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14:20
-10
A: Velocity of most distant galaxies

SDVThe fastest any galaxy could possibly be moving away from Earth is c. We can reason that if the big bang is receding at c then our recessional velocities should be fractions of c. I think this answers the question clearly. And a justification follows. There is nothing in relativity that allows fo...

Hi SDV, I'm sorry but this is very wrong. You're right that nothing can move through space faster than $c$, as predicted by special relativity. But the expansion of the Universe is governed by general relativity. Galaxies do not move through space (except for small velocities of 100–1000 km/s), but the expansion carries them away at velocities which are not at all bounded. The answers by James K and ProfRob are correct. I suggest you delete your answer before the downvotes start ticking in :)
SDV
SDV
I don't care much about a down vote. I think it is a rather weak argument that expansion velocity somehow does not count as velocity, and I think it is adhoc to accommodate the observations I mentioned, with no real justification. Has someone shown that the effects of relativity are negated when the velocity is due to a change in the spatial component of the spacetime metric? If so show me and I'll retract the comment. It is a bold claim that a distant galaxy is moving faster than c and it need.a thorough justification. It must exist somewhere.
SDV - please read the existing answers. They articulate the accepted physics. You may have a different theory, but it is not accepted as correct [ it is in fact wrong.
Of course it exists somewhere, e.g. here. This is not at all controversial or anything (today; it was in 1922). You just have a flawed, but very common perception of relativity. In GR, space is dynamic and may have any metric, as well as temporal evolution of this metric, that you prefer. There is nothing preventing two points in space receding from each other at arbitrarily high speed. There is, however, something preventing two points receding from each other faster than two photons receding from each other at the same location.
SDV
SDV
No, read in it's entirety. I elucidate on the reason the linear approximation is used rather than the relativistic expression. It is not wrong that the values presented by others here are obtained by the linear approximation. It is not wrong that Hubble's data is relativistic. I completed satisfactorily all the coursework and then some in a graduate physics program, including general relativity, before I was unable to continue to. I am amply qualified to make this contribution. Just because it is accepted, does not mean it is right. You know better, they can't receed faster than c.
Not at all pretentious of you Pela. Thank you for your sincere contribution to understanding. The Friedman equations do not demonstrate that recession is some kind of special velocity that negates relativity. They depend on the Hubble parameter, which in turn depends on which expression you choose for the recessional velocity. It is as though you just said, "it's true because I chose it to be true." Because that is the relationship between linear expansion and the Friedman equations.
And we are not talking about two abstract points. We are talking about a planet and a very distant galaxy. So while there may be nothing stopping two abstract points from receding faster than c, if that can be shown, there is something preventing a galaxy from receding faster than c. If this is an exception to relativity then relativity is completely wrong. You are saying that in all cases but galactic recession relativity is special, but in recession it magically becomes Galilean and the relative velocity does not count as relative velocity.
14:20
I think we probably won't agree, but as I see it, you think that the Hubble law, which is linear, is an approximation. It is not (modulo the peculiar velocities I mentioned above), it's a consequence of the expansion being "homologous". The law describes the velocities that galaxies have right now. However, the redshift that we measure is not straightforward, and is only approximately $z=v/c$ for the most nearby galaxies (e.g. those measured by Hubble). This is not true because I choose it, it's just mainstream physics, which is what this site is about.
If you agree that two abstract points can recede at $v>c$, but galaxies cannot, then it follows that two galaxies at those two points would be able to travel through space faster than $c$, which is not possible (if we believe SR).
Anyway, as I said, we probably won't agree, so I will just point to the fact that this site is for mainstream physics. There are other sites that are more open to challenging mainstream physics where I think you'll be more well received. Good luck :)
SDV
SDV
There is a non relativistic and a relativistic expression for recessional velocity. The Hubble law is linear by construction. It chooses the non-relativistic expression for v(z), which implies a linear relationship between z and d. If you chose the relativistic expression there can't be a simple proportionality between d and v, or d and z. z is a function of d, and d is calculated from z, but we have a assumed a linear relationship between d and z and it can't be because v /~ d or z in the relativistic case. We lack a relationship between z and d for the relativistic case, so we approximate.
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Centre of star $T\simeq 1.5\times 10^7$K. Thermal RMS velocity of a proton $\simeq$ 600 km/s. Time dilation effect: a factor of 1.000002.
SDV
SDV
Profrob, you are estimating at the suns surface, ~5000K, although you are reporting a temp much higher. That velocity is consistent with the surface. I spoke of the thermal velocities in the interior of the star. Try your calculation again with a temperature of 10-27 million K and then tell us the thermal velocities as fractions of c.
Stack exchange, turning truth into a popularity contest since 1998, or whenever. You all misunderstand, Just because there an article reports a result does not mean that this is the view of most physicists. It just means that the result was reported clearly within the constraints of the assumptions made. Go to a physics colloquium at your local university and you will hear many physicists voicing their displeasure at certain conclusions. This conclusion is just one that has proliferated around the internet, inside it is considered absurd and incomplete. Because relativity.
The fact that it is general relativity that governs the expansion doesn't exempt the motion from relativity It ensures the motion is governed by relativity. Special relativity is just a special case of general relativity.
@SDV $1.5\times 10^7$ K is of course the temperature in the centre of the Sun and the thermal velocities I quote are those in the centre. The fact that you don't understand that $1.5\times 10^7$ means 15 million makes me think you are wasting everybody's time, including your own.
SDV
SDV
14:20
Prof rob, you are trying to show that I am wrong by posting a fallacious response. Your comprehension is not sharp. I said you appear to have reported using the higher number in your calculation ie 1.5x10^7, but your answer of 600 km/s for the thermal velocity is more consistent with temps at the surface of the sun, ~ 5000K there are any number of articles and calculators available online to show this, it is you who are wasting my time.
You did not get a thermal velocity of 600 km/s by using 15 million K in your calculation. You made that up completely, or terribly misinterpreted the first result you found on Google, and then suggested I didn't know what I was talking about. A calculation using 10-27 million K will yield thermal velocities that are fractions of c.
 
6 hours later…
SDV
SDV
20:33
Idiot
Idiots
21:10
Show your calculation. Here is mine. The RMS velocity is the square root of 3kT/m, where m is the mass of a proton. This yields 600 km/s for a temperature of 15 million Kelvin. The RMS velocity for 5000 K is much lower,. That you can't do this very basic calculation, even when challenged on it, says much about the rest of your thesis.

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