# Can two galaxies receded from one another faster than C?

1. Mar 14, 2006

### Chaos' lil bro Order

Sorry if this is a lay question, here it goes:

Consider that on Earth we observed two galaxies, one directly above the north pole, the other directly above the south pole (aka. opposite directions). When we measured the redshifts of these galaxies we concluded that they had recessional velocities of 0.99C and 0.98C, respectively. The question then is simply this, are these two galaxies receding from one another with a combined recessional velocity of 1.97C?

If so, then consider this...
If we could put an observatory on the 0.99C galaxy, would we see the 0.98C galaxy as receding from us at 1.97C?

I'm pretty sure the answer is no, but I can't think of why its no. Am I simply thinking of the Universe's structure incorrectly?

Thanks

2. Mar 14, 2006

### pervect

Staff Emeritus
I don't have a formula that answers this specfic question at this time, though I'll do some more looking and as a last resort I might even try to figure it out from first principles.

It might be a good idea to move this question to the cosmology forum, as there is a good chance that SpaceTiger will be able to get you an answer more quickly. Part of the answer to this question depends on certain assumptions as to cosmological parameters.

There is however some very useful background information on your question at

http://www.astro.ucla.edu/~wright/cosmology_faq.html#FTL

however.

If we could multiply the redshifts together, it would be easy to answer the question from the above formula relating redshift to velocity, but I do not believe that this technique is justified for sevaral reasons - one reason being that the formula above is stated to work only for an empty universe. Presumably you'd want an answer that uses the current "concordance model" .

Note that this is an entirely different question than how velocities add together in SR, which is a much easier question. Not only is the geometry of the universe not flat, but as the FAQ above mentions, cosmologists tend to use different measures of distance (and velocity) than people in SR do. (For that matter, their defintion of simultaneity is different, too).

In case you were interested in the answer to the SR version of your question, that is given in many places, for instance

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html" [Broken]

Last edited by a moderator: May 2, 2017
3. Mar 14, 2006

### kmarinas86

Speed of light is measured as 299,792,458 meters / 1 second. Whose second is that? The key is that the length of the "second" varies. You might as well say that the speed of the observed galaxy is 299,792,458 meters * ??? Hz, where the Hz of the observed galaxy depends on the fraction of the speed of light they are receeding i.e. $v/c$

$??? Hz = 1 Hz * \frac{v}{c}\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}$

The velocity addition formula can be derived in the following way:

Let:

$\alpha=\frac{v_1}{c}$

$\beta=\frac{u}{c}$

$\gamma=\frac{v_2}{c}$

$f_1=f_0 \sqrt{\frac{1-\alpha}{1+\alpha}}$

$f_2=f_1 \sqrt{\frac{1-\beta}{1+\beta}}$

$f_2=f_0 \sqrt{\frac{1-\alpha}{1+\alpha}\frac{1-\beta}{1+\beta}}$

$f_2=f_0 \sqrt{\frac{1-\gamma}{1+\gamma}}$

$\sqrt{\frac{1-\gamma}{1+\gamma}}=\sqrt{\frac{1-\alpha}{1+\alpha}\frac{1-\beta}{1+\beta}}$

$\left(1-\gamma\right)\left(1+\alpha\right)\left(1+\beta\right)=\left(1-\alpha\right)\left(1-\beta}\right)\left(1+\gamma\right)$

$\left(1-\gamma\right)\left(1+\alpha+\beta+\alpha\beta\right)=\left(1-\alpha-\beta+\alpha\beta\right)\left(1+\gamma\right)$

$1+\alpha+\beta+\alpha\beta-\gamma-\gamma\alpha-\gamma\beta-\gamma\alpha\beta=1-\alpha-\beta+\alpha\beta+\gamma-\gamma\alpha-\gamma\beta+\gamma\alpha\beta$

$2\left(\alpha+\beta\right)=2\left(\gamma+\gamma\alpha\beta\right)$

$\alpha+\beta=\gamma+\gamma\alpha\beta$

$\frac{\alpha+\beta}{1+\alpha\beta}=\gamma$

Last edited: Mar 14, 2006
4. Mar 14, 2006

### Loren Booda

To all central observers, opposite horizons recede near the speed of light. Special relativity does not allow communication between these horizons, although we may observe them both individually.

