# Second derivative of a function in a manifold [closed]

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?

• Is $a^{i}=\dot\gamma^{k} \nabla_{k} \gamma^{i}$ or is it $a^{i}=\dot\gamma^{k} \nabla_{k} \dot{\gamma}^{i}$? – AFG Apr 18 at 7:23
• Actually in the book they have $a^{i}=\gamma^{k} \nabla_{k} \gamma^{i}$ but they called it accelaration am assuming they meant $^{i}=\dot\gamma^{k} \nabla_{k} \dot{\gamma}^{i}$ – amilton moreira Apr 18 at 7:28
• 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? – AFG Apr 18 at 7:41
• No $\dot \gamma^i=\frac{d x^i}{dt}$ – amilton moreira Apr 18 at 7:51
• Okay, then I think they meant what you have now in the question. – AFG Apr 18 at 7:54