# Killing vectors and momentum conservation

• jcap
In summary, the conversation discusses the implications of the flat FRW metric in Cartesian spatial co-moving co-ordinates, specifically the existence of Killing vectors in the x, y, and z directions and the conservation of 3-momentum for particles traveling on geodesics. It also touches on the difference between proper co-ordinates and co-moving co-ordinates, and the cooling of a gas of particles in a rigid box as the Universe expands. It is clarified that the temperature of the gas inside the box would remain unchanged unless it exchanges energy with its surroundings. The conversation also addresses the misconception of relative velocities between particles and the walls of the box. In conclusion, the gas inside the box will eventually reach equilibrium regardless of the initial conditions.
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##?

jcap said:
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 $1/a$.

bapowell said:
Yes, the proper velocity redshifts as $1/a$.

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

jcap said:
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 $1/a^3$ for nonrelativistic particles.

bapowell said:
Yes. The energy density declines by $1/a^3$ 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?

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).
jcap said:
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}##.

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?

timmdeeg said:
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.

Chalnoth said:
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?

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.

Chalnoth said:
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?

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

timmdeeg said:
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
Chalnoth said:
Regardless of the initial conditions, the gas inside the box will rapidly reach equilibrium.
Understand, thanks.

## 1. What are Killing vectors?

Killing vectors are a type of vector field in mathematics that describe the symmetries of a given space. In physics, they are often used to study the symmetries of physical systems, such as in general relativity where they represent the symmetries of spacetime.

## 2. How do Killing vectors relate to momentum conservation?

Killing vectors are closely related to momentum conservation because they are associated with the symmetries of a physical system. In particular, the components of a Killing vector correspond to conserved quantities, such as momentum, angular momentum, and energy, in a given system. This means that if a system has a Killing vector associated with a particular symmetry, then the corresponding conserved quantity will be conserved throughout the evolution of the system.

## 3. Can Killing vectors be used to solve physical problems?

Yes, Killing vectors can be used to solve physical problems in several ways. They can be used to simplify the equations of motion for a system by identifying symmetries and corresponding conserved quantities. They can also be used to find exact solutions to certain physical systems, such as in general relativity where Killing vectors are used to find solutions to Einstein's equations.

## 4. What is the relationship between Killing vectors and Noether's theorem?

Killing vectors are closely related to Noether's theorem, which states that for every symmetry of a physical system, there exists a corresponding conserved quantity. Killing vectors can be used to identify and classify the symmetries of a given system, and the components of a Killing vector correspond to the conserved quantities associated with these symmetries.

## 5. Are Killing vectors only applicable to spacetime?

No, Killing vectors can be applied to any type of space, not just spacetime. They are a mathematical concept that can be used to study the symmetries of any system, whether it is in classical mechanics, electromagnetism, or general relativity. In some cases, Killing vectors may have physical interpretations, such as in relativity, but they are not limited to only describing spacetime symmetries.

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