Differentiating with coordinate transformations

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The discussion focuses on the transformation equations T and X, aiming to demonstrate that ds² = -(1 + αx)² dt² + dx². Participants analyze the derivatives dT and dX to substitute into the equation ds² = -dT² + dX². Incorrect attempts at calculating dT are highlighted, emphasizing the need for proper application of the chain rule and distribution of derivatives. A key point is the clarification that t is treated as a coordinate, not a function, which affects the differentiation process. The conversation concludes with a reminder to include dt in the final expression for dT.
liu111111117
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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)
 
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Hello
Take care distribution of derivative
d(x \ cosh(\alpha t)) = dx \ cosh(\alpha t) + x\ d(cosh (\alpha t))=...
 
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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.
 
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Of course.

If y = f(x),

dy = \frac{dy}{dx} dx

Thank you.
 

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