# I Proportionality of the Redshift

1. Jun 22, 2016

### JohnnyGui

Good day all,

Sorry if this has been posted a lot before but I've been fiddling with the formulas for the redshift and I came up with a question regarding the proportionality of it. I'm new at this so please bear with me. For the sake of argument, I'm talking about redshift in an expansion with a constant velocity. Here it goes:

I was able to conclude that ΛObs / ΛEmit = D2 / D1 = z + 1 where D1 is the initial distance of a star when it emitted the light and D2 the distance of the star when the light is received.

The D2 can be rewritten as v(D1/c) + D1 and plugging that in the formule gives me:

(This formula can be dumbed down to v/c = z which is what z indeed is)

Here's my problem. It seems that when plotting this relationship of v D1 and z, using v and D1 as the x variable (since they're related through the Hubble value) and z as the y, that z is proportional to v and D1 even when v reaches near lightspeed. This is obvious since the formula is basically x / c = y after simplifying.
I have however read that z shouldn't be proportional to the velocity if it gets too high but my graph doesn't show this at all.

What am I doing wrong here? What exact circumstances makes the redshift not proportional to the velocity and why?

Last edited: Jun 22, 2016
2. Jun 23, 2016

### andrewkirk

1. What do you mean by a constant velocity? In particular, is the recession rate of a given (co-moving) galaxy constant in metres per second, or in metres per second per parsec? It would be very unusual to assume the former, as there is (so far as I am aware) no realistic cosmological model that describes that type of expansion.
2. What is the velocity v that is used above? There a number of different reference frames that are relevant to the consideration of redshift of a distant galaxy, some of which are inertial and some of which are not. Perhaps v is intended to be a component of the four-velocity as represented in one of those frames, but if so which frame?
I suspect that the problem may be that you are trying to analyse this problem as if both the Source and Observer are in the same inertial frame. They are not. But it will be clearer whether there is a problem, and if so what it is, if you can answer these two questions.

3. Jun 23, 2016

### JohnnyGui

1. I did mean the "stretching scenario" and thus a constant velocity in metres per second per parsec. But even so, if I write down D1 as a function of v / H (H being the value at the time the star was at D1) it would still give me a proportional relationship with z. The velocity v would be the x variable.

The formula would be:

Perhaps because the H goes down exponentially in case of a higher constant velocity of another star this function shouldn't be proportional to z? But even then, because H is in the numerator and the denominator they all cancel each other out so H changing with the v shouldn't matter.

2. Not sure what you mean with this but I was talking about the recession velocity of a galaxy/star with respect to us.

I think the whole problem is describing z + 1 as v/c in the first place since that's a proportional relationship. If you describe the relationship of z with the age ratio of the universe (the age of the universe when we received a star's light / the age when that star emitted the light = z +1), the distance ratio (the distance of a star when we received its light / the distance of that star when it emitted the light = z +1) while these relationships are based on z + 1 = v/c, you'd always get a proportional relationship (in case of a constant expanding universe).

But what is it exactly that makes z eventually not proportional anymore and why?

Last edited: Jun 23, 2016
4. Jun 23, 2016

### andrewkirk

Recession velocity can be defined in a number of ways. You appear to be defining it as
$$c(\frac{D2}{D1}-1)$$
where D1 and D2 are the 'distances' of the source from the observer at the time when the light is emitted and the 'time' when it is observed respectively.

I've used quotes around 'distance' and 'time' because they are ambiguous. We need to pick a reference frame in order to talk meaningfully about distance and time (including simultaneity). The reference frame usually employed for calculations at cosmological scale is the FLRW frame. That frame is non-inertial, so operations like dividing velocity by distance to get time are generally not valid.

I think that's the heart of the problem here. I get the impression that you are performing calculations as if the source and observer are in the same inertial reference frame. But they are not. The formula you derive above is valid for nearby receding objects but not for distant galaxies. This wiki section derives a formula equivalent to yours, which is valid for a local inertial frame, not for astronomical measurements. A famous experiment used to test this calculation (the Ives-Stilwell experiment) was done in a laboratory, not an observatory.

5. Jun 25, 2016

### Ken G

Yes, I think that's the problem, it's not clear to me why you assumed v/c = 1+z, that's just something you introduced with your expression for D2 and I can't see why it is justified. It sounds like what we could call the "first order Doppler shift", which is not even true in special relativity, let alone cosmology.
It sounds like you are trying to apply a special relativity expression to cosmology. But even if you do that, you have to use the full special relativity expression, not the first-order Doppler shift. The full expression is 1+z is the square root of the quantity 1+v/c divided by 1-v/c. Try that, but as pointed out above, don't reason too far from what you get, it only makes a connection between a particular z and a particular v, and implicit in that will be a rather unusual coordinate system that is generally not used in cosmology.

6. Jun 26, 2016

### JohnnyGui

Thanks for your explanation. Your given formula regarding velocity is indeed how I concluded it. So if I understand correctly, the z can not be proportional to velocity (and thus not for D1, D2 or even time) for far away objects because they don't share an inertial frame. Could you please elaborate what exactly the factor is that makes the universe have non-inertial frames among locations? I reckon it's maybe the curvature of space but how does this translate to non-inertial frames?

