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Time Dilation Q (2 Viewers)

astroman

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I think you have to add relativistic velocities, but idk how to do it lol.
 

InteGrand

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Can someone explain how to do it please.
There's a formula for addition of velocities in Special Relativity: https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

It's not as simple as in classical mechanics, but it ensures that adding two velocities with magnitude each less than c together in Special Relativity will never result in a velocity with magnitude greater than c (which is good, because that should be impossible, as we can't go faster than c as Special Relativity says).

And yes, this (addition of velocity vectors in Special Relativity) is beyond the syllabus.
 

astroman

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There's a formula for addition of vectors in Special Relativity: https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

It's not as simple as in classical mechanics, but it ensures that adding two velocities with magnitude each less than c together in Special Relativity will never result in a velocity with magnitude greater than c (which is good, because that should be impossible, as we can't go faster than c as Special Relativity says).

And yes, this is beyond the syllabus.
haha it's a Ruse question.
 

mrpotatoed

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let distance of the spaceship be x, use length dilation, it is 0.8x
let time taken for the ball to reach the end be t, use time dilation, it is 1.25t

v=d/t=0.64 * speed of ball from frame of reference of spaceship=0.064c from frame of reference of earth

add to 0.6 and you've got your answer, don't see why this couldn't be in the hsc
 

InteGrand

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let distance of the spaceship be x, use length dilation, it is 0.8x
let time taken for the ball to reach the end be t, use time dilation, it is 1.25t

v=d/t=0.64 * speed of ball from frame of reference of spaceship=0.064c from frame of reference of earth

add to 0.6 and you've got your answer, don't see why this couldn't be in the hsc
0.664c is not actually the correct answer. If the speed of the ball is 0.064c from the frame of reference of the Earth, why are we adding 0.6c to it? We shouldn't be doing anything to it, since that would be the answer, since what we are after is the ball's speed from the Earth's frame of reference.
 

mrpotatoed

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The speed of the ball moving through the spaceship is 0.064c, as worked out using relativity laws that we learn in HSC physics. But of course the reason we are using relativity laws is because the spaceship is also going at 0.6c away from us. So the spaceship is going away at 0.6c, the ball IN the spaceship is moving at 0.064c relative to earth. You can then use normal vector addition once you have accounted for the relativistic changes to the speed of the ball to get 0.664c which is the answer.

(and then obviously the MC choices are two sig fig so it gets rounded down)
 

InteGrand

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The speed of the ball moving through the spaceship is 0.064c, as worked out using relativity laws that we learn in HSC physics. But of course the reason we are using relativity laws is because the spaceship is also going at 0.6c away from us. So the spaceship is going away at 0.6c, the ball IN the spaceship is moving at 0.064c relative to earth. You can then use normal vector addition once you have accounted for the relativistic changes to the speed of the ball to get 0.664c which is the answer.
So what if the ball was magical and happened to be moving through the spaceship with a speed of 0.5c say...would the answer then be 1.1c ??

I don't think you mean 0.064c relative to the Earth, because that is exactly what we are trying to calculate.
 

mrpotatoed

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no... if the ball was moving at 0.5c, you would use what I did above and then you get the ball moving at 0.32c THROUGH THE SPACESHIP relative to earth. That's not accounting for the fact that the spaceship is also moving, you then need to add 0.6c to that and the ball is then moving at 0.92C AWAY FROM EARTH relative to earth.
 

InteGrand

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no... if the ball was moving at 0.5c, you would use what I did above and then you get the ball moving at 0.32c THROUGH THE SPACESHIP relative to earth. That's not accounting for the fact that the spaceship is also moving, you then need to add 0.6c to that and the ball is then moving at 0.92C AWAY FROM EARTH relative to earth.
I meant increase the ball's speed so that you would need to add say 0.5c to 0.6c. So say the ball was moving at (0.5 c)/(0.8) = 0.625c in the spaceship.

In special relativity, velocities don't add simply.
 

mrpotatoed

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I'm guessing what I did works as an approximation when the ball isn't going at relativistic speeds itself then? There is no other way within the course you could possibly do that question, and I'm guessing the ruse teachers intended it to be done this way
 

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