<p>The de Broglie relation suggests the &quot;wavelength of a particle&quot; is proportional to the reciprocal of its momentum.</p>
<p>Like all wavelengths, the de Broglie wavelength is effectively stretched by a factor <span class="math-container">$(1+z)$</span>, where <span class="math-container">$z$</span> is the redshift.</p>
<p>The net effect is to increase the wavelength and hence reduce the momentum when measured in the co-moving rest frame.</p>
<p>This, for example, is why the cosmic neutrino background should be very cold <span class="math-container">$\sim 2$</span>K, despite the neutrinos decoupling from the rest of the universe when they were ultra-relativistic. i.e. the temperature that characterises their momentum distribution is redshifted in a <em>similar</em> (but not identical) way to photons in the cosmic microwave background.</p>
<p>i.e. the redshifting <em>does</em> produce a (potentially) measurable effect on the momentum <em>distribution</em>.</p>

CC BY-SA 4.0
2021-04-26T22:51:37.913Z
@eshaya because it is a perfectly valid way of looking at the problem and yields the correct results.
ProfRob
CC BY-SA 4.0
2021-04-24T20:17:11.627Z
2021-04-26T23:15:35.917Z
2021-04-26T23:15:35.917Z
ProfRob
<p>Yes, there is a &quot;retardation of the co-moving velocity&quot; of particles.  It is important to take it into account to understand the time history of peculiar velocities of galaxies and for determining the energy distribution of cosmic rays coming from other galaxies.</p>
<p>Peculiar velocities, the difference between an object's velocity and the local rest velocity with respect to the microwave background, decay as 1/a(t), if there are no forces.  The global scale factor a(t) is the relative size of the universe compared to today.  So there is a &quot;redshift&quot; that applies to particles similar to that of the frequency of light.
The ratio of &quot;observed&quot; velocity to &quot;emitted&quot; velocity is <span class="math-container">$\frac{v_o}{v_e} = \frac{1}{1 + z}$</span>. (Peebles, Principles of Physical Cosmology). The <span class="math-container">$z = \frac{1}{a_e}-1$</span> here is different from the <span class="math-container">$z$</span> of the galaxy because the time of the particle's emission is different from the emission time of the light that we see now.</p>

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2021-04-24T00:27:38.393Z
But there's no way to know the intrinsic, emitted velocity of a cosmic ray particle, is there?  So even if the velocity does change, I'm not sure you could tell by how much.  So I'd be inclined to say that the answer to the original question is simply "no".
Eric Jensen
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2021-04-24T02:19:29.677Z
@EricJensen That's a good point; there aren't as many energy peak-producing process for particles as there are for photons, especially at high energy. For low energies there's [alpha decay](https://en.wikipedia.org/wiki/Alpha_decay) and [internal conversion](https://en.wikipedia.org/wiki/Internal_conversion). So I've gone ahead and asked [Astrophysical particle spectroscopy; narrow-line particle sources (charged or uncharged) for things other than photons? Have any been detected?](https://astronomy.stackexchange.com/q/43531/7982)
uhoh
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2021-04-26T22:03:27.227Z
@Eric Jensen.  It is a good observation, but the question asks if there is a physical effect of expansion on particle motion.  It is not asking if we are able to measure it.  As I see it, the answer is "yes, there is a physical effect, but it is hard to measure."    Nevertheless, it is an important physical effect, because one must take it into account when trying to ascertain the energy distribution of particles emanating from a galaxy.  I should note that the z(t) here is not the z of the galaxy since the particle, traveling slower, emanated before the light.
eshaya
CC BY-SA 4.0
2021-04-26T23:07:48.553Z
@eshaya your comment deserves to be added to the answer itself (perhaps with `@Eric's` comment reworded as a preface), and would improve the answer by providing not only the extra information, but the clarification on how you've approached the question's intent. :-)
Chappo Hasn't Forgotten
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2021-04-23T19:17:07.813Z
2021-04-27T16:00:40.230Z
2021-04-27T16:00:40.230Z
eshaya
<p>I had read somewhere that light from very distant sources can be measured to be increasingly red shifted the further away the object is (due to cosmic inflation?).</p>
<p>Suppose you had an object emitting cosmic rays or neutrinos other physical matter (not photons), is there an equivalent effect for these objects? ex: a red-shifted helium nuceli in a cosmic ray or red-shifted neutrinos coming from a neutrino source?</p>

<redshift><neutrinos><cosmic-ray>
Is there an equivalent of the red shift effect for cosmic rays?
Sidharth Ghoshal
# Is there an equivalent of the red shift effect for cosmic rays?

