B Cosmological Red Shift in a Perfectly Reflecting Box

[timmdeeg]

If one goes into details one should mention that if the mirrors initially are kept in a fixed proper distance and then are freed they would be free-floating but to become comoving would take a long time. So to have them comoving initially requires to adjust them accordingly.

 

[PAllen]

Aargh! Yes, you're right. I've been thinking about Milne's cosmology recently where this is possible, but not in a general FLRW universe.

Specifically, initially free floating mirrors that start with exactly 0 mutual spectral shift for a back and forth signal, will gain redshift and move apart over time if the second derivative of the scale factor is positive; the reverse if it is negative. (This idealizes that somehow local matter is irrelevant, and the mirrors are responding only to the the global space time geometry). Note, this means that if the scale factor is decreasing, but the decrease is decelerating, the mirrors will move apart rather than converge, because the second derivative of the scale factor will be positive. This is distinct from cosmologically comoving mirrors, because the initial conditions (their state of motion) give them a starting spectral shift, such that they follow the scale factor over time.

 

[PAllen]

If one goes into details one should mention that if the mirrors initially are kept in a fixed proper distance and then are freed they would be free-floating but to become comoving would take a long time. So to have them comoving initially requires to adjust them accordingly.

Actually, they would never become comoving. At any point, there are an infinite number of inertial trajectories, exactly one of which is cosmologically comoving. Thus, a geodesic that is not-comoving at one event will never become comoving. Otherwise there must exist a point with two comoving geodesics, which is impossible.

 

[timmdeeg]

Actually, they would never become comoving.

So I have been misinterpreting the statement "As shown in Fig. 3, the untethered galaxy asymptotically joins the Hubble flow in each cosmological model that expands forever." in https://arxiv.org/pdf/astro-ph/0104349.pdf
because "asymptotically" means "never".
In Fig. 3 the authors write: "In the accelerating universe (ΩM, ΩΛ) = (0.3, 0.7), the perturbed galaxy joins the Hubble flow more quickly than in the decelerating universes ... ".

Thanks.

 

[PAllen]

So I have been misinterpreting the statement "As shown in Fig. 3, the untethered galaxy asymptotically joins the Hubble flow in each cosmological model that expands forever." in https://arxiv.org/pdf/astro-ph/0104349.pdf
because "asymptotically" means "never".
In Fig. 3 the authors write: "In the accelerating universe (ΩM, ΩΛ) = (0.3, 0.7), the perturbed galaxy joins the Hubble flow more quickly than in the decelerating universes ... ".

Thanks.

Not necessarily. What the paper is saying is that ratio of peculiar velocity to recession rate goes to zero over time, for an untethered galaxy with some initial peculiar velocity (all as measured by some reference comoving observer). This asymptotic behavior does not imply that untethered galaxy is ever exactly comoving, any more than the the function 1/x ever has a slope of zero.

 

[timmdeeg]

This asymptotic behavior does not imply that untethered galaxy is ever exactly comoving, any more than the the function 1/x ever has a slope of zero.

Yes, understand, thanks.

 

[votingmachine]

Summary:: Does a photon in a box undergo cosmological red shift over time?

When a photon travels from a distant galaxy to us it undergoes an increase in wavelength due to the expansion of the universe during the time of flight. On the other hand, physical objects such as atoms and galaxies do not undergo a similar expansion because they are bound together by elctromagnetic and other forces. My question is this: suppose you put one or more photons into a box which has 100% perfectly reflecting walls. Will the photon(s) in the box experience a cosmological red shift over time or not? If so - why? and if not -why not?

I initially read this as a question about CMBR wavelength shifting. I has elements of that and elements of ordinary red-shift.

If it is strictly a question about CMBR, I think it has been answered that it won't shift. But the answers here confuse me whether they generalize to CMBR or not ... some mention velocity based red-shift. Can anyone specifically support that answer for CMBR?

 

[jbriggs444]

But the answers here confuse me whether they generalize to CMBR or not ... some mention velocity based red-shift. Can anyone specifically support that answer for CMBR?

There is no single way to split an observed wavelength into pieces so that "this piece comes from cosmological expansion" and "this piece comes from velocity". Such statements depend on the coordinates that one chooses. There is no one right choice of coordinates.

