Chain rule when taking vector derivatives

[Coffee_]

Consider a function of several variables $T=T(x_{1},...,x_{3N})$ Let's say I have N vectors of the form $\vec{r_{1}}=(x_1,x_{2},x_{3})$ and $x_j=x_j(q_1,...,q_n)$. Awkward inex usage but the point is just that the each variable is contained in exactly 1 vector.

Is it correct to in general use the chain rule in this way? :

$\frac{\partial T}{\partial q_j} = \sum\limits_{k=1}^N \frac{\partial T}{\partial \vec{r_k}}.\frac{\partial \vec{r_k}}{\partial q_j}$

Where the notation $\frac{\partial T}{\partial \vec{r_k}}$ is just $(\frac{\partial T}{\partial x_k},..., \frac{\partial T}{\partial x_{k+2}})$

 

[jedishrfu]

There's something wrong with your Latex, please edit your post to fix it.

 

[Coffee_]

There's something wrong with your Latex, please edit your post to fix it.

Done

 

[PeroK]

Consider a function of several variables $T=T(x_{1},...,x_{3n})$. Let's call vector $\vec{r_{k}}=(x_{k},x_{k+1},x_{k+2})$ in $R^{3}$

Is it correct to in general use the chain rule in this way? :

$\frac{\partial T}{\partial x_j} = \sum\limits_{k=1}^n \frac{\partial T}{\partial \vec{r_k}}.\frac{\partial \vec{r_k}}{\partial x_j}$

Where the notation $\frac{\partial T}{\partial \vec{r_k}}$ is just $(\frac{\partial T}{\partial x_k},..., \frac{\partial T}{\partial x_{k+2}})$

The way you've written this, I'd infer that the x's are all independent variables, so it makes no sense to differentiate one wrt another.

 

[Coffee_]

The way you've written this, I'd infer that the x's are all independent variables, so it makes no sense to differentiate one wrt another.

Yeah totally right. I've edited it, correctly this time I hope.

 

[Ray Vickson]

Consider a function of several variables $T=T(x_{1},...,x_{3N})$ Let's say I have N vectors of the form $\vec{r_{1}}=(x_1,x_{2},x_{3})$ and $x_j=x_j(q_1,...,q_n)$. Awkward inex usage but the point is just that the each variable is contained in exactly 1 vector.

Is it correct to in general use the chain rule in this way? :

$\frac{\partial T}{\partial q_j} = \sum\limits_{k=1}^N \frac{\partial T}{\partial \vec{r_k}}.\frac{\partial \vec{r_k}}{\partial q_j}$

Where the notation $\frac{\partial T}{\partial \vec{r_k}}$ is just $(\frac{\partial T}{\partial x_k},..., \frac{\partial T}{\partial x_{k+2}})$

Do you mean that you have $\vec{r}_1 = (x_1,x_2,x_3)$, $\vec{r}_2 = (x_4, x_5, x_6)$, etc? Using the notation $\xi_i(q_1,q_2, \ldots, q_n)$ instead of $x_i(q_1, q_2, \ldots, q_n)$ (just so as to keep straight the distinction between the variable $x_i$ and the function $\xi_i$ that delivers you the value of $x_i$), it seems you are asking for the derivative of
$$T = {\cal T}(\xi_1(q_1, q_2, \ldots, q_n), \xi_2(q_1, q_2, \ldots, q_n) , \dots, \xi_{3N}(q_1, q_2, \ldots, q_n)),$$
again being careful to distinguish between the variable $T$ and the function that delivers you $T$ (which I call ${\cal T}$).

 

[Coffee_]

Do you mean that you have $\vec{r}_1 = (x_1,x_2,x_3)$, $\vec{r}_2 = (x_4, x_5, x_6)$, etc? Using the notation $\xi_i(q_1,q_2, \ldots, q_n)$ instead of $x_i(q_1, q_2, \ldots, q_n)$ (just so as to keep straight the distinction between the variable $x_i$ and the function $\xi_i$ that delivers you the value of $x_i$), it seems you are asking for the derivative of
$$T = {\cal T}(\xi_1(q_1, q_2, \ldots, q_n), \xi_2(q_1, q_2, \ldots, q_n) , \dots, \xi_{3N}(q_1, q_2, \ldots, q_n)),$$
again being careful to distinguish between the variable $T$ and the function that delivers you $T$ (which I call ${\cal T}$).

Yeah. Maybe I should have included the physics context since this is a physics forum. Would have messed up less and wasted less of people's time by converting it into math format wrongly.

Basically T is supposed to be a kinetic energy function of a scleronomic system of particles. Where q's represent generalized coordinates and the r-vectors represent the positions of the particles in a carthesian frame.

I was wondering how to formulate the chain rule this way when taking the partial derivative of T wrt. of one of the generalized coordinates. In case you are not familiar with Lagrangian mechanics just nevermind what I said and yes you seem to have stated my problem very well.