General relativity, however, does allow points in the expanding universe to accelerate, thus enlarging the boundary and spacetime beyond where we are unable to observe light because v>c. In effect, the vast majority of spacetime regresses faster than the speed of light.

Remember that Einstein's postulate of physics equivalency in inertial frames does not necessarily apply to GR, where non-interial frames are considered.

5. Mar 15, 2006

### pervect

Staff Emeritus
[/quote]

This looks like it is equivalent to the SR velocity addition formula

$$v_tot = \frac{v1+v2}{1+\frac{v1 v2}{c^2}}$$

while this is correct for flat space-time (i.e. Special Relativity) it's not correct for the original problem.

6. Mar 15, 2006

### pervect

Staff Emeritus
I think I've dug up the correct way to proceed, though I could use a double-check by another SA who works with cosmology more than I do.

The relationship between velocity and $D_{now}$, the comoving radial distance, is NOT given by the standard SR doppler formula. (Ned Wright mentions this explicitly in his cosmology tutorial, which was my primary source for this post:)

http://www.astro.ucla.edu/~wright/cosmo_02.htm

All we really need to work this problem is Hubble's law, v = D H

We can thus determine D1 = v1/H, and D2 = v2/H

The co-moving distances add, so that the total distance between the two galaxies is just D1+D2 = (v1+v2)/H.

We multiply the result by H, to find that the velocities add linearly!

It should be noted, again, that these are NOT the velocities as defined in Special Relativity. To quote Ned Wright from the above link.

In particular, the "comoving radial distance" is measured along a curve that is not a geodesic, i.e. it's measured along a curve that's not the shortest distance between the two points. You can thus think of it as being larger than the distance that SR gives, which helps explain why the rate of change of this distance (which is the cosmological defintion of velocity) can be greater than 'c' - the cosmological distance scale is larger than the SR defintion, so the cosmological velocities (the rate of change of cosmological distance) are likeways larger.

Last edited: Mar 15, 2006
7. Mar 16, 2006

### Chaos' lil bro Order

In particular, the "comoving radial distance" is measured along a curve that is not a geodesic, i.e. it's measured along a curve that's not the shortest distance between the two points. You can thus think of it as being larger than the distance that SR gives, which helps explain why the rate of change of this distance (which is the cosmological defintion of velocity) can be greater than 'c' - the cosmological distance scale is larger than the SR defintion, so the cosmological velocities (the rate of change of cosmological distance) are likeways larger.

So basically, two comoving galaxies CAN recede from one another at a combined rate of over C and you say since space is curved, the distance is actually greater than the geodesic path between the two galaxies because of this curvature.

As for Hubble's Law, I thought that it predicts a maximum velocity of C. z=1000 equals something like .99999999999C, so doesn't velocity approach C as z approaches infinite?

Wouldn't: Recessional Velocity = Beta/ Hubble Constant imply that recessional velocity is constrained by C since Beta=v/C.

Since these two forumulas, the 1st cosmology, the 2nd Relativity are linked mathematically, I don't understand the comment saying:

The time and distance used in the Hubble law are not the same as the x and t used in special relativity, and this often leads to confusion. In particular, galaxies that are far enough away from us necessarily have velocities greater than the speed of light.

8. Mar 16, 2006

### Severian596

It looks like you destroyed pervect's well-spoken explanation. He didn't say that two galaxies can recede from one another at a combined rate over C. He said that we on earth can measure their apparent velocities relative to each other as being greater than C. But this is pretty meaningless...if the two galaxies measured their velocities relative to each other they would arrive at a v < c.

9. Mar 16, 2006

### DrChinese

Just as a reminder in case you are not familiar with Lineweaver & Davis. There are plenty of galaxies receding from us at velocities greater than 2c, including some which are experimentally verified as such via red shift.

10. Mar 16, 2006

### pervect

Staff Emeritus
That's not quite what I'm saying.

What I'm saying is that there are several distance measures used in cosmology, and that one has to read the "fine print' of what cosmologists are actually measuring, to make sure that one is using the right distance for the right application.

