Light Clock: Why Aren't Delta T and 2D/c Equal?

[morrobay]

TL;DR Summary

Why does delta t = gamma delta t' not equal to 2D /c

With this light clock delta t' in S' would be 2 Ls proper time. Since time measurements were done in same location delta x'=0 . Then from stationary S frame delta t = gamma delta t' . With gamma equal to 1.25 then that is 2.5 Ls. However the path the light takes as calculated from S frame in this example would be 2D/c = 2.33 light seconds . Why are they not equated ?

 

[Dale]

Your value for D is incorrect. Please show your calculation. Remember 1 ls is a distance and 0.6 c is a speed not a distance. The unlabeled side is not 0.6 ls long.

 

[morrobay]

Isn't .6c a distance of .6 light second ? Then just solving the right triangle .

 

[Ibix]

Isn't .6c a distance of .6 light second ? Then just solving the right triangle .

No. The point is that the ship and the light pulse travel for the same time duration. The ship travels 0.6 light seconds in the time it takes the light pulse to travel 1 light second. The light pulse does not travel 1 light second, so that distance cannot be 0.6 light seconds.

You need to solve for the lengths of the hypotenuse and the unlabelled side.

 

[Dale]

Isn't .6c a distance of .6 light second ? Then just solving the right triangle .

No. The time is not 1 s.

 

[morrobay]

Your value for D is incorrect. Please show your calculation. Remember 1 ls is a distance and 0.6 c is a speed not a distance. The unlabeled side is not 0.6 ls long.

No. The time is not 1 s.

The distance to mirror in light clock is one light second. So after light reaches mirror in S' the train at .6c seems to me to have traveled .6 light second (rate)(time) .Then it's just solving the right triangle. .6^2 + 1^ 2 = 1.166^2

 

[Dale]

The distance to mirror in light clock is one light second. So after light reaches mirror in S' the train at .6c seems to me to have traveled .6 light second (rate)(time) .Then it's just solving the right triangle. .6^2 + 1^ 2 = 1.166^2

So as I said above the time is not 1 s. By this calculation the time is 1.166 s. Do you recognize that?

So it travels 0.6 ls in 1.166 s. What is the speed?

 

[morrobay]

So as I said above the time is not 1 s. By this calculation the time is 1.166 s. Do you recognize that?

So it travels 0.6 ls in 1.166 s. What is the speed?

I don't know how you came to that conclusion from my quote, so how about you diagraming a light clock with mirror distance of 1 light second and v= .6c ?

 

[Dale]

I don't know how you came to that conclusion from my quote, so how about you diagraming a light clock with mirror distance of 1 light second and v= .6c ?

What path does the light take? Does the light travel on the vertical path or on the hypotenuse? Based on which path it takes, how long does it take to get there?

 

[Ibix]

I don't know how you came to that conclusion from my quote, so how about you diagraming a light clock with mirror distance of 1 light second and v= .6c ?

Your original post shows the light traveling 1.166 ls. The perpendicular distance is 1 ls. How long must the bottom of the triangle be? If the ship traveled that distance in 1.166s (the same time it took light to travel the length of the hypotenuse), how fast was it going?

 

[morrobay]

What path does the light take? Does the light travel on the vertical path or on the hypotenuse? Based on which path it takes, how long does it take to get there?

The light path on train is the vertical path . As seen from the observer in S the light path is taking the hypotenuse. Are you actually asking me to explain a light clock ??

 

[Dale]

As seen from the observer in S the light path is taking the hypotenuse.

Excellent, that is the observer of interest here. So how much time does it take for the light to travel a hypotenuse which is 1.166 ls distance?

If an object travels 0.6 ls distance in that same amount of time, then what is that object's speed?

Are you actually asking me to explain a light clock ??

Yes. Once you can explain it correctly then you will understand it

 

[morrobay]

Excellent, that is the observer of interest here. So how much time does it take for the light to travel a hypotenuse which is 1.166 ls distance?

If an object travels 0.6 ls distance in that same amount of time, then what is that object's speed?

Yes. Once you can explain it correctly then you will understand it

The 1.166 light seconds is based on the .6 light seconds. So how is .6/1.166 = .5145 c going to work ?

 

[Sagittarius A-Star]

Isn't .6c a distance of .6 light second ?

That argument is valid in reference frame S', because in this reference frame, the time of the light to reach the mirror is 1s. But in reference frame S', that is a moving distance, so the 0,6 LS in S' is a lengths-contracted 0.75 LS distance.

 

[Dale]

The 1.166 light seconds is based on the .6 light seconds. So how is .6/1.166 = .5145 c going to work ?

Exactly, it doesn't work. The length of the adjacent side is not equal to 0.6 ls.

You are treating the length of the adjacent side as a known, but it is an unknown that you need to solve for. You have two unknowns, the length of the hypotenuse and the length of the adjacent side.

Luckily, you have two equations, the Pythagorean theorem and the speed. But the speed is the adjacent divided by the hypoteneuse/c, not the adjacent divided by the opposite/c.

 

[PeroK]

The light path on train is the vertical path . As seen from the observer in S the light path is taking the hypotenuse. Are you actually asking me to explain a light clock ??

What you have done essentially is confused measurements from the two frames. You need to draw your diagram either:

1) In the rest frame of the clock, where you have a simple up and down of, I assume, $1$ light-second each way.

2) Where the clock is moving at $0.6c$. In which case the light-clock is still $1$ light second high, but the light (travelling along the hypoteneuse) takes longer than $1s$ to reach the top. In which case, the clock has moved more than $0.6$ light seconds horizontally (in that frame) while the light travels from the bottom to the top.

 

[morrobay]

That argument is valid in reference frame S', because in this reference frame, the time of the light to reach the mirror is 1s. But in reference frame S', that is a moving distance, so the 0,6 LS in S' is a lengths-contracted 0.75 LS distance.

From ∆t = gamma ∆t' = 2.50 . Then with hypotenuse 1.25 the distance is indeed .75 Ls

 

[robphy]

Possibly useful:
https://www.geogebra.org/m/XFXzXGTq
(in interactive 3D, with what should be everyone’s favorite [sub-light-speed relative-]velocity: v=0.6c)