PDA

View Full Version : Length Contraction, which way is it?


Living_Dog
Apr30-06, 06:59 PM
I thought I understood SR's time dilation and length contraction. But after reading the section on "simultameity" in Tipler I am just as confused as before. Here is my source of confusion.

[A] Muon Decay:

S frame = frame of the earth; S' frame = frame of the muon

A muon falls to earth at a speed of 0.998c ('c' of course is the speed of light). In its own frame it's lifetime is 2.0 μs, and the distance it travels is only 600 m.

But, in the earth's frame, the speed measured is the same, but the length is 9000 m and the lifetime is 30 μs. This makes perfect sense to me since, L = γL' and the distance (L' < L) is contracted for the moving particle.

[B] Rocket Ship:

A rocket ship flies past 2 points, A and B in S (the earth's frame) and is measured to be a length, L, when the front is over the point B and the back is over the point A. Tipler's explanation is that this is the contracted length of the ship in S', namely: L = L'/γ. Therefore the ship's length in S' is L' = γL. This means that the moving ship's length is longer than that measured in the earth's frame, or, L ' > L. And this is a direct contradiction to length contraction in [A]!

I know I the truth is that I don't understand this so I came here for an answer.

Thanks in advance!

-LD

Hurkyl
Apr30-06, 07:03 PM
In its own frame it's lifetime is 2.0 μs, and the distance it travels is only 600 m.
That's not true; in its own frame, it doesn't travel! It's the Earth that travelled 600 m.

Living_Dog
May1-06, 06:57 AM
That's not true; in its own frame, it doesn't travel! It's the Earth that travelled 600 m.

Doesn't the muon "measure" 600 m, but we measure 9000 m?

V = 0.998c = L'/\Deltat' = L/\Deltat

So, 0.998c = 600 m/ 2.0 \mus = 9000 m/ 30 \mus

No??

-LD

nrqed
May1-06, 07:04 AM
Doesn't the muon "measure" 600 m, but we measure 9000 m?

V = 0.998c = L'/\Deltat' = L/\Deltat

So, 0.998c = 600 m/ 2.0 \mus = 9000 m/ 30 \mus

No??

-LD

The correct way to understand the equation is this. A *proper distance* is a distance between two points as measured in a frame in which the two points are at rest. Then the equation says that the distance between the two points as measured in any other frame is the proper distance divided by gamma.
In th emuon example, the 9000 m is the proper distance. In the other example, L is the length of the spaceship measured from Earth (in which frame the spaceship is in motion) therefore L is not the proper length. The proper length of the spaceship is the length of the spaceship measured in the frame of the spaceship (where it is at rest).

Living_Dog
May1-06, 08:05 AM
The correct way to understand the equation is this. A *proper distance* is a distance between two points as measured in a frame in which the two points are at rest. Then the equation says that the distance between the two points as measured in any other frame is the proper distance divided by gamma.
In th emuon example, the 9000 m is the proper distance. In the other example, L is the length of the spaceship measured from Earth (in which frame the spaceship is in motion) therefore L is not the proper length. The proper length of the spaceship is the length of the spaceship measured in the frame of the spaceship (where it is at rest).

PERFECT! That was my confusion... now I have a fixed definition with which to apply to SR problems. Both proper time and proper LENGTH are the defintions which solve all such confusions! Dude, thx a million! :)

-LD

nrqed
May1-06, 08:35 AM
PERFECT! That was my confusion... now I have a fixed definition with which to apply to SR problems. Both proper time and proper LENGTH are the defintions which solve all such confusions! Dude, thx a million! :)

-LD
You are welcome.

It *is* very confusing when the authors (or the teacher) does not make it crystal clear when one can divide by gamma and when one must multiply by gamma. If this is not carefully explained, it makes things extremely confusing. I am glad I could help.

Patrick