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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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  • $\begingroup$ Possible. But, why would an atom become more energetic? Any explanation? $\endgroup$ – Schrödinger's Cat Apr 30 '14 at 22:02
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    $\begingroup$ 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? $\endgroup$ – Jerry Schirmer Apr 30 '14 at 22:06
  • $\begingroup$ 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. $\endgroup$ – John Eastmond May 1 '14 at 14:39
  • $\begingroup$ 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 $a(t) \propto t^{2/3}$. $\endgroup$ – John Eastmond May 1 '14 at 15:53
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    $\begingroup$ 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. $\endgroup$ – 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.

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  • $\begingroup$ Please see my comment above. $\endgroup$ – John Eastmond May 1 '14 at 15:37
  • $\begingroup$ 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 $m_0 = a(t) m$. $\endgroup$ – John Eastmond May 1 '14 at 20:44
  • $\begingroup$ But in that case, gravitational interactions that scale like $m^2$ will scale quadratically with $a$. $\endgroup$ – rob May 1 '14 at 21:41
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    $\begingroup$ 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. $\endgroup$ – rob May 5 '14 at 12:44
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    $\begingroup$ Ok - It's complicated! $\endgroup$ – John Eastmond May 6 '14 at 13:03
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The same interesting idea was brought recently by Prof Wetterich in his article "Universe without expansion":

https://www.sciencedirect.com/science/article/pii/S2212686413000332

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 main stream of physics yet, though is very intuitive comparing to 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 aceleration $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 aceleration should be constant, so one would rather expect that $G \propto 1/ M_{pl} $.

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