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Zula110100100
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My question is, which photon is the one that first hits the square in the following example
If I had a large platform that was 2777.78 meters long and at one end I had a laser that sent photons of increasing frequency at 1e-10second intervals(on a clock at rest relative to the platform on such a scale that it starts with 0 and counts up by 1 intensity.
On the other end of the straight 2777.78 meter platform is a mirror in a 45 degree angle and then an arm holding another mirror perpendicular to the platform and directly over the other mirror 1 meter away and sending the photon back to the 45, actually the 2777.78 meters is from the tip of the laser to the point the light hits to mirror. Then the 1m is from where it hits on the first mirror to where it hits the 2nd mirror.
I then have an extremely thin square that is little thicker than a photon moving at a relative speed of 277.778 meters/second(chosen since it's easily obtainable speed) in such a position and direction that it moved parallel to the platform and crosses VERY closely to the 2nd mirror and starting time 0 when the first end is even with the laser, it would take 10 seconds(on a clock at rest relative to the platform or at rest relative to the square with the platform moving by it at 277.778m/s) for the sheet to cross in front of the beam going from mirror to mirror.
My question is, what would the intensity of the first photon not returned to the laser, as in blocked by the square?
According to the clock at rest relative to the platform it takes (2777.78 meters divided by 300000meters per second) 0.0092592667 seconds and the time to the 2nd mirror from the first mirror is (1 meter divided by 3000000 meters per second) 0.0000033333 seconds for a combined total of .009262600 seconds
So the photon of 0 intensity would reach the 2nd mirror at .009262600 seconds elapsed from time zero relative to the station. and it continues at .0000000001 intervals until it counts up to 9.9999999999 seconds at that point the photon of intensity 99999999999(10 seconds worth of 1e-10 seconds minus one 1e-10 seconds) minus 0009262600(time to travel to 2nd mirror in 1e-10 seconds) and find that the photon to hit the mirror at 9.9999999999 seconds has intensity 99907373999.
The next photon 99907374000 would hit the square, should be the only possible answer.
From the p.o.v. of the square it takes 10 seconds to get to the light and block a photon according to relativity the square feels at rest and the platform is going by at 277.778 m/s
At time 0 it passes the laser and it begins, a photon is sent with 0 intensity it travels at 300000m/s relative to the square since that is what it MUST do. At 300000 m/s every 1e-10 seconds the light travels .0000300000 but the mirror on the platform is also moving towards the square at 277.778 meters per seconds or .0000000278 meters per 1e-10 seconds, for a combined .0000300278 m/1e-10s(Faster than light speed but only their relative speeds relative to me, but since light adjusts relative to each other still the speed of light) it takes 2777.78 divided by .0000300278 m/1e-10s which is 00092506944 1e-10 second intervals, and not considering the extremely small change from the lights change in direction another .000033333 seconds from the mirror to mirror.So from 0 seconds on the platform, as I observe from the square.
I see a photon of 0 intensity leaving the laser it takes 0.0092540277 seconds to reach the 2nd mirror, so when I reach 10 seconds on the square's clock, and finally interrupt the beam, the photon there should be of intensity 100000000000 - 000092540277 which equals 99907459722... which is 85722 intensities off, kinda a lot really... now it's probably a problem with my math or a factor I am not counting for so...any one explain please?
I had posted on a thread that I think was dead...and I cannot see how to solve this problem so I figured it's own thread might get more replies
If I had a large platform that was 2777.78 meters long and at one end I had a laser that sent photons of increasing frequency at 1e-10second intervals(on a clock at rest relative to the platform on such a scale that it starts with 0 and counts up by 1 intensity.
On the other end of the straight 2777.78 meter platform is a mirror in a 45 degree angle and then an arm holding another mirror perpendicular to the platform and directly over the other mirror 1 meter away and sending the photon back to the 45, actually the 2777.78 meters is from the tip of the laser to the point the light hits to mirror. Then the 1m is from where it hits on the first mirror to where it hits the 2nd mirror.
I then have an extremely thin square that is little thicker than a photon moving at a relative speed of 277.778 meters/second(chosen since it's easily obtainable speed) in such a position and direction that it moved parallel to the platform and crosses VERY closely to the 2nd mirror and starting time 0 when the first end is even with the laser, it would take 10 seconds(on a clock at rest relative to the platform or at rest relative to the square with the platform moving by it at 277.778m/s) for the sheet to cross in front of the beam going from mirror to mirror.
My question is, what would the intensity of the first photon not returned to the laser, as in blocked by the square?
According to the clock at rest relative to the platform it takes (2777.78 meters divided by 300000meters per second) 0.0092592667 seconds and the time to the 2nd mirror from the first mirror is (1 meter divided by 3000000 meters per second) 0.0000033333 seconds for a combined total of .009262600 seconds
So the photon of 0 intensity would reach the 2nd mirror at .009262600 seconds elapsed from time zero relative to the station. and it continues at .0000000001 intervals until it counts up to 9.9999999999 seconds at that point the photon of intensity 99999999999(10 seconds worth of 1e-10 seconds minus one 1e-10 seconds) minus 0009262600(time to travel to 2nd mirror in 1e-10 seconds) and find that the photon to hit the mirror at 9.9999999999 seconds has intensity 99907373999.
The next photon 99907374000 would hit the square, should be the only possible answer.
From the p.o.v. of the square it takes 10 seconds to get to the light and block a photon according to relativity the square feels at rest and the platform is going by at 277.778 m/s
At time 0 it passes the laser and it begins, a photon is sent with 0 intensity it travels at 300000m/s relative to the square since that is what it MUST do. At 300000 m/s every 1e-10 seconds the light travels .0000300000 but the mirror on the platform is also moving towards the square at 277.778 meters per seconds or .0000000278 meters per 1e-10 seconds, for a combined .0000300278 m/1e-10s(Faster than light speed but only their relative speeds relative to me, but since light adjusts relative to each other still the speed of light) it takes 2777.78 divided by .0000300278 m/1e-10s which is 00092506944 1e-10 second intervals, and not considering the extremely small change from the lights change in direction another .000033333 seconds from the mirror to mirror.So from 0 seconds on the platform, as I observe from the square.
I see a photon of 0 intensity leaving the laser it takes 0.0092540277 seconds to reach the 2nd mirror, so when I reach 10 seconds on the square's clock, and finally interrupt the beam, the photon there should be of intensity 100000000000 - 000092540277 which equals 99907459722... which is 85722 intensities off, kinda a lot really... now it's probably a problem with my math or a factor I am not counting for so...any one explain please?
I had posted on a thread that I think was dead...and I cannot see how to solve this problem so I figured it's own thread might get more replies
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