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Suppose we have a function curve $\gamma(t)$ on a manifold $M$ . Define the function $$f(t)=\frac{1}{2}g_{ab}\dot\gamma(t)^a\dot\gamma(t)^b$$

Introducing coordinates $x^i$ the first derivative of the function along the curve is $$\frac{df}{dt}=\frac{\partial f}{\partial x^i}\dot\gamma(t)^i$$

I am confused in how we calculate the second derivative .

In this lecture Lecture 10: Metric Manifolds to express the Euler-Lagrange equation it is calculated like this $$\frac{d^2f}{dt^2}=\frac{\partial^2 f}{\partial x^i\partial x^j}\dot\gamma(t)^i\dot\gamma(t)^j +\frac{\partial f}{\partial x^i}\ddot\gamma(t)^i$$

but in this book Relativity on Curved Manifolds page 278, in the taylor expansion of $f$ it is calculated like this $$\frac{d^2f}{dt^2}=\frac{\partial^2 f}{\partial x^i\partial x^j}\dot\gamma(t)^i\dot\gamma(t)^j +\frac{\partial f}{\partial x^i}a^i$$ where $a^{i}=\dot\gamma^{k} \nabla_{k} \dot{\gamma}^{i}$ with $\nabla$ the covariant derivative.

Which one is correct?

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  • $\begingroup$ Is $a^{i}=\dot\gamma^{k} \nabla_{k} \gamma^{i}$ or is it $a^{i}=\dot\gamma^{k} \nabla_{k} \dot{\gamma}^{i}$? $\endgroup$ – AFG Apr 18 at 7:23
  • $\begingroup$ Actually in the book they have $a^{i}=\gamma^{k} \nabla_{k} \gamma^{i}$ but they called it accelaration am assuming they meant $^{i}=\dot\gamma^{k} \nabla_{k} \dot{\gamma}^{i}$ $\endgroup$ – amilton moreira Apr 18 at 7:28
  • $\begingroup$ Maybe they have a parametrized curve $x^i(t)$ and $\gamma^i$ is the tangent vector to that curve, i.e. $$\gamma^i=\frac{d x^i}{dt},$$ and that's why $a^i=\gamma^k\nabla_k \gamma^i$. Could it be? $\endgroup$ – AFG Apr 18 at 7:41
  • $\begingroup$ No $\dot \gamma^i=\frac{d x^i}{dt}$ $\endgroup$ – amilton moreira Apr 18 at 7:51
  • $\begingroup$ Okay, then I think they meant what you have now in the question. $\endgroup$ – AFG Apr 18 at 7:54
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The derivative of a curve on a manifold is a vector field along the curve.

It's not possible to differentiate a vector field along a curve without using a connection. This is usually left implicit in the covariant derivative.

Hence the first derivative of a curve is the usual one but to take the second derivative we need to use the covariant derivative.

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