Projection Operator: Showing Orthogonality for Non-Null q

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The discussion centers on the projection operator defined by Schutz, specifically its application to non-null vectors. The original projection tensor, P_{\vec{q}}=g+\frac{\vec{q} \otimes \vec{q}}{\vec{q} \cdot \vec{q}}, is analyzed to show that the resulting vector V^{\alpha}{}_{\perp} is not generally orthogonal to \vec{q}. A comparison is made with Schutz's example using four-velocity, which successfully demonstrates orthogonality. The conclusion suggests that redefining the projection operator as P_{\vec{q}}=g-\frac{\vec{q} \otimes \vec{q}}{\vec{q} \cdot \vec{q}} would yield the desired orthogonality for arbitrary \vec{q}. This indicates a need for careful consideration of the projection operator's definition in different contexts.
Azrael84
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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?
 
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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}}
 
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