Lecture 2

Chasse_neige

Why Massless Spin-2?

Classical Field Theory

First, consider a massless spin-1 particle

$$\mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }-e{J}_{\mu }{A}^{\mu }$$$${\partial }_{\mu }{F}^{\mu \nu }=e{J}^{\nu }◻A^{\mu }=e{J}^{\mu }$$

for a static point source

$$\{\begin{array}{l}{J}^0=q_{\alpha }{\delta }^3(\overrightarrow{x}-{\overrightarrow{x}}_{\alpha })\\ \overrightarrow{J}=0\end{array}$$

So we can get the form of the vector potential

$$A^0=\frac{-e{q}_{\alpha }}{4\pi |\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}$$

And the total energy

$$E=\int {\mathrm{d}}^{}xe{J}_{b\mu }{A}^{\mu }=\int {\mathrm{d}}^{}xe(-q_b{\delta }^3(\overrightarrow{x}-{\overrightarrow{x}}_b))\left(\frac{-e{q}_{\alpha }}{4\pi |\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}\right)=\frac{\alpha q_a{q}_b}{|{\overrightarrow{x}}_{\alpha }-{\overrightarrow{x}}_b|}$$

Then consider a massive spin-0 (scalar field)

$$\mathcal{L}=-\frac{1}{2}({\partial }_{\mu }\varphi {)}^2-\frac{1}{2}{m}^2{\varphi }^2+g\varphi j\leftarrow \text{a scalar source}$$

and we can get

$$\begin{array}{c}(◻-m^2)\varphi =-gj\\ j_{\alpha }=q_{\alpha }{\delta }^3(\overrightarrow{x}-{\overrightarrow{x}}_{\alpha })\end{array}$$

the form of $\varphi$ is

$$\varphi =\int \frac{{\mathrm{d}}^{}k}{(2\pi {)}^3}\frac{g{q}_{\alpha }}{k^2+m^2}{e}^{i\overrightarrow{k}\cdot (\overrightarrow{x}-{\overrightarrow{x}}_{\alpha })}=\frac{g{q}_{\alpha }}{4\pi }\frac{e^{-m|\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}}{|\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}$$

The total energy

$$E=-g\int {\mathrm{d}}^{}x\varphi j_b=-\frac{g^2{q}_a{q}_b}{4\pi }\frac{e^{-m|\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}}{|\overrightarrow{x}-{\overrightarrow{x}}_{\alpha }|}$$

Lesson

1. Inverse square law -> massless mediator
2. Spin-0 mediates attractive force and Spin-1 mediates repulsive force

Gravity could be mediated by a massless spin-0 particle, but under this circumstance light won’t bend.

Field Theory for Spin-2

We guess

$$\mathcal{L}\propto -\frac{1}{2}{\partial }_{\lambda }{h}_{\mu \nu }{\partial }^{\lambda }{h}^{\mu \nu }-\kappa h_{\mu \nu }{T}^{\mu \nu }$$

This will lead to

$$◻h_{\mu \nu }=\kappa T_{\mu \nu }$$

where a static point particle is defined as

$$T_a^{00}=m_a{\delta }^3(\overrightarrow{x}-{\overrightarrow{x}}_a)$$$$h_{00}=-\frac{\kappa m_a}{4\pi |\overrightarrow{x}-{\overrightarrow{x}}_a|}$$

and the total energy

$$E=\kappa \int {\mathrm{d}}^{}x{h}_{00}{T}_b^{00}=-\frac{{\kappa }^2}{4\pi }\frac{m_a{m}_b}{|\overrightarrow{x}-{\overrightarrow{x}}_a|}$$

Massless Spinning Particles are peculiar!

Polarizations of the photon

for $\overrightarrow{k}=(0,0,k)$, we describe the polarization using the vector

$$\begin{array}{c}{e}_x=(1,0,0)\\ e_y=(0,1,0)\end{array}$$

or using the circular polarization

$$e_{\pm }=\frac{1}{\sqrt{2}}(e_x\pm i{e}_y)$$

for arbitrary momentum $\overrightarrow{p}$

$$e_{\pm }(\overrightarrow{p})\equiv R_z(\varphi )R_y(\theta )e_{\pm }(\overrightarrow{k})$$

is a SO(3) vector. But it is impossible to imbed $e_{\pm }$ into a Lorentz vector. (see the deduction from Feynman Lectures on Physics (3))

Guess

$$e_{\pm }^{\mu }(k)=\frac{1}{\sqrt{2}}(0,1,\pm i,0{)}^T$$

and we’ll get

$$R_y^{-1}(\theta )L_x({\eta }_2)L_z({\eta }_1)e_{\pm }^{\mu }(k)=e_{\pm }^{\mu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu }$$

So $e_{\pm }^{\mu }$ is a L-vec only up to an additive term $\propto k^{\mu }$.

3 types of L-trans will leave $k^{\mu }$ inv.

1. Rotation around $k$: $R_z(\theta )$
2. R $\times$ L-type L-trans: $R_y^{-1}{L}_x{L}_z$
3. R $\times$ L-type L-trans with $x\to yy\to -x$: $R_x^{-1}{L}_y{L}_z$

we call them LGT (little group transformation, ISO(2)). The generators of the LGT’s commutations are

$$\begin{array}{c}[R_z,T_x]=i{T}_y\\ [R_z,T_y]=-i{T}_x\\ [T_x,T_y]=0\end{array}$$

Polarization is a representation of not LGT, but SO(2).

The charge of SO(2) is called helicity: $R_z(\theta )X=e^{-ih\theta }X$, so the nonlinear term is price we have to pay for embedding $h=\pm 1$ into “$e_{\pm }^{\mu }$”.

The transformation $e\to e+\alpha k$ is traditionally called a gauge transformation. Gauge inv, is enforced by L-symmetry.

Graviton: Spin-2

Assume the momentum $k=(k,0,0,k)$, and the polarization tensor is defined

$$e_{\mu \nu }^{\pm }(k)=e_{\mu }^{\pm }(k)e_{\nu }^{\pm }(k)=\frac{1}{2}\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& \pm i& 0\\ 0& \pm i& -1& 0\\ 0& 0& 0& 0\end{array}\right)$$

for transformations within LGT, the polarization tensor will act as

$$e_{\pm }^{\mu \nu }(k)\to e_{\pm }^{\mu \nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}[e_{\pm }^{\mu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu }]k^{\nu }+\mu \leftrightarrow \nu$$$$e_{\pm }^{\mu \nu }(k)\to e_{\pm }^{\mu \nu }(k)+{\xi }^{\mu }{k}^{\nu }+{\xi }^{\nu }{k}^{\mu }$$

Spin-3 Particles

The polarization tensor $e_{\pm }^{\mu \nu \lambda }$ under LGT will be

$$e_{\pm }^{\mu \nu \lambda }(k)\to e_{\pm }^{\mu \nu \lambda }(k)+{\xi }^{\mu \nu }{k}^{\lambda }+\cdots$$

Summary

Polarization tensors are up to LGT $\subset$ L-transf.