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The standard explanation for the cosmological redshift is that photons emitted from far away galaxies have their wavelengths lengthened as they travel through the expanding Universe.

But perhaps the photons do not lose energy as they travel but rather the atoms in our detectors are more energetic in comparision with the atoms that emitted those photons a long time in the past leading to an apparent redshift effect?

Addition (having had a comment exchange with @rob, see below) : My hypothesis is that the Planck mass $M_{pl} \propto a(t)$ where $a(t)$ is the Universal scale factor.

Addition 2 Of course if the Planck mass $M_{pl}$ is changing then $G=1/M^2_{pl}$ is changing so that we no longer have standard GR!

I've asked this question before, see Cosmological redshift interpretation, but this time I'm including a little bit of theory to back up my hypothesis.

For simplicity let us assume a flat radial FRW metric:

$$ds^2=-dt^2 + a^2(t)\ dr^2$$

Consider the null geodesic path of a light beam with $ds=0$ so that we have:

$$dt = a(t)\ dr\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

Now at the present time $t_0$ we define the scale factor $a(t_0)=1$ so that we have:

$$dt_0 = dr\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Substituting equation (2) into equation (1) we have:

$$dt = a(t)\ dt_0$$

In order for the interval of time $dt$ to stay constant as the scale factor $a(t)$ increases we must have the corresponding interval of present time $dt_0$ varying inversely with the scale factor:

$$dt_0 \propto \frac{1}{a(t)}$$

Thus as cosmological time $t$ increases, and the Universe expands, equal intervals of cosmological time $dt$ correspond to smaller and smaller intervals of present time $dt_0$.

Now the energy of a system is proportional to the frequency of its oscillation which in turn is inversely proportional to its oscillation period:

$$E(t) \propto \frac{1}{dt}$$

The corresponding energy of the system in terms of the present epoch $t_0$ is given by

$$E(t_0) \propto \frac{1}{dt_0}$$

$$E(t_0) \propto a(t)$$

Thus an atom at time $t$ is a factor $a(t)$ times more energetic than the same atom at time $t_0$.

As the energy scale is ultimately set by the Planck mass then the Planck mass must be increasing as the Universe expands: $M_{pl} \propto a(t)$.

This effect alone would account for the gravitational redshift of distant galaxies without the assumption that photons travelling from those galaxies lose energy due to wavelength expansion.

Addition: I believe this hypothesis leads to a linear cosmological expansion $a(t)\propto t$ (see comments below).

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John Eastmond
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  • Possible. But, why would an atom become more energetic? Any explanation? – Earth is a Spoon Apr 30 '14 at 22:02
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    What do you mean by "energy"? That the fundamental energy of the transitions generating the photons has decreased over time? Or that the energy of the emitted photons themselves have decreased over time? Or something else? – Zo the Relativist Apr 30 '14 at 22:06
  • My hypothesis is that a unit time interval at time $t$ is equivalent to a time interval $1/a(t)$ at the present time $t_0$. As time and energy are reciprocally related this implies that a unit of energy at time $t$ is equivalent to $a(t)$ energy units at present time $t_0$. The change in energy is just due to different observers' perspectives analogous to the change in energy when one changes inertial frames in special relativity. – John Eastmond May 01 '14 at 14:39
  • My hypothesis has implications for cosmology. I believe the density of matter/radiation is always given by $\rho \propto a(t)/a^3(t) = 1/a^2(t)$. If you put this into the Friedmann equations (without cosmological constant) you derive a scale factor $a(t)$ that increases linearly with time. This is in better agreement with the Universe expansion data than the standard Einstein-de Sitter model for example where $a(t) \propto t^{2/3}$. – John Eastmond May 01 '14 at 15:53
  • @JohnEastmond: if you use current observational data with the assumptions of homogeniety and isotropy, and put this in the Friedmann equations, you are led inexoriably to a cosmological constant. – Zo the Relativist May 01 '14 at 16:28
  • The standard calculations fit to the current observational data assuming $\rho_{matter} \propto 1/a^3$ and $\rho_{radiation} \propto 1/a^4$. My alternative redshift interpretation implies $\rho \propto 1/a^2$ which leads to a linear cosmology. – John Eastmond May 01 '14 at 20:54
  • Astrophysicist Fulvio Melia has shown that a linear cosmology, in his terms $R_h = c t$, fits the current observational data surprisingly well. See his papers on the Arxiv: http://uk.arxiv.org/find/astro-ph/1/au:+Melia_F/0/1/0/all/0/1. – John Eastmond May 01 '14 at 21:08
  • Also perhaps $\rho \propto 1/a^2$ is consistent with the holographic principle in which a description of particles on the boundary area is equivalent to a description of particles in the enclosed volume. – John Eastmond May 01 '14 at 23:14
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    This hypothesis doesn't make any sense to me. If space is really expanding and there's a variable scale factor $a(t)$ in the metric, then it shows up in the null-geodesics equation, which implies that photons have wave number $k \propto 1/a(t)$. The photon wavelength have to vary as $\lambda \propto a(t)$. It's a direct consequence of the geodesic equation and a metric with a variable scale factor. – Cham Dec 12 '18 at 20:13

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This is an interesting idea. If I understand you correctly, you're suggesting that perhaps light emitted by very distant atoms has a different spectrum than light emitted by atoms in our cosmic neighborhood, and that a uniform shift in the energies of all atomic transitions would mimic the cosmological redshift.

However, the energies involved in atomic transitions depend on lots and lots of factors. The cosmological redshift has the theoretical advantage of simplicity: once the light is emitted and en route to us, all light is treated the same way. By contrast, the energy levels in an atoms depend on lots and lots of factors. In general the energies allowed in an atom depend on the value of ℏ, on the masses and charges of the constituents, on the length scales and speeds involved.