Each of these different distance measures has an associated "velocity", the rate of change of the distance with respect to some time parameter (said time parameter being usually, but perhaps not always, our time).

Thus it is ambiguous and potentially misleading to talk about "the" velocity of a distant galaxy without being more explicit as to _which sort_ of velocity one means.

Journalists do this all the time, unfortunately, which tends to lead to a confused public :-(.

Because you ARE talking about "the" velocity and "the" distance, one assumes that you have some particular model/application in mind, but it's unclear which of the many candidates for distance (or velocity) would be the closest match to your application.

Ned Wright's cosmology tutorial is a very useful reference, the Lineweaver and Davis paper that Dr Chinese posted a link to is also very useful.

I posted the link for Ned Wright's tutorial earlier, I'll repeat it here, in case you haven't at least skimmed it (there's quite a bit of material):

http://www.astro.ucla.edu/~wright/cosmolog.htm

(The main tutorial is under "enter tutorial').

Usually, we think of distance as being measured along the shortest path between two points. (Well, at least I do.)

The particular sort of distance used in the Hubble's law calculation (comoving distance) is very useful, but aside from not being the distance measured along a straight-line path (as one might intuitively think that distances should be), it is not even the only sort of distance that cosmologists use.

Among other sorts of distance in common use there are:

light travel time
angular size distance
luminosity distance

these are all conveniently computed in terms of z by a calculator program avaliable from the URL I quoted earlier.

This is not an exhaustive list of distance measures, one of my favorites, affine distance, is not even on the list. This concept of distance is closely related to light travel time distance, it could be regarded as counting the number of wavelengths of monochromatic light along the path of a lightbeam emitted by a visible object. It's probably more of theoretical interest than practical utiltiy, however.

As a conclusion to a rather long post, one can look to GR for guidance as to how to define distance, but all GR will tell you is that there isn't any unique way to determine the velocity of a distant observer when there is gravity involved (or when space-time is curved). There are some useful and general ways to proceed when one has a static metric - unfortunately, this doesn't help with cosmology, because the metric of the universe is not static, but is changing with time.

11. Mar 16, 2006

### Loren Booda

The fastest speed that can be measured is the speed of the measuring particle, the photon itself.

12. Mar 16, 2006

### pervect

Staff Emeritus
Let me post a simple example that explains part (not all) of the more complex cosmological situation. The good news is that the analogy is relatively simple - the bad news is that it is not exact.

Suppose that you are on Earth, somewhere above the equator (say 45 degrees lattitude to be specific).

Suppose you have two laboratories with a clear line of sight between them, and that the second laboratory is at exactly the same latitude, but 18.6 miles due east of the first.

Now, since light travels at 186,000 miles per second, you expect light to take .0001 seconds to travel from one laboratory to the other.

But you observe that the light reaches the laboratory sooner than this. Is light then moving faster than light?

The answer is no. The path that you've measured along a circle of constant lattitude by moving "due east" is actually not the shortest path. The shortest path, the one that light actually takes (ignoring height for the purposes of this analogy, you'd actually need a fairly tall tower to get an 18.6 mile LOS) follows a path that's a great circle.

It is not incorrect to say that the laboratory is 18.6 miles due east, it's just that one has chosen the wrong defintion of distance for the problem at hand - by appling plane geometry concepts to a geometry that's actually spherical.

The cosmological situation is more complicated, because it is not stationary like the simple example on the Earth. But co-moving coordinates are a lot like latitude and longitude in many respects - a galaxy always has a constant value for its cosmological coordinate, even as the universe expands.

For more details, I would again urge readers to read some of the references posted earlier in the thread. The devil is in the details, as they say.

Note that the cosmological problem is harder than the problem on Earth because the universe is not static - it's expanding. distance = time * velocity does one no good if one measures the wrong distance - and in an expanding universe, because the distance is constantly changing, picking the right distance and how to time-average it is even trickier than the simple example on Earth.

Last edited: Mar 17, 2006
13. Mar 18, 2006

### Chaos' lil bro Order

Thought I made it clear that I understood the difference between a geodesic path and a curved path. Cosmology deals with the curved path as space is curved.

