Recession Velocity near Cosmological Boundary: Hubble's Law?

[Asher Weinerman]

TL;DR

Question regarding my confusion between Hubble's Law and the freezing out of object velocities as they asymptotically approach the cosmological horizon

I'm confused whether Hubble's Law applies to objects near the cosmological horizon (CH). I'm told that objects asymptotically approach the CH and freeze there (v -> 0) in the same way that occurs during in-fall towards a black hole. But Hubble's Law says that velocity is proportional to distance. So which is right - Are objects at the CH moving fast with v proportional to d, or are they frozen out with v -> 0?

 

[Orodruin]

The basic error here is assuming that the object freezing out implies v->0.

You also need to be very careful when you talk about velocities in cosmology. Typically we are considering comoving objects, which are objects that by definition have zero comoving velocity. However, the proper separation speed between such objects is given in essence by Hubble’s law.

 

[Asher Weinerman]

So do objects just sail straight through the cosmological horizon or do they just asymptotically approach it as time here on Earth goes to infinity?

 

[Ibix]

It depends who is doing the describing and which coordinate system they are using. Using cosmological time, things move through the cosmological horizon without any problem. We will never see it happen, though, due to the redshift, which is probably where you've got the "things slow down" idea from.

 

[Asher Weinerman]

Hi Thanks Ibix. I also thought the Hubble law was pretty straightforward, not predicting any slowdown. But every once in a while I come across an expert talking about asymptotic behaviour at the CH. What do you make of the video below of Leonard Susskind. Check out the timestamp 39:32.



Is he wrong?
Thanks!

 

[PeroK]

Hi Thanks Ibix.
Is he wrong?
Thanks!

He seems to me to be saying exactly what @Ibix said.

 

[Asher Weinerman]

no, he isn't.

 

[PeterDonis]

no, he isn't.

You're going to need to give more detail than that. I have watched that part of the video and I think the same as @PeroK does. If you're having trouble seeing that, why?

 

[Ibix]

Is he wrong?
Thanks!

Note his repeated use of the words "see" and "through a telescope". He's talking about what I said in my last sentence. You see objects approaching but not crossing.

 

[Asher Weinerman]

A redshift makes objects appear redder but it doesn't make objects appear to slow down. For example in Special Relativity an object traveling away from you appears redder but obviously doesn't slow down (an object traveling away from you at near light speed "appears" visually to be traveling at half it's measured speed. However in the case of an event horizon such as a a Schwarzschild black hole objects do appear to freeze at the event horizon. The equation for that is dr/dt = -(1-2gm/r)sqrt(2gm/r). What is dr/dt for the Cosmic event horizon? I've only seen the derivation for the equation dr/dt = Hd. Thank you.

 

[Asher Weinerman]

(dr/dt above given for an object dropped from infinity at rest)

 

[Asher Weinerman]

What I should have asked at the end of my last statement is "If objects are "appearing" to slow down at the cosmic event horizon what is causing that? Can you point to an equation?

 

[Vanadium 50]

A redshift makes objects appear redder but it doesn't make objects appear to slow down.

Of course it does.

Imagine a light source blinking at once per second in its own frame. At z = 1 it's optical frequency is half what it was, and its blinking frequency is half what it was. If it moves a distance x in one second, an observer (who sees it at z = 1) sees it move that same distance in two seconds.

 

[Asher Weinerman]

If you are talking about the expansion of space then x = 0 locally. If you are talking about Special rel then velocity is symmetric between two objects. In the described case length x is contracted as well.

However I think I gave a clue to the answer of my own question when I said that objects traveling away at c appear to be moving at c/2. In the case of objects at the cosmic horizon moving outward the photons take longer and longer to arrive on Earth, making the objects motion appear to freeze out.

Strange because that is only apparent motion, not measured motion. So why does Leonard apply quantum mechanics to the thin layer of particles when the layer is not *measured* to be thin! Oh well, a question for another day :)

 

[Ibix]

A redshift makes objects appear redder but it doesn't make objects appear to slow down.

In this case it does - this is not kinematic redshift. The point here is that light emitted closer to the horizon takes longer to get here, with the horizon itself being the limiting case where light never does quite reach us. So each subsequent wave crest emitted as an object crosses the horizon takes longer and longer to get here - that is the source of the asymptotic behaviour in both the redshift and the apparent recession speed.

However in the case of an event horizon such as a a Schwarzschild black hole objects do appear to freeze at the event horizon. The equation for that is dr/dt = -(1-2gm/r)sqrt(2gm/r).

This is the rate of change of (Schwarzschild) coordinate radius as a function of coordinate time for an object free-falling from infinity, I think. It's not really a speed and I would be very surprised if it corresponds to any direct observable (I could be wrong - maybe stuff cancels out). I don't think there is an analogous concept in FLRW spacetime because there isn't a timelike Killing field to let you be sloppy about simultaneity.

I'm not sure exactly what you want to measure.

apparent motion, not measured motion

What's "measured motion" if it's not what we see?

 

[PeterDonis]

A redshift makes objects appear redder but it doesn't make objects appear to slow down.

Yes, it does.

in Special Relativity an object traveling away from you appears redder but obviously doesn't slow down

The coordinate speed you calculate for an object, after correcting for light travel time, is not the same as the apparent speed, the speed the object appears to have. An object moving away from you in flat spacetime appears to be moving slowly, because of the redshift, but when you correct for light travel time, you calculate its speed to be faster than the speed it appears to have.

I strongly suggest thinking very carefully about what that implies for the cases you are asking about (basically, the case of horizons in a curved spacetime, either cosmological horizons or event horizons).

 

[Asher Weinerman]

I think we all understand each other at this point and nothing further is being added.