(Asked by: username_0 on 2021-04-23)

I had read somewhere that light from very distant sources can be measured to be increasingly red shifted the further away the object is (due to cosmic inflation?).

Suppose you had an object emitting cosmic rays or neutrinos other physical matter (not photons), is there an equivalent effect for these objects? ex: a red-shifted helium nuceli in a cosmic ray or red-shifted neutrinos coming from a neutrino source?

**Tags:** redshift, neutrinos, cosmic-ray

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## 2 answers

### Answer by username_2 on 2021-04-23. Score: 6 | Accepted answer

Yes, there is a "retardation of the co-moving velocity" of particles. It is important to take it into account to understand the time history of peculiar velocities of galaxies and for determining the energy distribution of cosmic rays coming from other galaxies.

Peculiar velocities, the difference between an object's velocity and the local rest velocity with respect to the microwave background, decay as 1/a(t), if there are no forces. The global scale factor a(t) is the relative size of the universe compared to today. So there is a "redshift" that applies to particles similar to that of the frequency of light.
The ratio of "observed" velocity to "emitted" velocity is $\frac{v\_o}{v\_e} = \frac{1}{1 + z}$. (Peebles, Principles of Physical Cosmology). The $z = \frac{1}{a\_e}-1$ here is different from the $z$ of the galaxy because the time of the particle's emission is different from the emission time of the light that we see now.

> **Comments:**
> **username_3**: But there's no way to know the intrinsic, emitted velocity of a cosmic ray particle, is there?  So even if the velocity does change, I'm not sure you could tell by how much.  So I'd be inclined to say that the answer to the original question is simply "no".  
> **username_4**: @EricJensen That's a good point; there aren't as many energy peak-producing process for particles as there are for photons, especially at high energy. For low energies there's [alpha decay](https://en.wikipedia.org/wiki/Alpha_decay) and [internal conversion](https://en.wikipedia.org/wiki/Internal_conversion). So I've gone ahead and asked [Astrophysical particle spectroscopy; narrow-line particle sources (charged or uncharged) for things other than photons? Have any been detected?](https://astronomy.stackexchange.com/q/43531/7982)  
> **username_2**: @Eric Jensen.  It is a good observation, but the question asks if there is a physical effect of expansion on particle motion.  It is not asking if we are able to measure it.  As I see it, the answer is "yes, there is a physical effect, but it is hard to measure."    Nevertheless, it is an important physical effect, because one must take it into account when trying to ascertain the energy distribution of particles emanating from a galaxy.  I should note that the z(t) here is not the z of the galaxy since the particle, traveling slower, emanated before the light.  
> **username_5**: @eshaya your comment deserves to be added to the answer itself (perhaps with `@Eric's` comment reworded as a preface), and would improve the answer by providing not only the extra information, but the clarification on how you've approached the question's intent. :-)  

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### Answer by username_1 on 2021-04-24. Score: 3

The de Broglie relation suggests the "wavelength of a particle" is proportional to the reciprocal of its momentum.

Like all wavelengths, the de Broglie wavelength is effectively stretched by a factor $(1+z)$, where $z$ is the redshift.

The net effect is to increase the wavelength and hence reduce the momentum when measured in the co-moving rest frame.

This, for example, is why the cosmic neutrino background should be very cold $\sim 2$K, despite the neutrinos decoupling from the rest of the universe when they were ultra-relativistic. i.e. the temperature that characterises their momentum distribution is redshifted in a *similar* (but not identical) way to photons in the cosmic microwave background.

i.e. the redshifting *does* produce a (potentially) measurable effect on the momentum *distribution*.

> **Comments:**
> **username_1**: @eshaya because it is a perfectly valid way of looking at the problem and yields the correct results.  

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