# Differentiating with coordinate transformations

• liu111111117
In summary, the objective of the conversation was to show that ds^2 = -(1 + \alpha x)^2 dt^2 + dx^2 by finding dT and dX and inserting them into ds^2 = -dT^2 + dX^2. Incorrect attempts were made to find dT, but the correct expression is dT = dx sinh(\alpha t) + (\alpha x + 1) cosh(\alpha t) dt. The mistake was forgetting to include dt at the end. The conversation also discussed the use of the chain rule and clarified that t is a coordinate, not a function.
liu111111117
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:

$$dT= (\alpha x+1) cosh(\alpha t)$$

Hello
Take care distribution of derivative
$$d(x \ cosh(\alpha t)) = dx \ cosh(\alpha t) + x\ d(cosh (\alpha t))=...$$

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:
liu111111117
Of course.

If y = f(x),

$$dy = \frac{dy}{dx} dx$$

Thank you.

## What is a coordinate transformation?

A coordinate transformation is a mathematical process used to change the coordinates of points in a space to a new set of coordinates, often with the goal of simplifying or solving a problem.

## Why is differentiating with coordinate transformations important?

Differentiating with coordinate transformations allows us to solve problems in different coordinate systems, which can be more convenient or efficient for certain types of problems.

## How do you differentiate with coordinate transformations?

To differentiate with coordinate transformations, we use the chain rule and the transformation equations to express the derivative in terms of the new coordinates.

## What are some common coordinate transformations used in differentiation?

Some common coordinate transformations used in differentiation include polar coordinates, spherical coordinates, and cylindrical coordinates.

## What are the limitations of differentiating with coordinate transformations?

Some limitations of differentiating with coordinate transformations include the complexity of the transformation equations and the potential for introducing errors in the differentiation process.

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