Killing vectors and momentum conservation

[jcap]

Consider the flat FRW metric in Cartesian spatial co-moving co-ordinates:

##ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)##

As I understand it, since the metric does not depend on the the spatial co-ordinates, there exist Killing vectors in the ##x##,##y##,##z## directions.

Does this imply that the 3-momentum of particles traveling on geodesics is conserved (when one refers to the co-moving co-ordinates ##x##,##y##,##z##) ?

I have read elsewhere that the 3-momentum of free particles goes like ##1/a(t)## instead.

Is the difference due to proper co-ordinates being used instead of the co-moving co-ordinates ##x##,##y##,##z##?

 

[bapowell]

Is the difference due to proper co-ordinates being used instead of the co-moving co-ordinates ##x##,##y##,##z##?

Yes, the proper velocity redshifts as [itex]1/a[/itex].

 

[jcap]

Yes, the proper velocity redshifts as [itex]1/a[/itex].

If one has a gas of particles in a rigid box would that gas cool down as the Universe expands?

 

[bapowell]

If one has a gas of particles in a rigid box would that gas cool down as the Universe expands?

Yes. The energy density declines by [itex]1/a^3[/itex] for nonrelativistic particles.

 

[jcap]

Yes. The energy density declines by [itex]1/a^3[/itex] for nonrelativistic particles.

Are you assuming that the box expands with the Universe so that its volume is proportional to ##a^3##?

I'm assuming that the box is rigid - does the gas cool in that case?

 

[Chalnoth]

Yes, it cools. But to a very good approximation matter has a temperature of zero (because its temperature is usually much, much less than the rest mass of electrons, let alone protons).

Consider the flat FRW metric in Cartesian spatial co-moving co-ordinates:

##ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)##

As I understand it, since the metric does not depend on the the spatial co-ordinates, there exist Killing vectors in the ##x##,##y##,##z## directions.

Does this imply that the 3-momentum of particles traveling on geodesics is conserved (when one refers to the co-moving co-ordinates ##x##,##y##,##z##) ?

I have read elsewhere that the 3-momentum of free particles goes like ##1/a(t)## instead.

Is the difference due to proper co-ordinates being used instead of the co-moving co-ordinates ##x##,##y##,##z##?

I believe the conserved quantity in this case is ##a^2 \dot{x}##.

 

[timmdeeg]

Jcap is talking about a rigid box. So, the density of the gas within it should be constant over time. If correct, I don't understand, that the gas cools.

How about an initial state, where the gas particles are in rest to each other, means T = 0 K. In order to follow the Hubble flow, the distances should increase then and by bouncing against the rigid walls (which are not co-moving), the particles should gain kinetic energy. Is this reasoning correct?

 

[Chalnoth]

Jcap is talking about a rigid box. So, the density of the gas within it should be constant over time. If correct, I don't understand, that the gas cools.

How about an initial state, where the gas particles are in rest to each other, means T = 0 K. In order to follow the Hubble flow, the distances should increase then and by bouncing against the rigid walls (which are not co-moving), the particles should gain kinetic energy. Is this reasoning correct?

Yes, I was responding to the OP and didn't catch that part. Certainly the only change in temperature of a gas inside a rigid box will stem from an exchange of energy with its surroundings. So the temperature change would be related to the temperature of its surroundings and how efficiently it exchanges energy with them.

 

[timmdeeg]

Yes, I was responding to the OP and didn't catch that part. Certainly the only change in temperature of a gas inside a rigid box will stem from an exchange of energy with its surroundings. So the temperature change would be related to the temperature of its surroundings and how efficiently it exchanges energy with them.

Thanks. Would the temperature of the gas inside an adiabatically isolated box increase (without isochoric work being done on it) in expanding spacetime?

 

[Chalnoth]

If the box is isolated from exchanging heat with its surroundings, its temperature would, by definition, remain unchanged.

Basically, it's a rigid box. The expanding universe component is irrelevant when you have a rigid box. There is no way that an expanding universe can change any conclusions at all about the behavior of matter inside such a rigid box (except in terms of how the expansion impacts the temperature of the surroundings.

 

[timmdeeg]

The expanding universe component is irrelevant when you have a rigid box. There is no way that an expanding universe can change any conclusions at all about the behavior of matter inside such a rigid box (except in terms of how the expansion impacts the temperature of the surroundings.

Sorry, I still have a problem. Initially, the center of the box shall be co-moving and the matter particles inside too. So, locally close to the walls, there should be relative velocities between those and the particles, resulting in peculiar velocities of the particles after bouncing against the walls. After some time all particles except those in the center of the box should possesses peculiar velocities.
But it seems I am wrong. Could you kindly comment?

 

[timmdeeg]

But the kinetic energy in the frame of the box doesn't change during bouncing. I think I had a silly misconception.

 

[Chalnoth]

Sorry, I still have a problem. Initially, the center of the box shall be co-moving and the matter particles inside too. So, locally close to the walls, there should be relative velocities between those and the particles, resulting in peculiar velocities of the particles after bouncing against the walls. After some time all particles except those in the center of the box should possesses peculiar velocities.
But it seems I am wrong. Could you kindly comment?

Regardless of the initial conditions, the gas inside the box will rapidly reach equilibrium.

 

[timmdeeg]

Regardless of the initial conditions, the gas inside the box will rapidly reach equilibrium.

Understand, thanks.