Cosmological Redshift: Electron Energy Loss in Expanding Universe?

[Matterwave]

So I have been thinking. Light gets redshifted because of the cosmological expansion of the Universe. This would mean that other, material particles, should get "cosmologically redshifted" as well right? So, for example, if an electron were flying towards us from some distant galaxy (and we neglected all other effects), would this electron lose energy as it moved towards us simply due to the expansion of the universe? What is the rate at which it loses energy? It's been too long since I've taken a cosmology class for me to do this calculation myself with any confidence of correctness.

 

[Mordred]

yes other particles will lose energy. here is your relations

[tex]\frac{\Delta_f}{f} = \frac{\lambda}{\lambda_o} = \frac{v}{c}=\frac{E_o}{E}=\frac{hc}{\lambda_o} \frac{\lambda}{hc}[/tex]

 

[bapowell]

The redshift of matter particles is manifested in their coming to rest with respect to the comoving frame. From the geodesic equation of a matter particle, it is possible to show that the particle's proper velocity, [itex]{\bf u}[/itex], satisfies
[tex]|{\bf u}_0| = |{\bf u}_i|\frac{a(t_i)}{a(t_0)}[/tex]
where the subscript '0' refers to the present value, and a(t) is the scale factor. As the universe expands, [itex]{\bf u}[/itex] tends to zero. This is the same relation leading to the photon redshift -- just replace [itex]{\bf u}[/itex] with the momentum, [itex]{\bf p}[/itex].

 

[Matterwave]

The redshift of matter particles is manifested in their coming to rest with respect to the comoving frame. From the geodesic equation of a matter particle, it is possible to show that the particle's proper velocity, [itex]{\bf u}[/itex], satisfies
[tex]|{\bf u}_0| = |{\bf u}_i|\frac{a(t_i)}{a(t_0)}[/tex]
where the subscript '0' refers to the present value, and a(t) is the scale factor. As the universe expands, [itex]{\bf u}[/itex] tends to zero. This is the same relation leading to the photon redshift -- just replace [itex]{\bf u}[/itex] with the momentum, [itex]{\bf p}[/itex].

The four velocity is normalized to 1 (or -1) though right? So it can't actually turn to 0 can it? o.o

 

[bapowell]

[itex]{\bf u}[/itex] is the three-velocity.

 

[George Jones]

[itex]{\bf u}[/itex] is the three-velocity.

I think I have seen this terminology before, but it is much more standard (e.g., page 84 of Hartle's GR book) to write ##\bf{u} = \gamma \bf{v}##, where ##\bf{v}## is called the three-velocity.

In any case, it is a very nice result.

 

[George Jones]

if an electron were flying towards us from some distant galaxy (and we neglected all other effects), would this electron lose energy as it moved towards us simply due to the expansion of the universe? What is the rate at which it loses energy?

Use Killing vectors to do the calculation!

Let ##U## be the 4-velocity of a cricket ball that is tossed form one galaxy to another galaxy. From symmetry, we can take the motion to be on a 2-dimensional ##r-\chi## hypersurface of constant ##\theta## and ##\phi## (##\chi## is a comoving distance coordinate).

On this hypersurface, the FLRW metric induces the metric

$$ds^2 = -dt^2 + a \left(t\right)^2 d\chi^2 .$$

Since ##\chi## does not appear explicitly, ##\partial / \partial \chi## is a Killing vector, and ##k = g \left( U , \partial / \partial \chi \right)## is a conserved quantity on the ball's worldline.

To make contact with physically measured quantities, choose orthonormal bases for the comoving (with the Hubble flow, not the ball) observers that the ball passes, ##e_0 = \partial / \partial t## and ##e_1 = \left( 1/a \left(t\right) \right)\partial / \partial \chi##. Then, the constant

$$k = g \left( U , \partial / \partial \chi \right) = g \left( U^0 e_0 + U^1 e_1 ,a \left(t\right) e_1 \right) = -U^1 a \left(t\right).$$

Because of the orthonormal bases, ##U^1## takes the special relativistic form ##U^1= \gamma v##, and Brian's nice result follows.

 

[Mordred]

thanks for that explanation, helps me with the metrics in an article I just picked up the other day lol

 

[Matterwave]

Great, thanks guys. :D