Question on special relativity from "Basic Relativity"

In summary: No - this is incorrect. You should be looking for the time at which clock ##1## coincides with the front of the train, which is ##(4,L)##.
  • #36
PhysicsTruth said:
Ok, I panicked here and made a calculation mistake, it does turn out to be ##\gamma vL/c^{2}## using the space transformation as well.
That's right for part c). Did you notice that part b) was just the lack of synchronization of platform clocks in the train frame?
 
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  • #37
PeroK said:
That's right for part c). Did you notice that part b) was just the lack of synchronization of platform clocks in the train frame?
Yeah, I realized that. That is precisely what leads to the statement - "Simultaneity is Relative".
 
  • #38
PeroK said:
That's right for part c). Did you notice that part b) was just the lack of synchronization of platform clocks in the train frame?
So, when the clock 1 coincided with the front end, clock 2 read 4:00 for the train observer, right? It's just that the rear end doesn't coincide with clock 2 at that point of time in space.
 
  • #39
PhysicsTruth said:
So, when the clock 1 coincided with the front end, clock 2 read 4:00 for the train observer, right? It's just that the rear end doesn't coincide with clock 2 at that point of time in space.
Your problem which is common and is exemplified in this thread is that you are unable to imagine the scenario other than in the platform frame.

In the train frame: when clock 2 is at front of train, clock1 is only half way along the train and reads ##-\frac{Lv}{c^2}##. When that clock reaches the rear of the train it does indeed read 4:00, but those events are not simultaneous in that frame.

A second way to solve this problem was direct kinematics in the train frame, plus the leading clock lags rule. No need for Lorentz.

But, that requires you to reimagine the scenario in the train frame.

In fact, that's how I did it. I only used Lorentz to double check the answer!
 
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  • #40
PeroK said:
No need for Lorentz.

But, that requires you to reimagine the scenario in the train frame.
That's why I recommend the Lorentz transform method. No imagination needed, just grind through the maths. Once you've done that a few times you can learn to see a qualitative picture in a transformed frame, but based on a solid basis.

YMMV of course.
 
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  • #41
Ibix said:
That's why I recommend the Lorentz transform method. No imagination needed, just grind through the maths. Once you've done that a few times you can learn to see a qualitative picture in a transformed frame, but based on a solid basis.

YMMV of course.
I think it's important to be able to mentally change reference frames. Otherwise, you can be applying the LT blindly. In any case for part b) there was no given event to transform.
 
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  • #42
PeroK said:
I think it's important to be able to mentally change reference frames. Otherwise, you can be applying the LT blindly. In any case for part b) there was no given event to transform.
I agree with you. But I'm absolutely new to Special Relativity and Lorentz Transformations. I still can't figure out why it would read -Lv/c^2 on clock 2. I'm trying hard to figure out.
 
  • #43
PhysicsTruth said:
I agree with you. But I'm absolutely new to Special Relativity and Lorentz Transformations. I still can't figure out why it would read -Lv/c^2 on clock 2. I'm trying hard to figure out.
Have you seen a derivation of the "leading clock lags" rule? Which is effectively a statement of the relativity of simultaneity.
 
  • #44
PeroK said:
Have you seen a derivation of the "leading clock lags" rule? Which is effectively a statement of the relativity of simultaneity.
I'm checking that out right away.
 
  • #45
PeroK said:
Have you seen a derivation of the "leading clock lags" rule? Which is effectively a statement of the relativity of simultaneity.
Ok, I seemed to have confused myself a lot. But, I feel that it would read 4 - Lv/c^2, since the time on clock 1 was 4:00 and not 0.
 
  • #46
PhysicsTruth said:
Ok, I seemed to have confused myself a lot. But, I feel that it would read 4 - Lv/c^2, since the time on clock 1 was 4:00 and not 0.
Yes, of course.
 
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  • #47
PeroK said:
Yes, of course.
Thanks a lot for taking out time to guide me through this.
 
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  • #48
This is how I understand the problem. The two clocks on the left are synchronized, just as the two clocks on the right.
STR_Prob2b.jpg
 
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  • #49
kumusta said:
This is how I understand the problem. The two clocks on the left are synchronized, just as the two clocks on the right.
View attachment 288632
The train is longer than the platform. So, that's not right.
 
  • #50
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:
STR_Prob2c.jpg
 
  • #51
kumusta said:
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.
 
  • #52
Here is a graphically poor but physically correct version:

thumbnail_20210906_095725.jpg
 
  • #53
PeroK said:
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.
 
  • #54
PhysicsTruth said:
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!
 
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  • #55
PeroK said:
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!
 
  • #56
PeroK said:
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.
 
  • #57
PhysicsTruth said:
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.
 
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  • #58
kumusta said:
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!
 
  • #59
PeroK said:
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.
 
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  • #60
PhysicsTruth said:
*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!
 
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  • #61
[Sorry for the late reply]
PeroK said:
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.
 
  • #62
Ibix said:
[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.
 
  • #63
PeroK said:
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. 😁
 
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  • #64
Ibix said:
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?
 
  • #65
Chestermiller said:
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.
 
  • #66
Chestermiller said:
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.
 
  • #67
Ibix said:
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.
 
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  • #68
I wonder if this drawing of the moving platform seen fom the top would help solve the problem in this thread.
leng_con1a.jpg
 

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