- #1

Chenkel

- 482

- 109

Hello everyone, I was working on a thought experiment and came upon an inconsistent result that I thought maybe I could get some help with.

Firstly I am using natural units (c = unitless 1 i.e 1 light second per second) and I chose v = 3/5 to make gamma come out nicely (gamma equals 5/4 when v equals 3/5)

So the spaceship traveling at 3/5 light seconds per second has a pendulum at rest inside it with a period of 1 hertz (1 cycle per second) so I calculated the spacetime interval to be the squared distance between events minus the squared distance light travels between the events occurring, so that means the spacetime interval is equal to 0 - 1 = -1 light seconds.

In one second the spaceship travels 3/5 light seconds so relative to earth there is now a distance between the event of the pendulum coming back to its initial position of 3/5 light seconds.

So to discover the period of the pendulum relative to earth I solve the equation ##-1 = (\frac 3 5)^2 - (\Delta t)^2## So that means ##(\Delta t)^2 = \frac {9} {25} + \frac {25} {25} = \frac {34}{25}## so I have ##\Delta t = \sqrt {\frac {34} {25}}= 1.16##

So if the period of the pendulum in the spaceship is 1 second then the period of the pendulum that's on the spaceship relative to earth would be 1.16 seconds using my spacetime interval equations.

If I take the period of the pendulum on the spaceship (1 second) and multiply by gamma I should get the period of the pendulum on the spaceship relative to someone on earth which should be gamma, but gamma equals 1.25 using the time dilation formula, so these results seem inconsistent.

Any help would be appreciated, thanks in advance!

Firstly I am using natural units (c = unitless 1 i.e 1 light second per second) and I chose v = 3/5 to make gamma come out nicely (gamma equals 5/4 when v equals 3/5)

So the spaceship traveling at 3/5 light seconds per second has a pendulum at rest inside it with a period of 1 hertz (1 cycle per second) so I calculated the spacetime interval to be the squared distance between events minus the squared distance light travels between the events occurring, so that means the spacetime interval is equal to 0 - 1 = -1 light seconds.

In one second the spaceship travels 3/5 light seconds so relative to earth there is now a distance between the event of the pendulum coming back to its initial position of 3/5 light seconds.

So to discover the period of the pendulum relative to earth I solve the equation ##-1 = (\frac 3 5)^2 - (\Delta t)^2## So that means ##(\Delta t)^2 = \frac {9} {25} + \frac {25} {25} = \frac {34}{25}## so I have ##\Delta t = \sqrt {\frac {34} {25}}= 1.16##

So if the period of the pendulum in the spaceship is 1 second then the period of the pendulum that's on the spaceship relative to earth would be 1.16 seconds using my spacetime interval equations.

If I take the period of the pendulum on the spaceship (1 second) and multiply by gamma I should get the period of the pendulum on the spaceship relative to someone on earth which should be gamma, but gamma equals 1.25 using the time dilation formula, so these results seem inconsistent.

Any help would be appreciated, thanks in advance!

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