Samalkhaiat's challenge #001

[samalkhaiat]

Heh, heh, you should write a "Theoretical Physics Challenge" thread, similar to the "Math Challenges".

Heh heh, unfortunately I can’t do that. However, many of my posts in here do (sometimes) contain exercises. I will try to make it a habit in the future.

Here is one relevant for relativity forum:
Use the definition $$T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\delta (\sqrt{-g}\mathcal{L})}{\delta g_{\mu\nu}} ,$$ to show that $$T^{\mu\nu} = V^{\mu}V^{\nu} (\rho - p) - g^{\mu\nu}p ,$$ is derivable from the Lagrangian $$\mathcal{L} = 2 \sigma (x) \left(1 + \pi (\sigma) \right) ,$$ where $\sigma (x)$ is the density of an isotropic fluid in some space-time region, and $\pi (\sigma)$ is the potential energy per unit density $\sigma$, i.e., the elastic potential of the fluid.

I think the solution is in Pauli's lectures on QFT

Are these in English?

or also in Bogoliubov&Shirkov.

I learned QED from that book. In my opinion, it is the best book ever written on QED. I still use it whenever I get stuck on something.

 

[strangerep]

However, many of my posts in here do (sometimes) contain exercises. I will try to make it a habit in the future.

Here is one relevant for relativity forum: [...]

I've asked the moderators to move this exercise into its own "challenge" thread.

 

[samalkhaiat]

Let $j^{\mu}(x)$ be a conserved vector current, i.e., $$\frac{1}{\sqrt{-g}}\partial_{\mu} \left(\sqrt{-g} \ j^{\mu}\right) = 0 .$$ This means that the associated 1-form $\mathbf{J} = j^{\mu} \ g_{\mu\nu} \ \mbox{d}x^{\nu}$ is a co-closed field: $$\delta \mathbf{J} = ~^*\! \ \mbox{d} ~^*\! \ \mathbf{J} = 0,$$ and the flux of the “current” $\mathbf{J}$ through a space-like hyper-surface $\Sigma$ is identified with the associated charge: $$Q[\Sigma] = \int_{\Sigma} \ ~^*\! \ \mathbf{J} = \int_{\Sigma} \frac{\sqrt{-g}}{3!} \ j^{\mu}(x) \ \epsilon_{\mu\nu\rho\sigma} \ \mbox{d}x^{\nu} \wedge \mbox{d}x^{\rho} \wedge \mbox{d}x^{\sigma} \equiv \int_{\Sigma} d\sigma_{\mu}(x) \ j^{\mu}(x) .$$

Assuming that $\mathbf{J}$ vanishes sufficiently fast at spatial infinity, show that the charge $Q[\Sigma] = \int_{\Sigma} ~^*\! \mathbf{J}$, is independent of time and the (Lorentz) observer, i.e., you are asked to show that $Q$ is a conserved Lorentz-invariant scalar.

 

[ergospherical]

Here's a first attempt, I wouldn't be surprised if there's a few (or a lot) of issues...

$\dfrac{1}{\sqrt{-g}}\partial_{\mu} \left(\sqrt{-g} \ j^{\mu}\right) = \nabla_{\mu} j^{\mu} = 0$ implies that $d\star \mathbf{J} = 0$ therefore letting $\Omega$ be a "cylindrical-like" spacetime volume bounded by two spacelike hypersurfaces $\Sigma_{1}$ and $\Sigma_{2}$ and a timelike hypersurface $\Pi$,$$0 = \int_{\Omega} d \star \mathbf{J} = \int_{\partial \Omega} \star \mathbf{J} = \int_{\Sigma_1} \star \mathbf{J} - \int_{\Sigma_2} \star \mathbf{J} + \int_{\Pi} \star \mathbf{J}$$Assuming $\mathbf{J}$ vanishes sufficiently fast at spatial infinity then we can put $\mathbf{J} = 0$ on $\Pi$ leading to $0 = Q[\Sigma_1] - Q[\Sigma_2]$, therefore $Q[\Sigma]$ is time-independent.

For Lorentz invariance,\begin{align*}
\tilde{Q}[\Sigma] &= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} \tilde{j}^{\mu}(x) \epsilon_{\mu \nu \rho \sigma} d\tilde{x}^{\nu} \wedge d\tilde{x}^{\rho} \wedge d\tilde{x}^{\sigma} \\ \\

&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (\epsilon_{\mu \nu \rho \sigma} {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} {\Lambda^{\rho}}_{\gamma} {\Lambda^{\sigma}}_{\delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\

&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (|\Lambda|\epsilon_{\alpha \beta \gamma \delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\

&= Q[\Sigma]

\end{align*}since the determinant $|\Lambda|$ of a Lorentz transformation is unity.

