Why don't galaxies move faster then c?

[madmike159]

I understand Einstein’s theory of relativity quite well but I keep thinking about different things and getting confused. All galaxies are moving away from us and the further away are moving faster. I know nothing can move faster than c but if there was a galaxy far enough away should it not be moving faster than the speed of light when you look from a galaxy on the other side of the universe. You couldn’t see it though because the light would never reach you. Can some one explain why they don’t move faster than c?

 

[madmike159]

I just thought a better way of putting it. If there was a galaxy moving away from us (galaxy A) at 51% the speed of light, and another galaxy (B) moving the other way at 51% the speed of light. If you were at galaxy B galaxy A should be moving at 102% of c, but that can’t be possible.
Sorry for double posting.

 

[HallsofIvy]

In your first post you said "I understand Einstein’s theory of relativity quite well". If that is true then you should know why your second post is wrong. If one object, A, (whether it is a galaxy or not is irrelevant) is moving toward me at a speed, relative to me, of 0.51c, while a second object, B, is moving toward me from exactly the opposite direction, also at 0.51c, then the speed of B relative to A is
$$\frac{0.51c+ 0.51c}{1+ (0.51)(0.51)}$$
or about 81% the speed of light.

 

[madmike159]

Oh yea. I do that alot, start thinking about something and miss out really obvious things. Thanks for the help.

 

[buffordboy23]

The converse of the HallsOfIvy's statement is also equivalent; the speed of A relative to B is also 81%. This is just an application of the Relativistic Velocity Addition Theorem. From our reference frame on earth, we would actually observe the two objects to approach or recede from each other at 102% c. Motion is all relative and dependent upon the observer.

There are galaxies that do recede from us faster than c. This is due to the expansion of space, so the laws of relativity are not violated.

 

[neutralseer]

As buffordboy23 already said, it is okay for galaxies to "move" faster than the speed of light because it's not really motion, it's the expansion of the space between the galaxies. I think it's misleading to even talk about speeds or motion in the context of the cosmological expansion of space.

Also, I'm confused why HallsofIvy used a velocity addition formula that is only applicable to a flat spacetime (Minkowski space) for two objects that actually occupy a curved spacetime (the FLRW metric). Perhaps the take home message is that you cannot understand a General Relativistic effect-- http://en.wikipedia.org/wiki/Metric_expansion_of_space" -- within the context of special relativity.

Before I delve into trying to answer this question be warned: I'm no expert, I just want to try to solve the puzzle.

Let me try to reformulate madmike159's second question. Instead of talking of the speeds of galaxy A or B, I'll talk about the speed of coordinate A or B with respect to us. When I say B has a "speed" of 0.51c with respect to A, I just mean that if a spaceship at A were to move toward B at 0.51c, the separation between the spaceship and B would remain constant (neglecting the accelerating expansion of space). Finally I'll be referring to the speeds of these coordinates at a specific http://en.wikipedia.org/wiki/Cosmic_time" 't.'

Reformulated question: If A has a "speed" of .51c with respect to B, and B has a speed of .51c with respect to us, what is the speed of B with respect us?

My best guess: 1.02c.

 

[neutralseer]

Oops, let me change my reformulated question to be closer to madmike159's:
If A has a "speed" of .51c with respect us in one direction, and B has a "speed" of .51c with respect us in an opposite direction, what is the speed of A with respect to B? Here's the associated "picture" from our frame:

.51c <--A Us B-->.51c

My guess remains the same: B's velocity with respect to A is 1.02c.

 

[Fredrik]

The line element of the flat FLRW metric is just $ds^2=-dt^2+a(t)(dx^2+dy^2+dz^2)$, so the distance to a certain galaxy is the integral of $a(t)\sqrt{dx^2+dy^2+dz^2}$ along a path of constant t. a(t) can be taken outside the integral, so the distance at time t is just a(t)L, where L is a constant. When we say that the galaxy is moving away from us with speed v, we really mean that the derivative of the distance with respect to the time coordinate is v, i.e. that $\dot aL=v$. So if we see two galaxies, in opposite directions, each moving away from us at 0.51c, it means that they both are the same distance from us. If that distance is a(t)L, then the distance between them is just 2a(t)L, so the speed that one of them assigns to the other is 1.02c.

