Existence

Why does it have to be altered? The pen laser is directly above the second S when the light pulse strikes it.What you are failing to understand is that the point of emission is different in each coordinate system as defined by the inertial frames. In NoNuke's frame standing by the tracks the pen laser is moving. In the driver's frame of reference the pen laser is not moving. Therefore, in the driver's frame of reference the position of the pen laser and the point of emission are one in the same when the light pulse hits the second S. This is not so for NoNuke's frame of reference standing by the tracks. What the two frames of reference will agree on is the speed of light and that the light strikes the second S.If you can, please draw a diagram for the driver's frame of reference. Hopefully this will allow you to understand the differences between the frames of reference. That's just it, you have to define what vertical is and in relation to what, or else it doesn't make sense.Think of two guys throwing a ball back and forth to one another. Have them run sideways at the same speed and keep the same distance between them. IOW, have them run parallel paths. From the perspective of the guys throwing the ball they are throwing at a 90 degree angle with respect to the motion of the ground below them. From the perspective of someone in a hot air balloon above them the ball is being thrown at a less than 90 degree angle with respect to the guys motion with respect to the ground. This is exactly what is going on in our light experiments.

In the driver's inertial frame, the pen laser is not moving. How many times have I mentioned this? Why do you keep making this mistake? It doesn't matter what the angle is, it is still a straight line. You do understand that, don't you? You know, a straight line that connects the point of emission and the point of detection.Also, you need to specify which frame of reference that you are measuring the angle in. For this exercise, please tell me the angle measurement for the driver's frame of reference if the light strikes the second S. That is false. In the driver's frame of reference the car and pen laser are not moving, but the salt lake flats are moving at 0.5 c. Then why are you invoking an aether that is attached to the salt lake flats? It is also observed in the driver's frame of reference where the car and pen laser are motionless and the tracks, sensors, and detectors are moving at 0.5 c.If you can, please draw a diagram where the car and pen laser are motionless and the tracks, sensors, and detectors are moving. Is that because the vacuum tube was blocking the aether wind?

The pen laser is not moving in the driver's frame of reference. How many times have I told you this now? No, the detector is not at rest in the driver's frame of reference. It is moving. The pen laser is stationary. It would only hit the detector if the speed of light was instantaneous. It isn't. It travels at 3E8 m/s. By the time the light transits the space between the tip of the pen laser to the tracks the second S is right below the pen laser. The second S moves into position to be hit by the stationary pen laser. The light is in every possible frame of reference the entire time, including the driver's frame of reference. Frames of reference have no boundaries. They extend to infinity. Pedantic much? Let's place the driver directly above the pen laser. Please draw a diagram for this observer. The car is not moving in the driver's frame of reference. How many times have I mentioned this? Yes, there is a place where the ball is travelling at 90 degrees. In the boys' reference frame. All they need to do is throw the ball directly at each other and they will catch it. The ball will be directly between them for the entire flight of the ball. It is no different than throwing the ball to each other if they were stationary to the ground. The observer in the hot air balloon would observe the same sawtooth pattern that we have described for the light clock, would he not?

Found a really nice animation that ICAN'T should check out:http://galileoandeinstein.physics.virginia.edu/...tclock.swfAnd especially this one:http://einstein.byu.edu/~masong/htmstuff/Clock2.htmlEdited by Taq, : No reason given.

ICAN'T,If you are having trouble imagining what the ball looks like as it is tossed between the two boys, plug in 0.5 into the animation at the top of this page:http://webphysics.davidson.edu/...ativity/illustration4.html

D and S are further away than 1 attometer, so it wouldn't strike either of them. However, the laser pulse would be directly below the pen laser in both frames of reference, and it will be there for the entire flight of the photon. I linked several websites in my posts above that have nice animations showing just this. Also, this is exactly what Michelson and Morley saw in their experiment. This phenomenon has been directly observed in experiments.

Perhaps I misunderstood what ICAN'T was trying to say. When the light pulse is one attometer away from the end of the pen laser it is still directly below the pen laser, as it is the entire transit from the pen laser to the spot where it strikes the tracks.A light beam is nothing more than a collection of singular photons, so his description of a one attometer long pulse really doesn't make sense anyway. It would be much more productive to look at a single photon, wouldn't you agree?

