Derivative of a Vector with respect to its norm (special relativity) I came across an equation (related to special relativity) that requires me to to take a derivative of a vector with respect to to it's own norm. In a bit more detail, what I mean is, let:
$$\vec T(\tau) = \{T_1(\tau), T_2(\tau), \dotsb, T_t(\tau)\}$$
$$\vec X(\tau) = \{X_1(\tau), X_2(\tau), \dotsb, X_x(\tau)\}$$
With $\vec T$ and $\vec X$ having dimensions $t$ and $x$ respectively and both being vector functions of $\tau$.
Also let:
$$\tau^2 = w^2 - s^2 = \vec T \cdot \vec T - \vec X \cdot \vec X$$
Where:
$$w = \lVert \vec T \rVert \qquad s = \lVert \vec X \rVert$$
Knowing that:
$$d\tau^2 = dw^2 - ds^2= d\vec T \cdot d\vec T - d\vec X \cdot d\vec X$$
How then, can take the derivative $\frac {d\vec T}{dw}$?
It may be useful to know that $d\vec X = [\beta]\cdot d\vec T$, where $[\beta]$ is a constant matrix of dimension $x\times t$.
Thanks in advance.
 A: I'll trade your question for an easier one, perhaps my thoughts on this two-dimensional problem may give you some ideas. Consider $\mathbb{R}^2$ with time-space coordinates $(t,x)$ and the metric $\eta = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$. Note the interval $I((t_1,x_1),(t_2,x_2)) = (t_2-t_1)^2-(x_2-x_1)^2$ can be compactly expressed as $I(v,w) = (w-v)^T\eta (w-v)$. Getting back to the main point here, note the "norm" of $v=(v_1,v_2)$ is given by
$$ |v| = \sqrt{|I(v,v)|}  = \sqrt{|v_1^2-v_2^2|}. $$
Why the quotes? Well, as there are vectors like $(1,1)$ which are nonzero and yet have "norm" zero we should probably call it a pseudo-norm. That said, differentiation with respect to $|v|$ probably should be understood in terms of a choice of coordinates for which the psuedo-norm is one of the coordinates. The natural second choice in this hyperbolic context is that of the hyperbolic angle (rapidity in physics). In particular, define: $r \geq 0$ and
$$ x = r\cosh(\phi) \qquad \& \qquad y = r\sinh(\phi) $$
This suggests relations $x^2-y^2=r^2$ and $\tanh(\phi) = y/x$. Now, these coordinates are tricky, the formulas above only apply to $x,y >0$. To extend to other quadrants similar formulas can be written. Let's focus on quadrant I where the formulas given are reasonable. To convert a partial derivative with respect to $r$ into $x,y$ derivatives the standard technique is the chain rule:
$$ \partial_r=\frac{\partial}{\partial r} = \frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y} = \cosh(\phi) \partial_x + \sinh(\phi) \partial_y$$
I suppose that is a partial answer to your post, but I'll continue for a bit to show a bit more about this approach. We can also relate $\partial_{\phi}$ to the Cartesian coordinate derivations:
$$ \partial_{\phi}=\frac{\partial}{\partial \phi} = \frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y} = r\sinh(\phi) \partial_x + r\cosh(\phi) \partial_y$$
Consider, the psuedo-Laplacian 
$$ \partial_x^2-\partial_y^2= (\partial_x+\partial_y)(\partial_x-\partial_y)$$
A short calculation shows $\partial_x = \cosh (\phi) \partial_r-\frac{1}{r}\sinh(\phi)\partial_{\phi}$ and $\partial_y = -\sinh (\phi) \partial_r+\frac{1}{r}\cosh(\phi)\partial_{\phi}$. Note $\cosh(\phi) \pm \sinh(\phi) =e^{\pm \phi}$. This gives
$$ \partial_x^2-\partial_y^2= e^{-\phi}\left(\partial_r+\frac{1}{r}\partial_{\phi}\right)\left[ e^{\phi}\left(\partial_r-\frac{1}{r}\partial_{\phi} \right) \right] $$
I suppose this could be simplified some, but, I expect there will be a mixed derivative term. This is similar to the problem of writing the Laplacian in polar coordinates. There is a mixed-derivative term in that problem. Also, it seems to me that is similar to your question in the sense that differentiation with respect to the polar radius is in a sense differentiation with respect to the Euclidean norm. Similar comments apply to spherical coordinates and differentiation with respect to $\rho = \sqrt{x^2+y^2+z^2}$. Differentiation in curvelinear coordinates required some care. I hope this gives you some ideas about what perhaps you should do...
