How Do Conformal Transformations Extend Lorentz Symmetry in Physics?

[latentcorpse]

The group of four dimensional space time symmetries may be generalised to conformal transformations $x \rightarrow x'$ defined by the requirement

$dx'^2 = \Omega(x)^2 dx^2$

where $dx^2 = g_{\mu \nu} dx^\mu dx^\nu$ (recall that Lorentz invariance requires $\Omega=1$). For an infinitesimal transformation $x'^\mu = x^\mu + f^\mu(x), \Omega(x)^2=1+2 \sigma(x)$.

Show that $\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu} \Rightarrow \partial \cdot f = 4 \sigma$

That was easy enough - I just multiplied through by a metric.

The next bit is:

Hence obtain

$4 \partial_\sigma \partial_\mu f_\nu = g_{\mu \nu} \partial_\sigma \partial \cdot f + g_{\sigma \nu} \partial_\mu \partial \cdot f - g_{\sigma \mu} \partial_\nu \partial \cdot f$

I simply have no idea how to get this to work!

 

[fzero]

The easiest way to show that is to obtain a formula for $$g_{\mu\nu} \partial_\sigma \partial\cdot f$$ by differentiating the formula that you already obtained. This formula will let you obtain expressions for each term on the RHS. By simplifying, you will obtain the LHS.

 

[latentcorpse]

The easiest way to show that is to obtain a formula for $$g_{\mu\nu} \partial_\sigma \partial\cdot f$$ by differentiating the formula that you already obtained. This formula will let you obtain expressions for each term on the RHS. By simplifying, you will obtain the LHS.

ok thanks. i get:

$\partial_\sigma \partial \cdot f = 4 \partial_\sigma \sigma \Rightarrow g_{\mu \nu} \partial_\sigma \partial \cdot f=4 g_{\mu \nu} \partial_\sigma \sigma$
So,
RHS= $g_{\mu \nu} \partial_\sigma \partial \cdot f + g_{\sigma \nu} \partial_\mu \partial \cdot f - g_{\sigma \mu} \partial_\nu \partial \cdot f$
$=4 g_{\mu \nu} \partial_\sigma \sigma = 4 g_{\sigma \nu} \partial_\mu \sigma - 4 g_<br /> {\sigma \mu} \partial_\nu \sigma$

not sure how to simplify this - i can't get rid of the metric because the contracted term will need dummy indices such as $\partial \cdot f = \partial_\tau f^\tau$ and the metric cannot act on these since the indices won't match up, will they?

 

[fzero]

You need to use

$\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu}$

as well.

 

[latentcorpse]

You need to use

$\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu}$

as well.

great!

The next step is to show $2 \partial_\sigma \partial_\mu \partial \cdot f = -g_{\sigma \mu} \partial^2 \partial \cdot f$

so i tried multiplyign through my last expression by $\frac{1}{2} g_^{\mu \nu}$ to no avail as i keep getting contraction of my metrics and clearly i want to keep a metric in the RHS?

 

[fzero]

You should take an appropriate derivative of

$4 \partial_\sigma \partial_\mu f_\nu = g_{\mu \nu} \partial_\sigma \partial \cdot f + g_{\sigma \nu} \partial_\mu \partial \cdot f - g_{\sigma \mu} \partial_\nu \partial \cdot f$

to show that.

 

[latentcorpse]

You should take an appropriate derivative of

$4 \partial_\sigma \partial_\mu f_\nu = g_{\mu \nu} \partial_\sigma \partial \cdot f + g_{\sigma \nu} \partial_\mu \partial \cdot f - g_{\sigma \mu} \partial_\nu \partial \cdot f$

to show that.

ok. i used $\partial^\nu$ and that worked fine. Now I am supposed to show that $\partial_\sigma \partial_\mu \partial \cdot f=0$. Is this because $g_{\sigma \mu} = 0$ for $\sigma \neq \mu$ since the question said we are dealing with a Lorentzian metric? i.e. g is diag(1,-1,-1,-1) or (-1,1,1,1) depending on convention used.

Why does it then follow that $f_\mu(x)$ can only be quadratic in $x$?

