How Do Conformal Transformations Extend Lorentz Symmetry in Physics?

[latentcorpse]

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

[itex]dx'^2 = \Omega(x)^2 dx^2[/itex]

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

Show that [itex]\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu} \Rightarrow \partial \cdot f = 4 \sigma[/itex]

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

The next bit is:

Hence obtain

[itex]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[/itex]

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

 

[fzero]

The easiest way to show that is to obtain a formula for [tex]g_{\mu\nu} \partial_\sigma \partial\cdot f[/tex] 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 [tex]g_{\mu\nu} \partial_\sigma \partial\cdot f[/tex] 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:

[itex]\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[/itex]
So,
RHS= [itex] g_{\mu \nu} \partial_\sigma \partial \cdot f + g_{\sigma \nu} \partial_\mu \partial \cdot f - g_{\sigma \mu} \partial_\nu \partial \cdot f[/itex]
[itex]=4 g_{\mu \nu} \partial_\sigma \sigma = 4 g_{\sigma \nu} \partial_\mu \sigma - 4 g_
{\sigma \mu} \partial_\nu \sigma[/itex]

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

 

[fzero]

You need to use

[itex]
\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu}
[/itex]

as well.

 

[latentcorpse]

You need to use

[itex]
\partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu}
[/itex]

as well.

great!

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

so i tried multiplyign through my last expression by [itex]\frac{1}{2} g_^{\mu \nu}[/itex] 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

[itex]
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
[/itex]

to show that.

 

[latentcorpse]

You should take an appropriate derivative of

[itex]
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
[/itex]

to show that.

ok. i used [itex]\partial^\nu[/itex] and that worked fine. Now I am supposed to show that [itex]\partial_\sigma \partial_\mu \partial \cdot f=0[/itex]. Is this because [itex]g_{\sigma \mu} = 0[/itex] for [itex]\sigma \neq \mu[/itex] 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 [itex]f_\mu(x)[/itex] can only be quadratic in [itex]x[/itex]?

 

[fzero]

ok. i used [itex]\partial^\nu[/itex] and that worked fine. Now I am supposed to show that [itex]\partial_\sigma \partial_\mu \partial \cdot f=0[/itex]. Is this because [itex]g_{\sigma \mu} = 0[/itex] for [itex]\sigma \neq \mu[/itex] 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 [itex] \partial^2 \partial \cdot f=0[/itex] first.

Why does it then follow that [itex]f_\mu(x)[/itex] can only be quadratic in [itex]x[/itex]?

You can expand [itex]f_\mu(x)[/itex] in a power series if you want.

 

[latentcorpse]

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



You can expand [itex]f_\mu(x)[/itex] in a power series if you want.

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

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?
[itex]f_\mu(x)= \dots[/itex]
how do i organise the indices on the RHS of that?

 

[fzero]

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

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?
[itex]f_\mu(x)= \dots[/itex]
how do i organise the indices on the RHS of that?

Around the zero vector,

[tex]f_\mu(x) = \sum_n c_{\mu \mu_{1}\cdots\mu_{n} } x^{\mu_1} \cdots x^{\mu_n},[/tex]

where the [tex] c_{\mu \mu_{1}\cdots\mu_{n} }[/tex] are constant coefficients.

 

[latentcorpse]

Yes.



Around the zero vector,

[tex]f_\mu(x) = \sum_n c_{\mu \mu_{1}\cdots\mu_{n} } x^{\mu_1} \cdots x^{\mu_n},[/tex]

where the [tex] c_{\mu \mu_{1}\cdots\mu_{n} }[/tex] are constant coefficients.

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

so how do i justify this? can i just say

[itex]\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[/itex]
and then
[itex]\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[/itex]

how can i tell then that [itex]c_{\mu_i}=0 \forall i \geq 3[/itex]?

Thanks.

 

[fzero]

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

so how do i justify this? can i just say

[itex]\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[/itex]

The [tex]\mu_i[/tex] 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

[tex]\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},[/tex]

but written in a far more useful form.

and then
[itex]\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[/itex]

how can i tell then that [itex]c_{\mu_i}=0 \forall i \geq 3[/itex]?

Thanks.

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