 

[buffordboy23]

If A has a "speed" of .51c with respect us in one direction, and B has a "speed" of .51c with respect us in an opposite direction, what is the speed of A with respect to B? Here's the associated "picture" from our frame:

.51c <--A Us B-->.51c

My guess remains the same: B's velocity with respect to A is 1.02c.

From who's vantage point? From our vantage point, the answer is 1.02c. From A or B's perspective, use the velocity addition theorem; value is less than c.

Motion is relative, which means two objects are necessary to give a magnitude value for motion. Usually, when we analyze the motion of everyday objects, it is from our vantage point--for these cases, nothing can travel faster than c.

To explain the 1.02c answer, note from the discussion above that neither A or B alone can travel faster than c from our vantage point. If you are comparing the "relative motion" of two objects A and B from our vantage point, answers can be greater than c.

 

[buffordboy23]

So if we see two galaxies, in opposite directions, each moving away from us at 0.51c, it means that they both are the same distance from us. If that distance is a(t)L, then the distance between them is just 2a(t)L, so the speed that one of them assigns to the other is 1.02c.

No, from our stationary perspective the relative speed between the two galaxies is 1.02c.

If you instantaneously transported yourself to one of the galaxies, your vantage point is now one of the galaxies. (In actuality, you would say this new vantage point is stationary but since you transported yourself, you know that your traveling at .51c and that the other galaxy is traveling .51c in the opposite direction.) However, when you make your new velocity measurement, you would find the relative speed between your new vantage point and the other galaxy is less than c.

If what you claim is true, then the other galaxy is moving away from our new vantage point (within the galaxy) at 1.02c and violates special relativity (if we allow the expansion of space this value is now okay).

You must use the relative velocity addition theorem as demonstrated by HallsOfIvy.

Your right about the 2a(t)L from our original vantage point. When you change your perspective in the manner above, you now have to take length contraction into account, so the distance between the two galaxies is now less than 2a(t)L. Check this with relative velocity .81c and you will see that this is true.

 

[madmike159]

I think I understand this now. We measure the speed of A and B and get 0.51c for each, so we think that they most be moving away from each other at 1.02c. If you were at B looking at A you would be looking through spacetime being curved by B and spacetime being curved by A so they would be slower.

 

[sketchtrack]

Isn't the whole reason time is added as a dimension in GR to satisfy that paradox. The answer would be 1.02 C, but since their clocks slow down relative to the observer at rest, the idea that they aren't breaking the speed limit is protected.

Of coarse when you think about it, isn't an object then in different places at the same time according to different observers.

 

[madmike159]

Yes true. I think I understand it well enough to answer my question, there's always something in physics to complicate things just as you begin to understand.

 

[Fredrik]

Most of you are missing the point. The "velocities" here are not velocities in the sense of SR: "change of spatial coordinate"/"change of time coordinate". That's why the velocity addition formula from SR doesn't apply. The spatial coordinates of the galaxies are actually approximately constant here, so the "velocities" would be close to zero.

No galaxies are moving at relativistic speeds relative to the background (i.e. points of constant spatial FLRW coordinates), so I can only interpret the question in #2 to be a question about "velocities" that are due to the expansion of space. Those velocities are defined as the change of their distance from us per unit of time (with "distance" being the proper length along the shortest possible curve in a hypersurface of constant FLRW time, and "time" being the FLRW time coordinate).

So the answer is 1.02c because the distance from one of the galaxies to the other is the integral of $a(t)\sqrt{dx^2+dy^2+dz^2}$ along a path in a hypersurface of constant t. This means that if both galaxies are at the distance $aL$ from us (in opposite directions), the two galaxies are at distance $2aL$ from each other, so the velocity that one of them assigns to the other is just $2\dot aL=1.02c$.

This does not violate relativity in any way, and it has nothing to do with curvature.

 

[madmike159]

Ok I think I get that. Does that mean that beond the edge of the universe (1x10^26 m I think), there could be other galaxies which we can no longer see because they are moving faster than c (due to expansion). Or am I still getting this all wrong?

 

[Fredrik]

We can see galaxies that are moving away from us at superluminal speeds too. Here's something pretty funny that I didn't realize until I wrote my previous post: If we take off in a spaceship towards one of the galaxies that are moving away from us at 0.51c, accelerate to say 0.01c relative to Earth, and then switch off the engines, we will eventually catch up with that galaxy! The distance to that galaxy will be increasing at first, but we have to catch up with it since its spatial coordinates are constant and ours are changing at a constant rate.

