Relativity of Simultaneity: The Spreading Stain

[wsellers]

This is another (hopefully not tiresome) question related to the relativity of simultaneity! Suppose you have the usual train situation, but with this addition: suspended from one end of the car to the other is a long roll of paper towels. At each end there is a vial of red dye. From an the observer-on-the-embankment's perspective, two lightning flashes strike simultaneously at the front and rear of the moving train, breaking the vials. A stain spreads along each end of the paper towel strip, towards the middle. Will the observer on the embankment predict that the stains, when they meet, will meet in the exact middle of the train car? If not, why not? If so, is he correct in this prediction?

 

[Dale]

Interesting. You would need a relativistic description of Brownian motion and diffusion. I don't know if anyone has formulated such a description.

However, the answer is clearly "no". If you consider the rest-frame of the paper towel the lightning flashes were not simultaneous and therefore the stain meets at some point other than the middle of the train. Any correct relativistic description of diffusion would need to replicate that result in the rest frame and agree with the conclusion under any boost.

 

[cesiumfrog]

I don't see any reason why you should want a relativistic description of diffusion here; just calculate the spread in the towel's rest frame then transform the result as desired.

 

[Jorrie]

Interesting. You would need a relativistic description of Brownian motion and diffusion. I don't know if anyone has formulated such a description.

To avoid this, one can consider two identical guns, one at each end of the car, firing bullets at each other at identical speeds relative to the train. As said, the bullets ('stains') will not meet in the middle of the car, because they are not fired simultaneously in the train's frame of reference. And the middle of the car remains the middle for all inertial observers.

 

[Ich]

The resolutuion lies in the relativistic addition of velocities. The stain is supposed to propagate with speed u relative to the towel, which itself is moving at speed v. This does not transform to v+-u, which would yield the wrong result, but to (v+-u)/(1+vu/c²).
If you transform that result (incorrectly) using x'=x-vt, you would conclude that the stain propagates backwards faster than forwards.

 

[Jorrie]

Relativity of simultaneity needed

The resolutuion lies in the relativistic addition of velocities. The stain is supposed to propagate with speed u relative to the towel, which itself is moving at speed v. This does not transform to v+-u, which would yield the wrong result, but to (v+-u)/(1+vu/c²).

But what does transforming the stain velocity to the static observer's frame tell you? You don't need that to answer the OP's question - just plain old relativity of simultaneity. In the train (paper towel) inertial frame, the lightning strikes $vd/c^2$ seconds earlier at the front than at the rear, where $d$ is the distance between the two strikes in the [edit]train's frame.

 

[wsellers]

But what does transforming the stain velocity to the static observer's frame tell you? You don't need that to answer the OP's question - just plain old relativity of simultaneity. In the train (paper towel) inertial frame, the lightning strikes $vd/c^2$ seconds earlier at the front than at the rear, where $d$ is the distance between the two strikes in the static observer's frame.

Isn't this overstating things a bit? The observer in the middle of the moving car would have earlier seen the lightning flashes arrive at their eyeballs separated by a certain time delta [which, incidentally, I thought would have been 2dv/(c^2 - v^2) and not vd/c^2]. From this they would have concluded EITHER the lightning strikes were not simultaneous (in their frame of reference), OR that the frame of reference must be moving at v.

Similarly, when one stain (presumably) arrives at the middle of the paper towel a certain time delta before the other one [which can be calculated using the formula for the addition of relativistic velocities], the observer can conclude the same thing.

And presumably the value for "v" in both cases would be the same (but of course there would still not be enough evidence for the observer to know if the frame of reference was moving).

 

[Jorrie]

The observer in the middle of the moving car would have earlier seen the lightning flashes arrive at their eyeballs separated by a certain time delta [which, incidentally, I thought would have been 2dv/(c^2 - v^2) and not vd/c^2].

I do not know where you get your value from, but the standard synchronization offset between two frames in relative motion is vd/c^2. This can easily be established by drawing a Minkowski spacetime diagram and doing the relatively simple trig. [Edit: also straightforward from the Lorentz transformations.]

From this they would have concluded EITHER the lightning strikes were not simultaneous (in their frame of reference), OR that the frame of reference must be moving at v.

I have a problem with the second conclusion: "OR that the frame of reference must be moving at v." Moving relative to what? By simply observing two events one cannot say that you are moving... OK, maybe we are just misunderstanding each other. I thought that your 'speeding train' puzzle was designed to hammer home the principles of the relativity of simultaneity. Was there perhaps another educational aim?

 

[Ich]

OR that the frame of reference must be moving at v.

No, that he is moving with speed v relative to a frame where the strikes are simultaneous. There is nothing special about this frame that would justify the phrase "moving at v" without reference.

 

[Dale]

The resolutuion lies in the relativistic addition of velocities.

Doh! Of course. That is much simpler than trying to derive some relativistic diffusion law. But now that you mention it that is all that relativistic diffusion would be.

 

[wsellers]

I do not know where you get your value from, but the standard synchronization offset between two frames in relative motion is vd/c^2. This can easily be established by drawing a Minkowski spacetime diagram and doing the relatively simple trig. [Edit: also straightforward from the Lorentz transformations.]

Sorry, I must have mis-applied the analysis that Feynman gives in his lecture on SR (reprinted in Six Not So Easy Pieces, p. 56 of hardcover Perseus edition), where the time going in one direction is d/(c-v) and in the other it is d/(c+v).

 

[wsellers]

Belated Reply

...I have a problem with the second conclusion: "OR that the frame of reference must be moving at v." Moving relative to what? By simply observing two events one cannot say that you are moving...

Speaking just theoretically, could one say the following?

1. Consider a spherical impulse of light emitted from a point A at time t1, and the frame of reference that is "at rest" with respect to A. Wouldn't this frame of reference also be "at rest" with respect to ANY point at which ANY light source emanates? (I'm not talking about the light but the points.) If so, call this frame FR.

2. Consider another spherical impulse of light emitted from a point B at time t2 in FR. At some point, the edges of the two spheres will meet--call this point C. If C is equidistant from A and B, can't we conclude that the light was emitted from A and B simultaneously in FR?

 

[Jorrie]

1. Consider a spherical impulse of light emitted from a point A at time t1, and the frame of reference that is "at rest" with respect to A. Wouldn't this frame of reference also be "at rest" with respect to ANY point at which ANY light source emanates?

By definition, a reference frame is at rest relative to any point with fixed spatial coordinates in that frame, but you are possibly referring to event A and not 'point A', though. If so, a reference frame cannot be 'at rest' relative to an event.

2. Consider another spherical impulse of light emitted from a point B at time t2 in FR. At some point, the edges of the two spheres will meet--call this point C. If C is equidistant from A and B, can't we conclude that the light was emitted from A and B simultaneously in FR?

Yes, this can only happen if t1=t2 and given that C is equidistant from A and B in FR, not in some other frame.

If this does not answer your questions, please rephrase them more precisely.

 

[wsellers]

That does answer my questions--thank you very much for taking the time to respond.