One can use so-called "co-moving" coordinates, decide that the CMBR was emitted at such and such a wavelength, red shifted (massively) by cosmological expansion, then red or blue-shifted a bit due to the peculiar velocity of the Earth. This will account for the observed wavelength when the CMBR is measured here. But that's just one choice of coordinates. There is no underlying physical truth to that particular accounting.

 

[jartsa]

That's your objection? THAT'S your objection? That once the light is observed by someone outside of the box the light is no longer in the box? That's philosophical navel-gazing.

Possibilities:

1. The light is really, really bright and the boxes are a little transparent.
2. There is a gizmo where an outside light source is made to match the wavelength of the light in the box, as verified by a local observer.
3. You have billions of boxes and once a year a box opens sending its light to earth.

If you travel to the box to study the light, it's not redshifted or redshifting.

If the box travels to you, so that you can study the light, it's not redshifted or redshifting.

Seem that the box prevents any redshift happening to the light in the box.

 

[Demystifier]

In principle, the answer to the question is straightforward. One has to solve the D'Alembert wave equation in a time-dependent FLRW background and pick only those solutions which vanish on the boundary defined by the box. I haven't actually do that, but it seems pretty clear that before hitting the wall, a wave packet suffers a redshift expansion during the propagation. But when it hits the wall, it changes the packet in a way that I cannot guess easily without a calculation.

 

[PeterDonis]

it seems pretty clear that before hitting the wall, a wave packet suffers a redshift expansion during the propagation

It suffers a coordinate effect that many cosmologists misleadingly refer to as "redshift", but it suffers no actual effect at all unless it interacts with something.

To expand on "no actual effect at all": what seems "pretty clear" to me is that the only invariant thing you can say about the wave packet's propagation in the absence of interaction is that the wave vector is parallel transported along the null geodesic worldline of the light ray, and "parallel transport" along a geodesic translates to "unchanged".

 

[anuttarasammyak]

Expansion of the universe is derived from the equation where energy is distributed homogeneous. It is of super macro average scale. In our daily scale of planets, stars, and galaxies matters distribute inhomogeneous. I wonder it is inappropriate to apply the expansion law to such micro scale distance problems.

 

[Ibix]

I think it is inappropriate to apply the expansion to such micro scale worlds.

It isn't inappropriate. You can always find a state of motion in which you would see the CMB as isotropic, and someone with that state would be a co-moving observer. Two such people would move very slightly apart over time, absent other influences.

However, that last caveat is a huge one - on small scales, other gravitational and mechanical influences are more than enough to knock the observers out of their co-moving trajectories in extremely short order. That doesn't mean that using co-moving coordinates is wrong, but it's a bit like trying to use ground-fixed coordinates to calculate how to pick up your coffee on a train that's going over points. That is, it's complicated, and with an extra layer of disconnection between experiment and maths, but not wrong.

 

[anuttarasammyak]

It isn't inappropriate. You can always find a state of motion in which you would see the CMB as isotropic, and someone with that state would be a co-moving observer. Two such people would move very slightly apart over time, absent other influences.

Two such CMB-isotropic people near the Earth go down to the Earth or fall into the Sun by gravity, don't they? Such local geometry regarded as noise in universal average view is one of the concerns in my post with micro inhomogeneous distribution of matter.

 

[Ibix]

Two such CMB-isotropic people near the Earth go down to the Earth or falling into the Sun by gravity, don't they?

They're doing 600km/s with respect to Earth and escape velocity is only 11km/s, so they don't really fall. But their trajectories are affected, yes. That doesn't mean you can't analyse their trajectories with respect to observers who continue to see the CMB as isotropic.

 

[anuttarasammyak]

Two such people would move very slightly apart over time, absent other influences.

Thanks. I took absent here as "if" absent. I think the reality is that they may see on the Earth, fall into the sun or do something different from "very slightly apart" due to local inhomogeneous distribution of energy around them.

I watch the figure https://upload.wikimedia.org/wikipe...Hubble_constant.JPG/250px-Hubble_constant.JPG
and observe that around 1MPC is minimum distance threshold to show Hubble's law, so I assume one million light year is the minimum scale to show homogeneous distribution of energy.

We may need to prepare box of one million light year length with light mass enough to disregard its effect on spacetime geometry to show the result which OP expects.

 

[PAllen]

They're doing 600km/s with respect to Earth and escape velocity is only 11km/s, so they don't really fall. But their trajectories are affected, yes. That doesn't mean you can't analyse their trajectories with respect to observers who continue to see the CMB as isotropic.