After I wrote my previous post, I checked one of the other threads on this subject to make sure I didn't make a big blunder here. It was the thread that was started by "Chaos' lil bro Order". (See the "similar threads" links below). Pervect posted "my" answer in #6 and other competent posters (like Marcus) said equivalent things. The thread also contained a link to this article, which explains a lot, including why we can see galaxies that are moving away from us at superluminal speeds.

 

[buffordboy23]

so I can only interpret the question in #2 to be a question about "velocities" that are due to the expansion of space.

The article that you posted is interesting; I will explore it to more depth in the future. If you are interpreting the answer due to the expansion of space, then I can't comment.

I thought the original post and following discussion assumed no expansion. SR by itself can be a challenge to grapple with; GR adds to the complexity.

Let's change the scenario. If we change the galaxies to spaceships and so that only three objects (two spaceships and "stationary" observer) exist in a non-expanding universe, then my description would be correct according to SR. Would you agree Fredrik?

 

[Fredrik]

Yes, I agree.

 

[neutralseer]

Regarding a few posts by Fredrik:

I'm confused by his statement that:

"If we take off in a spaceship towards one of the galaxies that are moving away from us at 0.51c, accelerate to say 0.01c relative to Earth, and then switch off the engines, we will eventually catch up with that galaxy! The distance to that galaxy will be increasing at first, but we have to catch up with it since its spatial coordinates are constant and ours are changing at a constant rate."

I don't follow. Let's say I'm moving towards a spatial coordinate (say a galaxy is centered there with zero peculiar velocity) that has a "speed" (due to expansion) of .51c. I agree that since I'm moving through 3-space and the galaxy isn't, the galaxy's speed due expansion will be less than if I were to remain still with zero peculiar velocity. Nonetheless, I don't think that proves that I'd eventually "catch up" with the galaxy. Here's the mathematical form of my logic:

At t=0 a spaceship starts moving towards a galaxy a distance L(t=0) away at a constant speed u. In a time dt, the galaxy will move away from the spaceship by a distance dx1 due to expansion (also, a(t=0)=1):
$dx1=\dot a(t=0)L(t=0)dt$.
Also, in a time dt the spaceship will move towards the galaxy by a distance dx2:
$dx2=udt$.
Thus, the net change in distance between the spaceship and the galaxy is:
$dL=\dot a(t)L(0)dt-udt$.
Giving us the differential equation:
$\dot L(t)=\dot a(t)L(0)-u$.

Therefore, if, as Fredrik said, $v(t=0)= \dot a(t=0)L(0)=0.51c$ and u=0.1c, then $\dot L(t=0) >0$, meaning the separation between the spaceship and galaxy would be increasing initially. So would the spaceship ever reach the galaxy? This depends on if $\ddot a(t)$ is positive or negative. Since the data indicates cosmological expansion is accelerating, the spaceship wouldn't reach the galaxy. (Unless the dark energy density for some reason isn't constant and suddenly starts decreasing.)

Finally, at one point Fredrik said that Galaxy A and B having a relative "speed" of 1.02c "does not violate relativity in any way, and it has nothing to do with curvature." I think what he meant to say is that the "speeds" of the galaxies has nothing to do with spatial curvature (which is currently thought to be zero). It should be emphasized the expansion of the universe (which causes such "speeds") is a manifestation of spacetime curvature.

 

[Cvan]

Ok I think I get that. Does that mean that beond the edge of the universe (1x10^26 m I think), there could be other galaxies which we can no longer see because they are moving faster than c (due to expansion). Or am I still getting this all wrong?

The observable edge of our universe is nothing special.
Galaxies before it cannot be moving near the speed of light, because that would take an infinite amount of energy by e=mc^2.
Galaxies after it cannot be moving faster than the speed of light because nothing moves faster than the speed of light.

Thus, the logical conclusion is that they are not "speeding away from one another with velocities" as much as the "space between them is expanding in such a way as to distance the two galaxies.

The thing you're having trouble wrapping your head around is the mechanism of how space expands, if I understand everything correctly.

 

[Fredrik]

I don't think that proves that I'd eventually "catch up" with the galaxy.

