Something about calculating the Age of the Universe

[marcus]

Hubble law is defined in terms of proper distance (i.e. distance at a definite moment in time) and the v is the speed of distance change at that moment.

People often omit the time variable. But to really understand the law you have to have the concept of universe time also called Friedmann time.

this is the time that the standard cosmic model runs on and it is defined in terms of a family of observers all at CMB rest.

So whenever you write Hubble law, be sure to mentally include the time variable, or even make it explicit and write v(t) = H(t)D(t).

If you don't write it in explicitly at least THINK it's there and ask yourself WHEN you are applying the law.

the equations in 2. are exact
the equation in 3. is approximate for very small distances but typically it is wrong because most of the time we aren't talking about about small distances. so one can draw no conclusion from zxc=v
and what do you mean by "v". at what point in TIME is v evaluated? z is evaluated when the light comes in. but the speed at that moment ordinarily does NOT correspond to the speed that the thing has been receding.

the thing has been receding at many different speeds while the light has been traveling. there is no simple relation.

also you do not give a clear idea of WHEN you are applying the Hubble law, so writing the equation v = HD is meaningless. no mathematial conclusion can be drawn about D or about v. And H is changing constantly over time

So you ask what D is? Is D the first distance? or is D the second distance? Mathematically it is neither.

so your point 2. is exact AFAICS
but I can't make anything out of your point 3.

also your point 1. is exact. AFAICS

it is only point 3 that does not go anywhere.

 

[Jorrie]

My question here is, is D the D2 or the D1 of the star and how would one be able to tell?

D in the HD part is D2 in your point 4, because I see that you associated D2 with the present (proper) distance.

 

[JohnnyGui]

Hubble law is defined in terms of proper distance (i.e. distance at a definite moment in time) and the v is the speed of distance change at that moment.

People often omit the time variable. But to really understand the law you have to have the concept of universe time also called Friedmann time.

this is the time that the standard cosmic model runs on and it is defined in terms of a family of observers all at CMB rest.

So whenever you write Hubble law, be sure to mentally include the time variable, or even make it explicit and write v(t) = H(t)D(t).

If you don't write it in explicitly at least THINK it's there and ask yourself WHEN you are applying the law.

the equations in 2. are exact
the equation in 3. is approximate for very small distances but typically it is wrong because most of the time we aren't talking about about small distances. so one can draw no conclusion from zxc=v
and what do you mean by "v". at what point in TIME is v evaluated? z is evaluated when the light comes in. but the speed at that moment ordinarily does NOT correspond to the speed that the thing has been receding.

the thing has been receding at many different speeds while the light has been traveling. there is no simple relation.

also you do not give a clear idea of WHEN you are applying the Hubble law, so writing the equation v = HD is meaningless. no mathematial conclusion can be drawn about D or about v. And H is changing constantly over time

So you ask what D is? Is D the first distance? or is D the second distance? Mathematically it is neither.

so your point 2. is exact AFAICS
but I can't make anything out of your point 3.

also your point 1. is exact. AFAICS

it is only point 3 that does not go anywhere.

Hmm, let's see if I understand this correctly, please verify this for me: So the formula z x c = v is inacurrate in the case of large distances because by the time we measure z, the object has already changed velocities due to the expansion? So can I say that z x c = v is only accurate in the case of a constant velocity?

If that's the case, then let's say an object is traveling at roughly a constant velocity (I know this isn't the case due to H changing over time, but I'm trying this to understand the formula myself) or has a small distance. Let's associate two different time points, the time point when the object was at D1 being t1, and the time point when the object is at D2 being t2. By the time one measures z, the object would have traveled to D2 and thus we measure z at t2. However, if we use that same z to calculate v with the formula z x c = v, then my instinct would say that v was the velocity of the object when it was at D1 since that was the velocity for letting the object travel to D2 and giving the z we measure. So v was the velocity when the object was at D1 and thus it was the velocity at t1. Concluding from that I'd say that z(t2) x c = v(t1). Since the velocity was at t1, that means that v(t1) = H(t1) x D(t1). D at t1 was D1, so D = D1.

 

[marcus]

So can I say that z x c = v is only accurate in the case of a constant velocity?

If that's the case, then let's say an object is traveling at roughly a constant velocity...

No, z = v/c is fundamentally inaccurate. There is a more precise doppler formula one learns in Special Relativity. It involves z+1. You might be interested in learning it!
It just happens that z = v/c is a good approximation at low speeds. It is not "constancy" that is the issue so much as the range of speeds.

You will have fun if you start to learn SR and GR. But now you seem to be thinking of expansion redshift as a DOPPLER effect of a moving object.
As long as you think of this in terms of moving objects you will probably have a hard time understanding---at least this seems to be what I and other people have experienced.

