# The energy of redshifted light

1. Mar 17, 2009

Light travelling across the universe is red shifted, so that it loses energy. Is that energy destroyed? If not, where does it go? Could it be driving the expansion of the universe?
Could light be thermalised, i.e. could a low frequency light increase its energy by passing through a hot enough, although transparent, medium? Or vice versa...

2. Mar 17, 2009

### Staff: Mentor

There is no energy lost in a red shift, just as there is no energy lost if you are in a car accident where you hit another moving car from behind. Kinetic energy is simply frame of reference dependent.

3. Mar 18, 2009

### Chalnoth

There are multiple ways to answer this question. You can answer it by noting that energy need not be conserved in General Relativity: GR forces the conservation of the stress-energy tensor, which includes energy, momentum, pressure, and anisotropic stresses. Conservation of this tensor as a whole forces non-conservation of energy, under certain circumstances.

Another way of looking at it is to pay attention to the gravitational potential energy as well as the energy in matter fields, by looking at the Hamiltonian formalism. In that case, energy is always conserved by construction. For the case of photons, you'd see the energy loss in the photon field in a comoving volume as stemming from a gravitational potential energy that becomes less negative. For the case of a cosmological constant, you'd see the energy gain in the cosmological constant in a comoving volume as stemming from a gravitational potential energy that becomes more negative.

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

4. Mar 18, 2009

### cesiumfrog

Do you have an example where local conservation of stress-energy implies local non-conservation of energy and/or momentum?

Could you clarify that a bit? I would have thought such a potential energy would depend on the configuration of (a spacelike slice of) the spacetime, independent of whether the universe is in (cosmological constant driven) accelerating expansion or coasting expansion or even contraction?

5. Mar 18, 2009

### v2kkim

Simply looking at photon red shift there is energy loss obviously. But photon has traveled longer than expected compared to no space expansion case, which I can not relate to energy.

Last edited: Mar 18, 2009
6. Mar 18, 2009

### Chalnoth

The example in JuanCasado's post works here. Just take a uniform gas of photons. The number density in a local co-moving region stays the same (since it's uniform, the number going into the local co-moving region equals the number exiting it), but if space is expanding, then the photons are getting stretched along with space, and so the energy density in a local co-moving region drops with time.

Quick note on what co-moving means: the size of the region you're looking at expands along with the universe.

How does this work in the context of the conservation of the stress-energy tensor? Well, with a uniform radiation fluid, there are no off-diagonal elements. There's just energy density in the time-time component, and pressure along the space-space components. Since the pressure of a photon gas is 1/3rd its energy density, and since it's uniform in all directions, the diagonal elements of the space-space part are all $$rho/3$$, where $$rho$$ is the energy density of the photons. You can express the conservation of stress-energy in the following form:

$$\dot{\rho} = -3H\left(\rho + p\right)$$

Since $$p = \rho/3$$, and expanding the derivatives with respect to time:

$$\frac{d\rho}{dt} = -4H\rho$$

We can change all of our derivatives with respect to time to derivatives with respect to a by setting:

$$\frac{d}{dt} = \frac{da}{dt}\frac{d}{da} = a H \frac{d}{da}$$

So that we have:

$$aH\frac{d\rho}{da} = -4H\rho$$
$$\frac{d\rho}{da} = -\frac{4}{a}\rho$$
$$\frac{d\rho}{\rho} = -4 \frac{da}{a}$$
$$\ln\left(\rho\right) = -4 \ln\left(a\right) + C$$
$$\rho(a) = \rho(0) a^{-4}$$

Since the volume is increasing as $$a^3$$, but the energy density is falling as $$a^{-4}$$, this represents an energy loss per unit volume, taken directly from the conservation of stress-energy and making use of the fact that $$p = \rho/3$$ for photons.

Note that you can follow this process in the exact same way for any form of matter where $$p = w\rho$$ with $$w =$$ constant. For $$w = 0$$, energy is conserved in a comoving volume. For $$w < 0$$, energy grows with expansion. For $$w > 0$$, energy drops with expansion.

Last edited: Mar 18, 2009
7. Mar 18, 2009

### Chalnoth

Huh?

8. Mar 18, 2009

### v2kkim

Actually my wording was not good so I modified a little. I meant photon travel more than 'c' due to space expansion.

9. Mar 18, 2009

### Chalnoth

Oh, well, no. The only relative speed that has any meaning in general relativity is local relative speed. A photon always travels at speed c relative to any local observer, irrespective of the expansion.

10. Mar 18, 2009

### cesiumfrog

Oops, it seems you need a quick note on what "http://en.wikipedia.org/wiki/Local_reference_frame" [Broken]" means in general relativity. How about we drop this point (hint: locally is where GR conserves the stress-energy tensor, and it actually does require that energy and momentum are locally conserved individually) and we move straight on to you clarifying your claim regarding the balancing of potential energy?

Last edited by a moderator: May 4, 2017
11. Mar 18, 2009

### Chalnoth

Well, if you're going to reduce all the way to flat Minkowski space-time, then you can't talk about the effects of the expansion. So I took one step up from pure locality: a local comoving volume. This is localized in space, but not in time, so that the space-time curvature has an impact.

Last edited by a moderator: May 4, 2017
12. Mar 18, 2009

### cesiumfrog

You haven't proven that the stress-energy tensor is conserved (whatever you mean by that) for your comoving volume.

Last edited: Mar 18, 2009
13. Mar 18, 2009

### Chalnoth

No. I made use of the equation that can be derived from conservation of the stress-energy tensor:
$$\dot\rho = -3H\left(\rho + p\right)$$

You can go ahead and go through the covariant derivative of the stress energy tensor of an isotropic, homogeneous fluid if you'd like (it's a bit much to put down in tex for a forum post). But the equation above is what you get.

No.

14. Mar 30, 2009

As far as I have seen, the second part of my question has not been discused:
Could light be thermalised, i.e. could a low frequency light increase its energy by passing through a hot enough, although transparent, medium? Or vice versa...

15. Mar 30, 2009

### Chalnoth

Yes. This is the nature of the Sunyaev-Zel'dovich effect, which can be used to detect galaxy clusters (which contain a very hot gas that "warms up" the CMB photons).

16. Mar 30, 2009

Good. So, could CMB photons be (conversely) red shifted, i.e. "cooled down", due to interactions with very cold intergalactic media?

17. Mar 30, 2009

### Chalnoth

Perhaps. But there isn't much of anything that cool out there.

18. Mar 30, 2009

Well, any intergalactic particle should be in thermal equilibrium at about 2.7K, right?

19. Mar 30, 2009

### Chalnoth

Not necessarily. There's also starlight to consider, which tends to heat up the IGM above the CMB temperature.

20. Mar 30, 2009