Find the tensor that carries out a transformation

[Frostman]

Homework Statement:

Following the non-linear transformation of coordinates (conformal transformation)
$x'^{\mu}=\frac{x^{\mu}-x^2a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}$
with $a^{\mu}$ constant four-vector, the relativistic line element is transformed (obviously) in the following way
$g_{\mu \nu}dx'^{\mu}dx'^{\nu}=K_{\mu \nu}dx^{\mu}dx^{\nu}$

Relevant Equations:

Tensor calculus

I got stuck in this calculation, I can't collect everything in terms of $dx^{\mu}$.

$x'^{\mu}=\frac{x^{\mu}-x^2a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}$

$x'^{\mu}=\frac{x^{\mu}-g_{\alpha \beta}x^{\alpha}x^{\beta}a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}$

$dx'^{\mu}=\frac{dx^{\mu}-a^{\mu}g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}$

$dx'^{\mu}=\frac{dx^{\mu}-2dx\cdot xa^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a\cdot dx+2a^2dx \cdot x)}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}$

$dx'^{\mu}=\frac{(1-2x_{\mu}a^{\mu})dx^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-(2a_{\mu} -2a^2 x_{\mu})dx^{\mu})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}$

I stop here, because in the second fraction there is that 1 in the numerator that I don't know how to collect with respect to $dx^{\mu}$.

I suppose to express $1$ as $\frac{dx_{\mu}dx^{\mu}}{\sqrt{dx^2}}$ but I don't know if is it correct because I leave inside the $K_{\mu \nu}$ this term $\frac{dx_{\mu}}{\sqrt{dx^2}}$

 

[andrewkirk]

That 1 that is causing the problem should not be there.
It disappears in line 3 because the differential of the denominator is
$$-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})
$$
because the derivative of 1 with respect to any variable is 0.

 

[haushofer]

To add to andrewkirk: your expression contains mu both as free index and dummy index. In combination with the fact that the 1 without indices should ring alarmbells, you should probably go back to the basics of tensor calculus before considering conformal transformations.

 

[PeroK]

That 1 that is causing the problem should not be there.
It disappears in line 3 because the differential of the denominator is
$$-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})
$$
because the derivative of 1 with respect to any variable is 0.

Is there an assumption here that $g_{\alpha \beta} = \eta_{\alpha \beta}$?

 

[Frostman]

That 1 that is causing the problem should not be there.
It disappears in line 3 because the differential of the denominator is
$$-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})
$$
because the derivative of 1 with respect to any variable is 0.

Damn, you're right.

$dx'^{\mu}=\frac{(1-2x_{\mu}a^{\mu})dx^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}-\frac{(x^{\mu}-x^2a^{\mu})(2a^2 x_{\mu}-2a_{\mu})dx^{\mu}}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}$

$dx'^{\mu}=(\frac{1-2x_{\mu}a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}-\frac{(x^{\mu}-x^2a^{\mu})(2a^2 x_{\mu}-2a_{\mu})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2})dx^{\mu}$

Renaming

$Q = 1-2a_{\nu}x^{\nu}+a^2x^2$

$dx'^{\mu}=\Big[\frac{1-2x \cdot a}{Q}-\frac{4x^2a^2-2x \cdot a -2x^2 a^2 x \cdot a}{Q^2}\Big]dx^{\mu}$

$dx'^{\mu}=\frac{1-2 x \cdot a + x^2 a^2 -2 x \cdot a + 4(x \cdot a)^2 -2x^2 a^2 x \cdot a -4 x^2 a^2 +2 x \cdot a + 2 x^2 a^2 x \cdot a }{Q^2}dx^{\mu}$

$dx'^{\mu}=\frac{1-2 x \cdot a + 4(x \cdot a)^2 -3 x^2 a^2}{Q^2}dx^{\mu}$

$dx'^{\mu}=\frac{(1-2 x \cdot a)^2 -3 x^2 a^2}{Q^2}dx^{\mu} = \frac{(1-2 x \cdot a)^2 -3 x^2 a^2}{(1-2a \cdot x+a^2x^2)^2}dx^{\mu}$

Now I think I have done the right math. I just have to apply the initial definition and verify that:

$g_{\mu \nu}dx'^{\mu}dx'^{\nu}=K_{\mu \nu}dx^{\mu}dx^{\nu}$

So I have:

$K_{\mu \nu}=\Big[ \frac{(1-2 x \cdot a)^2 -3 x^2 a^2}{(1-2a \cdot x+a^2x^2)^2} \Big]^2g_{\mu \nu}$

Is it correct?

