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

<br /> \partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu} <br />

as well.

 

[latentcorpse]

You need to use

<br /> \partial_\mu f_\nu + \partial_\nu f_\mu = 2 \sigma g_{\mu \nu} <br />

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

<br /> 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<br />

to show that.

 

[latentcorpse]

You should take an appropriate derivative of

<br /> 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<br />

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.