Something about calculating the Age of the Universe

[JohnnyGui]

Thank you so much @wabbit and @marcus for the sources. I'll dive into them for now and see how much I'll be able to comprehend

Btw, one thing I'm wondering about for a while now, is how we are so sure that the expansion of U should be negatively influenced by gravity to the extent that we have concluded that there must be dark energy that counteracts the effects of gravity (preventing a Big Crunch). Isn't there a possibility that the expansion ISN'T influenced by gravity at all? Or are the effects of gravity on expansion merely concluded from the fact that H or expansion rate changes accordingly to the density over time? (i.e. the expansion rate being slower when the density of the U was high early on and faster when the density decreased)

@Chronos : I haven't studied anything like astronomy or cosmology but am really fascinated by them to the extent of reading and following lectures on YouTube. Any time now until I consider this as a hint to begin such a study ;).

 

[wabbit]

Thank you so much @wabbit and @marcus for the sources. I'll dive into them for now and see how much I'll be able to comprehend

Don't hesitate asking questions here or in a new thread if you start diving into one of these - if you don't have questions and general relativity seems natural and obvious, read them again, you must have missed something : )

Btw, one thing I'm wondering about for a while now, is how we are so sure that the expansion of U should be negatively influenced by gravity to the extent that we have concluded that there must be dark energy that counteracts the effects of gravity (preventing a Big Crunch). Isn't there a possibility that the expansion ISN'T influenced by gravity at all? Or are the effects of gravity on expansion merely concluded from the fact that H or expansion rate changes accordingly to the density over time? (i.e. the expansion rate being slower when the density of the U was high early on and faster when the density decreased)

No, and the reason is very simple : gravity is attractive. If two bodies are flying apart from each other, their mutual gravitational attraction will tend to slow them down (and possibly bring them back together in a big crunch). This does not depend on the details of the model, only on the assumption that gravity behaves at large scales broadly similarly to how it does at smaller scales.

 

[JohnnyGui]

Hey wabbit, I'm reading the PDF you've given me (General Relativity without Calculus) and so far, I have survived.
There's a random question that popped up in my head when watching yet another Yale lecture on YouTube about calculating the velocity of an recessing galaxy or star using redshift.

Here's a drawing I made about this:
https://www.dropbox.com/s/z5gn2k5eldik63g/Redshift.jpg?dl=0

λEmit is the wavelength of the lightray right when the star sent it out. λObs (Observed) is the redshifted lightray after the star has recessed to distance D2.

Here's what the lecture covered about calculating the velocity of the star. Please correct me if I'm wrong in these steps:

1. The lightray the star is sending out to the observer comes from distance D1 and while that lightray was traveling to us, the star recessed to distance D2. One would then be able to calculate D1 by using the luminosity formula for this star.
2. Now that D1 is calculated, you can calculate the time duration that took for the star to recess to distance D2. This is done by using D1 / c (speed of light)
3. Now, the lightray the observer sees is the redshifted λObs. Furthermore, I read that D2 / D1 = λObs / λEmit.

Here's where I'm stuck; how can one measure λEmit to be able to know the ratio and calculate D2?? Do they just assume the star is sending out a particular wavelength value? If so, what are these assumptions based on? If they're assuming the star belongs to a particular class of stars with a particular wavelength λEmit, how would one be so sure the concerning star belongs to that specific class without knowing its original λEmit?

 

[wabbit]

Hi again, glad you're surviving that read : )

I'll have to check these calculations you mention, usually I don't look at D2 (we don't observe anything at D2, it is a distance "now" and as we know "now" is a matter of convention in relativity - even though this particular convention is not arbitrary).

But about this emission wavelength, yes, astronomers measure spectra of stars (supernovae) and/or galaxies, and these have shapes and emission and absorption lines which can be identified - at least that's my understanding, maybe someone better versed in these things can chime in.

Those spectra (and their evolution over time, for supernovae) are then used to classify the type of source - notably, in the case of supernova cosmology, in order to retain only those which meet the criteria for being (likely) "standard candles".

Note that there are uncertainties and noise in these measurements, for instance galaxies have proper velocities relative to the Hubble flow and this affects their redshift - so at a given distance we see a range of velocities, and it is the average that is assumed to be representative of overall expansion.