Recession is the increase of distance between. It is not like ordinary motion thru space. Recession speeds are typically faster than light. That is, for most of the objects we can see with telescopes. anything with z bigger than 1.4, and most objects have z > 1.4

I'm short of time at the moment and can't discuss further but surely others here will discuss your analysis with you!
Your analysis seems to me to be off track for two reasons: the formula z = v/c is no good except in certain special limited cases (then it is a very handy approx)
and also the Hubble law v=HD is not a Doppler effect of motion thru space, and understanding it requires understanding the CMB rest criterion, and the idea of universe time. It holds for objects at rest, v is not a speed of motion, and the law holds at a given moment t in universe preferred time. It really should be written
v(t)=H(t)D(t)

 

[Jorrie]

So v was the velocity when the object was at D1 and thus it was the velocity at t1. Concluding from that I'd say that z(t2) x c = v(t1). Since the velocity was at t1, that means that v(t1) = H(t1) x D(t1). D at t1 was D1, so D = D1.

This formula V = H₀ D actually uses comoving distances and then it works for any distance, not only for short ones. The comoving distance of any galaxy remains constant forever, because it is defined as the proper distance of that galaxy today. Hence it is D2 in your example. The formula z c = V is an approximation for short distances, because cosmological redshift is not directly proportional to recession velocity.

As Marcus said before, recession velocity does not really mean a velocity - it is really only a recession rate that we can actually do without. All we need is the redshift and we can do most of the cosmology we require.

 

[JohnnyGui]

This formula V = H₀ D actually uses comoving distances and then it works for any distance, not only for short ones. The comoving distance of any galaxy remains constant forever, because it is defined as the proper distance of that galaxy today. Hence it is D2 in your example. The formula z c = V is an approximation for short distances, because cosmological redshift is not directly proportional to recession velocity.

As Marcus said before, recession velocity does not really mean a velocity - it is really only a recession rate that we can actually do without. All we need is the redshift and we can do most of the cosmology we require.

Thanks for your answer. I understand that it's not a real velocity. What I'm trying to do here is applying v = z x c to a scenario in which the formula IS accurate enough (in special limited cases as @marcus said) so that I'm sure I understand how and WHEN the formula should be used. I now indeed understand from Marcus that the formula couldn't be applied in the case of cosmological time.

I have thought of an example I'd like to give in which I think the formula v = z x c is accurate. Please correct me if I'm still using this formula the wrong way in this example:

Since I'm thinking the redshift as a doppler shift which is wrong, let's assume we're now measuring the change in a sound wave from a sound emitting device that is moving away from an observer in an expandable fasion with a constant velocity. What I mean in an expandable fasion is that there are other devices that are also moving away from the observer whom their speeds depend on their distances (such as in a stretching scenario) so that there's a H that can be calculated (and which is changing over time).

The formula combined is: λ_Obs / λ_Emit = D2 / D1 = z + 1 = (v = H x D) / s (= speed of sound which is 340 m/s)
The observer knows the initial distance D1 being 750m. The timepoint at which the device is at D1 is t0

At t1, the observer measures λ_Obs of the sound wave the device is emitting, already knowing what the initial λ_Emit was.
At the time he measures this, the device has moved to D2. Thus, the device reaching D2 occurs at t1.
The measured random values the observer gets are λ_Obs and λ_Emit being 500 and 300 respectively.
Thus, the formula is now: 500 / 300 = D2 / 750 = z + 1 = (H x D) / s

Calculating D2 gives a value of 1250m and z a value of 0,6667, thus 0,6667 = (H x D) / 340
Therefore, 0,6667 x 340 gives the speed of the object, being 226,667 m/s. Thus, 226,667 = H x D.

My question: When the device was at D1 = 750m (t0), what value did H have? Is it 226,667 / 750 = 0,302 m/s/m or is that at t1 when the device is at D2? If this H value is at t0 that means that calculating D gives 226,667 / 0,302 = 750m, thus D is D1. v being 226,667 m/s, can be taken at any time point (t0 and t1) since it's constant. It is therefore the timepoint at which H is measured (t0 or t1) that decides if D is D1 or D2! Even if z is calculated at t1. So if the calculated H is the H value at t0, then I'd say in general that z(t) would give an H at (t - Δt)

Am I making any sense here?

 

[Jorrie]

My question: When the device was at D1 = 750m (t0), what value did H have? Is it 226,667 / 750 = 0,302 m/s/m or is that at t1 when the device is at D2? If this H value is at t0 that means that calculating D gives 226,667 / 0,302 = 750m, thus D is D1. v being 226,667 m/s, can be taken at any time point (t0 and t1) since it's constant. It is therefore the timepoint at which H is measured (t0 or t1) that decides if D is D1 or D2! Even if z is calculated at t1.

