Light's redshift in a vacuum tube due to expansion

Question

If we have two bodies separated by 1 Mpc, but still connected with a tube then it's evident that these two bodies distance won't change because they are connected (If they're not connected their distance increase due to expansion of universe). if inside the tube is vacuum and a light ray is emitted inside the tube from one body to another; will that light be redshifted as a result of expansion of universe?

Answer 1

Yes, the light will be redshifted. However, assuming the tube is rigid${}^1$ - ie the proper distance between emitter and absorber is fixed - there'll also be additional Doppler (blue!-)shifting: Relative to the Hubble flow, the ends of the tube have a nonzero peculiar velocity.

Let's assume the tube has length $l$ and its center follows the Hubble flow. Then, the peculiar velocity of each end of the tube is given by

$$ \beta = \frac{Hl}{2c} $$

leading to a total redshift of

$$ 1+z = \frac a{a_0}\cdot\sqrt{\frac{1-\beta_0}{1+\beta_0}}\cdot\sqrt{\frac{1-\beta}{1+\beta}} $$ where the index $0$ denotes time of emission.

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For those unsure that this is the correct solution, here's the nitty gritty:

In terms of conformal time $\eta$ and comoving distance $\chi$ and ignoring the angular part, the FLRW metric reads $$ ds^2 = a^2(\eta)(d\eta^2 - d\chi^2) $$ That's conformally flat, so null geodesics will be manifestly straight lines.

Photon momentum will be given by $$ p = \epsilon(\eta)(\partial_\eta + \partial_\chi) $$ where $$ \epsilon\cdot a^2 = \text{const} $$ This follows from the parallel transport equation.

The emitting end of the tube follows the trajectory $$ \chi(\eta) = -\frac{l/2}{a(\eta)} $$ leading to a velocity $$ u_e = (\partial_\eta + \frac{la'}{2a^2} \partial_\chi)\big/\sqrt{a^2 - \left(\frac{la'}{2a}\right)^2} $$ where $a' = \partial a/\partial\eta$.

Similarly, for the absorbing end, we arrive at $$ u_a = (\partial_\eta - \frac{la'}{2a^2} \partial_\chi)\big/\sqrt{a^2 - \left(\frac{la'}{2a}\right)^2} $$

The redshift will be given by $$ 1+z = \frac{g(p_e,u_e)}{g(p_a,u_a)} $$ where $p_e$ and $p_a$ denote photon momentum at time of emission and absorption, respectively.

Plugging in the values is left as an exercise for the reader. It will reproduce the result I gave above, using $H=\dot a/a$ and $a'=\dot aa$.

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${}^1$ if the tube isn't perfectly rigid, metric expansion would deform the tube (remember, locally, gravitational effects including metric expansion will manifest as pseudo-forces)

Answer 2

The answer to your question is a big yes. Contrary to popular belief, space does expand everywhere because of dark energy (it is just that the effect is hardly measurable in places where the other forces dominate, and counteract dark energy). It is very confusing when you learn on this site that matter does not expand, because the other forces overcome space expansion. True, matter does not expand, but space does on all scales (the effect is different in different regions with different matter density).

For example, the solar system does expand due to cosmological expansion, but the effect is undetectably small. See Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," arxiv.org/abs/astro-ph/9803097v1 The strain on a bound system is proportional to (d/dt)(a¨/a), where a(t) is the cosmological scale factor. This quantity is not constant in realistic models, and can be nonzero even if the cosmological constant is zero. Also, it vanishes identically in a cosmology that consists only of dark energy (=cosmological constant).

Why does space expansion not expand matter?

Now to the that tube. Yes, there is tension on that tube.

So if you could tie two objects together with a string many light years long, to keep them at rest relative to each other, there really would be a tension in that string. That tension arises because you are forcing the objects to accelerate away from the geodesics they would otherwise follow, and it arises in the same way as the tension in the string if you suspend an object in Earth's gravity.

On the expansion of space on small distances

Now to that tricky photon inside the tube (in vacuum).

Yes, the photon would undergo comsological redshift. This redshift is because space itself expands inside the tube, outside the tube, everywhere.

It is only a question of to what extent the wavelength would be stretched, because the two objects are held by a rigid tube.

You are confused because the distance between the two objects is held tight by the tube. You would intuitively think this stops space from expanding between the two objects too.

In reality it does not. Nothing stops space expansion, not even on the smallest scale (true, though, that on the QM scale the effect is hardly detectable). It is only the matter of the tube that is withstanding the expansion (but there is tension on it). But this does not stop space from expanding between the two objects.

It is just that the objects relative distance is constant (near constant because the tube gets stretched a little bit), but space still expands inbetween. Tricky beast. You could say that the objects get even closer relative to the expanding space. Welcome to dark energy.