I think you're right, but I'm pretty confused right now because the results I'm getting seem to contradict what I've read: Here's my math:

Assume that the metric is $ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2)$. If the coordinate distance between the galaxies is L, then the proper distance along a path of constant t is L(t)=a(t)L, so the velocity (entirely due to expansion) is L'(t)=a'(t)L. If we say that our ship moves at constant velocity u, that means that the coordinate velocity dx'/dt'=u in a local inertial frame defined by the world line of a particle with constant spatial coordinates, and we have dt'=dt, dx'=a(t)dx, so u=a(t)dx/dt.

The distance D(t) between the ship and the destination galaxy at time t is L(t) minus the distance X(t) that the ship has traveled. D(t)=L(t)-X(t) implies that D(t) is decreasing when X'(t)>L'(t). X(t) is the integral a(t)dx along the ship's world line:

$$\int a(t)dx=\int_0^t a(t)\frac{u}{a(t)}dt=u\int_0^t dt=ut$$

so X'(t)=u and since L'(t)=a'(t)L, D(t) is decreasing when u>a'(t)L. If the rate of expansion is constant (yeah, I know), D(t) is either always increasing or always decreasing. If $L\geq 1/a'(t)$, i.e. if L'(t)>1, then even light won't ever reach the other galaxy.

OK, now I'm confused, because this seems to contradict that article and stuff I've read. An accelerating expansion seems to make things worse, not better. I think I'm going to have to read that article myself.

Finally, at one point Fredrik said that Galaxy A and B having a relative "speed" of 1.02c "does not violate relativity in any way, and it has nothing to do with curvature." I think what he meant to say is that the "speeds" of the galaxies has nothing to do with spatial curvature (which is currently thought to be zero). It should be emphasized the expansion of the universe (which causes such "speeds") is a manifestation of spacetime curvature.

Yes on the first part. Not sure about the second. Is $ds^2=-dt^2+a^2(t)dx^2$ curved?

 

[nwright]

The reason why we can't see galaxies beyond the observable edge of the Universe is because the light from them hasn't had time to reach us. Quite simply, speed of light = distance to edge of observable universe / time universe has existed for.

Also, someone mentioned earlier that velocities aren't the same if the space between two objects is expanding. Technically (since all velocities are relative) those objects do have velocities that exist and are measurable. The space between them is expanding (and the space they exist in) but it still pushes them apart.

 

[WarPhalange]

From who's vantage point? From our vantage point, the answer is 1.02c. From A or B's perspective, use the velocity addition theorem; value is less than c.

Okay, but then from our vantage point light from Galaxy A never reaches light from Galaxy B, correct? And that's not the case.

 

[neutralseer]

Hey Fredrik,
About what you said:

so X'(t)=u and since L'(t)=a'(t)L, D(t) is decreasing when u>a'(t)L. If the rate of expansion is constant (yeah, I know), D(t) is either always increasing or always decreasing. If $L\geq 1/a'(t)$, i.e. if L'(t)>1, then even light won't ever reach the other galaxy.

Okay now I'm confused, because this seems to contradict that article and stuff I've read. An accelerating expansion seems to make things worse, not better. I think I'm going to have to read that article myself.

Yeah, this stuff is simultaneously very confusing and very cool. Before trying to resolve this dilemma, let me state my understanding of it. We know that photons emitted from galaxies "receding" from us superluminally (at the time of emission) have reached us. Thus it must be possible for a slow spaceship to "catch up" with a galaxy with a high recessional velocity, or for a photon to "catch up" with a superluminally receding galaxy. However, our math shows that:

$D(t)=a(t)L-ut$, or
$\dot D(t)=\dot a(t)L-u$

Which for constant expansion or accelerating expansion implies we'll never reach the galaxy (even if we're a photon, i.e. u=c and recessional velocity > c).

Here's my guess at resolving the issue:
I think we're failing to take into account the fact that earlier in the universe,
$\ddot a(t)<0$. Take a look at any graph of a(t) versus time, (e.g. on http://www.astro.ucla.edu/~wright/cosmo_03.htm" , look at the graph with the magenta colored line) and you see that $\ddot a(t)<0$ for most of the universe's history. It's only recently that $\ddot a(t)>0$ (I believe cosmologists are a bit suspicious of the fact we are near the point of inflection of the a(t) vs. t graph). Thus, for photons being emitted from superluminal galaxies, D(t) would initially be increasing, that is, the recessional velocity of our galaxy, $\dot a(t)L$, would initially be greater than c. However, since, $\ddot a(t)<0$, at some later time t,
$\dot D(t)=\dot a(t)L-c<0$.

