# I Cosmological redshift

1. Jan 2, 2018

### exmarine

In studying the chapters on cosmology in Misner/Thorne/Wheeler (what a textbook!), I see that the cosmological redshift is different than Doppler. If I understand correctly, it is due to the expansion of the (Friedman, FLRW) universe during the photon’s long journey to us, rather than being due to some enormous retreating velocity during its emission. So my question is how do we distinguish between the two effects in astronomy? Or can we? Or do we need to tell the difference? If we have recently determined that the “expansion of the universe” is accelerating, is that one type or the other, or the sum of the two, etc.?

And where do the energy and momentum of a red-shifted photon go as it is reduced, either during emission or its journey to us?

Thanks.

2. Jan 2, 2018

### jbriggs444

The notions of energy and momentum conservation in an expanding universe are not as simple as one might prefer. The pithy version is that the energy does not have to "go" anywhere because "energy" in the simplistic form that we understand it is not a conserved quantity.

A quick search turns up: https://www.physicsforums.com/threads/is-energy-conserved.798114/#post-5011166

3. Jan 2, 2018

### Orodruin

Staff Emeritus
This is a truth with serious caveats. In the end, they both boil down to the same thing: parallel transport of the 4-frequency along the null geodesic and inner product with observer 4-velocity.

In fact, in local normal coordinates, comoving objects are actually moving apart and the local redshift is therefore exactly a Doppler shift. (Edit: This is a fun and meaningful exercise to show! I never felt I understood the relation between expansion and local velocity difference until I sat down and did it explicitly.)

Last edited: Jan 2, 2018
4. Jan 2, 2018

### Arman777

You can check here for answers, I think it explains your question perfectly.
http://astronomy.swin.edu.au/cosmos/c/cosmological+redshift

It's due to cosmological redshift, cause cosmological resdshift is caused by the expansion of the universe, meanwhile the doppler shift is caused by the motion of the objects at that instant while they emit light. Also the link will be helpful.

Just to space ?

5. Jan 2, 2018

### Orodruin

Staff Emeritus
As I stated above, the separation into cosmological, Doppler, and gravitational redshift really is an artificial one based on particular coordinate charts. In the end, the redshift depends on the parallel transport of the 4-frequency and the observer 4-velocity, which gives the coordinate independent statement.

6. Jan 2, 2018

### Arman777

Are you trying to mean that, in comoving coordinate system doppler redshift is equal to cosmological redshift.

So, they are the same thing but we use different names due to different coordinate system ?

But even thats the case, I am not sure. Maybe mathematically they are the same but do they mean same thing also physically ?

I find this,

"Practically speaking, the difference between the two (Doppler redshift and cosmological redshift) is this: in the case of a Doppler shift, the only thing that matters is the relative velocity of the emitting object when the light is emitted compared to that of the receiving object when the light is received. After the light is emitted, it doesn't matter what happens to the emitting object - it won't affect the wavelength of the light that is received. In the case of the cosmological redshift, however, the emitting object is expanding along with the rest of the universe, and if the rate of expansion changes between the time the light is emitted and the time it is received, that will affect the received wavelength. Basically, the cosmological redshift is a measure of the total "stretching" that the universe has undergone between the time the light was emitted and the time it was received."

7. Jan 2, 2018

### Arman777

I agree with you, same math but different description.

8. Jan 2, 2018

### Orodruin

Staff Emeritus
No. I am saying that frequency is one thing and one thing only, the product of the 4-frequency and the observer/emitter 4-velocities. This is the coordinate independent statement. There is no such thing as “the frequency of this/that wave” because frequency is observer dependent in much the same way as energy is. (You cannot say “the ball has x kinetic energy” without specifying the observer that makes the measurement)

Hence, there is only redshift, which depends on the ratio of the frequency of the signal for the receiver and the frequency of the signal for the observer.

Obviously. There is no measurement you can make of the signal itself that will tell you it was one or the other. It is just an effect of what frequency is to an observer and the geometry of spacetime.

To me, this sounds as if the writer has not thought about the issue enough and/or is making a very simplified and arbitrary division. It is easy to show that locally what is “Doppler shift” in one coordinate system is “cosmological redshift” in another. Any separation is arbitrary and coordinate system dependent.

I cannot see how you can separate the two. The math is the description.

9. Jan 2, 2018

### kimbyd

The issue there is that the curvature of space-time between the source and emitter makes a difference cosmologically.

One way to say it is that the simple version of a Doppler shift doesn't not work in cosmology because it doesn't take into account space-time curvature. As Orodruin correctly notes, however, it is possible to extend the classical notion of a Doppler shift to take the curvature (meaning: the expansion of the universe) into account. This more generalized notion of a Doppler shift does properly describe the redshift of cosmological objects.

However, this isn't usually what's done. What people working in cosmology usually do is separate the redshift into two pieces:
1) The local movement of the far-away galaxy relative to its surroundings, as compared to our local movement with respect to our surroundings. This component can be understood as a simple, classical Doppler shift.
2) The redshift which occurs due to the expansion of space over the time from when the light was emitted to when it was observed. This is the cosmological redshift.

I think it's this understanding of how redshift operates that MTW was trying to explain. The other one, using the generalized Doppler effect which uses General Relativity, will result in all of the same answers.

10. Jan 2, 2018

### Orodruin

Staff Emeritus
I used to think like this, then I did the computation of how comoving observers behave in normal coordinates. Locally (meaning loally enough to neglect curvature effects) the only difference between “Doppler” and “cosmological” shift is only the coordinate system used to describe it, which clearly is not physical in itself. I really suggest that people try this calculation. I might have to write it down in an Insight because it is something that is clearly not appreciated enough, even among cosmologists.

11. Jan 2, 2018

### Arman777

An insight would be great

I can try to calculate it but I dont think I can do that.

