# Light's redshift in a vacuum tube due to expansion

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?

• What does, "connected by a tube" mean? Are you imagining a rigid tube that is rigidly attached to each of the two bodies? Commented Jan 16, 2020 at 14:54
• Objects do not expand, gravitationally bound parts of the observable universe do not expand, the scale of space expands. Commented Jan 16, 2020 at 15:01
• Yes. rigid tube that is rigidly attached. Commented Jan 16, 2020 at 15:23
• Commented Jan 16, 2020 at 18:03

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.

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 $$a'=\dot aa$$.

$${}^1$$ 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)

• Why do i have the impression that space expansion as a "force" is very much weaker than even the gravitational force, and that is why we can measure expansion? If everything expanded we would have no way to measure it , no? The raisin bread analogy?. Another tack, would your calculations of redshift be larger that the heisenberg uncertainty for the energy of the photon? Commented Jan 16, 2020 at 19:18
• I think this is correct, have deleted my answer. There is a paper by Davis and Lineweaver about it. However, can't it be a blueshift? Commented Jan 16, 2020 at 19:44
• @annav: for simple order of magnitude estimates, you can probably just look at the $\Lambda c^2r/3$ term you get from the Einstein-de Sitter metric. At the surface of the earth relative to its center, that's $2\times10^{-29}m/s^2$ compared to $10m/s^2$ gravitational accceleration. It takes about 300ly for this effect to balance the sun's gravity, which I found a surprisingly 'short' distance compared to galactic scales (though this is really more of a testament to the weakness of gravity). Note this might not be the right way to think about things if you consider more realistic cosmologies Commented Jan 16, 2020 at 20:02
• @RobJeffries: It might be a blueshift: The Doppler contribution depends on spacetime dynamics via $H=\dot a/a$, and I don't see a reason why it couldn't become dominant. In fact, guesstimating things from some graphs, it does happen in our universe. Commented Jan 16, 2020 at 20:19
• Your answer is not correct. There is no redshift at all at scales smaller than galaxy clusters, because matter is clumped at those scale. Only a homogeneous distribution of matter and energy gives rise to the FLRW metric, which manifests as an expanding space. You cannot apply the FLRW metric to scales smaller than gravitationally bound systems. Commented Jan 17, 2020 at 20:00

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.

• it is just that the effect is hardly measurable in places where the other forces dominate - it's not just that: in swiss-cheese cosmologies, the metric in the holes can look rather different from FLRW; I would expect some indirect effect of the cosmological constant to be present via matching the boundaries, but I haven't read up on this... Commented Jan 16, 2020 at 19:02
• @Christoph correct thank you. Commented Jan 16, 2020 at 19:09