I think asking about small scale behavior of FLRW spacetime is, in fact, just a theoretical exercise useful for gaining insight. In practice, local geometry would completely dominate. Consider that the the FLRW geometry implies a specific stress energy tensor. The interior of the milky way is described by some completely different stress energy tensor. The cosmological stress energy tensor only arises as a large scale average, and its geometry arises only on such large scales.

So while you can talk about whether or not a body in the solar system is comoving in the sense of seeing CMBR isotropy, I don’t think there is any valid way to discuss small scale dynamics of bodies in the solar system as being affected by cosmological geometry. We have had threads here go nowhere trying to even properly formulate such questions unambiguously.

 

[Ibix]

I don’t think there is any valid way to discuss small scale dynamics of bodies in the solar system as being affected by cosmological geometry.

I completely agree. The only point I was making is that you ought to be able to use a (notional) family of non-inertial worldlines that see the CMB as isotropic as the beginnings of a coordinate system. And that it probably isn't a good idea.

OP's experiment I would do deep in intergalactic space, where inertial observers could be co-moving on long timescales.

 

[PAllen]

I completely agree. The only point I was making is that you ought to be able to use a (notional) family of non-inertial worldlines that see the CMB as isotropic as the beginnings of a coordinate system. And that it probably isn't a good idea.

OP's experiment I would do deep in intergalactic space, where inertial observers could be co-moving on long timescales.

And your last suggestion gets at how tricky this question is (in the real world, rather than as a mathematical exercise in properties of an idealized geometry).

Consider that region of deep intergalactic space. To me that suggests average total energy density is much less than the overall average of the observable universe. Thus one might approach it by suggesting that it ought to locally have a vacuum metric, possibly with cosmological constant. Considering first the case without cosmological constant, then you have a solution with at most pure Weyl curvature. Further, arguments based on the shell theorem suggest that it should, in fact, have no curvature to a good approximation. However, the curvature producing geodesic divergence for initially 'parallel' geodesics in FLRW metric is pure Ricci curvature. Thus, the phenomenology of 'expansion' cannot occur at all in a region described as essentially vacuum

Allowing for a cosmological constant only modifies the argument a bit - the region in question should be described by a vacuum solution with cosmological constant, which will certainly have different dynamics over the region than pretending the global geometry was present (which, by the EFE requires the energy density to match the global average).

By virtue of some of these other threads where this was discussed, I have come to believe the common statement "bound systems don't see expansion" is true but almost irrelevant. The real issue is that global total average energy density (including dark energy/cosmological constant) determine the large scale geometry of the manifold. This large scale geometry allows the initial post BB state to evolve as expected (there is "room" for the matter to move apart; expansion of space really means just this). However, the large scale geometry is 'emergent' and only relevant over regions approximating the overall universal energy density. Any regions, however large, for which the regional energy density is substantially different from the universal average are not governed at all by the global geometry. They must, instead, be analyzed with the stress energy tensor over the region.

Sorry for the long post, but the inadequacy of the traditional "bound systems exception" has been bugging me for a while. The more correct statement is that any region too small to look like the universal average is not described by the geometry of the universal average - specifically, including any dynamical effects of the expansion (except those due to cosmological constant).

 

[vanhees71]

In principle, the answer to the question is straightforward. One has to solve the D'Alembert wave equation in a time-dependent FLRW background and pick only those solutions which vanish on the boundary defined by the box. I haven't actually do that, but it seems pretty clear that before hitting the wall, a wave packet suffers a redshift expansion during the propagation. But when it hits the wall, it changes the packet in a way that I cannot guess easily without a calculation.

But these are standing-wave solutions. The corresponding cavity photons are not propagating. Cavity QED is in this sense qualitatively different from "vacuum QED".

There is nothing hitting the wall being "before" somewhere else. Since the box is bound by electromagnetic interactions it doesn't participate in the Hubble redshift.

 

[vanhees71]

It suffers a coordinate effect that many cosmologists misleadingly refer to as "redshift", but it suffers no actual effect at all unless it interacts with something.

To expand on "no actual effect at all": what seems "pretty clear" to me is that the only invariant thing you can say about the wave packet's propagation in the absence of interaction is that the wave vector is parallel transported along the null geodesic worldline of the light ray, and "parallel transport" along a geodesic translates to "unchanged".

