DarkMalice
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Hello,
I just started taking a Modern Physics course in University, and have a moderate understanding of time dilation. The problem is that I'm stuck with a problem I asked myself.
I understand the light clock thought experiment.
If you are within a rocket ship moving at a fast speed and a light pulse come from the floor vertically, hits the ceiling and returns to the floor, it take [tex]\Delta[/tex]t= [tex]\frac{2h}{c}[/tex], where h is the height of the ship.
For someone observing from Earth, the light does not move in a straight line but in a triangle, and it would take instead [tex]\Delta[/tex]t = ([tex]\frac{2h}{c}[/tex]) / [tex]\sqrt{1-(v^2/c^2)}[/tex]which is the time dilation formula.
My question was what would happen if the experiment was conducted on Earth? Within the reference frame of the Earth-dweller, [tex]\Delta[/tex]t= [tex]\frac{2h}{c}[/tex], and inside the reference frame of the moving rocket, [tex]\Delta[/tex]t =([tex]\frac{2h}{c}[/tex]) / [tex]\sqrt{1-(v^2/c^2)}[/tex] Does this mean that now instead of time moving slower in the rocket, does it move faster? I know it can't just move faster just because I decided to do the experiment somewhere else, but I'm still confused.
I just started taking a Modern Physics course in University, and have a moderate understanding of time dilation. The problem is that I'm stuck with a problem I asked myself.
I understand the light clock thought experiment.
If you are within a rocket ship moving at a fast speed and a light pulse come from the floor vertically, hits the ceiling and returns to the floor, it take [tex]\Delta[/tex]t= [tex]\frac{2h}{c}[/tex], where h is the height of the ship.
For someone observing from Earth, the light does not move in a straight line but in a triangle, and it would take instead [tex]\Delta[/tex]t = ([tex]\frac{2h}{c}[/tex]) / [tex]\sqrt{1-(v^2/c^2)}[/tex]which is the time dilation formula.
My question was what would happen if the experiment was conducted on Earth? Within the reference frame of the Earth-dweller, [tex]\Delta[/tex]t= [tex]\frac{2h}{c}[/tex], and inside the reference frame of the moving rocket, [tex]\Delta[/tex]t =([tex]\frac{2h}{c}[/tex]) / [tex]\sqrt{1-(v^2/c^2)}[/tex] Does this mean that now instead of time moving slower in the rocket, does it move faster? I know it can't just move faster just because I decided to do the experiment somewhere else, but I'm still confused.
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