# Observing supraluminally receding galaxies – is the “swimmer against variable current” analogy valid? [closed]

Davis and Lineweaver in Superluminal Recession Velocities state that “galaxies with distances greater than $$D=c/H$$ are receding from us with velocities greater than the speed of light and superluminal recession is a fundamental part of the general relativistic description of the expanding universe.” So how can light from superluminally moving galaxies be detectable here on Earth? Or, in Davis and Lineweaver's words, “How can a swimmer make headway against a current that is faster than she can swim?” At the end of their paper they say: “The swimming analogy fails because, unlike recession velocity which is smaller at smaller comoving distances, the current the swimmer has to face is the same at all comoving distances. Our swimmer has to battle an unrelenting current, while the photon constantly moves into regions with a slower “current” (slower $$v_{recessional}$$)”.

My question is in two parts: (a) is this second “swimming against a current” analogy valid as a means of explaining how we can observer supraluminally receding galaxies? And (b) have I successfully modelled this analogy?

My attempt to model this second swimming scenario (where the current velocity is – like recessional velocity – proportional to distance $$x$$ from origin) is to say $$v_{recessional}=H*x,$$ where $$H$$ is a constant. If the swimmer's velocity is $$c$$, her time to swim to the origin is then given by$$\frac{dx}{dt}=Hx-c$$

$$dt=\int\frac{1}{\left(Hx-c\right)}dx,$$which integrates to $$t=\frac{\lg\left(Hx-c\right)}{H}+K.$$But that doesn't make sense as if I set $$v_{recessional}=Hx=c$$, she will never reach us as $$\ln\left(0\right)$$ is not defined. So this “improved” swimming analogy fails as well. To repeat, is the swimmer analogy wrong or is my maths wrong?

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• The units of the first equation does not match. Also $v_{rec}=x$ ? I guess you forget to add $H$ – Reign Jan 17 at 17:22
• I've now included a constant of proportionality $H$. – Peter4075 Jan 17 at 18:00
• I didnt quite understand that what is your purpose here. To derive an equation which describes rhe swimming analogy ? – Reign Jan 17 at 18:25
• I'm asking a couple of things. First, is the tweaked swimmer analogy valid for understanding superluminal recession velocities - tweaked in the sense that I've attempted to assume not a fixed current velocity but one proportional to distance. Second, if it is a valid analogy, where have I gone wrong trying to model it. – Peter4075 Jan 17 at 18:51

## 1 Answer

Your equation is correct actually, but I think the problem is the Hubble constant is not actually a constant but changes with time. So,

As you said,

$$v_{net}=\dot {r}_p=Hr_p-c$$ When an object just at the Hubble radius emits a photon, it seems like the object will never reach us, however the missing point is $$H$$ is a function of $$t$$ so at next instant, $$H$$ will decrease (due to the expansion of the universe) and the photon can enter a region where $$v_{rec} Hence we will be able to see it.

• Afraid I don't understand your final "swimmer" derivation. Where did the $dt$s come from? Is $c$ the swimmer's velocity? And what does $dx=x+dx-x$ mean? – Peter4075 Jan 18 at 15:19
• I edited a bit, thered $dt$ cause $dx=vdt$, but again the same problem in this equation ($dx=vdt$) we are assuimg that $v$ is constant. Thats why its again "wrong" – Reign Jan 18 at 17:45
• Maybe, we can take some average velocity and calculate the "swimming" idea like that. – Reign Jan 18 at 17:46
• If its still unclear please write – Reign Jan 18 at 17:49
• I suspect the "improved" swimming analogy (speed of the current pushing back against the swimmer is proportional to distance from origin) is wrong as well. If the swimmer's velocity is even momentarily the same as the opposing current's velocity, she's just not going to make any headway. My groundbreaking discovery is that a swimmer is different to a photon! I await my Nobel prize. – Peter4075 Jan 18 at 18:58