Tagged Questions
5
votes
1answer
95 views
showing zero curvature implies a line
How can I show that a given (not necessarily unit-speed) parametrization $\gamma(t)$ of a curve in $\mathbb{R}^3$ which exhibits zero curvature is a line ? What I know is that zero curvature means ...
1
vote
1answer
62 views
curvature of space curve
I am slightly confused by the following curve $\gamma(t) = (e^t,0,0)$ in $\mathbb{R}^3$. Its curvature, defined as
$$
\kappa(t) = \frac{\|\dot \gamma(t) \times \ddot \gamma(t)\|}{\|\dot \gamma(t)\|^3}
...
7
votes
1answer
168 views
Directional derivative of vector field
I am trying to compute the directional derivative of a vector field $V$ along a direction $U$.
Actually, my vector field is initially only defined on a curve $\gamma(t)$ in a Riemannian manifold $(M, ...
1
vote
1answer
37 views
What are $\partial/\partial f^j$ in Jost's definition of the differential mapping?
Let $M$ be a $d$-manifold and $x_0=(x^1,x^2,\cdots, x^d)\in M$, Jost defines the tangent space at $x_0$ to be \begin{equation}\{x_0\}\times \operatorname{span}\left\{\frac{\partial}{\partial ...
2
votes
1answer
49 views
Equivalence of two definitions of path (in $\mathbb{R}^3$) length
In a previews question I asked here I used the following definition of path length:$\gamma=(x(t),y(t),z(t))$ : $L(\gamma)=\intop_{a}^{b}\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}$.
In the answer ...
6
votes
1answer
245 views
Coordinate-free differentiation techniques in Riemannian geometry
I encountered the following identities while reading this article on global calculus (p. 10):
$$
d(\|df\|^2)=2\mathop{\iota_{\mathop{\mathrm{grad}} f}} \mathop{\mathrm{Hess}} f,
$$
$$
...
