Differentiating with coordinate transformations

Click For Summary

Homework Help Overview

The discussion revolves around differentiating with coordinate transformations in the context of a mathematical physics problem. The original poster presents a transformation involving hyperbolic functions and aims to demonstrate a specific relationship involving the line element ds².

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of transformed coordinates and the application of the chain rule. There are attempts to express dT and dX, with some questioning the need for the chain rule in certain terms.

Discussion Status

Some participants have provided guidance on the distribution of derivatives and the proper handling of infinitesimals. There is an ongoing exploration of how to correctly express the terms involving dt and the implications of treating t as a coordinate.

Contextual Notes

There is a focus on ensuring the correct application of differentiation rules, particularly in the context of hyperbolic functions and their derivatives. The participants are navigating the complexities of the transformation without reaching a definitive conclusion.

liu111111117
Messages
4
Reaction score
0
Homework Statement
line element for Kottler-Møller coordinates
T = (x+\frac{1}{\alpha}) sinh(\alpha t)
X = (x+\frac{1}{\alpha}) cosh(\alpha t) - \frac{1}{\alpha}

Objective is to show that

ds^2 = -(1 +\alpha x)^2 dt^2 + dx^2

via finding dT and dX and inserting them into ds^2 = -dT^2 + dX^2

Incorrect attempt #1:

dT= (dx+\frac{1}{\alpha}) sinh(\alpha dt)

Incorrect attempt #2:

dT= (\alpha x+1) cosh(\alpha t)
 
Physics news on Phys.org
Hello
Take care distribution of derivative
d(x \ cosh(\alpha t)) = dx \ cosh(\alpha t) + x\ d(cosh (\alpha t))=...
 
  • Like
Likes   Reactions: liu111111117
Thus,

dT = dx sinh (\alpha t) + (\alpha x +1) cosh (\alpha t)

I find no way to yield a dt term.
 
Or does the second term need chain rule? I think not. t is coordinate, not function
 
liu111111117 said:
Thus,
dT = dx sinh (\alpha t) + (\alpha x +1) cosh (\alpha t)
I find no way to yield a dt term.
You forgot to put dt at the end. Thus
dT = \mathbf{dx} \ sinh (\alpha t) + (\alpha x +1) cosh (\alpha t) \mathbf{dt}
Both sides be infinitesimal including d(coordinate). t is coordinate. cosh at and sinh at here are its functions.
 
Last edited:
  • Like
Likes   Reactions: liu111111117
Of course.

If y = f(x),

dy = \frac{dy}{dx} dx

Thank you.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K