 

[vanhees71]

Are these in English?

The Pauli Lectures are in English and simply gems. It's Sommerfeld style just one generation younger and covers all the stuff about quantum theory Sommerfeld did not cover in his 6 volumes. Interestingly Pauli's 6 volumes don't include point-particle mechanics:

https://www.amazon.com/s?k=Pauli+Lectures&ref=nb_sb_noss_2

 

[samalkhaiat]

Here's a first attempt, I wouldn't be surprised if there's a few (or a lot) of issues...

See below.

$\dfrac{1}{\sqrt{-g}}\partial_{\mu} \left(\sqrt{-g} \ j^{\mu}\right) = \nabla_{\mu} j^{\mu} = 0$ implies that $d\star \mathbf{J} = 0$ therefore letting $\Omega$ be a "cylindrical-like" spacetime volume bounded by two spacelike hypersurfaces $\Sigma_{1}$ and $\Sigma_{2}$ and a timelike hypersurface $\Pi$,$$0 = \int_{\Omega} d \star \mathbf{J} = \int_{\partial \Omega} \star \mathbf{J} = \int_{\Sigma_1} \star \mathbf{J} - \int_{\Sigma_2} \star \mathbf{J} + \int_{\Pi} \star \mathbf{J}$$Assuming $\mathbf{J}$ vanishes sufficiently fast at spatial infinity then we can put $\mathbf{J} = 0$ on $\Pi$ leading to $0 = Q[\Sigma_1] - Q[\Sigma_2]$, therefore $Q[\Sigma]$ is time-independent.

That is correct and it is called “the fat-tube theorem”: A conserved current $\mathbf{J}$ that vanishes outside some “fat” world tube will produce the same flux $\int_{\Sigma} ~^*\!\mathbf{J}$ through any closed hypersurface $\Sigma$ intersected by the tube, i.e., the charge on a hypersurface will be the same for all hypersurfaces intersected by the tube.

For Lorentz invariance,\begin{align*}
\tilde{Q}[\Sigma] &= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} \tilde{j}^{\mu}(x) \epsilon_{\mu \nu \rho \sigma} d\tilde{x}^{\nu} \wedge d\tilde{x}^{\rho} \wedge d\tilde{x}^{\sigma} \\ \\

&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (\epsilon_{\mu \nu \rho \sigma} {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} {\Lambda^{\rho}}_{\gamma} {\Lambda^{\sigma}}_{\delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\

&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (|\Lambda|\epsilon_{\alpha \beta \gamma \delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\

&= Q[\Sigma]

\end{align*}since the determinant $|\Lambda|$ of a Lorentz transformation is unity.

If the current is not conserved, the charge will not be Lorentz invariant. So, in which of the above steps have you used current conservation?

To see that Lorentz invariance of $Q$ follows directly from $\partial_{\mu}j^{\mu} = 0$(or indirectly from the “tube theorem”), I will present you with 4 methods. To save space, I will drop the wedge signs, set $\sqrt{-g} = 1$ and write $$\begin{equation}Q[\Sigma_{1}] = \int_{\Sigma_{1}} \frac{1}{3!} \ j^{\mu}(x) \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} dx^{\rho} dx^{\sigma} , \end{equation}$$ in the first method.

In the second method, I write eq(1) as $$\begin{equation}Q[\Sigma] = \int_{\Sigma} d\sigma_{\mu}(x) \ j^{\mu}(x) .\end{equation}$$

In the third method, I will choose $\Sigma$ to be the hyperplane $x^{0} = t = \mbox{const.}$ which is possible because of the tube theorem. So, in the third method I put $$\begin{equation}Q(t) = \int d^{3}\mathbf{x} \ j^{0}(x) .\end{equation}$$.

And finally in the fourth method, I will start from the expression $$\begin{equation}Q = \int d^{3}\mathbf{x} \ j^{0}(0 , \mathbf{x}) .\end{equation}$$

In all methods, we will be using the following Lorentz transformation law for the vector current: $$\begin{equation}\bar{j}^{\mu}(x) = U^{\dagger}(\Lambda) j^{\mu}(x) U(\Lambda) = \Lambda^{\mu}{}_{\nu} j^{\nu}(\Lambda^{-1}x) , \end{equation}$$ where $$U(\Lambda) = e^{- i \epsilon M}, \ \ \ \ \ M = \omega_{\mu\nu}M^{\mu\nu} \in \mathfrak{so}(1,3),$$ and $\epsilon \ll 1$. In the second method we will use the following (Schwinger) identity $$\begin{equation}\int_{\Sigma} \ d\sigma_{\mu}(x) \ \partial^{\nu}T^{A}(x) = \int_{\Sigma} \ d\sigma^{\nu}(x) \ \partial_{\mu}T^{A}(x) ,\end{equation}$$ which can be proved for the well-behaved object $T^{A}(x)$