 

[JohnnyGui]

Hi again, glad you're surviving that read : )

I'll have to check these calculations you mention, usually I don't look at D2 (we don't observe anything at D2, it is a distance "now" and as we know "now" is a matter of convention in relativity - even though this particular convention is not arbitrary).

But about this emission wavelength, yes, astronomers measure spectra of stars (supernovae) and/or galaxies, and these have shapes and emission and absorption lines which can be identified - at least that's my understanding, maybe someone better versed in these things can chime in.

Those spectra (and their evolution over time, for supernovae) are then used to classify the type of source - notably, in the case of supernova cosmology, in order to retain only those which meet the criteria for being (likely) "standard candles".

Note that there are uncertainties and noise in these measurements, for instance galaxies have proper velocities relative to the Hubble flow and this affects their redshift - so at a given distance we see a range of velocities, and it is the average that is assumed to be representative of overall expansion.

Still have to finish the read though so I'm still yet to drown :P

I understand that D2 can't be seen, but it can't it be calculated?

About those spectra to classify a type, aren't these spectra also measured as λObs?? At the end, one would never really know the true emitted wavelength the moment those standard candles sent it out; it will always need time to reach us and therefore turn (even slightly) redshifted.

Regarding your last paragraph, I was exactly thinking about how they are able to distinguish the redshift caused by the expansion from the redshift by the proper velocites. It all sounds very roughly calculated.

 

[wabbit]

Yes offcourse D2 can be calculated, and it's used too - it's just that I don't pay attention, so I'm not sure about your formulas : )

As to the spectra, the spectrum at emission (which you do not see) is characterized by the chemical elements present, but the spectrum at reception (the one you measure) is just a shifted version which makes no sense as such - only when you shift it by the right amount does it match a meaningful spectrum, and how much you need to shift tells you the redshift.

To take a simple example say you re observing a nebula made of hydrogen alone - the hydrogen spectrum is well known, and you can tell by looking at all the lines together "this must be the 656 nm line", but it turns out you are seeing it at 800nm, so you know the shift is 144nm. I'm explaining this badly, it's hard without pictures - look at some spectra, check the wikipedia article about spectroscopy you'll see what I mean.

Disclaimer: I've never measured a redshift myself and I am just a layman with no qualification in the field, so take what I say with a grain of salt.

Edit: none of this is roughly calculated, these are precision measurements mades with great care, which are of course subject to various errors but those are modeled too and the measurements are given together with their uncertainty. Don't assume from a simplified broad description I give that those numbers are plucked out of thin air, they are not. Read some reference (*) articles about supernova cosmology, you'll see how detailed the analysis are, there are many more aspects covered than we discussed, studied in excruciating detail.

(*) I added "reference" because not all published articles are of the highest standard. But do visit http://supernova.lbl.gov if you haven't done so yet, it's a great resource.

 

[JohnnyGui]

Ah, reading your post and a wiki page on redshift (don't hit me for that) made me understand that they're looking at the pattern (i.e. the intervals between the absorbed spectral lines) of the observed wavelength. The only difference here is that it's redshifted but it's still the same pattern so they're able to compare it with known patterns from chemical elements to know what element the star is made out of. Is this correct?
There's apparently another way by looking if there's a same spectral line in both spectra but at a different wavelength and then calculate the redshift, only I don't know what they mean by "same spectral line" if it's not at the same wavelength. Doesn't a similar spectral line mean that it's at the same wavelength?

Regarding the formula I mentioned in post #103, I have compared its outcome with another redshift formula: v = c x (λΔ / λemit) and it gave me the same answer.

I will read the link you gave now.

 

[wabbit]

Yes that's what I was clumsily trying to explain : ) glad you found a better explanation in the wiki, I was struggling with how to formulate this.

A single line in isolation can't be used for redshift measurement, only a spectrum (pattern) can.

 

[JohnnyGui]

If it weren't for your many explanations in this thread, I would be still looking cluelessly at post #1 ;)

One thing I was wondering regarding the formulas:

- I read that redshift could be calculated by (λ_obs / λ_emit) - 1 = z.
- Since λ_obs / λ_emit = D2 / D1 (D1 being the distance of a star right when its light was emitted and D2 being the new distance after the redshift that we observe) I can say that (D2 / D1) - 1 = z
- For small distances, I read that z × c = v. I also know that v = H × D
- So I can say that ((D2 / D1) - 1) × c = H × D

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

 

[wabbit]

Where does this z×c=v come from ? Not saying it's wrong but I can't make sense of it.