I concluded it myself from scratch by using the standard Doppler shift yes. I have 2 questions for you if you don't mind:

1. I have seen that full special relativity expression of the formula you described, it looks like it's giving a ratio of the first order Doppler shift for a recessing object (z + 1) divided by the first order Doppler shift for an approaching object (z - 1) and then taking the square root of that ratio. It might be hard but is it possible to derive this equation you described from any simple special relativity formula?

2. You're saying that my formula doesn't go for special relativity nor cosmology. What are the exact factors in special relativity and cosmology that make my formula invalid?

Last edited: Jun 26, 2016
7. Jun 26, 2016

### Ken G

Actually, I don't think the inertial frame issue is the key there, it is simply that you are using a formula for 1+z that only applies for small z. So you'd have a problem even in special relativity, where there is no problem finding inertial frames.
Yes, it's gravity that does that. But again, that's not really the problem with your formula.
Yes, it is the special relativity formula that Einstein derived. The easiest way to derive it is to have two rockets in deep space, receding from each other a speed v, and have each one send out a light signal at a given frequency, which is received by the other. Then the stretch in the wavelengths they receive will equal 1+z, and will equal the square root of the ratio of 1+v/c divided by 1-v/c. To see this, you can choose a coordinate system where one rocket is stationary, and it sends the first wavefront of the signal when the two rockets coincide, so the time delay is zero. Then the second wavefront intercepts the moving rocket when c(t-P) = vt, where P is the period the wave is created with. That solves to t=P/(1-v/c), so the stationary rocket reckons that the other rocket receives a wavefront every P/(1-v/c). We can also calculate the period between waves received by the stationary rocket from the moving rocket, so if the moving rocket sends out a wavefront every P' in the frame of reference of the stationary rocket, the time between wavefronts received is P'+D/c, where D is the distance the other rocket travels in time P' and is D=P'/v. Hence the time between wavefronts received by the stationary rocket is P'(1+v/c), and the time between wavefronts received by the moving rocket is P/(1-v/c), where all these times are reckoned in the frame of the stationary rocket because that's the frame I used to calculate them. Since the light signals are the same frequency in their own frame, the only reason P is different from P' is time dilation, so let's account for that by saying P' = kP, where we don't know k yet. However, we also know that the two times just calculated must also differ only by the factor k, so we can say P/(1-v/c) = kP'(1+v/c) = k2P(1+v/c). That lets us solve for k, and we get the Lorentz factor. Then we say the time between received wavelengths is kP(1+v/c) and that is the same thing as P times the square root of the ratio of 1+v/c to 1-v/c, as claimed.
The above calculation.

8. Jun 26, 2016

### JohnnyGui

Thank you so much for this clarification. Only thing is, I got lost at P/(1-v/c) = kP'(1+v/c) = k2P(1+v/c). Where did that k squared come from? Could you perhaps show this step by step?

9. Jun 26, 2016

### andrewkirk

By the way, cosmological redshift can be written in a Doppler form, as:
$$1+z=\sqrt{\frac{1+v}{1-v}}$$
Here $v$ is the ratio of the distance to the time components, in the co-moving inertial frame of the Observer, of the vector $\vec u$ which is the (four-)velocity vector of the Source at the instant of emission, parallel transported along the path of the light ray to the observer.

The derivation of this formula uses the General Theory of Relativity.

10. Jun 27, 2016

### JohnnyGui

@Ken G Ï've got another question next to my other question in my post #8:

If P' = kP, then how can it be that t' = kt as well? Since t'= P'(1 + v/c) and t = P / (1 -v/c) the different notations between the 2 formulas which are (1+ v/c) versus 1 / (1-v/c) would cause a ratio other than k for t' / t.

Writing down P' / P in their t' and t functions respectively would give: t' / ( t (1-v/c)(1+v/c) ) which is not t' / t

11. Jul 2, 2016

### Ken G

The basic point is that if you use the nonrelativistic Doppler shift, like Doppler shift in sound, you get a different result if you take the source as moving or the receiver as moving. That's OK because sound has a medium, air, so you can say which one is "actually" moving through the air. But with light in a vacuum, that's not OK because it violates the basic symmetry that either person should be able to regard themselves as stationary. The only way to recover the symmetry is to also say that the other person's time is dilated. But when you do that, you also get that the period of the light they send out to you is increased. So then you can use the nonrelativistic Doppler shift for a moving source and a stationary receiver, coupled with the time dilation, and get the relativistic Doppler shift formula that we are talking about. You can also check that you get the same answer if you regard yourself as the source, send out a wave with the normal period, treat the receiver as moving in the nonrelativistic Doppler shift, and then account for the receiver time dilation to infer what they will perceive. Either path leads to the above square root formula, and that required symmetry is how you prove the necessity of time dilation.

