Suppose I have a hose pipe connected to a tap. I turn on the tap and water flows out of the other end of the pipe. Now imagine I install two valves at different points along the pipe. Water will flow normally if both valves are open, but not if either one of them is shut. No doubt you can now see where this is going. If the spacing of the valves along the pipe is sufficiently long, and the duration for which the valves are open is sufficiently short, we can always imagine a second reference frame in which at any given time at least one of the valves remains shut, so how can we account for the fact that water flows in that frame even when the valves are not both open? I imagine the resolution is somehow to do with the fact that the flow requires a pressure wave to propagate along the pipe when a valve is opened, but I haven't been able to find a convincing resolution. Has anyone come across an analogous problem before?
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1$\begingroup$ Well, real pipes are elastic and real water is slightly compressible. $\endgroup$– Jon CusterCommented Oct 17, 2023 at 12:17
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$\begingroup$ @JonCuster thanks, yes if you block a hose pipe, turn on the tap, turn off the tap and unblock it, water will flow because the pressure builds behind the blockage and then relaxes when the blockage is cleared. However, it is difficult to see how that could factor in to the explanation, since any stretching of the width of the pipe as a result of pressure build up would have to be the same in both frames. $\endgroup$– Professor SushingCommented Oct 17, 2023 at 13:01
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1$\begingroup$ I am struggling to work out what you hope to learn by cluttering up a rather dull problem with non-relativistic "bling", then labeling it as SR. $\endgroup$– m4r35n357Commented Oct 17, 2023 at 14:55
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$\begingroup$ @m4r35n357 what I'm trying to learn is how to explain why the water flows in a frame in which the two valves are not open simultaneously, given that if the two valves were not open simultaneously in the rest frame the water would not flow. How would you explain it? Incidentally, I've never hear of a hose pipe described as bling before! $\endgroup$– Professor SushingCommented Oct 17, 2023 at 15:01
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$\begingroup$ @m4r35n357 dictionary definition of bling: expensive, ostentatious clothing and jewellery. Ok, perhaps you were speaking figuratively, but why is a hose-pipe any more blingy than the usual train/platform scenario? Just asking, by the way, not arguing. $\endgroup$– Professor SushingCommented Oct 17, 2023 at 15:03
5 Answers
Even for the reference frame of the hose, the fact that water is passing through the tap B doesn't mean that the tap A is open at the same instant.
The 2 tap operators could arrange to open them at instant $t$ and close at $t + \Delta t$, so that $\Delta t$ is shorter even for a EM signal travel the distance AB. After a suitable time for the mechanical wave comes to B, the B operator opens the tap. The water pours in spite of A is already closed.
For another frame going from B to A, the conclusion is similar. Water flows through the tap B, while tap A is closed.
We can also imagine a long pipe with holes that can be opened or shut at A and B. If only one of them is open the pipe is dark when looking into the corresponding hole. If both are opened, each side see some light.
But if the holes are opened at the same time and shut quickly, one of them can open it later and see the light that travelled through the pipe. Again, the light can be seen without both holes are opened simultaneously.
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$\begingroup$ many thank- I should have guessed that an engineer would have some good insights here! However, I think the problem remains. Firstly I should be clear that I had assumed the entire length of the pipe was full of water, so that it was not the case that water could rush from A to B if B was closed. The set-up is similar to a simple electrical circuit with two switches wired in series, so that current flows only if they are both closed. If they are temporarily closed simultaneously current flows... $\endgroup$ Commented Oct 18, 2023 at 5:39
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1$\begingroup$ "we can always imagine a second reference frame in which at any given time at least one of the valves remains shut, so how can we account for the fact that water flows in that frame even when the valves are not both open?" a supposition with conclusion without demonstration? $\endgroup$ Commented Oct 18, 2023 at 7:26
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$\begingroup$ @TheTiler good point. Given the excellent feedback from you and Claudio, I realise there is too much ambiguity in my original version of the question, so at some point I will try to find the time to clarify it. $\endgroup$ Commented Oct 18, 2023 at 8:43
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$\begingroup$ @MarcoOcram There is another issue: If tap B is too high (more than $10.33$m) above tap A, some water flows through A even if B is closed, forming a vacuum behind. $\endgroup$ Commented Oct 18, 2023 at 12:59
Draw a spacetime diagram with one worldline for each small region of water. This should help you get clarity about the sequence of events and exactly what happens in one reference frame or another (if you know how to place lines of simultaneity correctly).
