What must be one's speed, relative to a frame S, in order....

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Discussion Overview

The discussion revolves around calculating the speed required for a moving clock to lose a specific amount of time as observed from a stationary frame S. The focus is on time dilation effects in special relativity, particularly in relation to clocks losing 1 second or 1 minute per day.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant poses a question regarding the necessary speed for a clock to lose 1 second or 1 minute per day as observed from frame S.
  • Another participant explains that for the one minute-per-day scenario, time dilation indicates that while 24*60 minutes pass on the stationary clock, only (24*60)-1 minutes pass on the moving clock, suggesting the use of the time dilation formula for calculations.
  • A participant calculates that for a clock losing 1 second per day, 86401 seconds will pass, leading to a value for ##\gamma## of ##\frac{86401}{86400}##, which can be used in the time dilation formula to solve for velocity.
  • Another participant suggests an approximate formula for velocity based on the close values of ##\gamma##, stating that ##v^2 ≈ 2c^2(γ - 1)##.
  • One participant corrects an earlier claim by clarifying that for a clock losing 1 second per day, it actually ticks 86399 times in a day as measured by a clock at rest in S, leading to a different calculation for ##\gamma## as ##\frac{86400}{86399}##.

Areas of Agreement / Disagreement

Participants express differing views on the calculations for ##\gamma## and the implications for the speed required, indicating that there is no consensus on the correct approach or final values.

Contextual Notes

Some calculations depend on assumptions about the definitions of time dilation and the specific conditions of the moving clock relative to the stationary frame. The discussion includes various interpretations of the time dilation formula and its application to the problem.

cosmos42
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... in order that one's clocks will lose:

(a) 1 second per day as observed from S?
(b) 1 minute per day as observed from S?

I was referencing this:( http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c2 ) hyperphysics page but I still can't seem to understand what I need to do to solve for velocity.
 
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In the one minute-per-day case, time dilation says that while 24*60 minutes elapse on the stationary clock, (24*60)-1 minutes elapse on the moving clock. The time dilation formula and some algebra will see you home from there.
 
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There are 86400 seconds in a day so on a clock that loses 1 second per day 86401 seconds will pass. Thus ##\gamma## equals ##\frac{86401}{86400}##. You can then use that value in the time dilation formula you linked to and solve for ##v##.
 
Or, since both of your γ are close to 1, you can use the following approximate formula:

v2 ≈ 2 c2(γ - 1)
 
Mister T said:
There are 86400 seconds in a day so on a clock that loses 1 second per day 86401 seconds will pass.
That's a clock gaining 1s. A slow clock will tick 86399 times in a day as measured by a clock at rest in S, which gives ##\gamma =\frac {86400}{86399}##.
 
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