# Projection operator

1. Oct 21, 2009

### Azrael84

Another question I have from Schutz (CH3, 31 (c)), where he defines the Projection tensor as

$$P_{\vec{q}}=g+\frac{\vec{q} \otimes \vec{q}}{\vec{q} \cdot \vec{q}}$$

This can be written in component form (or rather the associated (1 1) tensor can after operating a few times on it with the metric) as:

$$P^{\alpha}{}_{\beta}=\eta^{\alpha}{}_{\beta}+\frac{q^{\alpha}q_{\beta}}{q^{\gamma}q_{\gamma}}$$

This obviously takes a vector V and produces another vector , i.e.

$$V^{\alpha}{}_{\perp}=P^{\alpha}{}_{\beta} V^{\beta}=(\eta^{\alpha}{}_{\beta}+\frac{q^{\alpha}q_{\beta}}{q^{\gamma}q_{\gamma}})V^{\beta}=V^{\alpha}+\frac{q^{\alpha}q_{\beta}V^{\beta}}{q^{\gamma}q_{\gamma}}$$

The task is then to show that $$V^{\alpha}{}_{\perp}$$ is indeed orthoganal to $$\vec{q}$$, provided q is non-null.

So I start of by taking the dot product of $$V^{\alpha}{}_{\perp}$$ and $$\vec{q}$$:

$$\vec{q} \cdot \vec{V}_{\perp}=\eta_{\alpha \beta} q^{\alpha}V^{\beta}{}_{\perp}= \eta_{\alpha \beta}q^{\alpha}(V^{\beta}+\frac{q^{\beta}q_{\sigma}V^{\sigma}}{q^{\gamma}q_{\gamma}})=q_{\beta}V^{\beta}+\frac{\eta_{\alpha \beta}q^{\alpha}q^{\beta}q_{\sigma}V^{\sigma}}{q^{\gamma}q_{\gamma}}=q_{\beta}V^{\beta}+\frac{\vec{q} \cdot \vec{q} (q_{\sigma}V^{\sigma})}{\vec{q} \cdot \vec{q} }=q_{\beta}V^{\beta}+q_{\sigma}V^{\sigma}=2q_{\sigma}V^{\sigma}$$

Which is not generally equal to zero. In Schutz's previous example he used the four velocity as $$\vec{q}$$ which obviously has magnitude of -1. Also he used the projection operator as

$$P_{\vec{q}}=g+\vec{U} \otimes \vec{U}$$

Then everything works out fine, and you can easily show this does produce vectors orthoginal to $$\vec{U}$$ since following a similar derivation to above you end up with $$=U_{\beta}V^{\beta}+(\vec{U} \cdot \vec{U}) U_{\sigma}V^{\sigma}=U_{\beta}V^{\beta}- U_{\sigma}V^{\sigma}=0$$

So I don't believe this thing above really is the projection operator for arbitrary $$\vec{q}$$, although if we instead defined
$$P_{\vec{q}}=g-\frac{\vec{q} \otimes \vec{q}}{\vec{q} \cdot \vec{q}}$$
Then this would work I think?

Last edited by a moderator: Oct 21, 2009
2. Oct 21, 2009

### cristo

Staff Emeritus
I think you're right: the projection tensor should be $$P_{\alpha\beta}=\eta_{\alpha\beta}-\frac{q_\alpha q_\beta}{q^\gamma q_{\gamma}}$$