# Question on special relativity from "Basic Relativity"

PeroK
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I made an isometric view and not a side view. If only you will look carefully, you will see that the front end of the train coincides with the right edge of the platform while the rear end with that of the left edge. I made the distance between the two clocks on the platform equal to the length of the train at rest. The observer inside the moving train will not notice any change in the length of the train but he will see a decrease in the distance between clocks 1 and 2. Here is another illustration of the same problem, also in isometric view:
View attachment 288647
It still looks wrong to me. The rest length of the train is greater than the rest length of the platform.

PeroK
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Here is a graphically poor but physically correct version:

It still looks wrong to me. The rest length of the train is greater than the rest length of the platform.
Oh, but isn't the rest length of the train slightly greater than that of the platform in the isometric view in this attachment? I might be wrong.

PeroK
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Oh, but isn't the rest length of the train slightly greater than that of the platform in the isometric view in this attachment? I might be wrong.
I don't see how you could that sketch to help solve the problem. On the other hand, you definitely can use my diagram to solve the problem. Despite the lack of artistry!

PhysicsTruth
I don't see how you could that sketch to help solve the problem. On the other hand, you definitely can use my diagram to solve the problem. Despite the lack of artistry!
Sure, hand drawn diagrams on a piece of paper are way better on any day!

It still looks wrong to me. The rest length of the train is greater than the rest length of the platform.
When two objects having the same sizes are drawn in perspective, the one in front appears a little bigger than the one at the back in order to show depth of visual perception. I drew the stationary train in front of the platform, so naturally the train should appear a little larger than the slightly smaller platform behind it. A drawing of a flat surface made seen from the top, just like what you did, cannot show depth properly. Drawings are just visual representations and they do not determine the values of the quantities that must be used in the computations.

PeroK
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Sure, hand drawn diagrams on a piece of paper are way better on any day!
Yes, but the critical things in the train frame are:

1) The platform is (much) shorter than the train. Assuming ##v## is large.

2) Clock C2 lags behind clock C1.

That's what you needed to recognise to solve the problem.

If you want to use Lorentz Transformation, a diagram like that is still useful to establish the events you are interested in. For example, I would take C1 reading 04:00 as the common origin. (Note that that defines a unique spacetime point, which we take as the origin). The second event: clock C2 reading 04:00 has coordinates ##(t = 0, x = -L)## in the platform frame. Notice the way I dissociated the coordinates from the actual clock readings.

PhysicsTruth
PeroK
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When two objects having the same sizes are drawn in perspective, the one in front appears a little bigger than the one at the back in order to show depth of visual perception. I drew the stationary train in front of the platform, so naturally the train should appear a little larger than the slightly smaller platform behind it. A drawing of a flat surface made seen from the top, just like what you did, cannot show depth properly. Drawings are just visual representations and they do not determine the values of the quantities that must be used in the computations.
Thankfully for me, this isn't an art class!

Thankfully for me, this isn't an art class!
*An engineering drawing class. You don't get to see isometric projections often in an art class.

PeroK
PeroK
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*An engineering drawing class. You don't get to see isometric projections often in an art class.
I got an "A" in engineering drawing at school. A long time ago!

PhysicsTruth
Ibix
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[Sorry for the late reply]
In any case for part b) there was no given event to transform.
Sure there is - the event is the one on the worldline of clock 2 that is simultaneous in the train frame with clock 1 reading 4:00. Taking 4:00 on clock 1 to be the shared origin, we know that this event has the same ##t'##, so ##t'=0## and we know that the length ##L## platform is length contracted to ##L/\gamma## in this frame so ##x'=-L/\gamma##. We just need to calculate ##t## for this event and relate that to 4:00.

PeroK
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[Sorry for the late reply]

Sure there is - the event is the one on the worldline of clock 2 that is simultaneous in the train frame with clock 1 reading 4:00. Taking 4:00 on clock 1 to be the shared origin, we know that this event has the same ##t'##, so ##t'=0## and we know that the length ##L## platform is length contracted to ##L/\gamma## in this frame so ##x'=-L/\gamma##. We just need to calculate ##t## for this event and relate that to 4:00.
Isn't that equivalent to deriving the "leading clocks lag" rule in the first place? I don't see that every time you have the scenario of synchronised moving clocks, then you go back to the derivation from first principles.

Ibix
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Isn't that equivalent to deriving the "leading clocks lag" rule in the first place? I don't see that every time you have the scenario of synchronised moving clocks, then you go back to the derivation from first principles.
I find the Lorentz-every-time approach easier than remembering rules derived from it. I appreciate that not everyone agrees.

Motore
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I find the Lorentz-every-time approach easier than remembering rules derived from it. I appreciate that not everyone agrees.
I agree with you. BTW, doesn't the problem statement say that, in the platform frame of reference (and the clocks are at rest in this frame), the front of the train passes clock 1 at 4:00 and the rear of the train passes clock 2 at 4:00. Don't these same readings have to be on the ground clocks, as observed by train observers at the front and rear of the train, when the front of the train passes clock 1 and the rear of the train passes clock 2?

PeroK
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Don't these same readings have to be on the ground clocks, as observed by train observers at the front and rear of the train, when the front of the train passes clock 1 and the rear of the train passes clock 2?
These are spacetime events. They need no specific observer.

The physics is the same whether there is anyone on the train or not.

Ibix
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Don't these same readings have to be on the ground clocks, as observed by train observers at the front and rear of the train, when the front of the train passes clock 1 and the rear of the train passes clock 2?
Yes. But the question is what time does clock 2 show when the front of the train passes clock 1 according to the train observer, and "front passes 1" and "rear passes 2" are only simultaneous in the platform frame.

Chestermiller
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Yes. But the question is what time does clock 2 show when the front of the train passes clock 1 according to the train observer, and "front passes 1" and "rear passes 2" are only simultaneous in the platform frame.
Sorry. I guess I mis-read or mis-interpreted the question.

Ibix
I wonder if this drawing of the moving platform seen fom the top would help solve the problem in this thread.