Is momentum conserved in the entire observable universe?

[Arman777]

I know that in observable universe energy is not conserved.I don't know exactly why (it s possibly about GR and expanding universe but I don't know the equations)
In the observable universe...If we take a whole system like entire Observable universe, In this system is momentum conserved ?

Sorry,If its a stupid question

 

[BvU]

Yes: there is no external force

 

[Arman777]

Yes: there is no external force

Haha Thanks.

 

[Chalnoth]

Yes: there is no external force

I don't think this is accurate. Momentum suffers from many of the same problems as energy when it comes to conservation on a global scale.

As with energy, momentum conservation works quite well locally. But it starts to break down at very large scales where the expansion of the universe is important.

One way of thinking of this is that in General Relativity, the conserved quantity is not momentum or energy, but rather the stress-energy tensor. The stress-energy tensor includes components related to energy density, momentum, pressure, and twisting forces. All of these components are conserved together, which means that one of them can change, as long as it's counter-balanced by a related change elsewhere.

 

[BvU]

I don't think this is accurate. Momentum suffers from many of the same problems as energy when it comes to conservation on a global scale.

As with energy, momentum conservation works quite well locally. But it starts to break down at very large scales where the expansion of the universe is important.

One way of thinking of this is that in General Relativity, the conserved quantity is not momentum or energy, but rather the stress-energy tensor. The stress-energy tensor includes components related to energy density, momentum, pressure, and twisting forces. All of these components are conserved together, which means that one of them can change, as long as it's counter-balanced by a related change elsewhere.

I responded to Arman after we discussed $\Delta P_{\rm \;c.o.m.} =0$ (for a system with no external forces acting) in another thread. The jump to the whole universe is rather courageous and I don't think we should bring in GR & co in this stage.

 

[Arman777]

I responded to Arman after we discussed $\Delta P_{\rm \;c.o.m.} =0$ (for a system with no external forces acting) in another thread. The jump to the whole universe is rather courageous and I don't think we should bring in GR & co in this stage.

I understand the idea.yeah

 

[PeterDonis]

I don't think we should bring in GR & co in this stage.

I don't think you can avoid it if you're talking about the whole universe.

 

[Orodruin]

I responded to Arman after we discussed $\Delta P_{\rm \;c.o.m.} =0$ (for a system with no external forces acting) in another thread. The jump to the whole universe is rather courageous and I don't think we should bring in GR & co in this stage.

If you want to talk about the entire Universe it is unavoidable to bring in GR. Any conservation law is going to suffer from the same type of problem in defining a global conservation. Local conservation laws in terms of conserved currents work fine though and are essentially continuity equations.

 

[ShayanJ]

One way of thinking of this is that in General Relativity, the conserved quantity is not momentum or energy, but rather the stress-energy tensor.

But in GR, we have $\nabla_\mu T^{\mu \nu}=0$. This does give us a conservation law for an inertial observer in a small region of spacetime, but in general its not a conservation law!

 

[Orodruin]

But in GR, we have $\nabla_\mu T^{\mu \nu}=0$. This does give us a conservation law for an inertial observer in a small region of spacetime, but in general its not a conservation law!

It is a local conservation law. Many times "conservation law" is taken to be synonymous with "global conservation law". This does not mean local conservation laws do not exist.

 

[ShayanJ]

It is a local conservation law. Many times "conservation law" is taken to be synonymous with "global conservation law". This does not mean local conservation laws do not exist.

Yes, I know that. But I meant that equation can't give us a global conservation law, so I don't think the following sentence is true about the whole observable universe because at that scale, there is just no conserved quantity.(Except quantities like Komar mass or ADM mass but I don't think they're relevant here.)

All of these components are conserved together, which means that one of them can change, as long as it's counter-balanced by a related change elsewhere.

 

[Arman777]

Yes, I know that. But I meant that equation can't give us a global conservation law, so I don't think the following sentence is true about the whole observable universe because at that scale, there is just no conserved quantity.(Except quantities like Komar mass or ADM mass but I don't think they're relevant here.)

