# Redshift and Particle Momentum Decay

1. Dec 6, 2009

### Jorrie

Here is an interesting observation, which I would like to know the validity of.

The momentum of a relativistic 'cosmological' particle in a homogeneous universe can be written as $^{[1]}$

$$L = \gamma m a v_{pec} = K$$

where $\gamma=(1-v_{pec}^2/c^2)^{-0.5}$, $m$ the rest mass, $v_{pec}$ the peculiar velocity of the particle, $a$ the expansion factor, with K constant.

According to the de Broglie relations, the wavelength of the particle is

$$\lambda = \frac{h}{p} = \frac{h}{\gamma m v_{pec}}$$

where h is Planck's constant and p the local momentum of the particle.

From the above two equations, we can write

$$L = \frac{ah}{\lambda} = K$$

This is the same relationship as for the cosmological redshift of a photon. So, in a way, particle momenta do not 'decay', they simply 'redshift'.

Or, do I misinterpret something?

Ref: [1] http://arxiv.org/abs/astro-ph/0402278" [Broken] (section 3-2).

Last edited by a moderator: May 4, 2017
2. Dec 6, 2009

### marcus

Jorrie, I don't remember if we have discussed this before. Momentum decay in an expanding geometry seems an exact parallel to redshift. You may have already seen this papar by Hongbao Zhang.

http://arxiv.org/abs/0808.1552
Note on the thermal history of decoupled massive particles
Hongbao Zhang
4 pages, to appear in Classical and Quantum Gravity
(Submitted on 11 Aug 2008)
"This note provides an alternative approach to the momentum decay and thermal evolution of decoupled massive particles. Although the ingredients in our results have been addressed in Weinberg, Cosmology, the strategies employed here are simpler, and the results obtained here are more general."

You may be saying something more complicated, but let me recount the story simply and crudely: We know that expansion gradually slows relativistic massive particles down. This is what cooled the Dark Matter and allowed it to condense and begin the process of structure formation. This loss of momentum by particles is very important to our very existence. And it happens quite analogously to the redshift of light.

Now you have a way of understanding this based on DeBroglie waves. But also Steven Weinberg who wrote the current top book Cosmology proves this momentum decay in his book (says Hongbao, I didnt check.) And now Hongbao is offering us a slicker proof. He claims his proof is better than the one in Weinberg's textbook.

So it looks like everybody agrees on the basics: you, Steven Weinberg, and Hongbao Zhang. And also apparently your source which I think is the PhD thesis of Tamara Davis.
(though again I didn't check.)

I suspect you have a deeper message here and so far I have just responded superficially, but am I on the right track?

3. Dec 6, 2009

### Jorrie

Firstly, yes, it is the PhD thesis of Tamara Davis that I referenced.

Secondly, I haven't read the Hongbao Zhang paper before. I'm not sure I understand his/her "effective mass, chemical potential, and temperature" and how they relate to the velocity (or the de Broglie wavelength) of the particles.

Lastly, yes, I may have "a deeper message": if what I wrote is not wrong, there may be a very nice connection to your https://www.physicsforums.com/showthread.php?t=261161"

The equation I used:

$$L = \gamma m a v_{pec} = K$$

when used with a hyper-spherical model of radius a, can be viewed as a constant 'angular momentum' around the hyper-space origin. The peculiar velocity (or de Broglie frequency) of a massive particle, moving freely on the hyper-sphere, must then change inversely to a, taking SR time dilation into account. Likewise, the frequency of a photon must change inversely to a.

Not quite rigorous, but interesting...

Last edited by a moderator: Apr 24, 2017
4. Dec 6, 2009

### Jorrie

Marcus, just for the record: the de Broglie wave understanding is not a new idea...

Peacock 1998: "It is momentum that gets redshifted; particle de Broglie wavelengths thus scale with the expansion, a result that is independent of whether their rest mass is non-zero."

My humble contribution was to relate it to the 'cosmic balloon'.