Yes in a hypothetical universe with linear expansion, i.e. da/dt=constant, what you wrote is correct. Theory and observation favor an expansion curve that is not straight, but rather looks like this:

 

[JohnnyGui]

Yes in a hypothetical universe with linear expansion, i.e. da/dt=constant, what you wrote is correct. Theory and observation favor an expansion curve that is not straight, but rather looks like this:
View attachment 85171

I'm glad I got the linear expansion correct :). I gave the above example because you said that D in the v = H x D formula would be D2 but seeing that the calculated H in my example is the H value at t0, it would give a D that is D1 instead. That's what got me confused. z is calculated at t1 while the calculated H is the value from t0 so I would say that calculating z at t would give an H value that's from t - Δt. Please correct me if this is wrong.

What's weird about this though, is that suppose the device has now traveled a further distance and a totally new observer tries to calculate the speed of that same device. He would measure a bigger λObs / λEmit ratio (λObs has become larger because the device is further away, thus more time for the wavelength to become shifted while under its way to the observer) and thus a larger z. A larger z would consequently give a larger v (z x speed of sound) while we just said that the speed of that device is constant. How can this be corrected? How is one so sure that previous calculated v (266,667 m/s) is correct then??

 

[Jorrie]

What's weird about this though, is that suppose the device has now traveled a further distance and a totally new observer tries to calculate the speed of that same device. He would measure a bigger λObs / λEmit ratio (λObs has become larger because the device is further away, thus more time for the wavelength to become shifted while under its way to the observer) and thus a larger z.

I think you are confusing ordinary flat spacetime with the curved spacetime of the universe. Your example (which is flat spacetime) says the speed of recession remains constant for a given source, hence the Doppler shift never changes. The distance grows and your "H" is coming down in direct proportion, i.e the product remains the same.

A source that was originally farther away had to be given a greater (but constant) speed by your scenario, otherwise it would have had a lower "H". If you would put rockets on the sources so that the recession speed of each changes, then the z of each source would have become larger over time, but then v = HD does not hold anymore.

I think you are confusing yourself with these non-standard scenarios; why not look at the balloon analogy sticky at the top of this forum and learn real basic cosmology from that?

 

[JohnnyGui]

I think you are confusing ordinary flat spacetime with the curved spacetime of the universe. Your example (which is flat spacetime) says the speed of recession remains constant for a given source, hence the Doppler shift never changes. The distance grows and your "H" is coming down in direct proportion, i.e the product remains the same.

A source that was originally farther away had to be given a greater (but constant) speed by your scenario, otherwise it would have had a lower "H". If you would put rockets on the sources so that the recession speed of each changes, then the z of each source would have become larger over time, but then v = HD does not hold anymore.

This is exactly what I indeed concluded in my mind. What I didn't get was WHY the doppler shift never changes at a constant velocity. That is, until I pictured a simple scenario in my head which convinced me that at a constant velocity the λObs / λEmit ratio is always constant since λObs doesn't change in that case! So I guess at a constant v, this is the same case for a redshift of light being constant instead of sound?

You are right about me picturing non-standard scenarios. The thing is, I always like to imagine simple scenarios first and draw conclusions from that before going into more complex scenarios so that I can understand them better.
I'm aware of the balloon analogy and the accelerating expansion of the universe (i.e. changing velocities). Because of that knowledge in combination with what we discussed just now, I now understand why v = HxD doesn't hold at different velocities while it does at a constant one.

 

[JohnnyGui]

Guys, I need help

After looking at the formulas I encountered something that I keep stumbling upon.

After looking at marcus's explanation found here: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/ I concluded that when something is moving away while emitting a particular frequency in the other direction, the frequency would be smaller by a factor of ((c - v) / c) (let's call c here the speed of sound)
1. Inversely, the wavelength would be bigger by a factor of (c / (c - v)), correct?
2. Thus, the emitted wavelength λEmit would be multiplied by (c / (c - v)) to get λObs thus, λEmit x (c / (c - v)) = λObs
3. This means that λObs / λEmit = c / (c - v)
4. This also means that λEmit / λObs = (c - v) / c = 1 - (v / c)
5. We know that (v / c) = z thus, we can say that λEmit / λObs = 1 - z

Here's the weird part, I read on several sources (from which I got one of the previous mentioned formulae) that (λObs / λEmit) - 1 = z. This formula is clashing against my concluded formula λEmit / λObs = 1 - z since giving the λEmit and λObs random values doesn't give the same z.

What's wrong here?

