Which photon hits?

1. Jan 10, 2010

Zula110100100

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 alot 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

Last edited: Jan 10, 2010
2. Jan 10, 2010

Matterwave

I didn't read the whole thing, but each photon doesn't have a different "intensity". Intensity is really the NUMBER of photons. You can't send out individual photons of increasing amplitude, only increasing frequency. To increase the amplitude of the light waves, you'd send out more photons...

3. Jan 10, 2010

Zula110100100

I see...well yeah, I was using the increase in amplitude to represent the idea of knowing for sure which photon hit, a way to tell one photon from another in an experimental way.

If frequency would work then great, anyway that I could say the photon that hit was the nth one sent. Even if there isn't enough range for 100000000000 different frequencies you could use less and loop as long as it's enough of them to tell which hit.

I guess since we're theorizing we know which one we are talking about and it seems it'd be a different photon hitting from each perspective...which should be impossible since the photon hits the same spot in the light clock example the photon should hit the same spot(either mirror or square)

4. Jan 10, 2010

yuiop

Your mistake is in the above quoted sentence.

The square takes 10 seconds as measured by a clock at rest with the platform, to travel 2777.78 m at 277.778 m/s.

It will take less than 10 seconds as measured by the clock of an observer at rest with the square (the p.o.v. of the square), due to time dilation.

The platform will measure less than 2777.78 m from the p.o.v. of the square, due to length contraction.

Time dilation and length contraction are an intrinsic part of relativity and of course it will not make any sense if you ignore them.

You are saying the speed of light relative to the square is 300,000 m/s and yet SR makes it clear that that the speed of light relative to any observer is 299,792,458 m/s. Even in a crude Newtonian sense, the speed of light is 299,792,180 m/s faster than the square in the platform frame, so your numbers simply do not seem to add up.

Last edited: Jan 10, 2010
5. Jan 11, 2010

Zula110100100

Yeah, I look dumb now, was thinking it was 300,000 not 300,000,000. didn't have it in front of me....ooops...but as it was still a constant the misuse of time dilation and lorentz contraction aside, it's results being different would be a point. just valid if I used the correct lightspeed and 277778 m/s speed for the platform.

6. Jan 11, 2010

Zula110100100

It seems to me that even not factoring in time dilation and length contraction, I tried to do it with just variables(mostly).

If you and I at rest with the platform and measured it to be D, and there is a lightclock at rest with the platform of such dimension that the photon hits one side at 0 it would it hit other side at .0000000001 seconds(still at rest with platform) then the other at .0000000002, and so on.

Every time it hits either mirror a photon is sent of with increasing frequency. We would agree that after running this set up for an arbitrary amount of time, we'll call it Ta, the photon that is at the mirror at the time Ta could be ascertained with

(D/(D/Ta)-d/c)/photons per second, because (D/(D/Ta) reducing to Ta..(put it like that to show how it related later) So Ta because it's how long the experiment was ran, Minus D/c the time it takes a photon to go distance D and the speed of light.

So basically the total elapsed time, so the (Ta/photons per second) gives you the last photon sent at the time the experiment stopped, but that one hasn't made it to the mirror yet so the one that has traveled to the mirror has traveled D and it's rate was c, so the time it took was D/c and was the (D/(D/ta)-D/c)/photons per second one and had such a frequency.

Then we try it in motion, we agree that I will go at such a rate that the relative rate is Rv = D/Ta. But, I have no clock or measuring devices whatsoever, and then I go past the platform, or rather we go by each other, at Rv and you start the experiment when you see my go past. After I reach then end and block the laser, I can assume that your clock has ticked Ta seconds and since I don't know how far I went I could deduce it by knowing that I went Ta seconds at Rv I went TaRv distance, which since Rv = D/Ta RaTv = D, so i would know I went D for Ta seconds(in your point of view) and would agree the experiment stopped at D/(D/Ta)(from the perspective of whats sending the photons).

So from what I understand events themselves are supposed to be simultaneous, which is a major factor in SR since the light clocks and time dilation is based on the two postulates that Events are the same regardless of the observers velocities and The speed of Light is constant. If the photon hits the mirror in the light clock a certain number of times for someone at rest with it, it does for anyone else in any other inertial frame, and the sheet passes in front of the beam simultaneously for all observes they just may experience different amounts of time passed? SO if at rest you see the photon bounce Ta/photons per seconds times it would be so for all observers, just at different "t"s and would accordingly fire the same number of photons down the platform, but when you calculate which photon hit from my point of view moving Rv relative to the platform, I would use the equation

fequency of photon = (D/(D/Ta) - D/(Rv+c))/photons per second

The first portion is the same, but for the time it took the light to get there, from my p.o.v the photon should be moving c relative to me. And the mirror would be moving Rv in the opposite direction, relative to me. SO the gap between the photon and the mirror would close at Rv+c.

and finally, it should be fairly obvious that if Rv != 0:
(D/(D/Ta)-D/c)/photons per second != (D/(D/Ta) - D/(Rv+c))/photons per second
D/(D/Ta)-D/c != D/(D/Ta)-D/(Rv+c)
-D/c != -D/(Rv+c)
1/c != 1/(Rv+c)
c != Rv+c

So if that is still wrong due to time dilation and length contraction could anyone show me WHERE and HOW the time dilation/length contraction would be implemented to fix it?