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If I have a measure $d^4 x$ and I want to perform a conformal transformation $x^\mu \rightarrow \frac{x^\mu}{x^2}$, how do I get that the transformed measure is $\frac{d^4 x}{x^8}$?

I started by writing $d^4 x = \left| \text{det} \frac{\partial x'^\mu}{\partial x^\nu} \right| dx^\nu$, but it got somewhat crazy, and I would like to know if there is a more handy way, where that can be seen instantaneously.

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Use radial coordinates: $$ d^dx=r^{d-1}dr\, d\Omega_{d-1}, $$ where $d\Omega_{d-1}$ is the measure on $(d-1)$-dimensional sphere and contains the differentials of angles. In radial coordinates your transformation is simply $r\to 1/r$, which makes it a simple exercise.

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  • $\begingroup$ This is the best answer so far, since I also asked in my question for the most efficient way that one could come up with. Following your hint, I get the following: $d^4x’ = r’^3 dr’\ d\Omega’_3 = \left( \frac{1}{r} \right)^3 \frac{d (1/r)}{dr} dr\ d\Omega_3 = \frac{1}{r^3} \left( - \frac{1}{r^2} \right) dr\ d\Omega_3 = - \frac{1}{r^8} d^4 x$ which is the same as what I wrote in the OP, except for the $-$ sign. Am I still doing something wrong? 🙈 $\endgroup$ – Jxx Jun 12 '19 at 11:29
  • $\begingroup$ @Jxx, You also need to change the integration limits taking into account the orientation: $\int_0^\infty$ turns into $\int_\infty^0=-\int_0^\infty$ because $1/\infty=0$ etc. Alternatively you can use $|d(1/r)/dr|$ and ignore the orientation. $\endgroup$ – Peter Kravchuk Jun 12 '19 at 15:39
  • $\begingroup$ Ah yes, i didn’t think of that! Thanks a lot for your help! $\endgroup$ – Jxx Jun 12 '19 at 17:34
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There's no need to go crazy: you just have to work out

$$\frac{\partial}{\partial x^\nu} \frac{x^\mu}{|x|^2} = \frac{1}{|x|^4} \left[ x^\mu \frac{\partial |x|^2}{\partial x^\nu} - |x|^2 \frac{\partial x^\mu}{\partial x^\nu}\right] = \ldots$$ This is a concrete $4 \times 4$ matrix, and there's nothing strange about computing its determinant.

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    $\begingroup$ Thank you for your answer. Well, then I get $\partial^\nu \frac{x^\mu}{x^2} = \frac{1}{x^4} \left( 2 x^\mu x^\nu - x^2 \eta^{\mu\nu} \right)$. Taking the determinant would mean to multiply the diagonal elements together, so something like $\frac{1}{x^{16}} \left( 2x_{0}^2 - x^2 \right)...\left( -2x_{3}^2 - x^2 \right)$ and then add the off-diagonal contributions to the determinant, that is all the terms containing $x^0x^1$, $x^0x^2$ and so on... Is that the way you suggest? It does seem like heavy algebra just to get $\frac{d^4 x}{x^8}$. $\endgroup$ – Jxx Jun 11 '19 at 21:02
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    $\begingroup$ Determinant = product of eigenvalues. If you can guess the eigenvectors, you can get the eigenvalues. Also, the Jacobian should have one contravariant and one covariant index $\endgroup$ – Subhaneil Lahiri Jun 11 '19 at 22:30
  • $\begingroup$ Hint for guessing eigenvectors: what's the one special vector in the problem? For the rest, eigenvectors of symmetric matrices are... $\endgroup$ – Subhaneil Lahiri Jun 11 '19 at 22:37
  • $\begingroup$ (Hint 2: the special eigenvector is $x^\mu$, the other eigenvectors are orthogonal to $x^\mu$.) $\endgroup$ – Hans Moleman Jun 12 '19 at 12:59
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Note simply $d^4 x = dx^0 dx^1 dx^2 dx^3$ and perform the transformation for each component.

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  • $\begingroup$ That's a good idea, and thanks for your answer. Here is what I get for one component: $dx_0 = d\left(\frac{x_0}{x^2} \right)=\frac{dx_0}{x^2} + x_0 \frac{d(1/x^2)}{dx^2}\frac{dx^2}{dx_0} dx_0 = \left( \frac{1}{x^2} - 2 \frac{x_0^2}{x^4} \right) dx_0$. The first term is all I need, but why do I get this annoying second term? I guess I am doing something wrong somewhere. $\endgroup$ – Jxx Jun 11 '19 at 20:47
  • $\begingroup$ @Jxx why would it work with only doing it for one component? $\endgroup$ – lurscher Jun 11 '19 at 22:18
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    $\begingroup$ Since the change of variable does not factorize, you cannot compute the Jacobian separately for each component. In this form one really does need to compute the Jacobian. $\endgroup$ – Peter Kravchuk Jun 12 '19 at 0:42

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