Why are non-relativistic particles not redshifted?

[RedDwarf]

Hey!

I was reading some script and when it comes to the cosmological redshift, it says, that only relativistic particles are affected by cosmological redshift. This does feel quite natural, however, I haven't been able to come up with an explanation that shows it with proper physics and formulae.

Hope you can help!

 

[PeterDonis]

I was reading some script

Please give a specific reference.

 

[mathman]

Red shift usually refers to photons shift in frequency. Particles such as in cosmic rays are different.

 

[kimbyd]

Hey!

I was reading some script and when it comes to the cosmological redshift, it says, that only relativistic particles are affected by cosmological redshift. This does feel quite natural, however, I haven't been able to come up with an explanation that shows it with proper physics and formulae.

Hope you can help!

The reason is that non-relativistic particles have mass energy far greater than their kinetic energy (this is what is meant by the term non-relativistic). Redshift only reduces the kinetic energy, so the impact on non-relativistic particles is tiny. There is an impact, but it has so little effect on the overall energy of non-relativistic particles that can be safely ignored when looking at how their energy changes over time.

 

[Orodruin]

The geodesic equations for the FLRW universe are derived from maximising the expression
$$
S = \int g_{ab} \dot x^a \dot x^b d\tau = \int (\underbrace{\dot t^2 - a(t)^2 \dot{\vec x}^2}_{\equiv L}) d\tau.
$$
Since the integrand does not depend explicitly on the spatial coordinates, there are constants of motion
$$
k_i = \frac{\partial L}{\partial \dot x^i} = 2 a(t)^2 \dot x^i.
$$
Noting that the 4-momentum of a particle is its mass multiplied by its 4-velocity, it follows that it is given by $P^a = m \dot x^a$. The spatial momentum is therefore given by $\vec p^2 = - g_{ij} P^i P^j = m^2 a(t)^2 \dot{\vec x}^2 = m^2 \vec k^2 / a(t)^2 = \vec p(t_0)^2 a(t_0)^2/a(t)^2$. Hence, what is actually red-shifted according to the scale factor is the particle momentum.

For the particle energy, this means that
$$
E(t) = \sqrt{m^2 + p(t)^2} = \sqrt{m^2 + \frac{a(t_0)^2}{a(t)^2} p(t_0)^2}.
$$
In the case that $p(t_0) \gg m$ and $p(t) \gg m$, i.e., for relativistic particles, you would therefore find
$$
E(t) \simeq \frac{a(t_0)}{a(t)} E(t_0),
$$
whereas for $p(t_0) \ll m$ and $p(t) \ll m$, i.e., for non-relativistic particles, you instead obtain
$$
E(t) \simeq m \simeq E(t_0).
$$

 

[kimbyd]

The geodesic equations for the FLRW universe are derived from maximising the expression
$$
S = \int g_{ab} \dot x^a \dot x^b d\tau = \int (\underbrace{\dot t^2 - a(t)^2 \dot{\vec x}^2}_{\equiv L}) d\tau.
$$
Since the integrand does not depend explicitly on the spatial coordinates, there are constants of motion
$$
k_i = \frac{\partial L}{\partial \dot x^i} = 2 a(t)^2 \dot x^i.
$$
Noting that the 4-momentum of a particle is its mass multiplied by its 4-velocity, it follows that it is given by $P^a = m \dot x^a$. The spatial momentum is therefore given by $\vec p^2 = - g_{ij} P^i P^j = m^2 a(t)^2 \dot{\vec x}^2 = m^2 \vec k^2 / a(t)^2 = \vec p(t_0)^2 a(t_0)^2/a(t)^2$. Hence, what is actually red-shifted according to the scale factor is the particle momentum.

For the particle energy, this means that
$$
E(t) = \sqrt{m^2 + p(t)^2} = \sqrt{m^2 + \frac{a(t_0)^2}{a(t)^2} p(t_0)^2}.
$$
In the case that $p(t_0) \gg m$ and $p(t) \gg m$, i.e., for relativistic particles, you would therefore find
$$
E(t) \simeq \frac{a(t_0)}{a(t)} E(t_0),
$$
whereas for $p(t_0) \ll m$ and $p(t) \ll m$, i.e., for non-relativistic particles, you instead obtain
$$
E(t) \simeq m \simeq E(t_0).
$$

Small comment:
I think this kind of explanation is clearer if explained using the energy squared rather than the energy:

$$E(t)^2 = m^2 + p(t)^2 = m^2 + {a(t_0)^2 \over a(t)^2} p(t_0)^2$$

(note to others: both of our explanations are using units where the speed of light $c$ can be omitted)

 

[RedDwarf]

The reason is that non-relativistic particles have mass energy far greater than their kinetic energy (this is what is meant by the term non-relativistic). Redshift only reduces the kinetic energy, so the impact on non-relativistic particles is tiny. There is an impact, but it has so little effect on the overall energy of non-relativistic particles that can be safely ignored when looking at how their energy changes over time.

Thank you so much, this is exactly what my tired brain couldn't put into words!

 

[RedDwarf]

Small comment:
I think this kind of explanation is clearer if explained using the energy squared rather than the energy:

$$E(t)^2 = m^2 + p(t)^2 = m^2 + {a(t_0)^2 \over a(t)^2} p(t_0)^2$$

(note to others: both of our explanations are using units where the speed of light $c$ can be omitted)

Thank you, this helped very much!