As for semantics, velocity = recessional velocity
distance = Light Years (of course)
z = redshift (usually Lyman series for distant objects)

From the Lineman & Davis paper, its clear that galaxies can recede at superluminal velocities, in fact any with red shifts above ~1.46 meet the requirement.

Thanks for the ref. Dr. Chinese, now I won't need to read tons of half-true posts, I love a good Review paper.

14. Mar 18, 2006

### Chaos' lil bro Order

Severian, yup ok, you are just plain wrong.

15. Mar 19, 2006

### pervect

Staff Emeritus
So far so good, but.....

These are the sort of statements that make me think you are NOT understanding what I'm trying to say at all :-(.

After having spent a lot of time trying to explain that there are many ways of measuring distance, you talk about velocity as "recessional velocity" and the units of distance as being light years, as though that defined what velocity it is that you're talking about.

This gives me the feeling that real communication just isn't happening.

I rather suspect that you may be trying to understand cosmology without first understanding special relativity first. If my suspicion is correct this is a big mistake - you really need to understand SR first, before you tackle cosmology. (Understanding at least a little bit about GR as well as SR would be even better.)

Last edited: Mar 19, 2006
16. Mar 19, 2006

### Chaos' lil bro Order

I think I will reserve my cosmology questions to the cosmology forum, Space Tiger seems to know everything inside out.

But thanks.

17. Mar 30, 2006

### da_willem

Great analogy! Thanks!

18. Mar 30, 2006

### LnGrrrR

Pervect,
I'm trying to work out your dang signature...I'm at an area of trial and error though, and I'm trying to fight laziness to figure it out :)

19. Apr 3, 2006

### Conal

If, for example, a spaceship is flying away from earth at near the speed of light and communicating with us, we should percieve the signal to be travelling at the speed of light anyway. So if another spaceship is travelling in the opposite direction at near the speed of light, the signal should still be able to reach it too shouldnt it?

20. Apr 4, 2006

### marcus

The answer to the originally posted question is yes. One galaxy is receding from the other at 1.97 times the speed of light.

It is a popular mistake to think of recession speeds as limited by c. This was addressed in a Scientific American article last year. Also in the article cited by Dr Chinese in this thread.

A numerically simpler example would be the case where it is determined that galaxy A is receding from us at speed 2c and galaxy B is receding in the opposite direction at speed 3c. Then A is receding from B at speed 5c.

link to the SciAm article on popular misconceptions about the expansion of the universe is here
https://www.physicsforums.com/showpost.php?p=482299&postcount=76

It sounds like Chaos brother got discouraged and left this thread
Not a bad move. SpaceTiger would certainly be able to clear the matter up. (But it seems a pity if questions like this cannot be handled in the forum specifically set up to handle them.)

Last edited: Apr 4, 2006
21. Apr 4, 2006

### pervect

Staff Emeritus
I would say that the cosmology forum is probably best forum for cosmology questions, a better forum than the SR/GR forum. I would agree entirely that Space Tiger is a great reference, in fact I recommended him in #2.

If you scroll back over the past discussion, Chaos got a correct answer from me in post #6, and from Dr Chineese in #9. All the answers from science advisors were pretty much right on the mark, though there were a number of posts that missed the mark from non-SA's.

Thus it is possible to get good answers about cosmology in the GR forum, but it's probably easier and quicker to get good answers about cosmology in the cosmology forum.

I actually attempted to go into a litle more depth, and provide an explanation of how cosmologists do not use a coordinate system compatible with special relativity, citing Ned Wright's cosmology FAQ

http://www.astro.ucla.edu/~wright/cosmology_faq.html#FTL

I didn't cut and paste this quote in the original response, unforutnately, perhaps this would have been a good idea in retrospect for the convenience of lazy readers.

These different conventions (on the measure of basic things like distance and velocity) on the part of cosmologists are the reason this thread has generated so much confusion, and the source of many of the incorrect answers given by people who are familiar with SR, but not GR or cosmology.

In any event, questions on GR&SR are certainly welcome in the SR & GR forum. We do get a tiny bit tired about questions about FTL, though.

If marcus's point was that all "FTL" questions should be sent to the SR/GR forum as a sort of "dumping ground", I suppose I have to disagree. The cosmologists have to take their fair share of the load, when the issue actually does involve cosmology. Of course the vast majority of such questions do not involve cosmology, and are best addressed with the SR velocity addition formula.