What do you think?

 

[nwright]

Thus it must be possible for a slow spaceship to "catch up" with a galaxy with a high recessional velocity

Where did you get this idea from? I think you are quite mistaken here. If you are not traveling faster (colinearly) than the object you are pursuing, you will never catch it up. So if the "slow" spaceship is traveling slower than the galaxy (which is what you implying), it will never catch it up.

 

[Fredrik]

We know that photons emitted from galaxies "receding" from us superluminally (at the time of emission) have reached us. Thus it must be possible for a slow spaceship to "catch up" with a galaxy with a high recessional velocity, or for a photon to "catch up" with a superluminally receding galaxy.

This is my understanding too, except I'm still not sure I understand it. I think I do now that I've had to think about the things I'm writing in this post.

However, our math shows that:

$D(t)=a(t)L-ut$, or
$\dot D(t)=\dot a(t)L-u$

Which for constant expansion or accelerating expansion implies we'll never reach the galaxy (even if we're a photon, i.e. u=c and recessional velocity > c).

Here's my guess at resolving the issue:
I think we're failing to take into account the fact that earlier in the universe,
$\ddot a(t)<0$. Take a look at any graph of a(t) versus time, (e.g. on http://www.astro.ucla.edu/~wright/cosmo_03.htm" , look at the graph with the magenta colored line) and you see that $\ddot a(t)<0$ for most of the universe's history. It's only recently that $\ddot a(t)>0$ (I believe cosmologists are a bit suspicious of the fact we are near the point of inflection of the a(t) vs. t graph). Thus, for photons being emitted from superluminal galaxies, D(t) would initially be increasing, that is, the recessional velocity of our galaxy, $\dot a(t)L$, would initially be greater than c. However, since, $\ddot a(t)<0$, at some later time t,
$\dot D(t)=\dot a(t)L-c<0$.

What do you think?

I think you're right again. For me, the most interesting piece of information on that page is that in the model I've been considering (flat space), we have $a(t)=(t/t_0)^{2/3}$ where $t_0$ is a constant, so

$$\dot D(t)=\dot a(t)L-u=\frac{2L}{3}\Big(\frac{t_0}{t}\Big)^\frac{1}{3}-u$$

The first term goes to zero as t goes to infinity, so even if the distance between the ship and the destination galaxy is increasing at first, it will at some point begin to decrease. This is pretty cool. It means that in the flat model, we can move at a constant velocity of say 1 m/s towards a distant galaxy that's moving away from us at a million times the speed of light, and get there in a finite time!

If we want to explain why we can see galaxies that are moving away from us at speeds >c, we'd have to do these calculations for a more realistic model, but the flat model is good enough to explain the principles.

 

[nwright]

I think you are mistaken...

The first term goes to zero as t goes to infinity, so even if the distance between the ship and the destination galaxy is increasing at first, it will at some point begin to decrease. This is pretty cool. It means that in the flat model, we can move at a constant velocity of say 1 m/s towards a distant galaxy that's moving away from us at a million times the speed of light, and get there in a finite time!

I'm sorry, but I don't know where you've got this idea from so I don't know where your errors have arisen from. There are no galaxies moving faster than light, nothing can travel faster than light in any comparison of velocities.

If we want to explain why we can see galaxies that are moving away from us at speeds >c, we'd have to do these calculations for a more realistic model, but the flat model is good enough to explain the principles.

[/QUOTE]

Since no galaxy can be moving away from us at speeds >c we don't need to explain why we can see them. Could you please give a literature example of an observation of a galaxy moving faster than light? I don't think you'll find one!

 

[Fredrik]

I think I have explained pretty well why superluminal "speeds" are possible, and why they don't contradict GR. It's actually pretty simple (compared to other problems in GR). If the metric is e.g. $ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2)$ (a flat FLRW spacetime), the distance L(t) to another galaxy at time t is just $L(t)=a(t)L$ where L is the distance at the time when a(t)=1. The galaxy is "moving away from us" faster than the speed of light if L'(t)>1. (I expressed the metric in units such that c=1). We can make $L'(t)>1$ simply by choosing to consider a galaxy with a large enough L. ($L>1/a'(t)$).

Check out e.g. this thread to see what other posters here have said, or this entry at the physics FAQ.