0
$\begingroup$

How to prove that the expression $\varphi_{,ij}:=\frac{\partial^2\varphi}{\partial x_i\partial x_j}=\nabla\nabla\varphi$ is a tensor of second order where $\varphi$ is a scalar? Furthermore, how to prove that $a\times b:=a_i b_j\varepsilon_{ijk}e_k$ is a vector?

We can either prove it by definition or use the so-called "tensor recognition theorem" claiming that if $p_{i_1i_2\cdots i_mj_1j_2\cdots j_n}q_{j_1\cdots j_n} = r_{i_1\cdots i_m}$, then $p$ must be a tensor of order $m+n$, where $q_{j_1\cdots j_n}$ is a tensor of order $n$ and $r_{i_1\cdots i_m}$ a tensor of order $m$.

$\endgroup$
0
$\begingroup$

$1$) Suppose that $O X_1 X_2 X_3$ is the coordinate system corresponding to the given basis $e_1, e_2, e_3$ on $E_3$. The gradient of a scalar function $\varphi (X_1,X_2,X_3)$, using this coordinate system, is defined by $$\nabla \phi(X_1,X_2,X_3) = \frac{\partial \varphi}{\partial X_i} e_i.$$

Suppose that $\tilde{A} = \{ \tilde{e}_1,\tilde{e}_2,\tilde{e}_3 \}$ is another orthonormal basis with corresponding cartesian coordinates $O \tilde{X_1} \tilde{X_2} \tilde{X_3}$ given by $\tilde{X}_i = q_{ij} X_j$, where $Q = (q_{ij}) \in SO(3)$.

By the chain rule, $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = \frac{\partial \varphi}{\partial X_k} \frac{\partial X_k}{\partial \tilde{X}_i}.$$

Using $\tilde{X}_i = q_{ij} X_j$, we have $$\frac{\partial X_k}{\partial \tilde{X}_i} = q_{ik},$$ and hence $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = q_{ik} \frac{\partial \varphi}{\partial X_k}.$$

That is, by definition, $\nabla \varphi$ is a cartesian tensor of order $1$. Doing this again, but with $\nabla \varphi$ in place of $\varphi$, shows that ${\nabla}^{2} \varphi$ is a cartesian tensor of order $2$.

$2$) $a$ and $b$ are vectors so $a_i$ and $b_j$ are components of cartesian tensors of order $1$, and the alternating tensor $\varepsilon = \varepsilon_{klm}$ is a cartesian tensor of order $3$.

Taking their product we get that $a_i b_j \varepsilon_{klm}$ are the components of a cartesian tensor of order $5$.

Contracting indices (that is, setting two indices equal and thus effecting a sum) gives that $a_i b_j \varepsilon_{ilm}$ are the components of a cartesian tensor of order $4$, and contracting again gives that $a_i b_j \varepsilon_{ijm}$ are the components of a cartesian tensor of order $3$.

Hence, $a \times b$ is a cartesian tensor of order $3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.