The car is not travelling in the driver's frame of reference. It is the sensors and detectors that are travelling.Why do you keep making this same mistake over and over? The photons from the laser are directly between the pen laser and tracks the entire time. Take a look at this animation:http://einstein.byu.edu/~masong/htmstuff/Clock2.htmlThis is what happens in the real world.In your drawing, you will see that the pen laser is directly above the S when it strikes. This is completely consistent with what we have been telling you. The photon stays between the pen laser and the tracks the entire time.And I see that you still can not draw the same diagram from the driver's frame of reference. Why is that? What are you afraid of?

Since ICAN'T can not seem to draw the diagram from the reference frame of the driver of the car it seems that I will have to do it for him. As before, the sensor (S) will send a signal to pen laser (P) at c. The pen laser is aimed at an angle of 90 degrees with the Salt flats. There is a detector (D) between each sensor (S). We will be using the reference frame of the driver. I will include the position everything when the pen laser emits the first photon (1), when the photon is half way to the track (2), and when the photon strikes the tracks (3).(1)

D   ----driver looking down towards track
                         P   ----pen laser is motionless
.
.
.
.                   S----D----S----D    {----movement of track

(2)

D
                          P
                          |
                          |
.
.
                   S----D----S----D  {----movement of track

(3)

D
                          P
                          |
                          |
                          |
                          |
                S----D----S----D   <-----movement track

That's what happens from the drivers frame of reference. Can ICAN'T show where I made an error?At every point the pen laser is firing straight down from the driver's frame of reference, just as it should be since the pen laser is not moving in the driver's frame of reference. Only the tracks, sensors, and detectors are moving in the driver's frame of reference as shown in the diagram. At every moment the photons from the pen laser are directly between the driver/pen laser and the tracks. It does not travel at an angle in the driver's frame of reference.Edited by Taq, : No reason given.

Take a look at the username. ICAN'T . . . (add sentence here).

Everything. The observer is the one who plots the point of emission with the point of detection within their frame of reference. The distance between those two points determines how much time it takes for the light to move from the point of emission to the point of detection within that frame of reference.

Because that's how fast they are moving.If you are going down the freeway at 65 mph how fast do you see the road signs going by your window? 65 mph.All frames of reference are equal. The driver reporting the speed of the track as 0.5c is as accurate as someone standing by the tracks reporting 0.5c for the speed of the car.

Let's have some fun with ICAN'T physics.ICAN'T believes that the photon will strike the detector in our now infamous "car on the tracks" experiment. Why? Because the laser pen is aimed at the detector when the photon is released. The photon then enters the inertial frame of the detector causing it to strike the detector. So let's follow this a little further just to see how whacky the universe would act if ICAN'T is right.Let's launch a spacecraft and send it out to a distance of one light year and take up a stationary position with respect to another spacecraft here in the solar system. Once the spacecraft is in position we send off a radio transmission telling the captain to speed off in a random direction in 10 minutes. We follow this message with a laser beam [3 months later] aimed right at the spacecraft. According to ICAN'T, the photon released by the pen laser should follow the spacecraft wherever it goes, like a heat seeking missile following a jet fighter. If our aim is just a little bit off then the photon takes a nice straight path.What a strange world it would be if ICAN'T was actually right.Edited by Taq, : bracketed portion added in edit

Actually, we will need some clarification from ICAN'T on this one. You may only need to crack one of the windows to let in the Aether wind. Or perhaps the very existence of windows will allow the Salt Flats to see the experiment under way and adjust the results accordingly.

The car is not moving. The tracks are. Do I need to draw the diagram again? Yep, just as I showed in message 1092. When it travels in a straight line it strikes the second sensor. In my diagram in message 1092 the photon is travelling at c, and the emitting body is not moving. The tracks are. Look at message 1092. The photon travels at a 90 degree angle in the driver's inertial frame, and it strikes the second sensor. How so? You have the light travelling instantaneously. The photon I used in message 1092 is travelling at c. The only way that photon strikes the detector is if it travels instantaneously. Both the 26.57 and 90 degree angles would be straight lines. You do know how geometry works, don' t you?