Method 1: This is your method but properly done. The Lorentz transform of $Q[\Sigma_{1}]$ is obtained by conjugating it with the unitary operator $U(\Lambda)$. So, doing this to eq(1) and using eq(5), we obtain $$\bar{Q}[\Sigma_{1}] = U^{\dagger} Q[\Sigma_{1}] U = \int_{\Sigma_{1}} \frac{1}{3!} \ \Lambda^{\mu}{}_{\tau} j^{\tau}(\Lambda^{-1}x) \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu}dx^{\rho}dx^{\sigma} .$$ That is it. We have done the Lorentz transformation. Now, in the integral, we let $x \to \Lambda x$, which is a simple change of integration variables, not Lorentz transformation. After calculating the Jacobian and using $$\Lambda^{\mu}{}_{\tau}\Lambda^{\nu}{}_{\alpha}\Lambda^{\rho}{}_{\beta}\Lambda^{\sigma}{}_{\gamma} \ \epsilon_{\mu\nu\rho\sigma} = \epsilon_{\tau\alpha\beta\gamma},$$ we find (going to the new integration domain $\Sigma_{2}$) $$\bar{Q}[\Sigma_{1}] = \int_{\Sigma_{2}} \frac{1}{3!} \ j^{\tau}(x) \epsilon_{\tau\alpha\beta\gamma} \ dx^{\alpha}dx^{\beta}dx^{\gamma} \equiv Q[\Sigma_{2}].$$ But, according to the tube theorem, $Q[\Sigma_{2}] = Q[\Sigma_{1}]$. Thus $$\bar{Q}[\Sigma_{1}] = U^{\dagger}(\Lambda) Q[\Sigma_{1}] U(\Lambda) = Q[\Sigma_{1}] \ \ \ \ \ \ \ \mbox{qed}.$$
Method 2: In eq(5), put $$U = 1 - \epsilon \frac{i}{2}\omega_{\alpha\beta}M^{\alpha\beta},$$$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}_{\nu} + \epsilon \omega^{\mu}{}_{\nu},$$ then expand to first order in $\epsilon$ and factor out the parameters $\omega_{\alpha\beta}$ from both sides of eq(5). If you don’t make a mistake, you obtain the following commutator equation (the infinitesimal version of 5): $$\big[ iM^{\alpha\beta} ,j^{\mu}(x) \big] = x^{[\alpha}\partial^{\beta ]}j^{\mu} + \eta^{\mu [ \alpha}j^{\beta ]} ,$$ where $A^{\rho [ \mu}B^{\nu ]} = A^{\rho \mu}B^{\nu} - A^{\rho \nu}B^{\mu}$. Using $\partial^{\alpha}x^{\beta} = \eta^{\alpha\beta}$, we rewrite the above as $$\big[ iM^{\alpha\beta} ,j^{\mu}(x) \big] = \partial^{\beta} (x^{\alpha}j^{\mu}) - \partial^{\alpha}(x^{\beta}j^{\mu}) + \eta^{\mu [ \alpha}j^{\beta ]} .$$ We now operate with $\int_{\Sigma} d\sigma_{\mu}(x)$ on both sides, then we use eq(2) on the LHS, and apply Schwinger identity ,eq(6), to the first 2 terms on the RHS. This gives us $$\big[ iM^{\alpha\beta} ,Q\big] = \int d\sigma^{\beta} \partial_{\mu}(x^{\alpha}j^{\mu}) - \int d\sigma^{\alpha} \partial_{\mu}(x^{\beta}j^{\mu}) + \int d\sigma^{[ \alpha} j^{\beta ]} .$$ Using current conservation in the first two terms on the RHS, we get $$\big[ iM^{\alpha\beta} ,Q[\Sigma]\big] = \int_{\Sigma} d\sigma^{[\beta}j^{\alpha ]} + \int_{\Sigma} d\sigma^{[\alpha}j^{\beta ]} = 0.$$ This means that $Q$ is Lorentz invariant scalar (for it commutes with all the generators of the Lorentz group).

Now my time is up, so I leave remaining two methods for another day.

