What is the difference between time dilation and length contraction?

[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]

Welcome to PF!

Hello Evan! Welcome to PF!

(have a gamma: γ )

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.

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. I am 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, I've been thinking about it but i can't 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 doesn't sound right, but i can't 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!

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.