Nevertheless there is the observable Hubble redshift, i.e., an electromagnetic wave emitted very long ago from a far-distant galaxy has a higher frequency at emission than at observation, and you have a redshift-distance relation making the Hubble Law. It's just the free-em-field solution of Maxwell's equations in a FLRW spacetime.

 

[Demystifier]

But these are standing-wave solutions.

Yes, but their superposition does not need to be standing. You can have a wave packet moving to the right towards the wall and then reflecting from the wall to the left.

 

[PeterDonis]

there is the observable Hubble redshift

From a far-distant galaxy that is comoving (and assuming we ourselves are comoving), if we observe the light, yes. But that effect is due to the difference in 4-velocities of the emitter and the receiver; it is not due to anything that "happens" to the light during transit, at least not in any invariant sense, since, as I noted, the light's wave vector is parallel transported along its null geodesic worldline, and "parallel transport" means "unchanged".

 

[pervect]

I think it is not stressed enough in cosmology that "expanding space" is basically a coordinate dependent concept. Usually, cosmological coordinates are the most convenient, but covariance says that we can use any coordinated we like.

For this particular problem, Fermi Normal coordinates are (IMO) more convenient, at least to a good approximation. We use the usual approximation that we can write the metric in Fermi normal coordinates to second order knowing the curvature tensor. This is from memory, but I could probably find the relevant section in MTW if it became an issue. If we wanted to, in principle we could do a higher order analysis without the approximation, but this would make things a lost less straightforward and would get in the way of understanding the most significant physical effects. It's also something I don't want to do, so I'll leave it to anyone who may be interested.

Given this approximation of the metric. we can intuitively see that the major difference between the box in the FRLW spacetime vs the box in non-expanding Minkowskii space is the tidal forces that are present in the former.

Classically, this means that the photons measured frequency would depend where your measuring instrument is located. The center of the box will have a different expected frequency than the ends.

The effect will be minor for reasonably sized boxes though, we can calculate/estimae the magnitude of this shift knowing the size of the box. I haven't run any numbers, but I feel confident in saying that it can be considerd small. More to the point, we know that the effect must vanish in the limit as the box length becomes small, so that in the limit as the length approaches zero, there is no change in photon frequency with time. For a large box, it's less clear, and it depends on where we monitor the photon frequency. I suspect that the photon frequency measured in the center of the box would remain constant, while the photon frequency measured near the ends would vary slightly with time as the tidal forces, due to the changing curvature, change with time. Probably a more carfeful analysis would need to be done to confirm this,though.

 

[vanhees71]

From a far-distant galaxy that is comoving (and assuming we ourselves are comoving), if we observe the light, yes. But that effect is due to the difference in 4-velocities of the emitter and the receiver; it is not due to anything that "happens" to the light during transit, at least not in any invariant sense, since, as I noted, the light's wave vector is parallel transported along its null geodesic worldline, and "parallel transport" means "unchanged".

We measure a frequency of the light from a far-distant galaxy. That's a local observable. We infer from the known spectra of the involved atoms (of course under the assumption of the cosmological principle that the physics is everywhere and at any time the same) that this frequency is red-shifted.

I don't know which 4-velocities you compare. The standard red-shift equation is $1+z=a(t_{\text{obs}})/a(t_{\text{em}})$, i.e., the two frequencies by definition are defined to be measured by comoving observers (i.e., observers at rest wrt. the CMBR).

 

[PeterDonis]

We measure a frequency of the light from a far-distant galaxy. That's a local observable.

Yes.

We infer from the known spectra of the involved atoms (of course under the assumption of the cosmological principle that the physics is everywhere and at any time the same) that this frequency is red-shifted.

Yes. Or, to put it another way, we infer what the emitted wavelength was based on our knowledge of the spectra of atoms.

I don't know which 4-velocities you compare.

The emitted wavelength (or frequency, if you prefer to view it that way) is the inner product of the emitter's 4-velocity and the light's wave vector. The received wavelength (or frequency) is the inner product of the receiver's 4-velocity and the light's wave vector. Since the light's wave vector is parallel transported along the light's null geodesic worldline, it doesn't change from emitter to receiver; so if the received frequency is different from the emitted frequency, it means the receiver's 4-velocity is different from the emitter's 4-velocity.

The standard red-shift equation

Is derived from the 4-velocities of comoving observers, which in standard FRW coordinates are given by the coordinate vector field $\partial / \partial t$.

 

[vanhees71]

The four-velocities of the usually assumed observers are $(1,0,0,0)$ in the usual coordinates. As you say, it's $\partial_t$. It's just at different places, and the light has to travel from the one to the other, thereby undergoing the redshift due to the time-dependent scale factor, $1+z=a(t_{\text{obs}})/a(t_{\text{em}})$.

The derivation in the "naive photon picture" is simple. For radial light-like geodesics you have ($c=1$)
$$\mathrm{d} t^2-a^2(t) \mathrm{d} \chi^2=0$$
From this you have
$$\chi_{\text{obs}}-\chi_{\text{em}}=\int_{t_{\text{em}}}^{t_{\text{obs}}} \mathrm{d} t'/a(t').$$
Now let $T_{\text{em}}$ and $T_{\text{obs}}$ be the periods of the light at the emission and observation point. Then
$$\chi_{\text{obs}}-\chi_{\text{em}}=\int_{t_{\text{em}}+T_{\text{em}}}^{t_{\text{obs}}+T_{\text{obs}}} \mathrm{d}t'/a(t').$$
Subtracting both equations gives
$$0=\int_{t_{\text{obs}}}^{t_{\text{obs}}+T_{\text{obs}}} \mathrm{d} t'/a(t')-\int_{t_{\text{em}}}^{t_{\text{em}}+T_{\text{em}}} \mathrm{d} t'/a(t') \simeq T_{\text{obs}}/a_{\text{obs}} - T_{\text{em}}/a_{\text{em}},$$
and thus
$$\omega_{\text{obs}} a_{\text{obs}}=\omega_{\text{em}} a_{\text{em}} \Rightarrow \; \omega_{\text{obs}} = \omega_{\text{em}} \frac{a_{\text{em}}}{a_{\text{obs}}}.$$
In terms of wave lenghts
$$\lambda_{\text{obs}}=\lambda_{\text{em}} \frac{a_{\text{obs}}}{a_{\text{em}}}.$$
Both observers are assumed to be at rest, i.e., at $\chi_{\text{obs}}$ and $\chi_{\text{em}}$ during the entire time the light needs to travel.

 

[timmdeeg]

From a far-distant galaxy that is comoving (and assuming we ourselves are comoving), if we observe the light, yes. But that effect is due to the difference in 4-velocities of the emitter and the receiver; it is not due to anything that "happens" to the light during transit, at least not in any invariant sense, since, as I noted, the light's wave vector is parallel transported along its null geodesic worldline, and "parallel transport" means "unchanged".

Isn't it the point that the question what "happens" to light during transit is in the same sense meaningless as the question to the properties of a particle before measurement?

 

[timmdeeg]

Classically, this means that the photons measured frequency would depend where your measuring instrument is located. The center of the box will have a different expected frequency than the ends.

The effect will be minor for reasonably sized boxes though, we can calculate/estimae the magnitude of this shift knowing the size of the box. I haven't run any numbers, but I feel confident in saying that it can be considerd small. More to the point, we know that the effect must vanish in the limit as the box length becomes small, so that in the limit as the length approaches zero, there is no change in photon frequency with time. For a large box, it's less clear, and it depends on where we monitor the photon frequency. I suspect that the photon frequency measured in the center of the box would remain constant, while the photon frequency measured near the ends would vary slightly with time as the tidal forces, due to the changing curvature, change with time. Probably a more carfeful analysis would need to be done to confirm this,though.

Let's say the photon travels from one side of the box to the other side. During it's traveling measurements would reveal decreasing frequency corresponding to the increasing scale factor. Why do you distinguish the "center" from "near the ends"?

Do you say the tidal forces in curved spacetime offer another possibility to describe the frequency shift of the traveling photon?

 

[PAllen]

Let's say the photon travels from one side of the box to the other side. During it's traveling measurements would reveal decreasing frequency corresponding to the increasing scale factor. Why do you distinguish the "center" from "near the ends"?

Do you say the tidal forces in curved spacetime offer another possibility to describe the frequency shift of the traveling photon?

Any light, anywhere, in SR or GR only has a frequency only in relation to a particular measuring device or observer. Any light can be measured at any frequency. What distinguishes two light pulses at a given location is which observers measure which frequency.

