Calculus hangup in Gravitation

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    Calculus Gravitation
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SUMMARY

The forum discussion centers on the mathematical derivation of changing acceleration from the time coordinate \( t \) to the time coordinate \( T(t) \) as presented in MTW's "Gravitation." The user expresses confusion regarding the transition from the left-hand side (LHS) to the right-hand side (RHS) of the equation, particularly in understanding the derivative \( \frac{d}{dT}(\frac{dT}{dt}) = 0 \). The user provides specific examples and calculations, indicating a misunderstanding of derivatives with respect to functions rather than variables, which is crucial for grasping the underlying concepts of calculus in the context of gravitational physics.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and second derivatives.
  • Familiarity with the concepts of changing coordinates in physics.
  • Knowledge of the chain rule in differentiation.
  • Basic understanding of the mathematical framework presented in "Gravitation" by Misner, Thorne, and Wheeler.
NEXT STEPS
  • Study the chain rule in calculus to clarify the relationship between derivatives of different variables.
  • Review the concept of derivatives with respect to functions, particularly in the context of physics.
  • Examine the derivation of equations in "Gravitation" to understand the mathematical techniques used.
  • Practice solving problems involving changing coordinates and acceleration in physics.
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Students of physics, particularly those studying general relativity, mathematicians interested in calculus applications, and anyone seeking to deepen their understanding of the mathematical foundations of gravitational theory.

nonne
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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):

\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}}

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

\frac{dx}{dT}(\frac{d}{dt}(\frac{dT}{dt})) = \frac{{d}^{2}T}{{dt}^{2}}\frac{dx}{dT}

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

\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}}

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:

\frac{d}{dT}(\frac{dT}{dt}) = 0

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

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

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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nonne said:
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):

\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}}

Maybe this will help:

\frac{d^2x}{dt^2}=\frac{d}{dt}\left(\frac{dx}{dT}\frac{dT}{dt}\right)= \frac{dx}{dT}\frac{d^2T}{dt^2}+\frac{dT}{dt}\frac{d}{dt}\left(\frac{dx}{dT}\right)= \frac{dx}{dT}\frac{d^2T}{dt^2}+\frac{dT}{dt}\frac{dT}{dt}\frac{d}{dT} \left(\frac{dx}{dT}\right)= \frac{dx}{dT}\frac{d^2T}{dt^2}+\left(\frac{dT}{dt}\right)^2\frac{d^2x}{dT^2}

The "trick" is:

\frac{d}{dt}=\frac{dT}{dt}\frac{d}{dT}
 

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