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  Is the cosmological redshift caused by the Planck mass increasing?

+ 2 like - 0 dislike
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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).

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
asked Apr 30, 2014 in Astronomy by John Eastmond (55 points) [ no revision ]
Most voted comments show all comments
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?

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user Jerry Schirmer
@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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user Jerry Schirmer
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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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: uk.arxiv.org/find/astro-ph/1/au:+Melia_F/0/1/0/all/0/1.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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Possible. But, why would an atom become more energetic? Any explanation?

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user Sachin Shekhar
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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond

2 Answers

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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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user rob
answered Apr 30, 2014 by rob actually (90 points) [ no revision ]
Please see my comment above.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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 $m_0 = a(t) m$.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
But in that case, gravitational interactions that scale like $m^2$ will scale quadratically with $a$.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user rob
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 $a$.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user rob
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 $M_{pl} \propto a$.

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user John Eastmond
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 $ ?

This post imported from StackExchange Physics at 2014-05-04 11:22 (UCT), posted by SE-user rob
+ 2 like - 0 dislike

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?

Any interpretation of the redshift must explain how so efficiently it is used to measure relative velocities, for example of apparent slow rotating objects. It is the correlation of the spectrum variations with relative velocities known by other ways that had permited Hubble to interpret his observation and characterize an expansion.

However, one can freely theorize a sum of 2 effects.

Planck & Wmap collaborations and many searchers have investigated the "constancy of the constants", a kind of generalisation of your question on the mass. Other studies expected resulting ( old ) spectra and checked them against the satellites data. Even if their main purpose is to compare recombination-time constants to the currents, it contains interesting measures to test your hypothesis.

Using Planck 2013 data, around recombination, the electron mass $m_e$ was proven to coincide with the one obtained in the lab to within 1.6%. With Planck 2015 data $m_e/m_{e0} =$ 1.0056 +- 0.0080.

New constraints on time-dependent variations of fundamental constants using Planck data by Luke Hart & Jens Chluba

Exploring the fine-structure constant impact on matter spectra would be more meaningful...
But you will face the same constancy in the publications. Studies of the Lamb Shift are in the same sense.

Planck 2015 results. XIII. Cosmological parameters

Search other "Planck 2015 Results" on arxiv.

The question is still open but the scenario, if any, was probably more subtle.

Interesting pre Planck lecture on the constancy:
Uzan J.-P., 2011, Living Reviews in Relativity, 14, 2

answered Oct 21, 2017 by igael (360 points) [ no revision ]

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