One statement, which is rigorously true for any assumptions about the box, is that an observer whose 4 velocity is given by parallel transport of the 4velocity of the box side of last reflection along the light path (world line) after reflection, will measure the same frequency as that box side would measure for the reflected light.

This is true wether the box is considered bound or with free floating walls, and is actually independent of the spacetime. It is true for an arbitrary GR solution in all cases. Thus any change in measured frequency along the light path may be considered as due to nothing more than a state of motion of the observer being different from the the parallel transport of the emitter 4 velocity along the light path.

 

[pervect]

Let's say the photon travels from one side of the box to the other side. During it's traveling measurements would reveal decreasing frequency corresponding to the increasing scale factor. Why do you distinguish the "center" from "near the ends"?

Do you say the tidal forces in curved spacetime offer another possibility to describe the frequency shift of the traveling photon?

I do have to agree that if the tidal forces were changing rapidly enough, one would get a very very small drift in what we're calling the "photon frequency" (though it's a classical calculation).

 

[PeterDonis]

During it's traveling measurements by comoving observers, who will not be at rest relative to the box (they wil be moving outward towards the box walls), would reveal decreasing frequency corresponding to the increasing scale factor.

See my bolded insert above; it is the critical piece that you left out, and which, when left out, makes your claim false instead of true.

 

[PAllen]

See my bolded insert above; it is the critical piece that you left out, and which, when left out, makes your claim false instead of true.

I think I am missing something and coming to an odd conclusion. Prelude: we pretend that the the box and any instruments have no contributions to the stress energy tensor and that the perfect fluid assumed universally present in an FLRW solution (except for the Milne special case) is present throughout the box, and thus the total geometry is the FLRW geometry.

Assume the center of the box follows a comoving world line. Assume the walls of the box maintain fixed Fermi-Normal distance from the central world line (formalization of the box being a bound system). Now, the congruence of comoving observers includes the central world line, but otherwise moves isotropically away from the central world line towards the sides of the box. Then, when a signal is reflected by the box wall, at some frequency per the box wall, wouldn't it be seen as blue shifted by comoving observers before reaching the box center, then not shifted, then red shifted? (Symmetry argument suggests this would be true irrespective of curvature: how could center choose which wall not to match in an isotropic geometry?)

[edit: Now I think there is nothing odd about this, it is simply true. The comoving congruence will see decreasing frequency over the traversal, it just starts out blue shifted relative to the wall observer]

 

[PeterDonis]

when a signal is reflected by the box wall, at some frequency per the box wall, wouldn't it be seen as blue shifted by comoving observers before reaching the box center, then not shifted, then red shifted?

I think this is correct, yes. Right after the reflection, the frequency according to the box wall will be higher than the frequency according to a comoving observer co-located with the box wall at that event. Comoving observers inside the box observing the reflected signal as it traverses the box will see the shifts you describe relative to the frequency according to the box wall; but they will all see redshifts relative to the frequency according to the comoving observer co-located with the box wall at the reflection event.

However, that raises another question. Consider that comoving observer who is co-located with the box wall at the reflection event. According to that observer, the reflection event adds energy to the light pulse--i.e., the frequency the comoving observer observes just before the reflection event is lower than the frequency the comoving observer observers just after the reflection event. (For an observer moving with the box wall, these two frequencies are the same.) Where does that added energy come from?

 

[PAllen]

However, that raises another question. Consider that comoving observer who is co-located with the box wall at the reflection event. According to that observer, the reflection event adds energy to the light pulse--i.e., the frequency the comoving observer observes just before the reflection event is lower than the frequency the comoving observer observers just after the reflection event. (For an observer moving with the box wall, these two frequencies are the same.) Where does that added energy come from?

Well, to answer that we need to back off from too many idealizations. Consider first the congruence of Fermi observers centered on the box average center. Then, when the light hits one wall, it transfers a little momentum to the box. Similarly on the other side. Due to finite speed of sound, the box will be expanding and contracting with some frequency lower than than the traversal time of the light (and shifting a tiny bit around the center, but that would be a much smaller effect due to relative speed light and sound). Then, the box is exchanging energy with the light such that the light loses some energy on each reflection, then gains it at each reflection, at some resonant frequency of the box.

Now, express this using the comoving congruence, and there is no longer a contradiction. The comoving observers are seeing exchange of energy between the box and the light from a different perspective, but it exists for both congruences.