 

[JohnnyGui]

Believe me, I've been trying for half an hour to make sense out of that equation as well without result. There are several sources mentioning the formula, for example: http://astronomy.swin.edu.au/cosmos/c/cosmological+redshift as well as here: http://cas.sdss.org/dr5/en/proj/advanced/hubble/conclusion.asp

Apparently, the increase in wavelength by redshift compared to the emitted wavelength is the same fraction as the speed of the object compared to the speed of light but I don't get why.

EDIT: Ok, I found a thread from here where marcus discusses the formula, I'm reading it now: https://www.physicsforums.com/threads/prove-that-z-v-c.273160/

 

[wabbit]

Yes youre right it's the low velocity Doppler shift formula and it applies to recession velocities just fine as long as $v\ll c$

 

[JohnnyGui]

Knowing that, are you able to answer the question about D being D1 or D2 in post #109 and how one would be able to tell?

 

[wabbit]

Hmmm yes I was trying to avoid doing that

I was going to say D is neither D1 nor D2, it's the luminosity distance, but I'm not so sure now, I need to check, perhaps someone who knows this stuff better can help here.

 

[JohnnyGui]

Haha, nice tactic there ;)

Anyway, if D is based on the luminosity distance, then I would think that luminosity is based on the observed wavelength that was originally emitted when a star was in its first position and I would conclude that D is D1.

Perhaps @marcus could once again save the day here?

 

[Chronos]

It might be simpler to view z and D in terms of the scale factor of the universe - see http://www.physics.fsu.edu/users/ProsperH/AST3033/cosmology/ScaleFactor.htm for a brief discussion.

 

[JohnnyGui]

It might be simpler to view z and D in terms of the scale factor of the universe - see http://www.physics.fsu.edu/users/ProsperH/AST3033/cosmology/ScaleFactor.htm for a brief discussion.

Thanks, looks like a very good read to me. I'll dive into it. Isn't there a pm system on here so I can give @marcus a holler about my previous question?

 

[marcus]

Anyway, if D is based on the luminosity distance, then I would think that luminosity is based on the observed wavelength that was originally emitted when a star was in its first position and I would conclude that D is D1...

... Isn't there a pm system on here so I can give @marcus a holler about my previous question?

Hi JG, I haven't been following this thread and I'm not sure what D, and D1 and D2 are. I think luminosity is based on the RECEIVED wavelengths. I'm often mistaken at first and have to look things up. I'll look up "luminosity distance" and correct this if I'm mistaken.

You seem pretty well-informed and to be thinking clearly about this stuff so you probably realize already what I'm saying: distance reduces the received watts per cm^2 in two ways.
A. the energy is more spread out. It falls off as R^2 where R is the distance to the source by the time the light gets to us. Expansion contributes to the present distance to the source. So expansion contributes to R and reduces luminosity that way.
B. expansion also lengthens the wavelengths and so it drains energy from the light that way. You know the energy of a photon E =ħω so reducing the frequency (by wavestretch) reduces the energy of each photon AND expansion reduces the rate photons are arriving. So there is a kind of (z+1)² factor killing the energy FLUX.

So expansion reduces the watts per cm^2 on the detector surface TWO ways. By increasing the distance so things are more spread out, and by reducing the flux the other ways I mentioned.

I may be just confusing things by commenting like this without understanding what you are talking about : ^) I'll go look up "luminosity distance" and review that. Do you want to kind of summarize what your question is, so people just joining the thread can understand what it is about?

 

[marcus]

Damn! Astronomers have some pretty dumb conventions. Terminology is a proliferating accumulation of historical accidents.

