Differentiating with coordinate transformations

[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:

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

 

[mitochan]

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.

 

[liu111111117]

Or does the second term need chain rule? I think not. t is coordinate, not function

 

[mitochan]

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.

 

[liu111111117]

Of course.

If y = f(x),

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

Thank you.