You will find that the water is not and cannot be incompressible. The solution is that the water can and does stretch in one frame and get compressed in another. When it stretches one end can flow out while nothing flows in the other. When it is compressed water can flow in at one end while nothing flows out the other.
Things like this happen whenever a body of non-zero length changes its state of motion.
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$\begingroup$ classic advice, thank you. I will do as you suggest and report back. I suppose my lingering doubts are related to the idea that we might attach pressure gauges at various points along the hose in its rest frame, and whatever happens in the frame of the moving observer cannot be inconsistent with the pressure distribution along the pipe in the rest frame. $\endgroup$ Commented Oct 18, 2023 at 9:52
You've implicitly taken the speed of sound in water to be infinite. Make it finite and there's no paradox.
Consider what valves in a pipe actually do. They toggle-ably separate regions at different pressure. Opening a valve allows pressure to propagate through the water column. For our purposes this is a signal. It has the speed of sound in water $c_w$.
Suppose our system is set up left to right going uphill, with a high pressure reservoir on the left, a valve, a pipe of proper length $L$, a valve, and an opening.
If opening the right hand valve doesn't make water spurt out, then the initial pressure on the left of the right hand valve is atmospheric pressure or less. If it is exactly atmospheric pressure, there is no change in pressure at the valve, so no signal is transmitted.
If less, the pressure increases to atmospheric pressure and a pressure wave propagates right to left at $c_w$, with a tiny amount of water briefly moving (making room for a concave meniscus at the right hand end of the pipe). If we open the left hand valve before the signal arrives, superposition with the larger pressure wave propagating left to right will make the right to left wave negligible.
If we open the left hand valve, a pressure wave propagates left to right. At time $L/c_w$ the pressure at the right hand side of the pipe has increased. If the right hand valve is open, water begins to flow.
So we have an event (left valve opens) which transmits a signal which causes a second event separated in time (water flows), subject to a discriminator (right valve is open?).
Alternately, we could have the initial conditions such that water spurts out as soon as we open the right hand valve, by having larger than atmospheric pressure in the pipe at the right hand end. Then we have a negative Delta in pressure propagating right to left. If the left hand valve is opened before $L/c_w$ in the pipe frame, water continues to flow, else it stops after a brief spurt.
There are no initial conditions in any frame in which the water begins to flow when and only when both valves are open, except of course in the limit as $v \to c$ in which $L' \to 0$. Flow is either immediate and local, or separated in time from the event on the left by the ratio of $L'$ to the relativistically concatenated velocity of the wavefront.
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1$\begingroup$ +1 I was also thinking about sound speed. Moreover, even a possible shock wave (typically faster than sound) would be slower than light. It is possible to show that every physically reasonable hydrodynamic model in relativity propagates information only inside the future light-cone envelope (i.e. the fluid respects the "causality" requirement, arxiv.org/abs/2105.14621 ). $\endgroup$– QuilloCommented Oct 18, 2023 at 17:59
Any apparent "paradox" in special relativity can be resolved by reducing it to individual space-time events and Lorentz transformations (example).
When approaching your "paradox" with the similar analysis, one immediately sees the problem in its formulation. "Water flows normally in the pipe" is not a space-time event - it doesn't have a specific location.
If you do the work of actually describing what different localized observers will observe at different points of your experimental setup - then the "paradox" will dissolve itself. (Curiously, that resolution usually happens even before one gets to Lorentz transformation and any relativity suff.)
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$\begingroup$ Agreed. I have concluded that my initial question is far too vague, so when I get a chance I will clarify it. I suspect that I am handicapped by not having a sufficiently clear understanding of the dynamics of fluid flow in a hosepipe , so I will need to clear that up first. $\endgroup$ Commented Oct 18, 2023 at 10:20
Suppose we have two stars A and B which explode into supernovas at the same time (~taps), all observers in motion or at rest closer to A see the explosion of A into supernovas before B which is still a star (~ flowing water) (the invariance of the speed of propagation of electromagnetic waves in all frames of reference) except the case where it is on the line which passes in the middle of the segment [AB] (symmetry).
