Expansion of universe

• I
i have a question about the motion of galaxies and the speed limit of light.
using Hubble's law, it is theoretically found that after a certain distance(about 4200 mega parsec or something) from a observer, the galaxies are moving faster than the speed of light.
consider the following scenario:
let galaxy A be in any part of the universe. let galaxy B be at a distance greater than 4200 mega parsec from A. as per Hubble's law, B is moving away from A at speed faster than light and to B , A is moving away from A at speed faster than light . the movement of these two galaxies is observed by a observer in a absolute frame of reference F (Newtons's idea; assume such a frame exists)
here are the question:
1)for the observer in F , is the galaxy B and A moving by themselves , or are they moving due to the expanding space ?. because if galaxies were to move due to expansion of space, then aren't they violating the speed limit of light due to the following fact. let V be the velocity of the galaxy B and v be the velocity of the expanding space in which the galaxy is present . hence the total velocity of the galaxy B measured by the observer in F would be V+v ;where both V and v are comparable to the speed of light .
2) are the galaxies actually embedded in space and moving with space or are they moving in space and not embedded in the space itself?

Drakkith
Staff Emeritus
because if galaxies were to move due to expansion of space, then aren't they violating the speed limit of light

The speed of light is a local speed limit, not a global one. That means that two galaxies can recede from each other at a velocity greater than c thanks to the expansion of space, yet you will never be able to outrace a light-beam.

2) are the galaxies actually embedded in space and moving with space or are they moving in space and not embedded in the space itself?

There's no observable difference between the two that I know of. Keep in mind that we can't observe what space does. We can only observe what objects within space do. So there's no way to tell if space is "actually moving" or if objects are merely moving "through space". As far as I know at least.

The speed of light is a local speed limit, not a global one. That means that two galaxies can recede from each other at a velocity greater than c thanks to the expansion of space, yet you will never be able to outrace a light-beam.

There's no observable difference between the two that I know of. Keep in mind that we can't observe what space does. We can only observe what objects within space do. So there's no way to tell if space is "actually moving" or if objects are merely moving "through space". As far as I know at least.
then, would that mean its possible that galaxies have their own motion and the space itself has its own motion. and for both their motion the speed limit is c. hence the maximum possible observable velocity of a galaxy is 2c for a observer in a absolute frame of reference F (the frame F is at absolute rest - Newton proposed the idea of such a frame of reference) . hence the speed of light is not violated by the components in motion individually(space ; galaxy) but the speed of the galaxy is greater than the c . is this possible?

Drakkith
Staff Emeritus
then, would that mean its possible that galaxies have their own motion and the space itself has its own motion. and for both their motion the speed limit is c.

Keep in mind there is no limit to recession velocity, not merely 2c.

is this possible?

Nope. A preferred, absolute reference frame is ruled out by observations as far as I understand.

Keep in mind there is no limit to recession velocity, not merely 2c.

Nope. A preferred, absolute reference frame is ruled out by observations as far as I understand.
isn't a stationary inertial frame of reference similar to the absolute frame of reference? also if it is proved that something in the universe is at absolute rest, than would a absolute frame of reference be accepted?

Drakkith
Staff Emeritus
isn't an inertial frame of reference similar to the absolute frame of reference?

I don't believe so. If an absolute reference frame existed then an inertial frame of reference which was in motion with respect to the absolute frame should have differences in experiments run in that inertial frame. In other words, if you perform two identical experiments in each frame, there should be a difference in the results, and those differences should increase as the inertial frame moves with a higher and higher velocity relative to the absolute frame. This is not what we observe. All inertial frames behave identically, no matter which direction they are moving in nor how fast they are moving.

also if it is proved that something in the universe is at absolute rest, than would a absolute frame of reference be accepted?

Certainly. Absolute rest necessarily requires that an absolute reference frame exist.

I don't believe so. If an absolute reference frame existed then an inertial frame of reference which was in motion with respect to the absolute frame should have differences in experiments run in that inertial frame. In other words, if you perform two identical experiments in each frame, there should be a difference in the results, and those differences should increase as the inertial frame moves with a higher and higher velocity relative to the absolute frame. This is not what we observe. All inertial frames behave identically, no matter which direction they are moving in nor how fast they are moving.

