You're right. I jumped to a hasty conclusion when I set v<sup>.</sup>u'=0. It needn't be zero in general.
Hi CANGAS, If the photon continues at 0.99c in the direction of the ship's motion (which it does), as well as moving away from the ship (which is also does) then it is clearly moving at an angle, rather than perpendicular to the ship's motion... and it is at the angle that we would expect it to move at a speed of c. It's speed along the pipe in the Earth frame, on the other hand, would be well below c. The speed at which it moves along the pipe according to SR can be easily determined by time dilation. We know that in the ship frame it takes time T=9,999m/c to get from one end to the other, and because the two ends are at the same displacement in the direction of travel, we know that this time is divided by gamma to get the Earth-frame time. And since the length of the pipe is the same in the two frames, we can figure out that the speed of the photon along the pipe in the Earth frame is simply cdivided by gamma. So according to SR, we should expect one side of the triangle to be 0.99c (as you said), and the perpendicular side to be c/gamma = c.√(1-0.99² ) According to pythagorus, what does this make the hypotenuse? If SR is consistent, then it should be c.
I think we all missed the point of the opening post. CANGAS clearly meant for the tube to be perpendicular to the ship's axis of motion as designed and built (not as viewed from the earth frame during flight). CANGAS is testing whether the photon would be deflected to the side of the tube, thereby missing the detector altogether. I suppose that would happen if the test was run during a sufficently large acceleration, assuming the tube is sufficiently long and narrow. However, once inertial motion is acheived, the photon path should return to a straight line if the laws of physics hold true in all inertial frames. I believe it is the hypotenuse that is supposed to be defined as 1.0c, not one of the legs. Then the ratio of the hypotenuse to the undefined leg of the triangle is supposed to represent gamma. By edit: Whoops, issues already addressed by Pete.
Please don't include me in that "we all". Reread my post earlier in the thread. Please Register or Log in to view the hidden image!
Yes, in all fairness, you did have it right about the disagreement of the two different reference frames with respect to the angle between photon path and the spaceship path, and your reasoning made perfect sense. In fact it continues to make perfect sense regardless of whether CANGAS intended for the pipe to be mounted perpendicular to the direction of rocket-thrust, or at some other angle. I tried, but 'm afraid that I do not understand; perhaps you could give an example of another frame in which this agreement takes place. In my mind, the disagreement of the two different reference frames with respect to the angle between photon path and the spaceship path represents time dilation in the form of gamma. In a 3-dimensional cartesian coordinate system, it seems to me that the only way for the two frames to agree on this 3-dimensional angle is if they are co-moving (gamma=1) and therefore, essentiallially the same frame.
TomTom: You get a solid "F". Sherlock should be your nick. You found the attention getting gaffe, about as hard as finding a neon pink square hippopotamus standing across a corner of your living room. No credit for that. But then you failed to properly address the real issue, as I thought would be so, : the motion of a photon emitted perpendicular to its moving source. You don't really deserve an "F". But, there is no precedent for grading a "G".
Pete, if I had a shred of faith that you would honestly confront a science issue that I presented to you, it would be so. Let's see if you can do it. Tell us what a photon does, according to a stationary observer, when a moving object, going .99c according to the observer, emits a photon at a predetermined angle of perpendicular to the emitting object's path of motion. Now, gallant heckler, do we see you addressing the issue? Or picking out your next stupid question?
The stationary observer observes the photon to exude from the exit pupil of the Xaser. The photon is traveling at its normal velocity of c in its predetermined direction perpendicular to the direction of the starship motion. The starship is traveling at .99c, according to the stationary observer. The photon travels a short distance perpendicular to the starship motion when the pipe, travelling at .99c, runs into it from the side and absorbs it. As Special Relativists chronically reiterate, "Whatever happens in one frame happens in all the frames.". Therefore, since the photon is absorbed in the stationary observer frame, all the other photons in all the other frames were absorbed.
