I The Universe as an isolated system

[Trust10]

Can the universe be treated as a thermodynamic isolated system? And why?

 

[phinds]

What research have you done on this? What have you found?

 

[Trust10]

I have gone through some works relating to this topic and have found out that some people are still skeptical about the universe being thermodynamically isolated or closed. The universe being a closed system was based on the fact that, it could allow energy exchange between other universes in the multiverse system. But if this is true, why does the Friedman equation not give room for some extra energy parameter? And since it doesn't why can we treat the universe as an isolated system? On the other part some say since the universe is all there is, then it is Isolated.
I wrote something about it and sent to one of my lecturer, but he said i have no argument, and that i can treat it as an isolated system if i want to.

 

[phinds]

I am not aware of any valid argument that our universe is in causal contact with ANYTHING "outside" of itself. Personally, I'm doubtful that there IS anything "outside" of our universe but that's a different topic.

 

[Trust10]

I agree with you on that, i also personally don't think there is anything out there besides our universe. So do you think it can be treated as an isolated system?

 

[phinds]

I agree with you on that, i also personally don't think there is anything out there besides our universe. So do you think it can be treated as an isolated system?

Yes, but given that it is possibly infinite in extent, I'm not clear how that can even matter in any practical sense.

 

[PeterDonis]

I have gone through some works relating to this topic and have found out that some people are still skeptical about the universe being thermodynamically isolated or closed.

Can you give some specific references?

 

[PeterDonis]

The universe being a closed system was based on the fact that, it could allow energy exchange between other universes in the multiverse system.

Um, if the universe is exchanging energy with other universes, then it is not a closed system.

 

[Arman777]

Can the universe be treated as a thermodynamically isolated system? And why?

In the derivation of the Fluid equation, from thermodynamics,

$$dQ=dE+PdV$$ we take $dQ=0$. Which means that there's no heat flow from the universe (in or out). So this suggests that the universe is a closed system. If we take $dQ\neq 0$. Then it means that the universe can exchange heat or lose heat etc.

Now I think you are trying to ask that why we say $dQ=0$. I think the main reason is observation. If the universe was "gaining" heat then we could have seen this effect on the CMBR. Since CMBR is highly homogeneous we can treat the universe as a closed system.

 

[Trust10]

Thanks for the contributions. Although I further discussed this with some friends at school and we settled for this; Since there isn't any heat flow in or out from the universe, matter is not in anyway transferred out or into the universe, and work is not in any way done on the universe, why call it closed and not isolated system . For a closed system the condition to be satisfied is that only heat and not matter can be exchange between a system and a surrounding. But for an isolated system neither matter nor heat is exchanged with the surrounding and this satisfies the homogeneity of the CMB (as cited by Arman) and the homogeneity universe as a whole. Moreover dQ=0 is also true for an isolated system which is Adiabatic.

 

[PeterDonis]

why call it closed and not isolated system

Hmm...so your whole question here is about the choice of words? "Closed" vs. "isolated"?

For a closed system the condition to be satisfied is that only heat and not matter can be exchange between a system and a surrounding. But for an isolated system neither matter nor heat is exchanged with the surrounding

Perhaps this is how standard thermodynamics textbooks define those terms, but I don't know if cosmologists use those definitions. In any case, the physics doesn't depend on what words you use.

 

[PeterDonis]

If the universe was "gaining" heat then we could have seen this effect on the CMBR. Since CMBR is highly homogeneous

Why do you think that the CMBR being homogeneous rules out the universe gaining heat? Couldn't the heat gain be homogeneous?

 

[Trust10]

I guess you are right, "The physics does not depend on what words I use". Thanks.

 

[Arman777]

Why do you think that the CMBR being homogeneous rules out the universe gaining heat? Couldn't the heat gain be homogeneous?

I think by studying the recombination era and photon decoupling, we can deduce a theoretical temperature for the CMBR.

Or, we know the current CMBR temperature, 2.75K. We also know that scale factor is inversely proportional to temperature so we can deduce the temperature of the CMBR. And see if these two result match. If they match, it means that $dQ=0$

Also any heat exchange would change the photon-baryon ratio.

 

[Arman777]

Other crucial point that I want to make is that, any effect on the CMBR, that is done by "something" already would be a part of the universe. A universe which is apart from ours cannot effect the CMBR in the past.

 

[PeterDonis]

I think by studying the recombination era and photon decoupling, we can deduce a theoretical temperature for the CMBR.

Meaning the temperature of the CMBR at the time of decoupling? Yes, that can be deduced from the known properties of atoms.

we know the current CMBR temperature, 2.75K. We also know that scale factor is inversely proportional to temperature so we can deduce the temperature of the CMBR.

