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Length contraction??? (3 Viewers)
- Thread starter mlinger
- Start date
hysteria85
New Member
- Joined
- Mar 12, 2006
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- 13
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- Male
- HSC
- 2006
The answer is 0.3m which has already been explained. I'm going to explain one of the things that is supporting the .192 peoples argument.
Yorky, you said that it is impossible to tell whether an inertial frame of reference is still or moving at a constant velocity. Absolutely true. In this example, once the electron is moving, and seeing the tube as .24m long, it can't tell it's own velocity. And yes, when stationary, it sees the tube as .3m long. THIS DOESN'T MEAN IT CAN DIFFERENTIATE BETWEEN THE FRAMES. The change of length is only noticed by the electron when it is accelerating to .6c (ie. in a non-inertial frame).
So the electron can't tell if it's stationary or moving once it's already moving at .6c, which it is at the beginning of the question.
Also remember that the tube is in the same frame of reference as the lab - they are both stationary. The electron, moving at .6c, sees all other inertial frames contracted. This includes the lab and the tube. As the lab is in the same frame as the tube, the people in the lab see the real length of the tube, which is of course the longer of the two lengths.
If you say the two lengths are .24 and .192, the lab would see the tube as .24, which is incorrect because that's what the electron sees. Therefore the answer must be .3m.
Lets now work backwards from .3m to prove that it's correct. The lab sees a cathode ray tube of .3m in length. An electron stationary in the tube would also see the length as .3m. Now when the electron is moving at .6c, it sees the labs frame of reference (including the tube) contract.
Lv = Lo x (square root of .64)
Lv = .3 x .8
Lv = .24 which is the contracted length, the one seen by the electron as in the question.
Answer = .3m
Please feel free to attempt to prove the answer is .192. Good Luck
Yorky, you said that it is impossible to tell whether an inertial frame of reference is still or moving at a constant velocity. Absolutely true. In this example, once the electron is moving, and seeing the tube as .24m long, it can't tell it's own velocity. And yes, when stationary, it sees the tube as .3m long. THIS DOESN'T MEAN IT CAN DIFFERENTIATE BETWEEN THE FRAMES. The change of length is only noticed by the electron when it is accelerating to .6c (ie. in a non-inertial frame).
So the electron can't tell if it's stationary or moving once it's already moving at .6c, which it is at the beginning of the question.
Also remember that the tube is in the same frame of reference as the lab - they are both stationary. The electron, moving at .6c, sees all other inertial frames contracted. This includes the lab and the tube. As the lab is in the same frame as the tube, the people in the lab see the real length of the tube, which is of course the longer of the two lengths.
If you say the two lengths are .24 and .192, the lab would see the tube as .24, which is incorrect because that's what the electron sees. Therefore the answer must be .3m.
Lets now work backwards from .3m to prove that it's correct. The lab sees a cathode ray tube of .3m in length. An electron stationary in the tube would also see the length as .3m. Now when the electron is moving at .6c, it sees the labs frame of reference (including the tube) contract.
Lv = Lo x (square root of .64)
Lv = .3 x .8
Lv = .24 which is the contracted length, the one seen by the electron as in the question.
Answer = .3m
Please feel free to attempt to prove the answer is .192. Good Luck
Last edited:
MisterPresident
Member
It was 0.3. The lenght will be contracted from the frame of reference of the accelerated particles, therefore it will appear longer than 0.2 metres, or however long it was, to a stationary observer.
MisterPresident
Member
I could have also read hysteria's answer before posting a possibly superfluous comment.