 

[timmdeeg]

See my bolded insert above; it is the critical piece that you left out, and which, when left out, makes your claim false instead of true.

Yes I left it out, I implicitly presupposed it because otherwise the statement would't make sense.
Thanks for this hint.

I have been also assuming that the side of the box where the light was emitted is comoving all the time. But I see now it's much more meaningful to consider it's center comoving as @PAllen does.

 

[timmdeeg]

However, that raises another question. Consider that comoving observer who is co-located with the box wall at the reflection event. According to that observer, the reflection event adds energy to the light pulse--i.e., the frequency the comoving observer observes just before the reflection event is lower than the frequency the comoving observer observers just after the reflection event. (For an observer moving with the box wall, these two frequencies are the same.) Where does that added energy come from?

Wouldn't the comoving observer close to box wall interpret the increased energy of the light pulse after the reflection event as being due to the peculiar velocity of the wall towards him? This relative velocity doesn't change during reflection of the light pulse and hence the wall doesn't lose kinetic energy according to this observer because we think the box to be ideally glued.
In other words the added energy isn't created, it's due to the Doppler-effect.
Is this way of thinking too naive?

 

[vanhees71]

See my bolded insert above; it is the critical piece that you left out, and which, when left out, makes your claim false instead of true.

In this case you have both the gravitational/cosmlogical redshift and a Doppler effect due to the motion of the observer relative to the CMBR frame. The frequency for a given observer is always given by $u \cdot k$, where $u$ is the observer's four-velocity (normalized such that $g_{\mu \nu} u^{\mu} u^{\nu}=1$ in the west-coast sign convention).

 

[PeterDonis]

In this case you have both the gravitational/cosmlogical redshift and a Doppler effect due to the motion of the observer relative to the CMBR frame.

This is true if you adopt standard FRW coordinates. But what you are calling "the CMBR frame" (by which you appear to mean standard FRW coordinates) is not what people usually think of as a "reference frame" since different comoving observers are not at rest relative to each other.

 

[mitochan]

Summary:: Does a photon in a box undergo cosmological red shift over time?

My question is this: suppose you put one or more photons into a box which has 100% perfectly reflecting walls. Will the photon(s) in the box experience a cosmological red shift over time or not? If so - why? and if not -why not?

Say the box has side lengths a,b and c. QM says energy states of photon are
E_{mnl}=\frac{\hbar^2}{4}\{\frac{m^2}{a^2}+\frac{n^2}{b^2} +\frac{l^2}{c^2}\}
This holds for adiabatic change of a,b and c. Whether the box undertakes change of perimeters depends on the conditions of box we set. In the case that the box is the whole universe or its equiportional homogeneous part, CMB suggests that it holds.

 

[vanhees71]

This is true if you adopt standard FRW coordinates. But what you are calling "the CMBR frame" (by which you appear to mean standard FRW coordinates) is not what people usually think of as a "reference frame" since different comoving observers are not at rest relative to each other.

I indeed mean standard FRW coordinates, and I consider an "observer at rest" wrt. the corresponding reference frame if there world line is given by $(\chi,\vartheta,\varphi)=\text{const}$. Such an observer sees the CMBR as a homogeneous isotropic Planck spectrum, and that's what's usually called the rest frame of a thermal bath of photons.

 

[PeterDonis]

I consider an "observer at rest" wrt. the corresponding reference frame if there world line is given by $(\chi,\vartheta,\varphi)=\text{const}$.

And this usage of "at rest" is different from the usual meaning of that term, since comoving observers "at rest" by this definition are not at rest relative to each other; they are moving apart. Which is why you can't just help yourself to this definition of "at rest" and not expect it to cause confusion. Particularly not in a discussion which is explicitly considering a box whose walls are at rest relative to each other (and which therefore cannot all be comoving).

 

[anuttarasammyak]

And this usage of "at rest" is different from the usual meaning of that term, since comoving observers "at rest" by this definition are not at rest relative to each other; they are moving apart.

They become apart but we may say they keep rest in the sense that each of them observe all the bodies in inertial motion around them keep losing speed, i.e. proper speed or momentum, i.e. proper momentum though they keep zero proper momentum and stay at rest at constant $(\chi,\theta,\phi)$.

As an illustration of difficulty of "moving", let us see the north and south poles of an inflating sphere shell. They become apart on the 2D shell but how can we define their relative motion on the shell?

