I have questions regarding Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation ISBN 978-0-7167-0344-0. It is a book about Einstein's theory of gravitation. There is one exercise that I would test if I understood what it is about, and if someone could give me the solution. In page 234, the authors have presented the beginning of differential topology with the introduction of vectors and tensors to a metric-free space. And then they present the following exercise for beginners at differential topology:
Exercise 9.2. Practice Manipulating Tangent Vectors
Let $P_0$ be the point with coordinates $(x=0, y = 1, z=0)$ in a three-dimensional space; and define three curves through $P_0$ and define three curves through $P_0$ by: $$P(\lambda)=(\lambda,1,\lambda)$$ $$P(\xi)=(sin(\xi),cos(\xi),\xi)$$ $$P(\rho)=(sinh(\rho),cosh(\rho),\rho + \rho^3)$$ (a) Compute $(d/d\lambda)f$,$(d/d\xi)f$ and $(d/d\rho)f$ for the function $f=x^2 - y^2 +z^2$ at the point $P_0$ (b) Calculate the components of the tangent vectors $d/d\lambda$,$d/d\xi$ and $d/d\rho$ at $P_0$ using the basis $\{\partial/\partial x,\partial / \partial y , \partial / \partial z \}$.
$$(d/d\lambda)f = (\partial x / \partial \lambda) (\partial f/ \partial x) + (\partial y / \partial \lambda) (\partial f/ \partial y) + (\partial z / \partial \lambda) (\partial f/ \partial z)$$
From here I was kind of how to compute $(\partial x / \partial \lambda), (\partial y / \partial \lambda), (\partial z / \partial \lambda)$ on the expression given. However, the other components of the gradient of $f$ are obviously easy to compute. Or is this reasoning entirely wrong?
For (b), I compute the derivatives of the P's with respect to the variables:
$$P'_\lambda = (1,0,1)$$ $$P'_\xi = (cos(\xi)',-sin(\xi),1)$$ $$P'_\rho = (cosh(\rho),sinh(\rho),1+2\rho^2)$$
Then I need to express them in the basis $\{\partial/\partial x,\partial / \partial y , \partial / \partial z \}$ but I am not sure how to do that. The only thing that lacks me is the link with the greek letters variables and the latin letters variables so I must be close. The only link that I see between them is the components of the P's curves $P_\lambda \cdot e_x = \lambda$, $P_\lambda \cdot e_y = 1$ and $P_\lambda \cdot e_z = \lambda$ ( I am not even sure that this dot product is correct in differential topology).
- Is my reasoning correct?
- What am I missing?
- Can someone please provide a solution for both (a) and (b) please?
In this answer, I propose a detailed solution to your problem. I frame everything in the precise language of differential geometry (cfr. Lee, Smooth Manifolds, 2nd ed., 2013): please forgive the occasional notational changes motivated by this choice. My hope is that this approach will help you resolve some of your doubts.
(i) The standard tangent basis. We are working on the manifold $M \equiv \mathbb R^3$ with the (global) coordinate chart given by the identity map: $$\chi : M \to \mathbb R^3 \quad p \mapsto \chi(p)= (p^1,p^2,p^3), $$ where $p^j$ indicates the $j$-th component of the element $p\in M$.
Any coordinate chart induces a natural basis onto the tangent space $T_pM$, namely the set $$\left\{\frac{\partial}{\partial \chi^1}\Bigg{|}_p, \frac{\partial}{\partial \chi^2}\Bigg{|}_p, \frac{\partial}{\partial \chi^3}\Bigg{|}_p \right\}, \tag1$$ where, indicating by $\partial_i$ the partial derivative of a real function on $\mathbb R^n$ w.r.t. the $i$-th variable, $$\quad \forall g: M \to \mathbb R \quad \frac{\partial}{\partial\chi^i}\Bigg|_p[g] = \frac{\partial g}{\partial \chi^i}(p) := \partial_i(g \circ \chi^{-1} )(\chi(p)); \tag2$$ due to the simple form of $\chi$ in our situation (it sends points of $M \equiv \mathbb R^3$ to themselves), we have for all $i = 1,2,3$ that $$\frac{\partial g}{\partial \chi^i}(p) = \partial_i(g \circ \chi^{-1})(\chi(p)) = \partial_i g (p). \tag3$$ In other words, the basis $(1)$ reduces to the the basis of standard partial derivatives at $p$, which the authors of Gravitation simply call $\{\partial/\partial x, \partial /\partial y, \partial / \partial z\}$. Notice that this would not have been the case had $\chi$ been a different set of coordinates, e.g. affine coordinates.
