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Mar7-12, 05:23 AM
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Are you looking at the current English version of the article on Time Dilation? I cannot see much of what you say you are seeing.
Quote Quote by John232 View Post
Okay, it says the observer in motion observers the photon to travel a distance cΔt'.
I don't see any mention of the word "photon" in the article and I don't see any mention of the expression cΔt'. Tell me where you are getting this from.
Quote Quote by John232 View Post
I am saying that it doesn't. The observer at rest measures the hypotenus to be cΔt.
The observer at rest does not see a hypotenuse. Where are you getting this from?
Quote Quote by John232 View Post
The length of they hypotenus divided by the change in time would give the speed of light for the observer at rest and not the observer in motion.
The section starts off by saying that the speed of light is constant in all reference frames. It doesn't say anything about calculating the speed of light. Where are you getting this from?
Quote Quote by John232 View Post
The observer in motion would measure the vertical distance divided by time prime to be the speed of light, not the other way around. I think the error comes in when you say that the observer at rest see's the photon to travel a vertical distance, the observer in motion also see's this as well, but is being compared to what an observer at rest see's(the hypotenus).
You have to be looking at something different than what I'm seeing. None of what you are talking about is in the article that I'm looking at.
Quote Quote by John232 View Post
I am saying that the light clock doesn't measure ticks, but in fact that time is actually warped by the same amount as the relation between the sides of two light triangles. Where each side represents how the speed of light is measured from each frame of reference. So by putting cΔt' as the vertical distance it explains how time has to be adjusted to allow for the observer in motion to measure c even though he see's the photon travel a shorter distance. So then he has to experience less time to fill in this shorter side of the triangle. The observer at rest has to measure more time in order to account for the greater distance the photon is seen to travel. That is how it is possible to get the proper time out of the light clock example simply by assigning the time variables differently.

Also, Δt√(1-v^2/c^2) ≠ √(1-v^2/c^2)/Δt

v ≠ ΔL√(1-v^2/c^2)/(Δt/√(1-v^2/c^2)

v ≠ ΔL√(1-v^2/c^2)/(√(1-v^2/c^2)/Δt)

v = ΔL/τ the gamma cancels,

(v ≠ ΔL/Δt) So since this does not equal the velocity put into the equation, if you found a velocity with dialated spacetime using ΔL and Δt, the theory breaks down. You could then take that new velocity and then find new values for ΔL and Δt, and then find another new velocity and so on etc. This is why I think the time dialation and the length contraction equation should both be directly porportional to gamma.

Δt'=Δt√(1-v^2/c^2) solving for the time variables reversed

ΔL'=cΔt' I assume the distance traveled by the photon is a different value
since the speed of light is constant

ΔL'=cΔt√(1-v^2/c^2) substitute Δt' from the first equation

ΔL'=ΔL√(1-v^2/c^2) ΔL=cΔt, this also works for the direction of motion
ΔL'=vΔt' if you assume v=v'

Solving for length in the vertical dimension doesn't assume space contraction in that direction since both observers would agree that the photon reached the same position after being measured. Neither one says the photon traveled a greater vertical distance and is a requirement of this calculation to form the right triangle.

I then found an equation that describes time dialation for an object under constant acceleration. It assumes that because of the Michelson-Morely experiment that an object under acceleration still measures the photon to travel in a straight line since the Earth was accelerating and the effects of gravity where balanced out.

Δt'=Δt√(1-(vi+vo)^2/4c^2) I still need to look into how this differs from
constant motion
You're going to have to clue me in to where you are getting all this from. I have no idea.