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cianfa72
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- On the proof of chain rule applied to the definition of scalar product expression for the special case of gradient
Hi, I'm keep studying The Road to Reality book from R. Penrose.
In section 12.4 he asks to give a proof, by use of the chain rule, that the scalar product ##\alpha \cdot \xi=\alpha_1 \xi^1 + \alpha_2 \xi^2 + \dots \alpha_n \xi^n## is consistent with ##df \cdot \xi## in the particular case ##\alpha = df## for a scalar function ##f##.
My idea is to use the chain rule in a given chart around a point ##P## in the manifold ##M##. Pick a coordinate chart ##\phi: U \rightarrow \mathbb R^n## and a smooth curve ##\gamma (t)## in ##M##. Then in that chart the curve is ##\phi \circ \gamma (t)##. The class of equivalence of smooth curves with the same derivative $$\left. \frac {d(\phi \circ \gamma)} {dt} \right|_{\phi(P)}$$ at ##\phi(P)## defines the tangent vector ##\xi## in ##P##.
Now ##\xi(f)## is defined as the derivative of ##f## along one of the curves in the equivalence class. By definition in the given chart we get: $$\frac {d(f \circ {\phi}^{-1} \circ \phi \circ \gamma (t))} {dt}$$ By the chain rule $$\frac {\partial {(f \circ {\phi}^{-1})}} {\partial x^i} \cdot \left. \frac {d(\phi \circ \gamma(t))} {dt} \right|_{\phi(P)}$$The first term are the components ##(\alpha_1, \alpha_2 \dots \alpha_n)## of ##df## in the chart and the latter the components ##(\xi_1, \xi_2 \dots \xi_n)## of tangent vector ##\xi## in the chart. Hence the result holds.
Does it make sense ? Thank you.
In section 12.4 he asks to give a proof, by use of the chain rule, that the scalar product ##\alpha \cdot \xi=\alpha_1 \xi^1 + \alpha_2 \xi^2 + \dots \alpha_n \xi^n## is consistent with ##df \cdot \xi## in the particular case ##\alpha = df## for a scalar function ##f##.
My idea is to use the chain rule in a given chart around a point ##P## in the manifold ##M##. Pick a coordinate chart ##\phi: U \rightarrow \mathbb R^n## and a smooth curve ##\gamma (t)## in ##M##. Then in that chart the curve is ##\phi \circ \gamma (t)##. The class of equivalence of smooth curves with the same derivative $$\left. \frac {d(\phi \circ \gamma)} {dt} \right|_{\phi(P)}$$ at ##\phi(P)## defines the tangent vector ##\xi## in ##P##.
Now ##\xi(f)## is defined as the derivative of ##f## along one of the curves in the equivalence class. By definition in the given chart we get: $$\frac {d(f \circ {\phi}^{-1} \circ \phi \circ \gamma (t))} {dt}$$ By the chain rule $$\frac {\partial {(f \circ {\phi}^{-1})}} {\partial x^i} \cdot \left. \frac {d(\phi \circ \gamma(t))} {dt} \right|_{\phi(P)}$$The first term are the components ##(\alpha_1, \alpha_2 \dots \alpha_n)## of ##df## in the chart and the latter the components ##(\xi_1, \xi_2 \dots \xi_n)## of tangent vector ##\xi## in the chart. Hence the result holds.
Does it make sense ? Thank you.
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