For example, in the energy-time uncertainty relation $\Delta E\Delta t \ge \hbar/2$ we have an inverse relationship between energy and time, which suggests that if a global unit of $\Delta t$ is changing, the spectrum of virtual particle-antiparticle pairs that contribute to an interaction. This is called polarization of the vacuum and it contributes to changes in the electromagnetic coupling constant and the weak mixing angle as you look at interactions with different energy.

Similarly, for a massless photon the Einstein equation $E^2=p^2+m^2$ gives a total energy $E=hf$, where again $E$ and $t$ are inversely proportional to each other (though the time measurement is buried in the photon's frequency). But for a massive particle the total energy becomes $$ E = \gamma m c^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} mv^2 + \cdots $$ Now you start to see complications. Does your scale factor affect $c$ and $v$? If so, then for massive objects the energy varies like $1/t^2$, rather than like $1/t$. If not, then massive objects see no variation in energy as the scale factor changes. If your scale factor changes $v$ but not $c$, then you have a mess. Maybe it's rest masses that change inversely with $t$, but there's no theory that would support that. These are the energies that go into the computation of atomic excitations; you don't have the luxury of wishing them away.

As a real example of the sort of thing you're thinking of, there is evidence — not incontrovertible evidence, not universally accepted, but not convincingly refuted, either — that the electromagnetic fine structure constant $\alpha = e^2/\hbar c$ is different in the fifth decimal place in very distant galaxies. One of the strengths of this evidence is that a small shift if $\alpha$ causes some atomic transitions to become less energetic, and others to become more energetic, very different from an error in a redshift measurement. I think the best explanation of the physics was in one of the original papers, though the experimental situation has evolved since then.

Urb
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rob
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  • Please see my comment above. – John Eastmond May 01 '14 at 15:37
  • I would say that it is the rest mass $m$ that increases with the scale factor $a(t)$ when viewed from the present time $t_0$. The rest mass of an electron for example is due to constant interaction with the Higgs field. This produces an oscillatory motion with a characteristic time period $dt$ at some future cosmological time $t$. At the present time $t_0$ the corresponding time period is $dt_0 = dt/a(t)$. The corresponding mass/energy at the present time $t_0$ is then $m_0 = a(t) m$. – John Eastmond May 01 '14 at 20:44
  • But in that case, gravitational interactions that scale like $m^2$ will scale quadratically with $a$. – rob May 01 '14 at 21:41
  • I suppose a gravitational interaction will be of the form $Gm^2/r$. If $r \propto a$ as well as $m \propto a$ then the gravitational interaction will be proportional to $a$. – John Eastmond May 01 '14 at 23:48
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    If you don't allow $h$, $c$, or $G$ to change, then the Chandrasekhar limit will scale like $1/a^2$. Now you have to recover the Type Ia supernova data. – rob May 02 '14 at 01:24
  • Actually what I said above about $r \propto a$ was wrong - the scale $r$ of a bound system does not expand with the Universe. If we take $h=c=1$ then I think the Planck mass $M_{pl} \propto a$. – John Eastmond May 02 '14 at 08:57
  • But if it's not $r$, $\hbar$, or $c$ changing that cause gravitational energies $Gmm/r$ to scale with $a$, then what causes scaling with $a$ for electrical energy between unit charges, $\alpha \hbar c /r $ ? – rob May 02 '14 at 14:33
  • Ok. The proper distance at time $t$, $ds$, and the proper distance at the present time $t_0$, $ds_0$, is related by the equation $ds = a(t)\ ds_0$ (using the FRW metric with $dt=0$). Thus if $ds$ is fixed then we have $ds_0 \propto 1/a(t)$. Therefore relative to the present time, t_0, $\alpha \hbar c/ds_0 \propto a(t)$ (I'm using $ds_0$ for an interval of proper distance where you use $r$). – John Eastmond May 05 '14 at 12:06
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    Then $Gmm/r$ is no longer linear in $a$. I'm not interested in playing whack-a-mole with the details of your model; I just hoped I could get you to realized that it's complicated. – rob May 05 '14 at 12:44
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    Ok - It's complicated! – John Eastmond May 06 '14 at 13:03
  • if mass increase has the same effect as cosmological expansion, then shouldn't that be encodeable as a gauge symmetry? – lurscher Jan 22 '22 at 01:56
  • @lurscher I don’t know, and that’s too big of an “if” for me to think much about it. (I answered this question back before I really understood the community’s policy on non-mainstream physics.) You might ask a follow-up question. – rob Jan 22 '22 at 03:05
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The same interesting idea was brought recently by Prof Wetterich in his article "Universe without expansion":

https://doi.org/10.1016/j.dark.2013.10.002

He explains that the cosmological redshift can be understood as masses variation (growth) of all particles in Universe - as alternative to metrics expansion. So in this case scale factor a(t) is not needed to describe redshift and in some cases it can be replaced with m(t) as variation of energy in above post. As I understand this idea is out of the mainstream of physics yet, though is very intuitive compared to the current theory of the metric expansion of space.

Regarding your Addition 2. The variation of Gravitational constant is constrained by orbitaly motions of the planets which seem to be almost constant in time. The values of orbits are tighten by gravitational acceleration $a=\frac{Gm }{r^2} $ (see for example "Does the Newton’s gravitational constant vary sinusoidally with time? Orbital motions say no", Lorenzo Iorio). So in our case when mass varies in time the gravitational acceleration should be constant, so one would rather expect that $G \propto 1/ M_{pl} $.

Urb
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