 

[timmdeeg]

They become apart on the 2D shell but how can we define their relative motion on the shell?

Why not just say their relative motion is given by the increase of their proper distance per proper time unit?

 

[vanhees71]

Sure, one has to clearly say what one means with "at rest". The definition I'm used to from our standard cosmology lecture (and which I considered the standard meaning in cosmlogy) where usually "at rest relative to each other" means the world lines defined by constant spatial standard FLRW coordinates (i.e., $\chi$ or $r$, $\vartheta$, $\varphi$), i.e., the congruence of comoving "dust particles". It's clear that a "box" is bound by (dominantly em.) forces and thus the different parts of the box are not all co-moving.

 

[anuttarasammyak]

Why not just say their relative motion is given by the increase of their proper distance per proper time unit?

A distance, say $l_0$ increases to $l$ with time t is mentioned
l=\frac{a}{a_0} l_0
\frac{l-l_0}{t-t_0}=\frac{\frac{a}{a_0}-1}{t-t_0}l_0=[\dot{a}(t_0) + \frac{1}{2}\ddot{a}(t_0)(t-t_0)+...]\frac{l_0}{a_0}=V
I named it V and it has dimension of $LT^{-1}$ but I hesitate to call it "velocity" or "motion".

As for red shift well described in #77,
\omega a = \omega_0 a_0
\frac{\omega_0-\omega}{\omega_0}=1-\frac{a_0}{a}
In interpretation that same effect comes from Doppler effect in IFR the recessional velocity v is
\frac{\omega_0-\omega}{\omega_0}=\frac{v}{c}
Equating these two equatins
\frac{v}{c}=1-\frac{a_0}{a}=1-\frac{a_0}{a_0+\dot{a}(t_0) (t-t_0)+ \frac{1}{2}\ddot{a}(t_0)(t-t_0)^2+...}

I think v is not real velocity which takes place but a parameter in interpretation "as if" it is due to Doppler effect in IFR. As an example, in Doppler effect in IFR v is velocity of emitter wrt receiver at the time of emission $t=t_0$, however, I cannot read in the above equation for what time v is defined.

 

[timmdeeg]

I think v is not real velocity which takes place but a parameter in interpretation "as if" it is due to Doppler effect in IFR. As an example, Doppler effect in IFR, v is velocity of emitter at the time of emission $t=t_0$, however, I cannot read when v is defined in the above equation.

One can argue with the special relativistic Doppler formula, as Bunn & Hogg do in this paper, equation (6):

https://arxiv.org/pdf/0808.1081.pdf
Equation (6) can be derived by a straightforward calculation using the rules for parallel transport, but the derivation is easier if we recast the statement in more physical terms ...

 

[anuttarasammyak]

In the paper, the accumulation of many infinitesimal Doppler shifts, is a good alternative idea to understand the red shifts. Thank you.

 

[PeterDonis]

They become apart but we may say they keep rest in the sense that each of them observe all the bodies in inertial motion around them keep losing speed, i.e. proper speed or momentum, i.e. proper momentum though they keep zero proper momentum and stay at rest at constant $(\chi,\theta,\phi)$.

I have no idea what you mean by this.

As an illustration of difficulty of "moving", let us see the north and south poles of an inflating sphere shell. They become apart on the 2D shell but how can we define their relative motion on the shell?

I have no idea what point you are trying to make here.

 

[PeterDonis]

The definition I'm used to from our standard cosmology lecture (and which I considered the standard meaning in cosmlogy) where usually "at rest relative to each other" means the world lines defined by constant spatial standard FLRW coordinates

No, that is not what at rest relative to each other (my emphasis) means, even in cosmology. That's why I used that specific term, with the specific qualifier I just bolded; because that specific term means, specifically, at rest in the ordinary lay person's sense--the proper distance between the objects stays constant. If we wanted an explicit physical test for this, we would have the two objects exchange repeated round-trip light signals; they are at rest relative to each other if and only if the round-trip light travel time stays constant.

The obvious conflict between "at rest relative to each other" (which, as I just noted, matches the ordinary lay person's understanding of what "at rest" means) and "at rest in FRW coordinates" is why I do not think using the term "at rest" to mean "at rest in FRW coordinates" is a good idea. Particularly, as I said, in this thread, where the discussion is explicitly focused on a case where not all objects are comoving. The walls of the box are at rest relative to each other, even though they are not at rest in FRW coordinates.