How? We don't directly measure the scale factor. We deduce it from other observations. But those other observations include assumptions, one of which is, basically, that the universe is not gaining heat from somewhere we can't observe.

Also, the argument you are making here is not that the universe gaining heat from somewhere we can't observe would make it not homogeneous. It is that the universe gaining heat from somewhere we can't observe would change the temperature of the CMBR--which could happen in a homogeneous fashion. So you appear to have shifted your ground from the post of yours that I originally responded to.

any heat exchange would change the photon-baryon ratio.

If the heat exchange were in the form of photons, yes.

 

[Arman777]

How? We don't directly measure the scale factor. We deduce it from other observations. But those other observations include assumptions, one of which is, basically, that the universe is not gaining heat from somewhere we can't observe.

Yes, you are right. The scale factor comes from the fluid equation which in the derivation we use $dQ=0$

We can derive the $a∝T^{-1}$, by using the thermodynamics. So if see that, theoretical CMBR temperature derived from the at the time of photon decoupling (using particle physics etc.) and the temperature derived from the $a∝T^{-1}$ (that is derived by setting $dQ=0$) matches then it means that there cannot be any heat exchange.

But I did not understand why you said scale factor is observable. We can deduce it from the Friedmann Equations. Oh you mean the density parameters ?

Also, the argument you are making here is not that the universe gaining heat from somewhere we can't observe would make it not homogeneous. It is that the universe gaining heat from somewhere we can't observe would change the temperature of the CMBR--which could happen in a homogeneous fashion. So you appear to have shifted your ground from the post of yours that I originally responded to.

But then I remembered that, It does not matter if the heat distrubition is homogeneous or not, If something affects the CMBR its already part of the our universe.So we cannot say there can be heat flow.

I discussed this question before on this site.

https://www.physicsforums.com/threads/observable-universe-tempature-and-cmb.935554/#post-5912337

You can look post #4 and #7

 

[PeterDonis]

The scale factor comes from the fluid equation

What do you mean by this?

We can derive the $a∝T^{-1}$, by using the thermodynamics

Can you be more specific?

if see that, theoretical CMBR temperature derived from the at the time of photon decoupling (using particle physics etc.) and the temperature derived from the $a∝T^{-1}$ (that is derived by setting $dQ=0$) matches

Matches what? How can you tell, observationally, that it matches?

I did not understand why you said scale factor is observable

I said the scale factor is not observable.

If something affects the CMBR its already part of the our universe.

This is basically defining the "universe" to be an isolated system. But you can't change physics by defining words.

I discussed this question before on this site.

The points @Bandersnatch makes in that discussion are valid, but they're not the same as the claims you're trying to make here. Also, the points @Bandersnatch makes in that discussion do not imply that the universe must be an isolated system. He is just making the obvious point that heat transfer, like any other process, can only take place at or below the speed of light.

 

[kimbyd]

Can the universe be treated as a thermodynamic isolated system? And why?

With the assumptions of homogeneity and isotropy, yes, definitely. Every co-moving volume will have as much heat leaving the volume as entering it, so it can be treated as isolated. Though this is often done by considering it to have periodic boundary conditions instead.

However, the thermodynamics of the universe depends critically upon the thermodynamics of gravity, and we don't have a good way of understanding how gravity behaves with respect to thermodynamics. We know a few edge cases (black holes, de Sitter space), but that's it.

 

[PeterDonis]

With the assumptions of homogeneity and isotropy, yes, definitely. Every co-moving volume will have as much heat leaving the volume as entering it

And also, every co-moving volume will have no net matter crossing its boundary.

 

[Arman777]

What do you mean by this?

To Solve the Friedmann Equations (or to find the scale factor), we need a fluid equation, acceleration equation and the normal Friedmann equation that is derived from GR.

I mean that when we are deriving the fluid equation we are setting $dQ=0$ hence in the process of finding the scale factor ( by solving these 3 equations) we are already assuming that universe is a closed system.

Can you be more specific?

Okay well,

$$dQ=dE+PdV$$
we set $dQ=0$ so we have,
$$dE/dt=-P(t)dV/dt$$
By setting $E=εV$ where $ε$ is the energy density of the photons. And we also know that $ε=σT^4$ and the pressure of the photons $P=σT^4/3$ (since ($P=wε$ ,and for photons $w=1/3$)

If we solve this we would get,

$$\frac {1} {T}\frac {dT} {dt}=-\frac {1} {3V}\frac {dV} {dt}$$

for $V∝a(t)^{-3}$ we see that $$T(t)∝a(t)^{-1}$$

Matches what? How can you tell, observationally, that it matches?

So you agreed with me that we can calculate the CMBR temperature by using the particle physics etc. So here we get a Temperature value.

