# Orthogonal Projection Tensor

Hi my question is about this orthogonal projection tensor $P^\sigma_\nu\equiv\delta^\sigma_\nu+U^\sigma U_\nu$.

It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.

$P^\sigma_\nu V^\nu_\parallel ~~= ~~0$

$P^\sigma_\nu W^\nu_\perp ~~=~~ W^\sigma_\perp$

However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.

$P^\sigma_\nu V^\nu_\parallel ~~=~~\delta^\sigma_\nu V^\nu_\parallel+U^\sigma U_\nu V^\nu_\parallel$

If $\sigma=\nu$ then $\delta^\sigma_\nu=1$.

$P^\sigma_\nu V^\nu_\parallel~~=~~ V^\sigma_\parallel +U^\sigma U_\sigma V^\sigma_\parallel \qquad \qquad [U^\sigma U_\sigma=-1]\\ \quad\qquad=~~ V^\sigma_\parallel - V^\sigma_\parallel \\ \quad\qquad=~~0 \quad\checkmark$

That looks good, now let's check when $\sigma\neq\nu$ and $\delta^\sigma_\nu=0$.

$P^\sigma_\nu V^\nu_\parallel~~= ~~0 +U^\sigma U_\nu V^\nu_\parallel \qquad \qquad~~~ [V\parallel U\implies U_\nu V^\nu\neq0]\\ \quad\qquad~\neq~~ 0 \qquad \qquad \qquad \qquad~~~~~~ [\mathrm{contradicts~Eq.}(2)]$

It looks like I should totally disregard the case where $\sigma\neq\nu$ but I don't see why.

Talking about $\sigma=\nu$ and $\neq \nu$ are ill defined. $\nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^\nu U_\nu$, because $\delta^\sigma_\nu V^\nu = V^\sigma$ always.

• Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $\delta^\sigma_\nu V^\nu=V^\sigma$ and ALSO all it changes all the other instances of $V^\nu$? That is hard for me to wrap my head around. Aug 30, 2016 at 20:27
• Nope. The other $\nu$ is contracted with the $U$. So, it works like this:\begin{align}P_\nu^\sigma &= \delta_\nu^\sigma + U^\sigma U_\nu\\ P_\nu^\sigma V^\nu &= (\delta_\nu^\sigma + U^\sigma U_\nu)V^\nu\\ & = \delta_\nu^\sigma V^\nu + U^\sigma U_\nu V^\nu \\ & = V^\sigma + U^\sigma U_\nu V^\nu. \end{align} So, the only question is how $V$ is related to $U$. Aug 30, 2016 at 21:18
• Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^\sigma_\nu V^\nu=V^\sigma$? Aug 30, 2016 at 21:41
• The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{\perp},$ where the defining equation of $V_{||}$ is $V_{||}^\mu = a U^\mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{\perp}$? What happens to the equation when we plug this broken down version of $V$ into it? Aug 30, 2016 at 21:51
• LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$. Aug 30, 2016 at 22:00

I think what you are missing is the following: by applying the operator $P^{\sigma}_{\nu}$ to the vector V this is what you are actually doing: \begin{equation*} \sum_{\nu=0}^{3}P^{\sigma}_{\nu}V^{\nu}=\sum_{\nu=0}^{3}(\delta^{\sigma}_{\nu}+U^{\sigma}U_{\nu})V^{\nu} \end{equation*} From which follows: \begin{equation*} P^{\sigma}_{\nu}V^{\nu}=V^{\sigma}-V^{\sigma} =0\end{equation*} As you correctly stated in the beginning of your question. The point is that $\nu$ becomes a dummy index only $\textbf{after}$ the operator is contracted on the vector.

$$P^{\sigma}_{\nu} V^{\nu} = V^{\sigma} + U^{\sigma} U_{\nu} V^{\nu}$$ holds generally for a vector $$V^{\nu}$$.

Now, if we additionally assume that $$V^{\nu}$$ is parallel to $$U^{\nu}$$, then its components are just those of $$U^{\nu}$$ but scaled by some number $$\beta$$, i.e: $$V^{\nu} = \beta U^{\nu}$$.

Therefore, $$U^{\sigma} U_{\nu} V^{\nu} = \beta U^{\sigma} U_{\nu} U^{\nu} = - \beta U^{\sigma} = - V^{\sigma}$$, using the fact that the norm of the 4-velocity is -1 ($$U_{\nu} U^{\nu} = -1$$).

It follows that $$P^{\sigma}_{\nu} V^{\nu} = V^{\sigma} - V^{\sigma} = 0$$, if $$V^{\mu}$$ is parallel to $$U^{\mu}$$.