(ii) The curves. Now let $p_0$ be the point corresponding to the coordinates $\chi(p_0) = (0,1,0)$, and let $\gamma_1, \gamma_2, \gamma_3 : (-\varepsilon,\varepsilon) \to M$, for some $\varepsilon > 0$, be three curves through the point $p_0$ such that, in coordinates, $$\begin{split} (\chi \circ \gamma_1)(\lambda) &= (\lambda,1,\lambda), \\ (\chi \circ \gamma_2)(\xi) &= (\sin\xi, \cos\xi, \xi), \\ (\chi \circ \gamma_3)(\rho) &= (\sinh \rho, \cosh \rho, \rho + \rho^3), \end{split}$$ for $\lambda,\xi,\rho \in (-\varepsilon,\varepsilon)$.
All three curves pass through $p_0$ when the parameter takes the value $0$, and at that point they define a tangent vector which we shall call $\gamma_i'(0)$ for $i=1,2,3$. This vector, which is what the authors call $d/d\lambda$ when $i=1$, $d/d\xi$ when $i=2$, and $d/d\rho$ when $i=3$, is obviously an element of the tangent space $T_{p_0}M$ for all three curves, and can therefore be expressed in terms of the standard basis of vectors given in $(1)$.
(iii) The tangent vectors $\gamma’_i(0)$. To see how we can do so, let's fix an arbitrary smooth function $g : M \to \mathbb R$ and $i$, and let's calculate the derivative of the function $(g \circ \gamma_i) : (-\varepsilon,\varepsilon) \to \mathbb R$: this will yield the value of $\gamma_i'(0)$ applied to $g$. By the chain rule, indicating $\partial_t \equiv \partial_1$ (where $t$ is the parameter in $(-\varepsilon,\varepsilon)$),
$$\partial_t(g \circ \gamma_i)(0) = \partial_t(g \circ \chi^{-1} \circ \chi \circ \gamma_i)(0) = \sum_{k=1}^3 \partial_k(g \circ \chi^{-1})((\chi \circ \gamma_i)(0)) \cdot \partial_t(\chi \circ \gamma_i^k)(0); $$ here, we recognize the definition of the $k$-th vector in the standard basis of $T_{p_0}M$, as given in $(2)$, so that we may rewrite $$\partial_t(g \circ \gamma_i)(0) = \partial_t(\chi \circ \gamma_i^k)(0) \frac{\partial g}{\partial \chi^k}(\gamma_i(0)). \tag4$$ Erasing the arbitrary function $g$ from the equation, we transform an equality between real numbers into an equality between tangent vectors: $$\gamma'_i(0) = \partial_t(\chi \circ \gamma_i^k)(0) \frac{\partial }{\partial \chi^k}\Bigg|_{\gamma_i(0)}. \tag5 $$
For example, if $i=1$, eq. $(5)$ yields $$ \gamma'_1(0) = \sum_{k=1}^3 \partial_\lambda(\chi \circ \gamma_1^k)(0) \frac{\partial}{\partial \chi^k} \Bigg|_{\gamma_1(0)} = (1)\frac{\partial}{\partial \chi^1}\Bigg|_{p_0} + (1) \frac{\partial}{\partial \chi^3}\Bigg|_{p_0}, $$ or, if $i = 3$, $$\gamma'_3(0) = \sum_{k=1}^3 \partial_\rho(\chi \circ \gamma_3^k)(0) \frac{\partial}{\partial \chi^k} \Bigg|_{\gamma_3(0)} = \cosh(0)\frac{\partial}{\partial \chi^1}\Bigg|_{p_0} + \sinh(0) \frac{\partial}{\partial \chi^2}\Bigg|_{p_0} + (1 + 3(0)^2)\frac{\partial}{\partial \chi^3}\Bigg|_{p_0}.$$
(iv) The function $f$. Now let $f : M \to \mathbb R$ be the specific smooth function such that $$p \mapsto f(p) = (p^1)^2 - (p^2)^2 + (p^3)^2; $$ the exercise asks us to apply the tangent vectors $\gamma_i'(0)$ to this $f$ and compute the result. Pick e.g. $i=2$: then we may simply substitute $f$ in place of the arbitrary $g$ in eq. $(4)$ and simplify using $(3)$: $$\begin{split} \gamma_2'(0) [f] &= \sum_{k=1}^3 \partial_\xi(\chi \circ \gamma_2^k)(0) \frac{\partial f}{\partial \chi^k}(p_0) = \sum_{k=1}^3 \partial_\xi(\chi \circ \gamma_2^k)(0)\cdot \partial_k f (p_0) \\ &= \cos(0) (2 p_0^1) - \sin(0) (-2p^2_0) + (1) (2p^3_0) = 0. \end{split} $$ The same procedure can be applied to the cases $i=1$ and $i=3$.