On the idea of time in physics-relativity

Well then think about this: According to the platform frame, the proper length of the platform just happens to be the same as the length-contracted train car. The lightning strikes occur when the endpoints of the length-contracted train car line up with the endpoints of the un-contracted platform. But according to the train frame, the proper length of the train is larger than the length-contracted platform. So it would be impossible for both endpoints of the train car to line up with the both endpoints of the platform at the same time.

Indeed it does, as explained above. The effects do not cancel out. Have a look at post #432:

http://www.sciforums.com/showthread.php?133989-On-the-idea-of-time-in-physics-relativity&p=3055673&viewfull=1#post3055673

 

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Your reasoning doesn't seem to have any natural logic to it. The train, the platform, and the speed can vary from place to place. Not all trains are the same length, not all platforms are the same length, and the speed of the train is never defined.
To make the train and the platform length in someway relate is possible but in most situations there would be no relationship at all.

What you might be saying is as lightning struck the ends of the train it struck the platform at the same time. so the marks on the ground represent the actual length of the train to an observer if the lightning struck simultaneous to an observer.

 

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Pete defined the speed of the train to be:

\(v = 0.6c\)

Which makes the Lorentz factor:

\(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.25\)

Pete defined the proper length of the train to be:

\(L'_0 = 25\)

Which makes the length-contracted length of the train, according to the platform:

\(L' = \frac{L'_0}{\gamma} = \frac{25}{1.25} = 20\)

Pete defined the proper length of the platform to be:

\(L_0 = 20\)

Which means the endpoints align simultaneously in the platform frame, because:

\(L' = L_0\)

\(20 = 20\)

But in the train frame, the platform is length contracted to:

\(L = \frac{L_0}{\gamma} = \frac{20}{1.25} = 16\)

Which means the endpoints cannot align simultaneously in the train frame, because:

\(L \neq L'_0\)

\(16 \neq 25\)

 

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Fair enough, and as I said you can select lengths and speed where there is a nice relationship. The lightning strikes are only simultaneous to the platform observer. From the train the lightning wasn't simultaneous so as you have found (shown the marks) are more to one end than the other (I presume). It would be interesting to see where the marks actually were on the platform so both observers could go back and check. For there were only two lighting strikes so I imagine there are only two marks on the ground. So if the platform is shorter than the train the front strike has to be earlier than the second in the train frame and that is what was found to have happened.

 

Yes, that is it exactly. I thank you!

 

If there was only length contraction in one frame then wouldn't everything look bigger from that frame? Like in Honey I Shrunk the Kids, everything looks real big to the kids. But, if they shrunk everything then it could look to be the same size. One frame wouldn't see everything as getting all huge, lol.

 

Read what he said again, Layman. Length contraction is in both frames, not just one.

Are you going to respond to my post?

 

What did you mean by that?

 

he means that according to platform's frame,train is contracted and according to train's frame,platform is contracted.

 

But in this case I think PL is correct. From the observers frame everything seems normal but the moving frame measures length contracted, so they seem small and the observer rather large.
Even if it is really reversed for the moving frame when it is their turn at being the one at rest.

 

This is in violation of Theory of relativity. Every reference frame is not unique.. You are giving preference to platform's frame.
Principle of relativity: If,
relative to K, K
1
is a uniformly moving co-ordinate system devoid of rotation, then natural
phenomena run their course with respect to K
1
according to exactly the same general laws
as with respect to K. This statement is called the principle of relativity (in the restricted
sense).

 

You cannot make that type of conclusion. You are mixing observer's reference frame with train's reference frame.

 

moving frame is contracted with respect to observer. Length contraction is relative and Time dilation too..

 

Well that is sad. PL appears to have a his PhD in trolling. His is purposely saying nonsenes to annoy people; it is obvious from his posting. Agreeing with him is like believing that the Onion is a real news outlet.

 

What? The Onion is a real news network. They portrayed Obama as what he really is.

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You can be either of the two observers. You could be on the platform as the train passed through. Or you could be on the train. Wherever you put yourself your frame is at rest. So you give your own frame preferred status.
I seem to think you have still missed a point about relativity.

 

What you say is like the the saying even a broken watch is right twice a day.
I agree he is not often right. but this time I think he hit the nail on the head.

 

obviously this conversation has some regional meaning.

 

Almost every interpretation of the MME in popular science says that they arrive at the same time in both frames. If it took longer for one beam to reach the same detector as another beam, then they wouldn't match up when they came back together when you compared them to each other. One beam would have more or less waves than the other in its journey. This is one reason why quantum mechanics is so strange, and it doesn't work like everyday real world stuff. You want to apply Newtonian logic to how this is possible, but it can't really be done. There is gauge invarience of frames observing each other in relative motion in the MME.

 

Pete is talking about TE, not MME. You still don't get the difference, do you?