It is pretty straight forward how light is redshifted in an expanding universe, yet I still can't understand why the De'Broglie wavelength of a massive particle isn't redshifted in an expanding universe. There is no proper notion of conserved mass energy in the expanding universe (without considering gravitational energy that is). anything that doesn't involve handwaving would be great (I'm not afraid of getting my hands dirty in the math). thanks!!


3 Answers 3


The de Broglie wavelength of a massive particle is redshifted in an expanding universe.

The de Broglie wavelength is given by:

$$ \lambda = \frac{h}{p} $$

so a red shift of the de Broglie wavelength simply means that the momentum is decreasing, which for a massive particle means that its velocity relative to us is decreasing.

And that is exactly what we see. Suppose someone on a distant galaxy fires a particle towards us with an initial velocity (relative to us) of $v$. As the particle crosses the space towards us the space expands under its feet so the particle slows down. We would see the particle slow down and in the absence of dark energy eventually come to a halt - in the presence of dark energy the particle can reverse direction and then accelerate away from us.

The result is that we observe the de Broglie wavelength of the particle to increase as the universe expands.

  • $\begingroup$ @John_Rennie Thank you. That is precisely what I had thought intuitively, but hadn't come across any mentions. Thanks again! $\endgroup$
    – R. Rankin
    Commented Dec 4, 2016 at 8:32
  • $\begingroup$ Looking a little deeper, shouldn't this redshift should hold for any initial particle momentum (in a non-bound system). Then from that standpoint, wouldn't the zero-point fluctuations of a massive particle, which manifest as momentum fluctuations, necessarily redshift as well? This is actually what I'd been looking into. Thanks again! $\endgroup$
    – R. Rankin
    Commented Dec 4, 2016 at 20:49
  • $\begingroup$ @R.Rankin a free particle has no zero state energy. In any case zero point energy doesn't cause fluctuations. The fluctuation is in the measuring process not the quantum state. See here for more on this. $\endgroup$ Commented Dec 5, 2016 at 6:43
  • $\begingroup$ Great answer Rennie. Hindsight being 20/20 I was wondering: if a particle is fired towards us at initial velocity $v$ and distance $D$, wouldn't it also accelerate away from us if $v<DH$? Without dark energy $\endgroup$
    – R. Rankin
    Commented Feb 11, 2023 at 5:17

The difference between massless and massive particles in the expanding Universe is in their energy densities. Since the energy of heavy particle comes almost entirely from its mass, the energy density behaves as $\frac{1}{r^3}$, while the energy of the massless one comes from its wavelength, which changes as the Universe expands, so the energy density of these particles is $\propto \frac{1}{r^4}$.

  • 2
    $\begingroup$ This is true but doesn't answer the question. $\endgroup$
    – user4552
    Commented Dec 3, 2016 at 18:59

This answer (the most upvoted answer) is incorrect. The redshift is NOT related to the decreasing momentum of the particle. This hypothesis (known as the 'tired particle' hypothesis) has been disproven experimentally in favor of the Doppler hypothesis.

If it were true that the redshift was a result of a loss of particle momentum, it would imply inelastic collisions between the particle and interstellar medium. These collisions would also produce a 'smearing' effect, in addition to slowing the particle down. No such smearing has ever been found. This is the basis for my answer.

The only explanation for the redshift is the well known Doppler effect - as applied to the expanding Universe (not to the particle itself).

Reference - Michael Berry - Principles of Cosmology and Gravitation, pg 21 (chapter 2, section 3 - 2.3) If you think this answer is incorrect, be prepared to cite a reference to support your claim. I have listed a reference above - but ANY modern cosmology textbook will support my claim.

  • 5
    $\begingroup$ The answer you refer to is correct. The momentum of massive particles (e.g. neutrinos from the big bang) is much lower in the present day universe. No collisions are required, just as photons from the CMB have not interacted with anything but now have microwave energies. The Doppler effect is not responsible for cosmological redshift - this is a common misconception that causes a number of difficulties. $\endgroup$
    – ProfRob
    Commented Jun 18, 2017 at 11:51
  • $\begingroup$ I think you are the one with the misconception. The doppler effect IS the cause of redshift. Look up any modern textbook - or not. If you have a reference to support your claim, please cite it. Here is my Reference - Michael Berry - Principles of Cosmology and Gravitation, pg 21 (chapter 2, section 3 - 2.3) $\endgroup$ Commented Jul 9, 2019 at 18:25
  • $\begingroup$ none of the other theories have any experimental validation. There is no smearing observed - which would be observed if any kind of momentum change was taking place and causing the shift. So far, no such smearing has been detected. If you can cite a counter theory - please also cite any experimental validation of it. $\endgroup$ Commented Jul 9, 2019 at 18:32
  • $\begingroup$ You're right that the redshift can be understood as a Doppler shift, but the peculiar velocity of nonrelativistic particles does also decrease with time, for similar reasons. It happens even in a Newtonian universe with no gravity and a uniformly expanding sphere of particles. A particle with a large peculiar velocity ends up closest to the Hubble particles with the most similar velocities, and relative to them its peculiar velocity is small. $\endgroup$
    – benrg
    Commented Aug 13, 2020 at 5:49

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