But from the recent posts, I understand the idea, thanks.

12. Jan 2, 2018

### kimbyd

When I read this, what I understand amounts to:
"When taking space-time curvature into account, you need to modify the Doppler shift equations in a particular way. But you can make a specific coordinate choice that sets that modification to zero, so that you can still use the old Doppler equations as long as you're careful."

This fact can be read either way, really. You can read it one way and say that the cosmological redshift and the Doppler shift are basically the same thing, because they are the same thing in some coordinates. Or you can read it as stating that cosmological redshift and Doppler shift are two similar aspects of a different property: the redshift, which is what fundamental. Distinguishing one kind of redshift from another comes down to arbitrary choices that will change the answer depending upon said choices.

One of the things that my GR professor emphasized is that when you want to do GR correctly, what you need to focus on are observables. In this case, the redshift is an observable. What's not observable is how much of the redshift is down to Doppler effects and how much is down to cosmological expansion.

Orodruin shows, quite correctly, that you can describe the whole redshift as Doppler. But it's also to describe it as some coming from Doppler effects, and some coming from the intervening space-time curvature. Either description is accurate, and they both have merit in teaching something useful about how the redshift occurs. Neither description, however, should be taken as some sort of fundamental description of reality. The fundamental description is only that there was some specific amount of redshift (the amount observed).

13. Jan 2, 2018

### Orodruin

Staff Emeritus
You do not even need a curved space-time. The most pathological example is a 1+1-dimensional space-time with a scale factor $a(t) = kt$ for some constant $k$. Clearly, as $a(t)$ grows in time, there will be a “cosmological” redshift due to the expansion.

Now here is the surprise to many: what was just described above is the future light cone of an event in Minkowski space. Should we instead impose regular Minkowski coordinates (which happen to be normal everywhere) then this “cosmological” redshift is a regular SR “Doppler” shift - there is no expansion.

I agree with your GR professor. The only things that are physical are invariants and coordinate independent objects such as tensors (as contrasted to tensor components, which are coordinate dependent). Coordinate dependent quantities such as the scale factor (which is directly tied to comoving coordinates) or the split in cosmological vs Doppler or gravitational vs Doppler shift may give you a particular heuristic and aid thinking but should not be taken as dogma.

In particular, one should be aware that anything related to comoving coordinates, including comoving distances, utterly depend on a simultaneity convention that does not agree locally with the one we would be used to from SR.

To me, redshift is the same question as asking what angle a geodesic makes to two vectors at different points it goes through. Just in a Lorentzian manifold instead of a Riemannian one. Simply because this is the invariant physics that lie underneath.

14. Jan 2, 2018

### Orodruin

Staff Emeritus
Just to emphasize that this is the key point of the matter. You will find several accounts online where such an arbitrary choice is made and imposed as truth without any qualifications regarding the assumptions made to do it or their arbitrariness. The article linked to in #6 being an example of this.

15. Jan 2, 2018

### strangerep

16. Jan 3, 2018

### PeroK

I think you are allowing the complexities of the phenomena to obsure a very simple, but important point:

Suppose a ball is thrown down from a building with an initial speed and is accelerated by gravity on the way down. You measure its speed as it reaches the ground.

Part of the final speed is due to the initial speed. Part of the final speed is due to the acceleration. And, if you know enough about the experiment, you can calculate these proportions. All you have on the ground, however, is a single measurable speed. You cannot directly measure these components of the final speed. There is only one speed - it doesn't come with measurable initial and additional components.

It's not that the maths is the same, it's that you have a single measurable quantity.

In the same way, a light wave that has reached you will have a wavelength. A single measurable wavelength. You cannot tell from that measurement alone how the light wave got to have that wavelength. You can't look at the wavelength and see direct evidence that it was emitted by the source as $\lambda_0$, was redshifted (or blueshifted) to $\lambda_1$ due to the initial velocity of the source relative to you;, and was further redshifted to $\lambda_2$ due spatial expansion during its travels.

There is only one wavelength to measure and there is no way to measure directly how it got to be that wavelength. It's not to do with maths, but with what you can measure.

PS In fact, I would say in both these cases the maths is different. The mathematics of the Doppler redshift and expansion redshift are different. But, they do not produce physically, measurably different types of wavelength. You can only tell that something has been Doppler shifted, say, by additional measurements, and deduction or inference.

Last edited: Jan 3, 2018
17. Jan 3, 2018

### Orodruin

Staff Emeritus
Again, they are not different at all. Any redshift calculation boils down to parallel transport and inner product with a 4-velocity. It is only a question of how you complicate your life to actually do it. The same actually goes for your example with the ball. How much is ”initial” and how much is ”accelerated” velocity depends on the frame you consider, but the maths to describe it is still classical mechanics.

18. Jan 3, 2018

### PeroK

I wasn't thinking of "maths" as general as "classical mechanics". I was thinking more specifically the maths used to calcuate the phenomena.

True, the maths used to make the measurement is the same. But, the maths to explain the phenomena are different.

In the same way, changing your frame of reference changes the maths but not the phenomena.

19. Jan 3, 2018

### Arman777

20. Jan 3, 2018

### Orodruin

Staff Emeritus
21. Jan 3, 2018

### PeroK

Ah, I think I see your point. We can also always generalize the maths so that it is the same. We don't have to do the maths differently.

That's a good point and beyond what I was thinking about.

You didn't win that Physics Award for nothing!

22. Jan 5, 2018

### Orodruin

Staff Emeritus
Fine. I started. One Insight coming up. I make no promises as to the publication date though ...

23. Jan 5, 2018

### Arman777

Thanks

24. Jan 6, 2018

### Orodruin

Staff Emeritus
25. Jan 6, 2018