What Does Shankar Mean in Prof Shankar's Introduction to Relativity?

Click For Summary

Discussion Overview

The discussion revolves around the interpretation of a segment from Prof Shankar's introduction to relativity, specifically regarding the Lorentz transformation and the variables involved in the scenario of two trains crossing at the origin. Participants explore the meaning of the distance "u" and its relation to the time taken for a light pulse to reach a detector.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the distance "u" mentioned by Prof Shankar, questioning whether it represents the distance the moving train travels while the light beam reaches the detector.
  • Another participant clarifies that "u" is the speed of the moving train relative to the stationary train, suggesting the equation should include time: x' = x - ut.
  • A subsequent participant confirms that the time in question is indeed the time taken for the light pulse to reach the detector.
  • Further clarification is provided that both trains agree on the emission point of the light pulse at (0, 0), but perceive the event at different coordinates.
  • One participant reflects on the overall presentation style of Shankar, finding it somewhat disorganized and expressing a separate query about forces in inertial frames.
  • Another participant responds to the force question, stating that objects can still accelerate in an inertial frame, and that F=ma is typically taught using inertial frames.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of "u" and the relationship between the distances and time, but there is a lack of consensus on the clarity of Prof Shankar's presentation style and its implications for understanding the concepts discussed.

Contextual Notes

Some participants express uncertainty about the clarity of the explanations provided by Prof Shankar, indicating potential limitations in understanding due to the presentation style rather than the underlying physics.

abrogard
Messages
99
Reaction score
3
Hope this is a valid question.

I can't quite understand this very simple thing.

Prof Shankar's introduction to relativity on Youtube starting at about 50 minutes has a quick calculation that comes up with the Lorentz transform.

I follow the maths, no problem.

But he confuses me a bit when he's setting the stage.

He says two trains, mine and yours, cross at the origin. Yours is moving. As they cross a beam of light is sent forward to the detector, distance x from the origin for me. distance x prime for you.

And he says there is a distance u for you. So that x prime = x - u.

That's what confuses me. Where is the distance u from? Is that the distance you will travel during the time it takes for the light beam to reach the detector?

I am thinking it must be but I'm so much a novice at all this I'd love confirmation or correction if I'm wrong.
 
Physics news on Phys.org
abrogard said:
And he says there is a distance u for you. So that x prime = x - u.
"u" is the speed of the primed train with respect to the unprimed (stationary) train. So that equation should be x' = x - ut. (You left out the time.)
 
yes, thank you. and that time is the time taken for the light pulse to reach the detector?
 
abrogard said:
yes, thank you. and that time is the time taken for the light pulse to reach the detector?
That's right. According to the stationary train the light pulse hits the detector at (x, t). But the other train views that same event as happening at (x', t').

(Both trains agree that the pulse was emitted at point (0, 0).)
 
Thank you. I thought so but was a little uncertain.

I find Shankar a bit confusing sometimes. In that talk there he seems to be all over the place - talking of firecrackers and bullets and this and that - where the meat of the matter is a mere 6 minutes right at the end. And that's a Yale education!

Lots of talk of F=MA. And inertial frames of reference. I was just thinking, that'd mean there's no Force in an inertial frame of reference, wouldn't it? Seeing as there can be no acceleration.

different question. sorry.

thanks for the help.

:)
 
abrogard said:
Lots of talk of F=MA. And inertial frames of reference. I was just thinking, that'd mean there's no Force in an inertial frame of reference, wouldn't it? Seeing as there can be no acceleration.
No. Just because the frame isn't accelerating doesn't mean that objects cannot accelerate with respect to that frame. In fact, F = ma is generally first taught using only inertial reference frames.
 
Ah, yes. Thank you.
 

Similar threads

High School Shankar on Lorentz Transformation: Does x' = ct'?
  • · Replies 16 ·
Replies
16
Views
2K
Undergrad A question about static spacetime in GR
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Arguments leading to the speed of light as a dimensionless constant
  • · Replies 64 ·
3
Replies
64
Views
5K
Undergrad Galilean relativity for 2 frames
  • · Replies 52 ·
2
Replies
52
Views
4K
Undergrad Has anyone measured the speed of light in one direction?
  • · Replies 45 ·
2
Replies
45
Views
7K
Undergrad Variation of the lightning train thought experiment
  • · Replies 26 ·
Replies
26
Views
5K
Undergrad Revisiting the Light Clock
  • · Replies 25 ·
Replies
25
Views
2K
Undergrad Understanding Relativity of Simultaneity in Resnick, Part II
  • · Replies 20 ·
Replies
20
Views
3K
Undergrad Understanding Relativity of Simultaneity in Resnick, Part I
  • · Replies 16 ·
Replies
16
Views
2K
High School Solving Doubts about Simultaneity Exp. in Special Relativity
  • · Replies 13 ·
Replies
13
Views
3K