UPDATE: I think I found the problem, the thing is that (λObs / λEmit) - 1 = z is used for when the observed wavelength is SHORTER than the emitted wavelength (i.e. the object is moving in the direction of its emitting soundwave) while my concluded formula (λObs / λEmit) - 1 = z is used when the observed wavelength is LONGER than the emitted wavelength, thus the object is moving in the opposite direction from the soundwave that it's emitting. Is this correct?

 

[marcus]

Hi JG, I noticed your link to an earlier (2008 actually!) post of mine which got me curious

...After looking at marcus's explanation found here: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/ I concluded that when something is moving away while emitting a particular frequency in the other direction, the frequency would be smaller by a factor of ((c - v) / c) (let's call c here the speed of sound)...

so I went back and had a look. I think the main point of that post was that cosmological redshift is not Doppler shift.
Not in any simple sense we normally think of when we say Doppler effect---the effect associated with some particular speed. Cosmo redshift is not the effect of any particular speed---it is a cumulative effect that builds up over million, sometimes billions, of years.

So it is potentially a bit confusing to have a discussion of Doppler effect in Cosmology forum. You might have more success asking in either General Physics or in the (SR and GR) Relativity forum. There is a nice relativistic Doppler formula in Special Relativity which you might like if you haven't encountered it yet.

But none of that usual Doppler stuff, based on particular speeds at emission and reception, applies to cosmological redshift. That is "thing one" the most important thing to realize. Anyway you got me curious about what I said when you cited that old 2008 post so I will quote it and have a look:

(BTW I hope it's clear that distance growth, i.e. recession, is not like ordinary motion because nobody gets anywhere by it everybody just becomes farther apart. So you should never think of expansion cosmology in terms of familiar motion thru surrounding space--it leads to pretty bad confusion.)
==quote from that old post==
You aren't really talking about the cosmological redshift, because that is not the doppler effect of the current recession speed, or the recession speed at the time of emission, or at any other one particular time. the cosmo redshift is determined by the factor by which distances have expanded during the light's travel time. The formula they give you for it, on day one of cosmo class, is 1+z = a(now)/a(then), the ratio of the metric scalefactor now compared to what it was then, when the light was emitted.

So if you were talking about the cosmo redshift you would have totally the wrong formula. But I think what you are really asking about is the DOPPLER EFFECT shift. If z is defined as the fractional increase in wavelength, and v is actual motion away, of the observer from the source, then you could say z = v/c.

That would have nothing much to do with universe expansion, but it could apply to some random motions of neighboring galaxies relative to each other, and stars within galaxies, and stuff like that.
========quote from other guy's earlier========
https://www.physicsforums.com/threads/prove-that-z-v-c.273160/goto/post?id=1965349#post-1965349
... but can someone remind me how you prove mathematically z = v/c? (z is redshift, v is recessional velocity, c is speed of light.)
I realize that this equation only works in a non-relativistic Universe, but nevertheless I'd like to see it.
==========endquote============
As you point out, it isn't true that z = v/c. But for unrelativistic speeds it is nearly right (if we are clear that it is not cosmo redshift, but some small random motion doppler effect that we are talking about)
...
...
...You asked for a nonrelativistic picture. So we can interpret the formula for sound. It's more intuitive, quicker to understand, thinking of frequency. So let c be the speed of sound. Let v be your speed, towards. The frequency you hear as you go towards will be the emitted frequency increased by a factor of (1 + v/c)
You will be meeting the peaks of the waves that much faster, because you are going towards the source.

Frequency higher by a factor of ((c+v)/c) means wavelength shorter by a factor (c/(c+v))
But for small velocities (a small percentage of c) that number is about the same as 1 - v/c.

You know, 1/(1+x) is about the same as 1 - x, for small x.

So for example 5% higher frequency corresponds approximately to 5% shorter wavelength. The reciprocal of 1.05 is not exactly the same as 0.95, but pretty close.

What I've described is why the nonrelativistic doppler shift is v/c, where you the receiver are moving towards the source. And v/c applies both to the fractional increase in frequency and the fractional decrease in wavelength (approximately.)

The story is the same when you are moving away from the source---I just happened to imagine it going towards.
...
Given that intuitive framework can you attach algebraic symbols to the various key quantities and construct a proof with equations that you are happy with? If not, let us know. I or someone will help translate into equations.

I guess you know that the real formula, for the relativistic doppler, is 1+z = sqrt( (c+v)/(c-v))
To tie up loose ends I guess one should notice that for small v/c that is almost the same number as 1 + v/c
==endquote==

 

[JohnnyGui]

Hey Marcus!

I'm sorry, I'll try and shut up about the doppler shift right now. Thing is, I do understand that it's not cosmological redshift but since I got distracted by the doppler shift, I was tempted to at least understand it correctly. Hence my questions about it. :)