
042806, 12:47 AM #1
 Posts
 1,613
A light speed gedanken
Foreword: there is no intention here to present a trick question or anything of the sort. This is intended to be an analysis of photon motion on a path perpendicular to the motion of its source on theoretical physics grounds.
This gedanken thought science experiment consists of a straight and practically rigid pipe .99 meter in diameter and 9,999 meters long, coated inside with material that absorbs any photon and capped on one end with a photon detector and counter. On the other end is a Xaser ( a hard Xray laser ) adjusted to fire single photons. During manufacture the Xaser is aimed dead center on the end cap. The pipe is tightly packed full of pure vacuum. The entire assembly is painted with special pigment that completely absorbs any external photon or any substance that could create a photon inside.
The assembly is mounted on an Enterprise class starship with its major axis perfectly perpendicular to the fore and aft axis of the starship. It looks like the starship has a long skinny pipe for a wing.
The science experiment modus operandi is for the starship to execute a high speed fly by, on a straight path, past the Earth at a closest approach of 99,999 kilometers while stationary obversation station area observers on the ground observe it. The starship will fly on impulse only but will activate the forward asteroid deflector shield to ward off any and all disturbance from oncoming photons, cosmic rays, or anything else. It will build up speed and make a long fly by at .9999c, according to the observers. The experiment has been rehersed repeatedly to obtain timing parameters and the team is confident that that programming of automatic controllers on the starship will give perfect coordination.
During the fly by, when the starship is nearest the Earth, the controllers will trigger the Xaser to fire one photon. To be doubly and unnecessarily repetetively redundant, the pipe is exactly perpendicular to the starship flight path.
When the starship lands at the stationary observer station area, the stationary observers will probably rush over to to it at an average, or drift, velocity of about 9.9 kilometers per hour and anxiously examine the photon detector and counter on the end cap.
How many photon strikes will the counter show for the result of this experiment and why?

042806, 03:55 AM #2
One, assuming everything works as planned.
Because the photon strikes the end cap.

042806, 09:01 AM #3
 Posts
 727
Yes, if you fire one photon then one photon strikes the detector. This is true in every frame.
Originally Posted by CANGAS

042806, 09:25 AM #4Originally Posted by Tom2

042806, 09:33 AM #5
 Posts
 727
Yes, that would work too. Another way to do it would be to abandon the Earth frame and try to find a frame in which the motion of the ship and the photon are at right angles. I was suggesting the fix that would be the least work.
But no matter how you slice it, there's only one photon hit.

042806, 09:41 AM #6Originally Posted by Tom2

042806, 01:41 PM #7
 Posts
 727
Originally Posted by Pete
u=[γv+u'+((γ1)/v<sup>2</sup>)(v<sup>.</sup>u')v]/[γ+(γ/c<sup>2</sup>)v<sup>.</sup>u')
Let u= velocity of the photon in the new frame (call it S).
Let u'= velocity of the photon in the ship frame (call it S').
Let v= velocity of S' relative to S.
Let the ship axis lie along the y' axis and the photon tube lie along the x' axis. Then u'=ci. The problem is to find v such that v is perpendicular to u, that is, such that v<sup>.</sup>u=0.
Take the dot product of both sides of the velocity addition formula with v, set the left side equal to zero, and plug in u'. You will get the following.
0=v<sup>.</sup>v > v<sup>2</sup>=0 > v=0
In other words the condition u<sup>.</sup>v=0 cannot be satisfied unless you are in the frame of the ship.Last edited by Tom2; 042806 at 04:58 PM.

042806, 01:57 PM #8
 Posts
 727
And just because I'm SUCH a sweetheart ...
Originally Posted by Pete
Let S= Earth frame and S'=ship frame.
Let u= velocity of photon with respect to Earth = ci
Let u'=velocity of photon with respect to the ship = c cos(θ )i+c sin(θ )j
Let v=velocity of the ship with respect to the Earth=vj
Plugging the above vectors into the velocity addition formula and equate coefficients. You will find the following.
xcomponent
(1): [c cos(θ )]/[γ (1+(v/c)sin(θ )]=c
ycomponent
(2): [v+c sin(θ )]/[1+(v/c)sin(θ )]=0
From (2) we have that sin(θ )=v/c. The pythagorean theorem tells us then that the cos(θ )=(+ or )1/γ. The restriction that u=+ci tells us that we need the (+) sign.Last edited by Tom2; 042806 at 02:14 PM.

042906, 12:17 AM #9
Something feels wrong. If we can set the pipe at an angle so that the photon moves at perpendicular to the ship in Earth's frame, then it seems that we should be able to find a frame in which the motion of the ship and the photon are perpendicular...
Because rather than moving the pipe, we could adjust the angle of the whole ship, right? But then, rather than adjusting the angle of the ship, why not adjust our frame?
I'll work through the numbers and see what I can figure.

042906, 01:26 AM #10
 Posts
 993
The problem is that earthbound observer and the shipboard observer will disagree on the angle of emision of the photon, even though they do agree on the fact that the pipe is perpendicular to the line of travel of the starship. As viewed by the earhbound observer, the emiter is moving, and will have a forward angle emission, even though it is not angled forward.

042906, 09:19 AM #11
 Posts
 1,723
Originally Posted by Pete
Dale

042906, 01:42 PM #12
 Posts
 727
Originally Posted by DaleSpam
But the axis of the ship is not necessarily related to the motion of the ship, so it can be that the "perpendicular" frame is the rest frame of the ship where the ship has no motion.