Wikipedia on "luminosity distance" (D_L)indicates that the effect of redshift is NOT ALLOWED FOR in the definition of D_L
==quote==
Another way to express the luminosity distance is through the flux-luminosity relationship. Since,

where F is flux (W·cm−2), and L is luminosity (W),... From this the luminosity distance can be expressed as:

The luminosity distance is related to the "comoving transverse distance"

by the Etherington's reciprocity relation[citation needed]:

where z is the redshift.

is a factor that allows you to calculate the comoving distance between two objects with the same redshift but at different positions of the sky; if the two objects are separated by an angle [PLAIN]https://upload.wikimedia.org/math/6/7/7/677514f5ab1c654e3006964219831e7f.png, the comoving distance between them would be [PLAIN]https://upload.wikimedia.org/math/0/c/d/0cdb2ceca920b31a61e0ca85e430672e.png. In a spatially flat universe, the comoving transverse distance [PLAIN]https://upload.wikimedia.org/math/6/b/4/6b4edd602a37988d06321ebcc59d310e.png[B] is exactly equal to the radial comoving distance[/B] [PLAIN]https://upload.wikimedia.org/math/a/1/6/a16cff4686b360b3ddc1ab645852a051.png, i.e. the comoving distance from ourselves to the object.[1]
==endquote==
So you take the actual Watts output of the star, that's L. And you picture this in static Euclidean space and you naively say what radius R would be needed to spread this power out to get the observed Watts per cm^2. that is the first formula. It does not allow for redshift!

Now this obviously over estimates the distance. The absolute luminosity should not be L it should be L/(z+1)². so when you get it under the square root sign you get a correction factor of 1/(z+1)

So finally in the third equation they say "Oh drat we over estimated by a factor of (z+1)! So let's define a new distance D_M."

And that turns out to be the good, spatial flat, approximation of the right thing---the comoving or NOW distance that is actually the distance.

Astronomers can't go back and correct bad definitions that are time-honored and enshrined in the scholar literature because that would make the professional literature of the past unreadable. Words in the journals would mean different things before and after the reform/correction of the bad definitions. So we are stuck with something called "Luminosity distance" which is bigger than it should be by a factor of (z+1).

That is my dos centavos Señores. It may or may not be relevant and it may of course be incorrect. : ^)

 

[JohnnyGui]

Hi JG, I haven't been following this thread and I'm not sure what D, and D1 and D2 are. I think luminosity is based on the RECEIVED wavelengths. I'm often mistaken at first and have to look things up. I'll look up "luminosity distance" and correct this if I'm mistaken.

You seem pretty well-informed and to be thinking clearly about this stuff so you probably realize already what I'm saying: distance reduces the received watts per cm^2 in two ways.
A. the energy is more spread out. It falls off as R^2 where R is the distance to the source by the time the light gets to us. Expansion contributes to the present distance to the source. So expansion contributes to R and reduces luminosity that way.
B. expansion also lengthens the wavelengths and so it drains energy from the light that way. You know the energy of a photon E =ħω so reducing the frequency (by wavestretch) reduces the energy of each photon AND expansion reduces the rate photons are arriving. So there is a kind of (z+1)² factor killing the energy FLUX.

So expansion reduces the watts per cm^2 on the detector surface TWO ways. By increasing the distance so things are more spread out, and by reducing the flux the other ways I mentioned.

I may be just confusing things by commenting like this without understanding what you are talking about : ^) I'll go look up "luminosity distance" and review that. Do you want to kind of summarize what your question is, so people just joining the thread can understand what it is about?

Hey Marcus, I am indeed aware of this info but thanks a lot for the effort nonetheless :). Sure, I'll summarize my question for you:

I was talking with @wabbit about the following drawing that I made: https://www.dropbox.com/s/z5gn2k5eldik63g/Redshift.jpg?dl=0
λEmit is the wavelength of the lightray right when the star sent it out when it was at distance D1. λObs (Observed) is the redshifted lightray that we observe after the star has recessed to distance D2.

I concluded the following:
1. I saw a lecture from Yale and it covered the following formula: D2 / D1 = λObs / λEmit. I also read that redshift could be calculated by (λobs / λemit) - 1 = z.
2. Since λObs / λEmit = D2 / D1 (D1 being the distance of a star right when its light was emitted and D2 being the new distance after the observed redshift) I can say that (D2 / D1) - 1 = z as well.
3. Furthermore, for small distances, I read that z × c = v. I also know that v = H × D
4. So I can say that ((D2 / D1) - 1) × c = H × D

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

EDIT: Just noticed your second reply, but I'm not sure (yet) if this has answered my question or not. Sorry for that.

 

[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. :)