Why are non-relativistic particles not redshifted?

In summary, the cosmological redshift only affects relativistic particles and reduces their kinetic energy. This has a negligible effect on the overall energy of non-relativistic particles and can be safely ignored.
  • #1
RedDwarf
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0
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!
 
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  • #2
RedDwarf said:
I was reading some script

Please give a specific reference.
 
  • #3
Red shift usually refers to photons shift in frequency. Particles such as in cosmic rays are different.
 
  • #4
RedDwarf said:
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.
 
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  • #5
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
$$
E(t) \simeq m \simeq E(t_0).
$$
 
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  • #6
Orodruin said:
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)
 
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  • #7
kimbyd said:
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!
 
  • #8
kimbyd said:
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!
 

1. Why do non-relativistic particles not experience redshift?

Non-relativistic particles do not experience redshift because they are not moving at speeds close to the speed of light. Redshift is a phenomenon that occurs when an object is moving away from an observer at speeds close to the speed of light, causing the wavelength of light to appear longer and the frequency to appear lower. Since non-relativistic particles are not moving at these high speeds, they do not experience redshift.

2. How does the speed of a particle affect its redshift?

The speed of a particle determines its redshift because redshift is caused by the Doppler effect, which is the shift in frequency and wavelength of light due to the motion of the source. The faster an object is moving away from an observer, the greater the redshift will be. Non-relativistic particles, which are not moving at high speeds, will not experience a significant redshift.

3. Are all particles redshifted in the same way?

No, not all particles are redshifted in the same way. The amount of redshift experienced by a particle depends on its speed and the distance between the particle and the observer. Relativistic particles, which are moving at speeds close to the speed of light, will experience a greater redshift than non-relativistic particles.

4. Can non-relativistic particles ever experience redshift?

Yes, non-relativistic particles can experience redshift, but it will be much smaller compared to relativistic particles. The redshift experienced by non-relativistic particles is usually negligible and not detectable without precise instruments.

5. How does redshift affect our understanding of the universe?

Redshift plays a crucial role in our understanding of the universe. It is one of the key pieces of evidence for the expansion of the universe and the Big Bang theory. By measuring the redshift of distant objects, scientists can determine their distance from Earth and the rate at which the universe is expanding. This information helps us understand the history and evolution of the universe.

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