Certainly. Absolute rest necessarily requires that an absolute reference frame exist.
how is it possible that measurement made from a inertial frame is correct irrespective of the velocity of the inertial frame, which itself is not possible to know without an absolute frame . hence any measurement made from a inertial frame is relative , isn't it? then how is the measured value same in all inertial frames of reference? [ I seem to have deviated from the original topic :-) ]

Drakkith
Staff Emeritus
how is it possible that measurement made from a inertial frame is correct irrespective of the velocity of the inertial frame, which itself is not possible to know without an absolute frame .

It's not that they are "correct", it's that they are the same. And to clarify, I'm talking about experiments you can perform in a single reference frame, not observations of external events outside of your reference frame.

It's not that they are "correct", it's that they are the same. And to clarify, I'm talking about experiments you can perform in a single reference frame, not observations of external events outside of your reference frame.
so is it impossible to determine whether any measurement like velocity made by an observer(in a inertial frame) of an event occurring in the same frame of the observer or any other frame(inertial or non inertial) is the correct value of the quantity , since the velocity of the frame of the observer is unknown ?

Drakkith
Staff Emeritus
so is it impossible to determine whether any measurement like velocity made by an observer(in a inertial frame) of an event occurring in the same frame of the observer or any other frame(inertial or non inertial) is the correct value of the quantity , since the velocity of the frame of the observer is unknown ?

The velocity of the observer isn't unknown, it's relative. A spaceship flying by the Earth at 100 km/s is viewed by an observer on the Earth as traveling at 100 km/s, yet an observer on the ship views the Earth as passing by at 100 km/s. And both are correct.

Dale
Mentor
2021 Award
Newton proposed the idea of such a frame of reference
Newtonian gravity works on the scale of the earth and moon, but it does not work at the scale of the universe.

Newtonian gravity works on the scale of the earth and moon, but it does not work at the scale of the universe.
For the most part it does:

Also see this, this, this and this reference on the subject. For the gravity in cosmology Newton does most of the job since it is rather weak; the relativistic equations give the same result like Newtons, with the little difference that the energy of light is inversely proportional to the scalefactor. But you don't need hardcore relativity in standard cosmology like you would in black hole astronomy, the formalism is completely newtonian for the most part.

then, would that mean its possible that galaxies have their own motion and the space itself has its own motion.
That's right. The first is the local and the second the comoving recessional velocity. We observe a combination of both.

and for both their motion the speed limit is c.
Only for the local velocity. The recessional velocity has no limit.

hence the maximum possible observable velocity of a galaxy is 2c
UDFy-38135539 as one of the farthest observable galaxies with a redshift of 8.55 was receding with 2.08 c when it emitted its light and is receding with 3.62 c today.The CMB surface had 63.12 c relative to us when it emitted the light which we receive today 400 million years after the big bang, and is receding with 3.11 c today.

Newton proposed the idea of such a frame of reference
Later he rejected it.

,

so is it impossible to determine whether any measurement like velocity made by an observer(in a inertial frame) of an event occurring in the same frame of the observer or any other frame(inertial or non inertial) is the correct value of the quantity , since the velocity of the frame of the observer is unknown ?
All comoving observers, even when having a relative recessional velocity to each other, are at rest relative to the CMB. If we want to translate the measurements made from earth to that frame we first have to substract the CMB dipole and then add the proper distance times the Hubble constant to the equation if we want to transform into the system of another Robertson-Walker observer. The kinetic energy of an object in that frame is the energy it would have relative to an observer at rest to the CMB when passing by right next to him.

PeterDonis
Mentor
For the most part it does

He said the "equations of an expanding universe" are "Newton's equations", but he does not mean the fundamental equations of Newtonian gravity. He only means the equations from one particular solution in Newtonian gravity. The fact that one particular GR solution happens to yield the same equations as one particular solution in Newtonian gravity does not mean Newtonian gravity explains the expansion of the universe. It just means the same equations happen to come out of GR in this particular case. You certainly can't use the fundamental equations of Newtonian gravity to derive this particular solution for the universe.