 

[samalkhaiat]

So I have:
$$K_{\mu \nu}=\Big[ \frac{(1-2 x \cdot a)^2 -3 x^2 a^2}{(1-2a \cdot x+a^2x^2)^2} \Big]^2g_{\mu \nu}$$
Is it correct?

No, it is not correct. You have $$\bar{x}^{\mu} = K(x) (x^{\mu} + b^{\mu}x^{2}), \ \ \ \ \ (1a)$$ where $$K(x) = \frac{1}{1 + 2 b \cdot x + b^{2}x^{2}} . \ \ \ \ \ \ (1b)$$

So, $$g_{\mu\nu} \ d\bar{x}^{\mu} \ d\bar{x}^{\nu} = g_{\mu\nu} \frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}}\frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} \ dx^{\rho} \ dx^{\sigma} .$$ Thus $$K_{\rho\sigma} = g_{\mu\nu} \ \frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}}\frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} . \ \ \ \ \ \ \ \ (2)$$ Now, from (1) you find $$\frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}} = K(x) \left( \delta^{\mu}_{\rho} + 2b^{\mu}x_{\rho} - \frac{2(x^{\mu} + b^{\mu}x^{2})(b_{\rho} + b^{2}x_{\rho})}{1 + 2 b \cdot x + b^{2}x^{2}}\right) .$$ If we define the matrix $$\Lambda^{\mu}{}_{\rho} (x,b) = \delta^{\mu}_{\rho} + 2b^{\mu}x_{\rho} - \frac{2(x^{\mu} + b^{\mu}x^{2})(b_{\rho} + b^{2}x_{\rho})}{1 + 2 b \cdot x + b^{2}x^{2}} , \ \ \ (3)$$ then $$\frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}} = K(x) \Lambda^{\mu}{}_{\rho} (x,b).$$ Thus, Eq.(2) becomes $$K_{\rho\sigma} = K^{2}(x) \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ g_{\mu\sigma} . \ \ \ \ (4)$$ So, the problem reduced to finding the properties of the local matrix $\Lambda (x,b)$. To do this, it is sufficient to take the parameters $b^{\mu}$ to be very small, i.e., in Eq.(3), set $b^{2} \approx b^{3} \approx \cdots \approx 0$: $$\Lambda^{\mu}{}_{\rho} \approx \delta^{\mu}_{\rho} + 2 b^{\mu} \ x_{\rho} - 2 b_{\rho} \ x^{\mu} .$$ So, to first order in $b^{\mu}$ we have $$\Lambda^{\mu}{}_{\rho} (x , b) = \delta^{\mu}_{\rho} + 2 g_{\rho \tau} (b^{\mu}x^{\tau} - b^{\tau}x^{\mu}) .$$ Now, if you define $$\omega^{\mu\tau} \equiv 2 (b^{\mu}x^{\tau} - b^{\tau}x^{\mu}) = - \omega^{\tau \mu} ,$$ then $$\Lambda^{\mu}{}_{\rho} = \delta^{\mu}_{\rho} + g_{\rho\tau}\omega^{\mu\tau} = \delta^{\mu}_{\rho} + \omega^{\mu}{}_{\rho} .$$ Using this expression for $\Lambda$, you can easily show that $$g_{\mu\nu}\Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} = g_{\rho\sigma} .$$ Finally, substituting this result in Eq.(4), you get $$K_{\rho\sigma} = K^{2}(x) \ g_{\rho\sigma} .$$