# Find the tensor that carries out a transformation

• Frostman
In summary, the problem reduces to finding the properties of the local matrix ##\Lambda(x,b)## which is defined as ##\Lambda^{\mu}{}_{\rho} = \delta^{\mu}_{\rho} + 2 g_{\rho \tau} (b^{\mu}x^{\tau} - b^{\tau}x^{\mu})##. By setting the parameters ##b^{\mu}## to be very small, we can approximate ##\Lambda^{\mu}{}_{\rho}## to be equal to ##\delta^{\mu}_{\rho} + 2 g_{\rho\tau}\omega^{\mu\tau}##, where ##\
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}}##

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.

Frostman and jim mcnamara
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.

andrewkirk said:
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}##?

andrewkirk said:
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?

Frostman said:
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 $$K_{\rho\sigma} = K^{2}(x) \ g_{\rho\sigma} .$$

Last edited:

## 1. What is a tensor and how does it relate to transformations?

A tensor is a mathematical object that represents a linear transformation from one vector space to another. It is used to describe the relationship between different coordinate systems and how objects change under different transformations.

## 2. How do you find the tensor that carries out a specific transformation?

To find the tensor that carries out a transformation, you can use the matrix representation of the transformation. The elements of the matrix will correspond to the components of the tensor, and by rearranging the elements, you can find the tensor that represents the transformation.

## 3. What properties should a tensor have to effectively carry out a transformation?

A tensor should have the properties of linearity and covariance to effectively carry out a transformation. Linearity means that the transformation should be able to be broken down into smaller, simpler transformations, while covariance ensures that the transformation is consistent regardless of the coordinate system used.

## 4. Can a tensor carry out multiple transformations simultaneously?

Yes, a tensor can carry out multiple transformations simultaneously. This is because tensors can be multiplied together to form a new tensor that represents the combined transformation. This is known as tensor multiplication.

## 5. Are there any real-world applications of tensors and transformation?

Yes, tensors and transformations have many real-world applications, especially in fields such as physics, engineering, and computer graphics. For example, tensors are used to describe the stress and strain of materials, the motion of objects in space, and the deformation of images in computer graphics.

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