- #1

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Also define an arbitrary (dummy) scalar field ##f = C^{\infty}(M)##, and $$\bar{\gamma} = \phi_\alpha \circ \gamma, \qquad \qquad \bar{f} = f \ \circ \phi_\alpha^{-1}.$$ Here ##\phi_\alpha(\mathbf{P}')## gives the coordinates of the point ##\mathbf{P'}## on the manifold, and ##\phi_\alpha^{-1}## is its inverse. The tangent vector of ##\gamma## at ##\mathbf{P}## is then given as follows, $$\dot{\gamma}(0)f = \frac{d}{dt}(f \ \circ \gamma)|_{t=0} =\frac{d}{dt}(\bar{f} \circ \phi_\alpha \circ \phi_\alpha^{-1} \circ \bar{\gamma})|_{t=0} = \frac{d}{dt}(\bar{f} \circ \bar{\gamma})|_{t=0} = \dot{x}^i(0) \frac{\partial\bar{f}}{\partial x^i}|_{x(0)}.$$

Here ##\dot{x}^i(t) = \bar{\dot{\gamma}}^i(t)## and ##x^i(t) = \bar{\gamma}^i(t)## are coordinates. I understand everything up to this point. Next however the following step is introduced, $$\dot{x}^i(0) \frac{\partial\bar{f}}{\partial x^i}|_{x(0)} = \dot{x}^i(0) \frac{\partial f}{\partial x^i}|_{\gamma(0)}.$$

I cannot understand or derive why this holds. The lhs makes sense to me, since ##\bar{f}## takes coordinates as its arguments, and thus differentiating to coordinates seems logical. However on the rhs we are suddenly differentiating ##f##, which takes a point on the manifold ##\mathbf{P}## as its argument. How can we still derive this to the coordinates? Trying to fill in the definitions of ##\bar{f}## and ##\bar{\gamma}## on the lhs also doesn't seem to take me anywhere. Any help would be greatly appreciated.