B Can we travel faster than the speed of light?

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1. Nov 6, 2016

Stephanus

Dear PF Forum,
Before I get a NO answer, I'd like to ask a few question.
https://en.wikipedia.org/wiki/Hubble's_law#Observed_values
As of 13th July, 2016 Hubble flow is 67.6 km/s per mega parsec.
Or 67km/s per 31 trillion * 1 million km or per 3.26 million light year.
Say a galaxy 11.68 giga light year away from us. This galaxy is traveling at 80% the speed of light away from us.
What I want to know. Is it possible for an object at that galaxy to travels say, 30% the speed of light "away" from us relative to the galaxy?
If it is, then wouldn't the object travel at 1.1c?
And one more thing. Any object as far away as 13.8 giga light year from us can never reach us because it has already traveled more than the speed of light.
And I read somewhere (do I have to make a citation here?) that any object less than 13.8 gly from us can reach us, because it haven't traveled at the speed of light.
So the object at 11.68 gly can reach us. But can an object at that galaxy which travels 30% the speed of light away from us, then stop then travel back to us, say 0.99c reach us?
Thank you very much.

2. Nov 6, 2016

Jorrie

Galaxies at that distance do not "travel away from us at at 80% the speed of light". We say they a have a recession rate of 0.8c, because we cannot measure (SR-like) speed over the curved spacetime of those distances. As you probably know, the recession rate is the result of the metric expansion and does not have an upper limit.

Also, I do not think you can generally add recession rates like you did - perhaps at the lower ranges it may be approximately right.

Last edited: Nov 6, 2016
3. Nov 6, 2016

4. Nov 6, 2016

Jorrie

No, this is SR and does not work for cosmological distances and also not for any curved spacetime.

5. Nov 6, 2016

6. Nov 6, 2016

Charles Kottler

Sorry, I guess you are correct that it would not apply in the context of this question, where the 'velocity' is a result of the metric expansion. Would it be correct to say that SR would apply locally over every region between the remote galaxy and us, and that we can then think of the expansion as being similar to a long piece of elastic being stretched: the ends can move apart very fast relative to each other while in any small section the expansion is hardly noticeable?
67.6 km/s per mega parsec sounds a lot until you convert it back to more common units - turn it around and see how long it takes for one Km to 'grow' by 1mm. Regardless of the distance (and therefore speed) between the endpoints it will always be possible to get a signal or to travel from one point to another given enough time.

7. Nov 6, 2016

Jorrie

Actually, in comoving coordinates, we can add recession speeds normally. It is implied by the way comoving distance is defined: D = H0 Vrec/c.
As an example, at the Hubble radius (RH=14.4 Glyr), the recession speed is c. For a comoving observer at our Hubble radius, a galaxy that lies in the same direction, but on that observer's RH, will also have a recession rate of c to them. That galaxy will be 28.8 Glyr from us, at a recession rate of 2c.

8. Nov 6, 2016

Stephanus

Thanks for your reply . Some mentor said (in my previous post), you can't use that for hubble flow

9. Nov 6, 2016

Stephanus

Yes, I was afraid that we have to you SR velocity addition. Because if I recall correctly some mentor/advisor in PF said that we can't use that.

10. Nov 6, 2016

Jorrie

Yes, locally SR will apply everywhere.

With the present accelerating expansion, there is actually a cosmological event horizon (at some 16.5 Glyr comoving) from beyond which no present emission can ever reach us. We can observe particles originally from areas that are presently more than 45 Glyr away, due to the fact that accelerated expansion only started "recently" (actually some billions of years ago).

Last edited: Nov 6, 2016
11. Nov 6, 2016

Stephanus

Everywhere?? What about very distant galaxy that travels recedes more than the supposed speed of light?

Last edited: Nov 6, 2016
12. Nov 6, 2016

Jorrie

Did you notice the word "locally"? It means in the vicinity of the observer in the galaxy, or wherever...

13. Nov 6, 2016

Stephanus

Sorry, I have corrected my post. But thanks anyway. You beat me to it

14. Nov 6, 2016

Jorrie

No problem, feel free to ask more clarification.
This has gone into cosmology now, so maybe further questions should be asked in that sub forum.

15. Nov 6, 2016

Lautaro

16. Nov 6, 2016

Jorrie

Naturally, because expansion does not have a speed - but the recession speed of a specific galaxy/cluster is well defined. It is the rate of change of its proper (or physical) distance per unit cosmic time and it can far exceed c. For expansion, we do not have a specific distance that changes over time; we express it as the Hubble parameter, which is a fractional rate of increase per unit distance.

17. Nov 7, 2016

vanhees71

http://www.edu-observatory.org/physics-faq/Relativity/GR/hubble.html

The bottom line: the Hubble redshift is not a Doppler effect in the strict sense, and the recession velocity is not a proper relative velocity between (local!) events!

18. Nov 8, 2016

Chris Miller

If you were approaching a 100 light year distant (by earth measurement) star at nearly c, due to length foreshortening, it might be only a short distance away by your measurements. Then, if you were to decelerate very rapidly to 0 v, it would "suddenly" retreat 100 light years, for a relative velocity of many times c.

19. Nov 8, 2016

Ibix

The star doesn't move in that case. And how rapidly the distance to it "changes" is entirely a matter of definition.

20. Nov 8, 2016

Staff: Mentor

There is no inertial coordinate system in which this statement is true.

You could, of course, make a non-inertial coordinate system where that is true, but speeds are not limited to c in non-inertial coordinate systems.

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