Do we need very hot and density stuff in the beginning of the universe like $10^{-20}s-10^{-42}s$ Since energy and momentum and such things do not conserved in general scales ?

In general I wonder from this non-conservative concepts (energy,momentum...), what can we conclude about universe ?

 

[Mordred]

Yes, I know that. But I meant that equation can't give us a global conservation law, so I don't think the following sentence is true about the whole observable universe because at that scale, there is just no conserved quantity.(Except quantities like Komar mass or ADM mass but I don't think they're relevant here.)

How does one set a global conservation law when you have the added complexity of observer influence when measuring energy and momentum? This becomes a rather complex question. Different observers will measure different momentum values for the same object but in the objects own frame of reference its momentum is unchanged. So why would you apply a conservation law to describe the observer influence portion.? when locally ie object locality it is following the conservation law

 

[nikkkom]

How does one set a global conservation law when you have the added complexity of observer influence when measuring energy and momentum?

In SR, they are also observer-dependent, but nevertheless they are globally conserved.

 

[Mordred]

SR isn't particularly useful in this instance particularly under a commoving volume over time

 

[nikkkom]

SR isn't particularly useful in this instance particularly under a commoving volume over time

I only said that SR is a counterexample to your claim that dependence of momentum on chosen coordinate system poses serious problem for defining global conservation laws.

 

[Mordred]

Not if your following the arguments as per Sean Carroll on preposterous universe.

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

ie in GR energy is not conserved.

There is a particular important line that applies to my original post.
" If that spacetime is standing completely still, the total energy is constant; if it’s evolving, the energy changes in a completely unambiguous way."

 

[nikkkom]

Not if your following the arguments as per Sean Carroll on preposterous universe.

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

ie in GR energy is not conserved.

I've seen this gazillion times and I agree with it.

You evidently do not understand what I'm saying. I'm saying that this quote of you:

"Different observers will measure different momentum values for the same object but in the objects own frame of reference its momentum is unchanged. So why would you apply a conservation law to describe the observer influence portion.?"

is applicable to SR too. In SR, "different observers will measure different momentum values for the same object." And yet, in SR, global momentum conservation is easy to define. Ergo, "different observers will measure different momentum values" is not a problem (otherwise, SR would have the problem too, which it does not).

Problem is somewhere else. (For example, when it's impossible to even define global momentum, then it's kinda hard to define that it is conserved).

 

[Mordred]

My question deals with more.

Why would you need to do so? We all know the object doesn't gain momentum due to observer. So all your doing by applying observer corrections under GR and SR is finding the invarient quantities ie momentum all observers can agree to after all observer influence is removed

Isn't that what the conservation law applies to? the proper momentum not the observed momentum?

 

[nikkkom]

momemtum all observers can agree to after all observer influence is removed

Good luck :D

 

[Mordred]

I agree good luck with that but I think its easiest to simply define the conservation law in terms of action. Its easier to separate the observer influence due to GR.

The observer effects don't cause action so it doesn't make sense to apply the conservation law to it. Another way of describing the above from

"Introduction to Quantum fields from Classical backgrounds" by Muchanov quote below from my hardcopy.

"For example In a flat spacetime, the laws of energy and momentum conservation follow from the invariance of the action under spacetime translations. In the presence of gravitation the spacetime is curved, so in general the spacetime translations are not a physical symmetry any more"

 

[Chronos]

The relativity of simultaneity makes it difficult to unambiguously count things such as mass, energy, stress energy content, etc., over any sufficiently large volume of space [i.e., beyond the local inertial reference frame where an accounting is attempted]. The very notion of global conservation becomes slippery when your first hurdle is to achieve a consensus on how to meaningfully quantify whatever it is you intend to count at any arbitrary moment of time.

 

[nikkkom]

The relativity of simultaneity makes it difficult to unambiguously count things such as mass, energy, stress energy content, etc., over any sufficiently large volume of space [i.e., beyond the local inertial reference frame where an accounting is attempted]. The very notion of global conservation becomes slippery

No, it is not difficult and it does not become slippery in SR. You can choose any inertial frame, and intergal of 4-momentum over the entire space-like slice of the spacetime is conserved.