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

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Discussion Overview

The discussion centers around the discrepancies in time measurements and distances in a light clock scenario, particularly examining the relationship between proper time, light travel distance, and the effects of relativistic speeds. Participants explore concepts related to time dilation, the geometry of light paths, and the implications of different reference frames.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant claims that the time measurement in the stationary frame (S) should be gamma times the proper time in the moving frame (S'), leading to a calculated time of 2.5 light seconds.
  • Another participant challenges the value of D, asserting that 0.6c is a speed, not a distance, and questions the calculations based on this misunderstanding.
  • Some participants argue that the light pulse and the moving object (the ship) travel for the same duration, complicating the interpretation of distances in different frames.
  • There is a repeated assertion that the light travels along the hypotenuse in the stationary frame, while the vertical path is taken in the moving frame, leading to confusion about the distances involved.
  • One participant suggests that the time taken for light to travel the hypotenuse is longer than 1 second, proposing a calculation that results in 1.166 seconds.
  • Another participant points out that the adjacent side of the triangle formed by the light clock setup is not simply 0.6 light seconds, indicating the need for further calculations to determine its actual length.
  • Some participants discuss the implications of length contraction in different reference frames, suggesting that the perceived distances change based on the observer's frame of reference.
  • A link to an interactive 3D model is provided as a potentially useful resource for visualizing the scenario.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the calculations and interpretations of distances and times in the light clock scenario. There is no consensus on the correct values or methods to resolve the discrepancies presented.

Contextual Notes

Participants highlight limitations in understanding the geometry of the light clock, particularly regarding the relationships between distances in different frames and the assumptions made about time and speed. The discussion reveals unresolved mathematical steps and dependencies on specific definitions of distances and speeds.

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TL;DR
Why does delta t = gamma delta t' not equal to 2D /c
IMG20201218191508.jpg
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 ?
 
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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.
 
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Isn't .6c a distance of .6 light second ? Then just solving the right triangle .
 
morrobay said:
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.
 
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morrobay said:
Isn't .6c a distance of .6 light second ? Then just solving the right triangle .
No. The time is not 1 s.
 
Dale said:
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.
Dale said:
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
 
morrobay said:
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?
 
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Dale said:
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 ?
 
morrobay said:
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?
 
  • #10
morrobay said:
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?
 
  • #11
Dale said:
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 ??
 
  • #12
morrobay said:
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?

morrobay said:
Are you actually asking me to explain a light clock ??
Yes. Once you can explain it correctly then you will understand it
 
  • #13
Dale said:
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 ?
 
  • #14
morrobay said:
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.
 
  • #15
morrobay said:
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.
 
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  • #16
morrobay said:
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.
 
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  • #17
Sagittarius A-Star said:
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
 
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