22. Apr 4, 2006

### Chaos' lil bro Order

Thanks Marcus, best response yet!

23. Apr 4, 2006

### marcus

You are welcome!
There is another calculator besides Wright's that might be useful to you.
http://www.earth.uni.edu/~morgan/ajjar/Cosmology/cosmos.html [Broken]

You asked a question back in post #7 about something involving z = 1000. This calculator would tell you corresponding speeds and also would tell the light travel time. So I mention z = 1000 as an illustration. But that is a CMB size redshift and the numbers are a bit stupendous. So you might also want to try redshift 1.5 or 2 or 3.

Morgan's calculator has the same basic functions as Ned Wright's -----you put in parameters like 71 for Hubble and 0.27 for matter and 0.73 for dark energy (the famous "73%") and once that is taken care of then you can put in anything for the redshift z and it will tell you the Hubble-law distance

it does NOT have some other features that Wright does, but it does have a couple of features his doesnt. Morgan's will give you the distance, say for redshift z = 1000 (essentially same as Wright's) but it will also give you the PRESENT recession SPEED of the matter that emitted the light that we are now receiving that has redshift 1000.

And also interesting it will give you the speed that the matter had when it emitted the light.

In Morgan's simplified terminology "Omega" means matter density Omegamatter. So you type in 0.27----that is a common estimate (27%) comprising ordinary matter plus dark matter. Again in Morgan's terms, "Lambda" means OmegaLambda-----the dark energy or cosmol. const. part of total.

so to use the calculator you have to type in 0.27 for Omega ("matter density") and 0.73 for Lambda ("cosmological constant") and 71 for the Hubble parameter (71 km/s per Megaparsec). then you are ready.

then if you put in z = 3 or whatever redshift you want to see distance for (or age when the light was emitted, light travel time, or present/past recession speeds)
you should get that......

If you don't want to try it, or prefer to stick with Ned Wright's distance-from-redshift it is OK my feelings wont be hurt

but just as a check, in case you do try it: you should get that for z =3 the present distance is 21.07 billion LY.

that is where you put in, for "matter density", "cosmological constant", Hubble parameter, and z the numbers 0.27, 0.73, 71, 3

have to go help with supper.

Back now. Basically the calculator gives you a simple hands-on encounter with GENERAL relativity. what the calculator is using is a certain SOLUTION TO THE EINSTEIN EQUATION associated with names like Friedman, Lemaitre, Robertson, Walker. the metric is called the FRW metric (for Friedman-Robertson-Walker).

In the example I suggested you try----with 0.27 + 0.73 = 1-----the spatial slices are FLAT
so the Hubble-law distance that we are talking about is a STRAIGHT LINE DISTANCE. It is the shortest distance between two points barring excursions into the past or the future.
(It is misleading to attribute superluminal recesssion speeds to the Hubble-law distance not being "along geodesics".)

Anyway, if you are at all interested in General Relativity the spatially flat case of the FRW metric is a simple example to play around with and you have the basics of it on Morgan's calculator.

Bear in mind that individual solutions of Gen Rel do not have to have Lorentz symmetry---they don't need to obey Special Rel (except as a local approximation). So in a particular solution of Einstein's Equation like the FRW metric HAS a idea of rest----it makes sense to say a galaxy is sitting still in some space that is receding from another galaxy----and there is an idea of SIMULTANEOUS. Whereas in Special Rel you do NOT have a notion of rest or simultaneity.

some history: Special Rel was 1905. Gen Rel was 1915.
The earlier theory is the one that has no rest, no simultaneity, and no idea of a distance increasing faster than light.

I don't know if anyone here is particularly interested in Gen Rel or this kind of introduction to it. So I will stop and see if there is any response. First one example. The microwave background comes from z=1100. So think about the MATTER that emitted the light that we are now seeing as microwaves. That matter had a certain RECESSION SPEED at the time it emitted the light.
What was the recession speed of the matter that emitted the CMB photons that we are now receiving?
Anybody?