12. Jul 4, 2016

### JohnnyGui

Thanks for your explanation. After reading a few articles on this I indeed understand what you say and I fully agree on this.

What I noticed after reading however, is that the factor by which the P changes (as you described) based on the difference in the formulas for whether an observer is moving or the source is moving isn't the Lorentz factor but the following:

For an observer who is moving, he would receive a signal every P time which is c / (fc - fv) where f is the frequency. For a source that is moving, the observer would receive a signal every P time which is (1 + v/c) / f. If you divide those to formulas by each other to see the ratio among the different P's for these 2 different scenarios, you would get a ratio of 1 / ((1 - v/c)(1+v/c)) which isn't the Lorentz factor by which the P dilates. This is actually the factor that is caused by the Doppler asymmetry of whether the source of an emitting frequency is moving, or the observer itself.

It seems that the difference in time duration (P) because of that Doppler asymmetry, is something else than time dilation itself that cause the P's to differ. This https://en.wikipedia.org/wiki/Relativistic_Doppler_effect Wiki Article for example (see "Motion along the line of sight") shows that even if both observer and source agree that the observer is moving (thus throwing out Doppler asymmetry), there'd still be time dilation in the form of the Lorentzfactor.
Why do I say even if both the observer and the source agree that the observer is moving? Because the Wiki mentions the formula λ / (c - v) = P which is the formula for a moving observer plus the fact that it shows that the gamma factor dilates the time for the observer as well thus meaning that the observer is moving with respect to the source.

My problem is, if Special Relativity is all about a matter of perspective, why doesn't the Relativistic Doppler formula not count in the Doppler asymmetry factor as well next to the Lorentz factor since both of those factors cause the P's to differ and not just only the Lorentz factor?

13. Jul 4, 2016

### Ken G

I don't think it is helpful to think in terms of "Doppler asymmetry." The need for time dilation is to preserve a symmetry, so you can think of that as a need to get rid of an asymmetry, but the ratio that you describe is not particularly useful because you are ratioing results from two different reference frames. Best to choose a frame and stick with it, but apply a time dilation factor to the period that is either emitted or received in the other frame. All that is necessary to derive the result is to choose a reference frame, regard the other person as moving through a medium through which light moves at c. and assert that the moving time is dilated. Then you set the degree of dilation to retain the symmetry that either the receiver or the source can be taken as the moving frame, and you must get the same result for the observed period. All you ever need is time dilation, the rest is all just like the Doppler effect in sound, so you encounter the Lorentz factor but never the ratio you describe.

14. Jul 5, 2016

### JohnnyGui

I'm not sure how one can do this, could you please provide an example? How can one cancel out the Doppler asymmetry so that only the Lorentz factor is the factor that dilates the time (or P for that matter)?
If the observer is the one who actually moves, he'd still think that the source is moving (from his perspective) and he'd therefore use the formula (1 + v/c) / f to calculate his own P. Regarding the f in the formula, the observer would have reduce that f with the Lorentzfactor to get the frequency that he is receiving himself (the obsever thinks he's standing still so his time is faster than the source's, a faster time means a larger P and thus a smaller f for the observer than the source's).

On the other side, the source is thinking that the observer moving so the source would use the formula c / (fc - fv) to calculate the observer's P. Then, the source would use that P with the Lorentz factor to calculate the dilated P that the observer is receiving according to the source (since the source thinks the observer is moving).

As you see, the observer and the source are both using different formulas to calculate the observer's P (because of the Doppler asymmetry) so it's inevitable that they'd eventually calculate a different P for the observer, even if they both use the Lorentz factor. What am I missing here?

Last edited: Jul 5, 2016
15. Jul 14, 2016

### JohnnyGui

UPDATE: So apparently after rewriting all this, I came to the conclusion that both the source as well as the observer would calculate the same P for the observer, despite the difference in formula caused by the Doppler asymmetry or whether the time dilates for the source (observer sees the source moving away) or that the time dilates for the observer (source sees the observer moving away). So at the end, both the observer and the source would agree on the time P for the observer in Special Relativity with Doppler asymmetry.

Furthermore, I concluded that the factor by which the formulas for the Doppler asymmetry differ, which is the ratio of c / (fc - fv) and (1 + v/c) / f, is actually the Lorentz factor squared.

Is this statement true?

Last edited: Jul 14, 2016
16. Jul 25, 2016

### JohnnyGui

Could someone please verify my statements in my previous post #15? I'd appreciate it a lot.

17. Jul 25, 2016

### Ken G

I think the whole concept of "Doppler asymmetry" is awkward and difficult to parse-- I don't think that concept is useful here, it is the symmetry that is the key to the result, not some asymmetry.

18. Aug 11, 2016

### JohnnyGui

My problem is that I think that if one has to throw out Doppler asymmetry, one of the two (the observer or the source) would have to agree that he himself is moving with respect to the other so that they both use the same Doppler formula for P which then can be multplied by the Lorentz factor based on who is moving. But someone agreeing that he himself is moving is clashing with the rules of Special Relativity since in SR everyone is saying that the other is moving with respect to himself.