 

[fzero]

ok. i used $\partial^\nu$ and that worked fine. Now I am supposed to show that $\partial_\sigma \partial_\mu \partial \cdot f=0$. Is this because $g_{\sigma \mu} = 0$ for $\sigma \neq \mu$ since the question said we are dealing with a Lorentzian metric? i.e. g is diag(1,-1,-1,-1) or (-1,1,1,1) depending on convention used.

It's simpler than that. Try to use one of the identities to show that $\partial^2 \partial \cdot f=0$ first.

Why does it then follow that $f_\mu(x)$ can only be quadratic in $x$?

You can expand $f_\mu(x)$ in a power series if you want.

 

[latentcorpse]

It's simpler than that. Try to use one of the identities to show that $\partial^2 \partial \cdot f=0$ first.



You can expand $f_\mu(x)$ in a power series if you want.

is it as simple as taking
$2 \partial_\sigma \partial_\mu \partial \cdot f = - g_{\sigma \mu} \partial^2 \partial \cdot f$
and multiplying through by a $g^{\sigma \mu}$

to get 2 \partial^2 \partial \cdot f = -4 \partial^2 \partial \cdot f \Rightarrow 6 \partial^2 \partial \cdot f =0[/itex] hence result?

i am not sure what to do for the power series bit?
$f_\mu(x)= \dots$
how do i organise the indices on the RHS of that?

 

[fzero]

is it as simple as taking
$2 \partial_\sigma \partial_\mu \partial \cdot f = - g_{\sigma \mu} \partial^2 \partial \cdot f$
and multiplying through by a $g^{\sigma \mu}$

to get 2 \partial^2 \partial \cdot f = -4 \partial^2 \partial \cdot f \Rightarrow 6 \partial^2 \partial \cdot f =0[/itex] hence result?

Yes.

i am not sure what to do for the power series bit?
$f_\mu(x)= \dots$
how do i organise the indices on the RHS of that?

Around the zero vector,

$$f_\mu(x) = \sum_n c_{\mu \mu_{1}\cdots\mu_{n} } x^{\mu_1} \cdots x^{\mu_n},$$

where the $$c_{\mu \mu_{1}\cdots\mu_{n} }$$ are constant coefficients.

 

[latentcorpse]

Yes.



Around the zero vector,

$$f_\mu(x) = \sum_n c_{\mu \mu_{1}\cdots\mu_{n} } x^{\mu_1} \cdots x^{\mu_n},$$

where the $$c_{\mu \mu_{1}\cdots\mu_{n} }$$ are constant coefficients.

don't you mean $f_\mu(x) = c_\mu + c_{\mu_1} x^{\mu_1} + c_{\mu_2} x^{\mu_2} + \dots$?

so how do i justify this? can i just say

$\partial \cdot f = \mu_1 c_{\mu_1} + \mu_2 c_{\mu_2} x^{\mu_1} + \mu_3 c_{\mu_3} x^{\mu_2} + \dots$
and then
$\partial^2 \partial \cdot f = \mu_1 \mu_2 \mu_3 c_{\mu_3} + \mu_1 \mu_2 \mu_3 \mu_4 c_{\mu_4} x^{\mu_1} + \dots = 0$

how can i tell then that $c_{\mu_i}=0 \forall i \geq 3$?

Thanks.

 

[fzero]

don't you mean $f_\mu(x) = c_\mu + c_{\mu_1} x^{\mu_1} + c_{\mu_2} x^{\mu_2} + \dots$?

so how do i justify this? can i just say

$\partial \cdot f = \mu_1 c_{\mu_1} + \mu_2 c_{\mu_2} x^{\mu_1} + \mu_3 c_{\mu_3} x^{\mu_2} + \dots$

The $$\mu_i$$ in the series I wrote down are spacetime indices, not exponents. This is a series expansion in the spacetime coordinates written covariantly. It's equivalent to a series like

$$\sum_{(n_t, n_x,n_y,n_z)} A_{n_t n_x n_y n_z} t^{n_t} x^{n_x} y^{n_y} z^{n_z},$$

but written in a far more useful form.

and then
$\partial^2 \partial \cdot f = \mu_1 \mu_2 \mu_3 c_{\mu_3} + \mu_1 \mu_2 \mu_3 \mu_4 c_{\mu_4} x^{\mu_1} + \dots = 0$

how can i tell then that $c_{\mu_i}=0 \forall i \geq 3$?

Thanks.

It should be clearer once you figure out what the series means.