Proper Length, Time

[Piyu]

Hello! This is my first post on this forums, always been here since a year back through google while looking for some answers for questions. Anyways i have to say, the threads were extremely helpful in my learning :)

I got abit of confusion over time dilation and length contraction. Apparently, when we try to use time dilation for say the time experienced by a person on a rocket flying away, we use t0 where t=gamma t0. However, when we do megnth contraction, to find the length perceived by the moving observer, we use t0 as the stationary observer.

I understand that t0 has to be the person who goes to the 2 points(ei, their foot is standing on the spots where we take the time taken.). and l0 is the person who can take the length of the object at the same time. However, when i try to draw a connection between these 2 statements, i get confused.

Evan

[tiny-tim]

Hello Evan! Welcome to PF! :smile:

(have a gamma: γ :wink:)
However, when we do length contraction, to find the length perceived by the moving observer, we use t0 as the stationary observer.

I don't follow you. :confused:

If the moving observer goes past a stationary rod of length L in time t on a stationary clock,

then his speed is v = L/t.

Using his clock, his time is only t/γ, and his length is only L/γ, so the speed (of the rod now) is still v.

[Piyu]

Ah. Im sorry, that is my fault. I was referring to 2 different scenarios. In one case, t0 is the moving person. and in the other, L0 is to the stationary observer. That's whant causing my confusion

Actually, the more i type about it the more i think i get it. For the time dilation scenario, the person in the rocket is actually the stationary observer and hence we use t0 for him. Is that right?

[ghwellsjr]

The time experienced by a "moving" observer is just normal, he cannot perceive any dilation. The length experienced by a moving observer is just normal, he cannot perceive any contraction.

But a "stationary" observer, or any other observer moving with respect to the first "moving" observer, will perceive the first "moving" observer's times dilated and lengths contracted.

[Piyu]

So what about this scenario, ive been thinking about it but i cant seem to figure it out. Since gamma is always greater or equals to 1.

For a stationary observer A, the length of a ruler is L. So to the moving observer B, the length is L/gamma.

Then now we can use the moving observer A as the stationary 1 and say its the stationary observer B who is moving. We know that to B, the length of the ruler is L/gamma. Therefore, to A, the length will be L/gamma^2?.

This doesnt sound right, but i cant figure out the mistake in the thinking.

Edit: upon typing once again, could it be that the reason for this is that L0 has to always be measured by a person who is in stationary frame to the object?

[tiny-tim]

Hello Evan! :smile:
For a stationary observer A, the length of a ruler is L. So to the moving observer B, the length is L/gamma.

Then now we can use the moving observer A as the stationary 1 and say its the stationary observer B who is moving. We know that to B, the length of the ruler is L/gamma. Therefore, to A, the length will be L/gamma^2?.

It's because the observer measures the distance between two points at the same time on his clock …

that will not be at the same time on the other observer's clock …

they are measuring the distance between two different pairs of events. :wink: