1
$\begingroup$

Suppose we have two coordinate systems (Cartesian and spherical)

$$x^{\mu} = (t,x,y,z)$$

$$x'^{\mu'} = (t',r,\theta,\phi)$$

where $r= \sqrt{x^2 + y^2 + z^2} , \theta = \cos^{-1}(z/r), \phi = \tan^{-1} (y/x)$. My question is, in general, what are the components of a vector $A_{\mu} = (A_t,A_x,A_y,A_z)_{\mu}$ in the primed coordinates? From GR, I believe the answer is $A'_{\mu'} = (A_{t'},A_{r},A_{\theta},A_{\phi})_{\mu'} = \frac{\partial x^{\mu}}{\partial x'^{\mu'}} A_{\mu}$, with the inverse matrix used for upper-index vectors.

If this is the case, in particular it should work for position vectors. That is, $x'^{\mu'} = \frac{\partial x'^{\mu'}}{\partial x^{\mu}} x^{\mu}$. However, applying this transformation gives $x'^{\mu'} = (t',r,0,0)$, not $(t',r,\theta,\phi)$. Am I doing something wrong?

Edit: The second paragraph incorrectly applies the formula I've cited, as pointed out by mike stone.

As for the first question, since we have $x'_r = \sqrt{x_1^2 +x_2^2 + x_3^2}, x'_{\theta} =\cos^{-1}(x_3/x'_r)$,$x'_{\phi} = \tan^{-1}(x_2/ x_1)$, does it follow for any vector $A'_{\mu}$ (for instance, the EM gauge field) that $A'_r = \sqrt{A_1^2 + A_2^2 + A_3^2}$, $A'_{\theta} = \cos^{-1}(A_3/ A'_r)$, and $A'_{\phi} = \tan^{-1}(A_2/A_1)$?

$\endgroup$
  • 1
    $\begingroup$ You don't need to list an edit history. This is automatically done. Just make the appropriate edits to your equations. $\endgroup$ – Aaron Stevens Sep 21 '18 at 17:32
1
$\begingroup$

Your transformation matrix:

I will ignore the "t" coordinate

\begin{align*} &\text{The position vector for a sphere is: } \\ &\vec{R_s}= \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \left[ \begin {array}{c} r\cos \left( \vartheta \right) \cos \left( \varphi \right) \\ r\cos \left( \vartheta \right) \sin \left( \varphi \right) \\ r\sin \left( \vartheta \right) \end {array} \right]&(1) \\\\ &\text{we can now calculate the transformation matrix $R$:}\\\\ &R=J\,H^{-1}\\ &\text{$J$ is the Jakobi matrix }\quad\,, J=\frac{\partial\vec{R_s}}{\partial\vec{q}}\quad \text{with:}\\ &\vec{q}=\begin{bmatrix} r \\ \varphi \\ \vartheta \\ \end{bmatrix}\quad, H=\sqrt{G_{ii}}\,,H_{ij}=0\quad\text{and } G=J^{T}\,J\quad\text{the metric.}\\\\ &\Rightarrow\\\\ &R=\left[ \begin {array}{ccc} \cos \left( \varphi \right) \cos \left( \vartheta \right) &-\sin \left( \varphi \right) &-\cos \left( \varphi \right) \sin \left( \vartheta \right) \\ \sin \left( \varphi \right) \cos \left( \vartheta \right) &\cos \left( \varphi \right) &-\sin \left( \varphi \right) \sin \left( \vartheta \right) \\ \sin \left( \vartheta \right) &0&\cos \left( \vartheta \right) \end {array} \right]&(2) \end{align*} \begin{align*} &\text{We can solve equation (1) for $r\,,\varphi$ and $\vartheta$}\\\\ &r=\sqrt{x^2+y^2+z^2}\\ &\varphi=\arctan\left(\frac{y}{x}\right)\\ &\vartheta=\arctan\left(\frac{z}{\sqrt{x^2+y^2}}\right)\\\\ &\text{and with equation (2):}\\\\ &R= \left[ \begin {array}{ccc} {\frac {xz}{\sqrt {{y}^{2}+{x}^{2}}r}}&-{ \frac {y}{\sqrt {{y}^{2}+{x}^{2}}}}&-{\frac {x}{r}} \\ {\frac {yz}{\sqrt {{y}^{2}+{x}^{2}}r}}&{\frac {x} {\sqrt {{y}^{2}+{x}^{2}}}}&-{\frac {y}{r}}\\ {\frac {\sqrt {{y}^{2}+{x}^{2}}}{r}}&0&{\frac {z}{r}}\end {array} \right] &(3)\\\\ &\text{The components of a vector can transformed either with equation (2) or with equation (3) } \end{align*}

$\endgroup$
0
$\begingroup$

The GR vector transformation you cite applies to elements of the tangent space at a point. Positions are not vectors in any tangent space, so coordinates $x^\mu$ do not transform as vectors.

$\endgroup$
  • $\begingroup$ Thanks. That answers the second part. I've edited; could you take a look? $\endgroup$ – Dwagg Sep 21 '18 at 17:28

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.