Also, we can look at the CMBR and we can see that it has a temperature of 2.75K. Now by using the Friedmann Equations, we can derive the scale factor and then using the $$T(t)∝a(t)^{-1}$$ we can have another temperature value (by using their ratios). If these two values are nearly the same then we can say that $dQ=0$. Since while we were deriving the scale factor, we used the fluid equation for $dQ=0$.

My point is in this second way we are never using any particle physics related to the CMBR temperature. So these are two different ways of obtaining the temperature of the CMBR.

If they were give us different results, then I think we can say that $dQ\neq 0$

It's like this; using the particle physics gives us the theoretical value of the CMBR Temp. and the scale factor gives us the experimental value of the CMBR Temp.

I said the scale factor is not observable.

Oh sorry, I misread it I think.

 

[PeterDonis]

So you agreed with me that we can calculate the CMBR temperature by using the particle physics etc.

I agreed that we can calculate the temperature of the CMBR when it is emitted (i.e., at the time of recombination) using particle physics. You keep referring to this as "the CMBR temperature", but that's really not precise enough; the CMBR always has a temperature, but it's not always the one it had when it was emitted.

we can look at the CMBR and we can see that it has a temperature of 2.75K

It has a temperature of 2.75 K now, yes. Again, you really need to be more precise in your language.

by using the Friedmann Equations, we can derive the scale factor

No, we can't. We can, as you did, derive a relationship between the temperature of the CMBR and the scale factor, assuming that dQ = 0. But that relationship is a ratio; it doesn't tell you the actual scale factor now or at any time. It just tells you that, if dQ = 0, then the ratio of the scale factor now to the scale factor when the CMBR was emitted is the inverse of the ratio of the temperature of the CMBR now to the temperature at emission that we calculated from particle physics. But, again, all of this assumes that dQ = 0.

these are two different ways of obtaining the temperature of the CMBR.

No, they aren't. You have things backwards. Here's how the actual logic goes:

(1) We measure the temperature of the CMBR now.

(2) We calculate what the temperature of the CMBR was when it was emitted based on the physics of atoms.

(3) We calculate the relationship between the CMBR temperature and the scale factor of the universe, assuming dQ = 0.

(4) We use the relationship in (3) plus the temperatures from (1) and (2) to calculate the ratio between the scale factor of the universe now and the scale factor when the CMBR was emitted. This calculation depends on the assumption dQ = 0, since (3) does.

And that is all we can do. We have no way of actually measuring the scale factor now or the scale factor when the CMBR was emitted, so we have no way of using the above to check the assumption that dQ = 0. Cosmologists routinely quote the above numbers because they believe that the assumption dQ = 0 is extremely unlikely to be wrong, not because they have actually measured the scale factor and checked it.

 

[Arman777]

@PeterDonis I see your point. So there's no another way of deriving the CMBR temperature at the last scattering beside the particle physics ?

 

[PeterDonis]

there's no another way of deriving the CMBR temperature at the last scattering beside the particle physics ?

No.

 

[Arman777]

No.

Hmm okay then, thanks

 

[kimbyd]

We have no way of actually measuring the scale factor now or the scale factor when the CMBR was emitted,

To be a little bit more precise, the scale factor itself is not a physical quantity at all. The scale factor at anyone time is whatever we choose it to be. Usually it's convenient to define the scale factor today as being equal to 1.

Ratios in scale factors at different times are measurable, however. The scale factor today is measured at roughly 1090 times larger than it was at CMBR emission.

To Arman777:
Yes, the temperature at last scattering is determined by the temperature at which the hydrogen-helium plasma in the early universe cools to a transparent gas. This temperature is estimated by models of this gas which include inputs from both lab experiments and theoretical calculations using quantum electrodynamics.

 

[PeterDonis]

The scale factor today is measured at roughly 1090 times larger than it was at CMBR emission.

Based on the temperature ratio, yes. Strictly speaking, though, this temperature ratio assumes that no other heat source has produced homogeneous black-body radiation since the CMBR was emitted. I agree that is extremely likely to be the case, but it is still, strictly speaking, an assumption that has to be made in order to deduce the scale factor ratio.

 

[kimbyd]

Based on the temperature ratio, yes. Strictly speaking, though, this temperature ratio assumes that no other heat source has produced homogeneous black-body radiation since the CMBR was emitted. I agree that is extremely likely to be the case, but it is still, strictly speaking, an assumption that has to be made in order to deduce the scale factor ratio.

Assumptions have to be made with any measurement. It's the nature of the beast. The CMBR is on pretty solid footing with regard to that, especially as the detailed nature of the statistics of the hot and cold spots provides such rich, testable information about the nature of our universe (such as the ratio of dark matter to normal matter).