0
$\begingroup$

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.

$\endgroup$

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Aug 30, 2016 at 20:27
  • $\begingroup$ 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 $U$. $\endgroup$
    – Sean Lake
    Aug 30, 2016 at 21:18
  • $\begingroup$ Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^\sigma_\nu V^\nu=V^\sigma$? $\endgroup$ Aug 30, 2016 at 21:41
  • $\begingroup$ 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? $\endgroup$
    – Sean Lake
    Aug 30, 2016 at 21:51
  • $\begingroup$ 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 $PV=V$. $\endgroup$ Aug 30, 2016 at 22:00
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$P^{\sigma}_{\nu} V^{\nu} = V^{\sigma} + U^{\sigma} U_{\nu} V^{\nu}$ 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 $U^{\mu}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.