So, I'm going over this bit in the beginning of MTW's Gravitation where they give an equation for changing acceleration from time coordinate t to time coordinate T(t):(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{{d}^{2}x}{{dt}^{2}} = \frac{d}{dt}(\frac{dx}{dT}\frac{dT}{dt}) = \frac{{d}^{2}T}{{dt}^{2}}\frac{dx}{dT} + {(\frac{dT}{dt})}^{2}\frac{{d}^{2}x}{{dT}^{2}}[/tex]

The first term on the RHS is easy enough to figure out:

[tex]\frac{dx}{dT}(\frac{d}{dt}(\frac{dT}{dt})) = \frac{{d}^{2}T}{{dt}^{2}}\frac{dx}{dT}[/tex]

but for the second term I'm not sure if my method was the most elegant thing possible:

[tex]\frac{dT}{dt}(\frac{d}{dt}(\frac{dx}{dT})) = \frac{dT}{dt}(\frac{d}{dt}\frac{d}{dT}(x)) = \frac{dT}{dt}(\frac{d}{dT}(\frac{dx}{dt})) = \frac{dT}{dt}(\frac{d}{dT}(\frac{dx}{dT}\frac{dT}{dt})) = {(\frac{dT}{dt})}^{2}\frac{{d}^{2}x}{{dT}^{2}} + \frac{dT}{dt}\frac{dx}{dT}\frac{d}{dT}(\frac{dT}{dt}) = {(\frac{dT}{dt})}^{2}\frac{{d}^{2}x}{{dT}^{2}} + \frac{dT}{dt}\frac{dx}{dT}\frac{d}{dt}(\frac{dT}{dT}) = {(\frac{dT}{dt})}^{2}\frac{{d}^{2}x}{{dT}^{2}}[/tex]

Now, like I said I bet there's a better way to get from the LHS to the RHS, but my real problem is with the following equation, which kind of falls out of the last couple of steps I put up there:

[tex]\frac{d}{dT}(\frac{dT}{dt}) = 0[/tex]

because it seems like if, say, T(t) = (t + 1)^2, then:

[tex]\frac{d}{dT}(\frac{dT}{dt}) = \frac{d}{dT}(2(t+1)) = \frac{d}{dT}(2\sqrt{T}) = \frac{1}{\sqrt{T}} [/tex]

and not zero.

So I'm sure that what I'm missing is something about exactly what it means to take a derivative with respect to a function instead of a variable, but I can't remember what. I'm super hung up on this, so if anyone could help me out that would be wonderful.

(also, sorry for only posting half of my post earlier, but PF is acting kind of weird)

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# Calculus hangup in Gravitation

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