Recent content by liu111111117

Of course. If y = f(x), dy = \frac{dy}{dx} dx Thank you.

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

Thus, dT = dx sinh (\alpha t) + (\alpha x +1) cosh (\alpha t) I find no way to yield a dt term.

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})...