Last edited by a moderator: May 2, 2017
24. Apr 5, 2006

### pervect

Staff Emeritus
One first needs to define how the recession speed is measured. I gather cosmologists usually mean the rate of change of the comoving distance, as per Ned Wright - a coordinate dependent defintion.

Even given the precise coordinate dependent defintion of "velocity" that cosmologists use, the answer to the above question is highly model dependent. Plugging in z=1100 into Ned Wright's cosmological calculator and picking the model parameters of the cosmology that one believes are accurate, allows one to calculate the comoving distance and hence the "recessional velocity".

For example, for z=1100 with

H0=71, omegam=.27, and omegavac=.73 the calculator computes the comoving distance as about [correction] 14000 Mpc.

Multiply 14000 Mpc by H0, and one gets V_now 994,000 km/sec

But with different parameters, for example H0=71, omegam=1, omegav=0 the distance is only about 8200 Mpc, significantly lower. The velocity changes correspondingly.

Hence the remarks about the model dependence of this velocity.

I should note that in cases with a static metric, I would greatly prefer to define a notion of velocity in a coordinate independent manner, rather than in the coordinate dependent way in which cosmologists measure velocity.

For instance, if one has a black hole, I would (and have)! suggest defining the idea of a "velocity relative to the black hole" of a test object X as taking an object "at rest" with respect to the black hole, and measuring the velocity between this object and the test object X when the two objects are at the same point in space-time.

Note that I would not suggest using this concept of velocity without explaining it - the very idea of the relative velocity of two objects at different points in space-time is ambiguous without further explanation. But I feel it is a good defintion, once explained, because it does not depend on the usage of any specific coordiante system.

This coordinate independence is possible because there is a unique object that is "stationary" with respect to the static metric of the black hole at any given location in space.

Unfortunately, this coordinate independent technique will not work for FRW cosmologies, because they do not have static metrics. Hence, we are stuck with the rather ugly (and potentially confusing) facts that not only can velocities as defined by cosmologists be greater than 'c', but that the point where the objects recessional velocity becomes equal to 'c' is not the same as either the event horizon or the particle horizon!

Last edited: Apr 5, 2006
25. Apr 5, 2006

### marcus

Hi pervect, how nice that someone wants to try calculating!
It is after midnight here and I am going off to bed soon but before doing so I will calculate for z = 1100. Then if you do so we can compare results!

Remember that this is all done with the FRW metric which is one particular solution to the Einst. equation. Remember also that the Hubble parameter CHANGES. So the figure of 71 applies only to the present moment!

However the model is able to extrapolate back and find the Hubble that prevailed at any given time in the past.

OK, so we plug in z = 1100, the redshift of the Cosmic Microwave Background radiation.

I get that the recession speed of the emitter at the time it radiated the light which we are now picking up was 57c.

The recession speed at emission was FIFTY-SEVEN TIMES C. Is this what you get? I hope so.

The present distance of the matter that emitted that light is now 45.5 billion LY and it is receding at a speed of 3.3 c. Did you get any of these numbers? If not, you may have made a mistake.
Perhaps someone else will use one of the available online calculators and check my answers.

I will compare and see what Ned Wright's gets. z=1100 is really pushing it! So I wouldnt be surprised if there were some minor discrepancy. The distance should come out within say 10 percent 45.5 billion LY but maybe not exact due to round-off error and stuff.

=====================

WOW NED WRIGHT'S GIVES VERY NEARLY THE SAME ANSWER!
It gives the distance as 45.655 billion LY
And Morgan's gave it as 45.5 billion

This is when I put in the same numbers: 71, 0.73, 0.27 and say z=1100.

GREAT! they are probably using the same formulas and very similar numerical integration routines.
Ned's is just a few more decimal places accurate.

Ned's gives the age of universe at the time the CMB light was emitted as 372,000 years. That is close to what people usually give for the time of last scattering.

The light travel time is given by the calculator to be 13.655 billion years. Almost the present age of the universe.

OK that all seems fairly consistent. Anybody trying it, prevect or others, you should get that the emitter is now some 45 billion LY away
and the light has been traveling for some 13.6 billion years
and the emitting matter, when it radiated the light, was receding at around 57c.

If you have any trouble getting these answers, give me a brief description of what you did and i will try to help.

Last edited: Apr 5, 2006