Azrael84
Oct21-09, 01:02 PM
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^{\al pha}+\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_{\sigm a}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_{\sig ma}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?
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^{\al pha}+\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_{\sigm a}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_{\sig ma}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?