The velocity of the observer isn't unknown, it's relative. A spaceship flying by the Earth at 100 km/s is viewed by an observer on the Earth as traveling at 100 km/s, yet an observer on the ship views the Earth as passing by at 100 km/s. And both are correct.
but if both of the observers (on the earth and space ship) are already in motion such that both earth and spaceship are moving at 5 km/s , and 100km/s is the measured relative value. but the actual value is 105 km/s . hence the 5 km/s value is not known unless a absolute frame of reference is defined. isn't this correct?

Newtonian gravity works on the scale of the earth and moon, but it does not work at the scale of the universe.
but that does not have anything to do with the absolute frame of reference.

PeterDonis
Mentor
if both of the observers (on the earth and space ship) are already in motion

Relative to what?

such that both earth and spaceship are moving at 5 km/s , and 100km/s is the measured relative value. but the actual value is 105 km/s

Relative to what?

the 5 km/s value is not known unless a absolute frame of reference is defined

No, it's not known unless you specify what it is relative to. There is no absolute frame of reference.

UDFy-38135539 as one of the farthest observable galaxies with a redshift of 8.55 was receding with 2.08 c when it emitted its light and is receding with 3.62 c today.The CMB surface had 63.12 c relative to us when it emitted the light which we receive today 400 million years after the big bang, and is receding with 3.11 c today.

so if the galaxy is moving with the resulting speed of 3.11 c , then aren't the particles in that galaxy violating the local speed limit. (because the recession velocity is added to the local velocity of the particle.).

also in calculating the recession velocity using Hubble law, is the recession of the observer taken into account?

No, it's not known unless you specify what it is relative to. There is no absolute frame of reference.
not all motion is relative. only the measured value is relative. whether the motion is observed or not , the particle or whatever being observed is in motion because of the expansion of the universe and also due to its own energy. hence it would be impossible to find the correct value of the observed value without defining a absolute frame of reference.

PeterDonis
Mentor
the recession velocity is added to the local velocity of the particle

No, it isn't. The recession velocity is a coordinate velocity in FRW coordinates. The local velocity of the particle is the relative velocity that would be measured by a comoving observer at the same spatial location. You can't add these two things; the result is physically meaningless.

No, it isn't. The recession velocity is a coordinate velocity in FRW coordinates. The local velocity of the particle is the relative velocity that would be measured by a comoving observer at the same spatial location. You can't add these two things; the result is physically meaningless.
but we have actually observed the galaxy moving with the said 3.11 c velocity. if the two velocities are not added, how is it possible.

PeterDonis
Mentor
not all motion is relative

Yes, it is. You are coming very close to receiving a warning for misinformation.

it would be impossible to find the correct value of the observed value without defining a absolute frame of reference

This is not correct. Again, you are very close to receiving a misinformation warning. I strongly suggest that you take some time to learn more about relativity and cosmology and what our best current theories actually say. You are making a number of incorrect statements which appear to be based on a highly mistaken understanding.

we have actually observed the galaxy moving with the said 3.11 c velocity

No, we haven't. We have only observed the galaxy's redshift. Observed redshift only equates to relative velocity within a single local inertial frame. There is no local inertial frame that covers both us here on Earth and a faraway galaxy.

Don't be misled by the fact that cosmologists routinely quote "recession velocity" numbers for galaxies instead of redshifts. This is a convention and doesn't say anything about the physics; all it really says is that, for whatever reason, cosmologists prefer velocity units to redshift units. The "recession velocity" of 3.11c is just the redshift of 3.11 (i.e., the lines we observe in the galaxy's spectrum are at wavelengths 3.11 times the wavelengths of the same spectral lines observed in the laboratory) times the speed of light. [Edit: This last is not correct, see follow-up posts later in this thread.]