Good day

 

[samalkhaiat]

Method 3: Again, we write $U(\Lambda) = 1 - i \epsilon M , \ \ \ \Lambda = 1 + \epsilon \omega$ and expand eq(5) to first order in $\epsilon$. This time though, we will not factor out the Lorentz parameter $\omega$ because we will do some business with it. With simple algebra, we get $$\begin{equation}\big[iM , j^{\mu}(x)\big] = \omega^{\mu}{}_{\nu}j^{\nu}(x) - \omega^{\rho}{}_{\sigma}x^{\sigma} \partial_{\rho}j^{\mu}(x).\end{equation}$$

Using $\partial_{\rho}j^{\rho} = 0$, allows us to rewrite the first term on the RHS as total divergence $$\omega^{\mu}{}_{\nu}j^{\nu} = \partial_{\rho} \left(\omega^{\mu}{}_{\nu} x^{\nu} j^{\rho}\right).$$ Similarly, the condition $\omega^{\rho}{}_{\rho} = 0$ transforms the second term in eq(7) into a total divergence $$\omega^{\rho}{}_{\sigma}x^{\sigma} \partial_{\rho}j^{\mu} = \partial_{\rho} \left(\omega^{\rho}{}_{\sigma}x^{\sigma} j^{\mu} \right).$$ Thus, eq(7) becomes $$\begin{equation}\big[iM , j^{\mu}(x)\big] = \partial_{\rho} \Omega^{\mu\rho}(x),\end{equation}$$ where $$\begin{equation}\Omega^{\mu\rho} = - \Omega^{\rho\mu} = \omega^{\mu}{}_{\sigma}x^{\sigma} \ j^{\rho} - \omega^{\rho}{}_{\sigma}x^{\sigma} \ j^{\mu} .\end{equation}$$ We are almost done. Setting $\mu = 0$ and integrating over $\mathbf{x}$ gives us $$\big[iM , \int d^{3}\mathbf{x} \ j^{0}(x) \big] = \int d^{3}\mathbf{x} \ \partial_{k} \Omega^{0k}(x) = \oint dS_{k} \ \Omega^{0k}(x) .$$ The surface integral vanishes because the current satisfies the boundary condition: $\mathbf{x}^{2}x^{\sigma}j^{\mu}(x) \to 0, \ \ \mbox{as} \ \ |\mathbf{x}| \to \infty$. Thus $[iM , Q] = 0$, and this complete the proof of the invariance.Method 4: As we said before, we start from $$Q = \int d^{3}\mathbf{x} \ j^{0}(0 , \mathbf{x}),$$ and rewrite it as $$Q = \int d^{3}\mathbf{x} dx^{0} \ \delta (x^{0}) \ j^{0}(x^{0},\mathbf{x}).$$ If we introduce the time-like unit vector $n = (1,0,0,0)$, the above becomes $$Q = \int d^{4}x \ \delta (n \cdot x) \ n_{\mu}j^{\mu}(x) .$$ Conjugating with $U(\Lambda)$ and using the transformation law for the current, we find $$U^{\dagger}(\Lambda) Q U(\Lambda) = \int d^{4}x \ \delta (n \cdot x) \ n_{\mu} \Lambda^{\mu}{}_{\nu}j^{\nu}(\Lambda^{-1}x) .$$ Now we change integration variables as $x \to \Lambda x$. So $d^{4}x \to d^{4}x$, $j^{\mu}(\Lambda^{-1}x) \to j^{\mu}(x)$, $$n \cdot x \to n \cdot (\Lambda x) = (\Lambda^{-1}n) \cdot x ,$$ and $$n_{\mu}\Lambda^{\mu}{}_{\nu}j^{\nu}(x) \equiv n_{\mu} (\Lambda j(x))^{\mu} = (\Lambda^{-1}n)_{\mu} j^{\mu}(x) .$$ So, if we define the new unit vector $e = \Lambda^{-1}n$, we find $$U^{\dagger}QU = \int d^{4}x \ e_{\mu} \delta (e \cdot x) \ j^{\mu}(x).$$ Now, we need to use current conservation. The trick is to consider the difference $$U^{\dagger}QU - Q = \int d^{4}x \left( e_{\mu} \delta (e \cdot x) - n_{\mu}\delta (n \cdot x)\right) j^{\mu}(x) ,$$ and use the identity $m_{\mu} \delta (m \cdot x) = \partial_{\mu}\theta (m \cdot x)$. Thus $$U^{\dagger}QU - Q = \int d^{4}x \partial_{\mu}\left(\Theta (x) j^{\mu}(x)\right) = 0,$$ because $$\Theta (x) \equiv \left( \theta (e \cdot x) - \theta (n \cdot x) \right) \to 0 \ \ \mbox{as} \ \ x^{0} \to \pm \infty,$$ and $$\mathbf{x}^{2} j^{k}(x) \to 0, \ \ \mbox{as} \ \ |\mathbf{x}| \to \infty .$$ qed.