- #211
geistkiesel
- 540
- 1
Hurkyl said:I take it you're only considering the case when both photons came from the same direction, then? (I.E. you are not considering the case when the actual sequence of events goes something like "Pass A, see photon A, see photon B, pass B")
(P.S. I'm not faulting you for doing this; it is a fairly messy task to account properly for all the possibilities)
So the criterion for simultaneous emission, according to the moving observer, is that:
c * (y - z) = v (x - w)
Note that I think you have one of the sides reversed; if the moving observer passed A before B, then he should expect to get B's photon first if emission was simultaneous, making both sides positive. (And similarly if B was passed before A)
As an aside, this is a nifty equation, because it renders calculation of time dilation irrelevant; any sort of dilation of times cancels out!
So let's remember this criterion and go back into the stationary frame. We have an experimental setup consisting of two stationary lamps, A and B. We watch a moving observer pass by both lamps with velocity v, and then, by some means, we trigger A and B simultaneously.
Notice that, in particular, this corresponds to the previous setup; the moving observer will pass A, then pass B, then get B's photon, then get A's photon. Thus, let us compute the times these occur.
For simplicity, the stationary observer sets his clock to 0 when the moving observer passes A. Let's suppose A was at position 0, and B was at position p.
Then, the moving observer passes B at time p / v.
Let's trigger the lamps at time 2p/v. The moving observer is at position 2p when this happens...
Thus, it takes time p / (c - v) for the photon from A to reach him, and time 2p / (c - v) for the photon from A to reach him.
So we have the following: In our frame,
He passes A at time 0.
He passes B at time p/v.
He receives the photon from B at time 2p/v + p/(c-v)
He receives the photon from A at time 2p/v + 2p/(c-v)
Remembering from our aside that dilation is irrelevant for the criterion that the moving observer computes the emission to be simultaneous. Let's substitute the times into the equation:
c * (y - z) = v (x - w)
yielding
c * ((2p/v + 2p/(c-v)) - (2p/v + p/(c-v))) = v ((p/v) - 0)
simplifying...
c * p / (c-v) = p
cancel out p...
c / (c - v) = 1
et cetera...
c = c - v
v = 0
Since the observer is moving, this equation clearly cannot hold, thus we conclude the moving observer does not compute that A and B were activated simultaneously.
Your hypothetical to me was confusing. You had the frame moving, then the observer. Any way all that is your problem. I am standing by the hyopothetical that have the photons observed to be emitted simultaneously. When you tire of your trivial calculations and want to discuss the experimental result, just say so.
edit added: If your source of photons is the same as the hypothetical arrangement that /b/measured[/b] the photons being emitted simultaneously then what ever your conclusions are merely adds a few more centimeters to the depth of the grave hole that is being dug for SR theory. Every SR analysis conflictingwith the experuimental results proves my point why continue? Your analysis above is just another shovel of dirt thrown out by that eternal and tireless worker, Dellawsa Fizzicks.
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