042906, 11:52 PM #13
 Posts
 1,613
Gentleman:
How plainly does it need to be stated that we have stationary observers on Eath who are keeping tabs on these goingson? Did the thread starter say anything about observers on the starship? It could be a ballistic lump of metal.
The pipe was not postulated to be on a swivel. Any reasonable person would have to assume that it was welded in place during manufacture. Is there a little bit of desperation being shown by someone who fears that there may be a connection here with something inimical to SR?
Can a photon source which is traveling at very near c, according to a stationary observer, emit a photon at a right angle to its motion, according to a stationary observer, which then moves at its usual velocity of c in its emitted direction and also at a velocity of .99c in the direction of its source's velocity, all according to the stationary observers?
If so, what is the photon's resultant velocity, according to a stationary observer?
If we have a right triangle, with one side .99c and the other side 1.oc, what does the hypoteneuse calculate to be? Less than c? Or FTL?Last edited by CANGAS; 043006 at 02:20 AM.

043006, 07:29 AM #14
 Posts
 1,723
Originally Posted by CANGAS
The rest of the discussion is about your mistaken setup. If "This is intended to be an analysis of photon motion on a path perpendicular to the motion of its source" then you did not accomplish your goal. We are discussing how you could have accomplished it. How is that desperation?
Dale

043006, 11:11 AM #15
 Posts
 727
Originally Posted by CANGAS
Did the thread starter say anything about observers on the starship? It could be a ballistic lump of metal.
Furthermore, you are the one who specified the problem from the point of view of the ship. The 9999 meter length, the 0.99 meter diameter, and the perpendicularity of the pipe and ship axis are all determined from the ship's frame.
The pipe was not postulated to be on a swivel. Any reasonable person would have to assume that it was welded in place during manufacture.
Is there a little bit of desperation being shown by someone who fears that there may be a connection here with something inimical to SR?
Can a photon source which is traveling at very near c, according to a stationary observer, emit a photon at a right angle to its motion, according to a stationary observer, which then moves at its usual velocity of c in its emitted direction and also at a velocity of .99c in the direction of its source's velocity, all according to the stationary observers?
If so, what is the photon's resultant velocity, according to a stationary observer?
u= (c/γ )i+vj
I'll leave it to you to verify that the magnitude of this velocity vector is in fact c.
If we have a right triangle, with one side .99c and the other side 1.oc, what does the hypoteneuse calculate to be? Less than c? Or FTL?
1.) SR assumes that the speed of light is invariant, and
2.) SR is deductively valid, and therefore internally inconsistent.
It should come as no surprise then that in any gedanken involving velocity addition that you get out what was put in: the speed of light postulate.

043006, 09:01 PM #16For any "mounting angle" you can always find a frame where the motion of the photon is perpendicular to the axis of the ship.Originally Posted by Tom2
The point is that if you can orient the ship so that the motion of the photon is perpendicular to the ship's motion, then you can also orient your reference frame so that the motion of the photon is perpendicular to the ship's motion.
If the velocity of the ship as measured from Earth is v=vi, then an observer in any frame whose velocity (again, as measured from Earth) is w=vi+wk (doesn't matter what w is) is going to say that the photon moves at right angles to the ship's axis. But he won't say that the photon moves at right angles to the ship's motion. That's because both the ship and the photon have a velocity component in the same direction.
Originally Posted by Tom2
v.u' + v² = 0
v<sub>x</sub>c = v²
Giving us two frames for any given velocity, and an infinite number of frames altogether (even without considering the z direction).
* Not that I'm doubting you (much ), I just haven't seen it before (so I thank you), and havent bothered verifying or deriving it.

043006, 10:04 PM #17
This experiment seems to lack any mirrors. The usual SR examples rely upon a reflection in order to complete one full cycle of the timeclock (symmetrically). Consider two halfmirrors at each end of the tube, and a required roundtrip for the beam of light, and then it would have to be postulated that the mirrors are parallel to each other in every frame.
This discussion about maintaining a 90 degree angle between lightray and lineofmotion is leading people to consider a tube which is tilted toward the rear of the moving spacecraft. In this case, the lightray would be at a 90 degree angle to the lineofmotion, but only in one direction of the light beam's travel. Had there been parallel mirror, the reflected beam would be at an angle in the 'other' direction which would be very different from perpendicluar.Last edited by Neddy Bate; 043006 at 10:13 PM.

043006, 10:30 PM #18
 Posts
 727
Originally Posted by Pete
The point is that if you can orient the ship so that the motion of the photon is perpendicular to the ship's motion, then you can also orient your reference frame so that the motion of the photon is perpendicular to the ship's motion.
Why place that restriction on the other observer's velocity? What if w = ui + wk (u not equal to v)?
One of us is making a mistake. Assuming your equation is true*, I get:
v.u' + v² = 0
v<sub>x</sub>c = v²
By the way, the equation is true. I'll refer you to the following online textbook: Modern Relativity  SR. The 3D boost found in Equations 1.1.3 (this is the inverse of Equations 11.19 in Jackson's Classical Electrodynamics, 2ed). Put the boost in differential form, then divide dr by dt. Then on the left side divide the top and bottom by dt'. Set u=dr/dt and set u'=dr'/dt'. I simplified the result by substituting in the definition β=v/c.Last edited by Tom2; 043006 at 11:05 PM.

050106, 04:37 AM #19
Thanks, Tom!
Well, I did look for that frame and I showed you what I found...
The dot product v.u' is equal to zero.
So, there is no frame with velocity parallel to the ship's axis in which the photon is moving at right angles to the ship's motion.
But, I think that there are two frames with velocity in the shipEarth (xy) plane in which the photon is moving at right angles to the ship's motion, and infinite frames with arbitrary zaxis velocity in which the photon is moving at right angles to the ship's motion.

050106, 04:38 AM #20
I don't think it needs to get any more complicated than it is, Neddy... at least not until we sort out with CANGAS what's happening. Maybe later.
Bookmarks