Last edited:
No, it isn't. The recession velocity is a coordinate velocity in FRW coordinates. The local velocity of the particle is the relative velocity that would be measured by a comoving observer at the same spatial location. You can't add these two things; the result is physically meaningless.
lets use the expanding dough and raisin analogy. here the particle observed in the galaxy is inside the raisin moving at some velocity v close to c. the dough itself is expanding with any velocity V . hence, would it mean, to a observer on some other part of the dough in some other raisin , the total velocity of the particle would be v+V .

PeterDonis
Mentor
lets use the expanding dough and raisin analogy

You can't use this analogy to draw the conclusion you are trying to draw. It is a very limited analogy and you should not use it if you really want to learn the correct physics. The "velocity of the dough" you are imagining here has no physical meaning.

Once again, I strongly suggest that you take some time to learn correct relativity and cosmology before posting again. Ned Wright's cosmology FAQ and tutorial would be a good start:

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

Dale
Mentor
2021 Award
but that does not have anything to do with the absolute frame of reference.
Newton's absolute frame of reference idea can be patched on to Newtonian gravity, but it is incompatible with general relativity. So it won't work on cosmological scales where Newtonian gravity fails.

The "recession velocity" of 3.11c is just the redshift of 3.11
Sorry, but this is wrong. The redshift of the CMB surface of last scattering which is today receding away with 3.11c and was receding away with 63.12 when it emitted it's light is z=1089. You are mixing up a few things here.

so if the galaxy is moving with the resulting speed of 3.11 c , then aren't the particles in that galaxy violating the local speed limit.
No, the 3.11 c are not the local but the coordinate velocity. We are assuming galaxies to have local velocities of only a few 100 km/s so the local component is neglible at such distances.

because the recession velocity is added to the local velocity of the particle.
When you have a photon at the hubble horizon which is heading towards you the distance stays the same since c-c=0. If it is going in the other direction the distance increases with 2c (and accelerates with increasing distance). So there is no violation since you always add a local component of limit 1c to the arbitrarily high recessional velocity. The redshifts from local and recessional velocities just multiply: z_total=(1+z_rec)(1+z_pec)-1

PeterDonis
Mentor
The redshift of the CMB surface of last scattering which is today receding away with 3.11c

Do you have a reference for this number? And more generally for your numbers given in an earlier post here:

Fy-38135539 as one of the farthest observable galaxies with a redshift of 8.55 was receding with 2.08 c when it emitted its light and is receding with 3.62 c today.The CMB surface had 63.12 c relative to us when it emitted the light which we receive today 400 million years after the big bang, and is receding with 3.11 c today.

In particular, I don't understand how the CMB last scattering surface recession velocity today (redshift over 1000) can be less than the recession velocity today of a galaxy whose light we are now seeing with a redshift of only 8.55. With the obvious definition of "CMB last scattering surface" in this context (which is the comoving observer whose worldline, extended back to the time of last scattering, was co-located with the emission point of CMB light we are observing today), the recession velocity of the CMB last scattering surface today should be larger than that of any galaxy we can see.

Do you have a reference for this number?
I did the calculations here - the velocity at emission is calculated via the past light cone distance for the corresponding light travel time, see this plot. The velocity at absorption is that divided by the scale factor of the corresponding emission time. The whole calculation is a little bit longer, but you will quickly find that the CMB surface is in the rough distance of the particle horizon, which is about 46 GLyr away, which is 3 times the hubble radius. You can also find with Google that the redshift is 1089. So the present velocity is roughly 3 c and the redshift roughly 1000 (because the scale factor was 1000 times smaller then than it is today)

In particular, I don't understand how the CMB last scattering surface recession velocity today (redshift over 1000) can be less than the recession velocity today of a galaxy whose light we are now seeing with a redshift of only 8.55.
This is explained in the first chapter of Davis & Lineweaver: expanding confusion - the universe's expansion rate was slowing down and is now converging to a constant value (therefore the accelerated expansion). The galaxy was farther away when it emitted the light we receive today, and the CMB surface nearer so that compensates for the higher velocity.

Bandersnatch