Please Register or Log in to view the hidden image! Allow me to repeat myself: One, assuming everything works as planned. Because the photon strikes the end cap. If the photon continues at 0.99c in the direction of the ship's motion (which it does), as well as moving away from the ship (which is also does) then it is clearly moving at an angle, rather than perpendicular to the ship's motion... and it is at the angle that we would expect it to move at a speed of c. It's speed along the pipe in the Earth frame, on the other hand, would be well below c. The speed at which it moves along the pipe according to SR can be easily determined by time dilation. We know that in the ship frame it takes time T=9,999m/c to get from one end to the other, and because the two ends are at the same displacement in the direction of travel, we know that this time is divided by gamma to get the Earth-frame time. And since the length of the pipe is the same in the two frames, we can figure out that the speed of the photon along the pipe in the Earth frame is simply c divided by gamma. So according to SR, we should expect one side of the triangle to be 0.99c (as you said), and the perpendicular side to be c/gamma = c.√(1-0.99² ) According to pythagorus, what does this make the hypotenuse? If SR is consistent, then it should be c.
No. The stationary observer sees the photon emitted at the appropriate leading angle such that it travels down the center of the tube and is absorbed by the detector. This is in accord with Maxwell's equations for a moving emitter. >>edit Pete beat me to it.
Pete, if you divide T=9,999m/c by gamma for the Earth frame, the Earth clock beats slower than the ship clock. You are working the gedankin from a ship observer's perspective, letting the ship observer predict what an Earth observer would 'see'. Now, how do you get the ship clock to beat slower than the Earth clock from Earth observer's frame of reference?
Thanks, 2inq - you're correct. I mistyped. Time in the ship frame is indeed multiplied by gamma to get time in the Earth frame. The remainder of the post is correct (the error was just in the transcription of that part) - the speed of the photon along the pipe in the Earth frame is divided by gamma, as I said.
Why wouldn't you divide the APPARENT path of the photon by gamma in the Earth's frame? (the angle) That is supposed to be what the Earth observer sees.
Yes, that would give you the total speed of the photon in the Earth frame, if you knew the length of the photon's path. You might like to do this yourself, to check that the photon's speed is c, as it should be according to SR. By "the speed of the photon along the pipe in the Earth frame" I mean the component of the photon's speed parallel to the pipe. It's the length in the Earth frame of the pipe divided by the time in the Earth frame it takes the photon to traverse the pipe. The photon will also have a speed component in the Earth frame of 0.99c in the direction perpendicular to the pipe, because that's how fast the pipe is going in that direction. You can get the total speed of the photon in the Earth frame from the two components using pythagorus's theorem, and the answer (no surprise) is c.
Yes, Pete, I know exactly what you have done. You explained the gedankin from the point of view of the spaceship observer. When you divide the length of the tube by gamma, that is what the spaceship observer predicts the Earth observer would see. The photon does not travel at 'c' in the Earth's frame when this is done. This is the 'deformed' moving frame discussed in the absolute time thread. For the photon to travel at 'c' in the 'resting' Earth frame, you must divide the APPARENT path of the photon by gamma.
I didn't divide the length of the tube by gamma, 2inq. The tube is perpendicular to the motion of the ship, so it's length is not contracted in the Earth frame.
I did say that, but it was a mistranscription of my notes. I actually multiplied the time by gamma, which (because the pipe length is constant) meant that the speed of the photon along the pipe must be divided by gamma.
(Emphasis added) We did address the issue directly. The whole point is that in your set up here the photon does not travel in a direction perpendicular to the ships motion in any frame. We even suggested ways that you could get the velocity vector to be perpendicular to the ships motion. You seem to think that velocity vectors are made out of some sort of perfectly rigid adamantium that you can stick onto the ship instead of simply being mathematical lines drawn between events. I think that it is pretty amusing for you to be complaining of the indirectness of others. Particularly when you are wrong and you are the one who consistently writes the most rambling indirect posts on this forum. (Except for those with language barriers) -Dale
