say? Based on what? In what way?
All the math I've seen so far that successfully predicts the delay kinda relies on the light not slowing down.
You mean something like this answer?

Let's think of light from some distant star. There's an extra delay in how long it takes to reach us when the light happens to pass near the sun on its way. How come? We can describe it in a particular choice of coordinates, the Schwarzschild coordinates. Two things happen to the light as it goes near the sun:

1) Close to the sun, the effective rate at which time passes is slowed. According to local clocks there, the light is traveling at the usual speed, c, but we think those clocks are slow so from our point of view the light is going slower.
2. As it approaches and departs from the vicinity of the sun, the light travels extra distance, more than what you would calculate if you drew a big circle around the sun and took the diameter to be its circumference over 2pi. Space isn't Euclidean- the diameter of that circle is bigger than it should be based on the circumference. So the light has farther to go (as measured by local rulers) than it would if it weren't going near the sun.

These effects add up to give the Shapiro delay.

Should you say that under these circumstances the light travels farther? In our coordinate choice, that does account for half the effect. However, there are lots of different coordinate choices. Whichever one you like, the effect is real.

... maybe this is a topic for another thread?

13. May 9, 2013

### pervect

Staff Emeritus
A lot depends on seemingly minor, but ultimately crucial, details. There are a myriad of different clocks, ticking at different rates, one might choose, If one chooses proper time (as kept by actual, physical clocks) and proper distances, then there is *no* change in the speed of light. (Or, in modern terms, there is no change in the measured length of a proptotype meter standard that one carries around from place to place, by measuring the proper time it takes for light to make a 2-way trip across the meter standard and back).

Modern emphasis is mostly on the "no change" approach, I think. This ties in with the modern trend to de-emphasize coordinates.

The modern approach is also the one that appears in the NIST definition of the second, and the meter:

http://physics.nist.gov/cuu/Units/second.html
Note that is no mention of "where you are" in the defintion of the SI second. Or the SI defintion of the meter (which defines the speed of light as a constant).

If you take the "speed of light slowing down" really seriously, you'd have to start replacing the NIST defintions with different ones, because the NIST defintions do not define the speed of light as changing depending on your location.

With the modern interpretation, the rate of change of the PPN position coordinate with respect to the PPN time is not really a velocity. To make it an actual velocity, one needs to adjust it by the local metric coordinates, so that it reflects what a local observer would measure, not what a PPN observer would measure.

Last edited: May 9, 2013