Edit: Attached is a graph (using the above principles) for a particle shot towards us when the concordant cosmic model age was ~0.2 Gy, from a redshift distance z=19 (or a=0.05), at peculiar velocity V_pec = -0.999c. The particle would not have reached us yet - if my modeling is right, it will only reach our galaxy in another ~4 billion years and then travel at a local speed of v ~ -0.64c, due to peculiar velocity decay.

-J

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Last edited: Dec 7, 2009
5. Dec 7, 2009

### Ich

Yes, the balloon has some descriptive power, even quantitatively.
I explored these things a while ago, you can find https://www.physicsforums.com/showthread.php?p=2089458#post2089458"what I've written. I think there is also a paper somewhere that describes the same thing, at least in Newtonian approximation.

Momentum decay as well as photon redshift is consistent with de Brogie's approach, because it's in the first place the wavelength that changes with expansion, not the frequency.

Last edited by a moderator: Apr 24, 2017
6. Dec 7, 2009

### Wallace

Note that there is a very simple way to understand this 'decay' of momentum. Remember that if we want to define the momentum of a particle, we need to declare the reference frame we are using. Put simply, what does the particle have a momentum relative to?

When dealing with an expanding universe, it is conventional and convenient to define velocities (and momenta) with respect to the locally comoving reference frame, that is to say, how fast is that particle moving with respect to material that is simply partaking in the general expansion, at the particles current location?

Let's look at the consuqence of this. In an expanding universe, things further from a chose origin with be receeding more rapidly than things that are closer, by Hubbles law, v = H d.

So, we shoot off a particle from the origin at a particular velocity. Some time later it will reach the location of some point a distance d from the origin. We known from Hubbles law that this particle is moving away from the origin, so in terms of the reference frame of the origin, this point has a velocity in the same direction as the test particle. If we then ask what the velcoty of the test particle is with respect to that point, we find it is less than it has with respect to the origin.

As the particle gets further and further from the origin, it will reach points that are themselves moving faster and faster from the origin. Therefore the velocity and momentum of the test particle is constantly reducing with respect the the local Hubble flow.

Last edited: Dec 7, 2009
7. Dec 7, 2009

### Jorrie

Thanks for the link - I've missed that one of yours. I haven't seen it linked to the balloon analogy before, so maybe you were the first. :)

Last edited by a moderator: Apr 24, 2017
8. Dec 7, 2009

### Jorrie

Agreed. My first reaction was that it won't work for photons, but then I realized that if we use a relativistic subtraction of the 'origin speed' (c) and the Hubble flow (V_H) of the distant point, we get c anyway. I realize that this needs to be done over small increments (actually infinitesimal) in order to model it correctly.

Last edited: Dec 7, 2009
9. Dec 9, 2009

### Chronos

A photon can do as it pleases with respect to any observer as long as its energy is conserved wrt to that observer. I have an issue with any spacetime model that does not respect this simple courtesy.

10. Dec 9, 2009

### Wallace

Except for in some extreme cases, you can't measure the energy conservation of a photon with respect to a particular observer, because an observer only gets to see a particular photon once, so you can't compare energies at two different times to determine if it is conserved. I'm not sure what you are getting at?

Note that if we are talking about two observers in different locations seeing the same photon and comparing the energies they measure, then the energy (the 3D energy) of a photon is not conserved in an expanding universe. In particular, since this is an observation, it is invariant under co-ordinate transformations, it doesn't matter how we choose to conceptualise the expansion, we'll get the same non-conservation of energy as the answer (as long as we do the sum correctly, however we choose to do them).

On the other hand the photons 4-Momentum is always conserved, but that's a trivial axiom of the theory of General Relativity, so not particularly illuminating.

11. Dec 10, 2009

### Chronos

A photon can only be seen by one observer. A second observer sees another photon emitted by the same source at roughly the same time. The two should agree on whether both photons have the same energy. We can do this on earth by observing photons emitted by say the crab nebula at six month intervals. If there is a seasonal difference in redshift, after accounting for local velocity, then energy is not conserved, assuming sufficient instrument sensitivity.

Last edited: Dec 10, 2009