Faddeev-Popov Determinant of Chern-Simons Theory

I am asking this question in order to figure out the expression of the Faddeev-Popov determinant given by Edward Witten is his paper "Quantum Field Theory and Jones Polynomial".

Starting from the action

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right)$$

the variation gives equation of motion

$$F=dA+A\wedge A=0$$

Denoting the solutions to the equation of motion by $a$, then one expands a generic connection $A$ around a flat connection

$$A=a+B$$

Then the action splits into three pieces:

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(a\wedge da+\frac{2}{3}a\wedge a\wedge a\wedge a\right)+\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(B\wedge D_{a}B\right)+$$

$$+\frac{k}{6\pi}\int_{M}\mathrm{Tr}\left(B\wedge B\wedge B\right)$$

where the covariant derivative in the second term is defined as $D_{a}=d+[a,\,\,\,\,]$.

The gauge transformation $A[U]=U^{-1}dU+U^{-1}AU$ splits into two parts:

$$a[U]=U^{-1}dU+U^{-1}aU,\quad B[U]=U^{-1}BU$$

so that the flat connections remain flat and the perturbation part $B$ lives in the adjoint representation.

Then, the path-integral takes the form

$$Z=\int\mathcal{D}a\,e^{iS[a]}\int\mathcal{D}B\,\exp\left\{\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)\right\}$$

It's easy to check that the last two terms

$$S[a;B]=\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)$$

are indeed invariant under the gauge transformations $a[U]=U^{-1}dU+U^{-1}aU$ and $B[U]=U^{-1}BU$. Assuming the spacetime manifold $M$ is closed, and $k\in\mathbb{Z}$, then the first part

$$S[a]==\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(a\wedge da+\frac{2}{3}a\wedge a\wedge a\right)$$

is also gauge invariant up to a shift by $2\pi\mathbb{Z}$ from the Wess-Zumino-Witten term under large gauge transformations.

Assuming that the moduli space of flat connections $\mathcal{M}=\mathrm{Hom}(\pi_{1}(M),G)/G$ is a discrete set, then the partition function really is

$$Z=\sum_{m\in\mathcal{M}}e^{iS[a_{m}]}\frac{1}{\mathrm{Vol}}\int\mathcal{D}B\,\exp\left\{\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)\right\}$$

To fix the gauge, the author imposes the covariant gauge

$$\mathcal{F}[a;B]=(D_{a})_{\mu}B^{\mu}=0$$

The Faddeev-Popov determinant is then

$$\Delta[a;B]=\mathrm{Det}\left(\frac{\delta\mathcal{F}[a[U];B[U]]}{\delta U}\right)\Bigg|_{U=\mathrm{id}}=\mathrm{Det}(M)$$

Using chain rule, one has

$$M(x-y)=\int d^{3}z\left\{\frac{\delta\mathcal{F}(x)}{\delta B(z)}\frac{\delta B(z)}{\delta U(y)}+\frac{\delta\mathcal{F}(x)}{\delta a(z)}\frac{\delta a(z)}{\delta U(y)}\right\}$$

In Witten's paper, the final expression of the Faddeev-Popov is quite simple, which is $$M=(D_{a})_{\mu}(D_{a})^{\mu}$$

However, carrying on the calculation of functional derivatives, I obtained a quite different result.

To be specific, the functional derivatives are given by

$$\frac{\delta\mathcal{F}(x)}{\delta B(z)}=D_{a}(x)\delta(x-z),\quad \frac{\delta\mathcal{F}(x)}{\delta a(z)}=[\delta(x-z),B(z)]$$

$$\frac{\delta B(z)}{\delta U(y)}=[B(z),\delta(z-y)],\quad \frac{\delta a(z)}{\delta U(y)}=D_{a}(z)\delta(z-y)$$

where $(D_{a})_{\mu}(x)\delta(x-y)=\frac{\partial}{\partial x^{\mu}}\delta(x-y)+[a_{\mu}(x),\delta(x-y)]$, and the commutator here carries Lie-algebra indices and so is symmetric. i.e. $[A,B]=AB+BA$.

Plugging the above functional derivatives back into the determinant, what I obtained in the end is $$M(x)=4(D_{a})_{\mu}B^{\mu}(x)$$

This is obviously incorrect.

What am I mistaken in the above procedure?

(I also posted this question here)

Answer

Witten wants to evaluate the path integral for a given background $a$, he doesn't want to vary $a$ inside the $B$ action. If he had wanted to do so he wouldn't have chosen the gauge fixing $(D_{a})_{\mu}B^{\mu}=0$, because this gauge fixing is invariant under the combined gauge transformation of $a$ and $B$ given in the question.

The action for the fluctuation field has the following gauge freedom: $$ a[U] = a$$ $$B[U] = U^{-1}BU + U^{-1}D_aU$$ (Infinitesimally: $\delta B[U] = D_a \delta U + [ B, \delta U] = D_A \delta U$)

Please notice that the Chern-Simons action for $B$ is gauge invariant with respect to the above transformation, since $D_a$ is nilpotent due to the flatness of $a$: $$D_a^2 = F_a = 0$$ Thus we have a gauge theory of the non-Abelian $B$ gauge potential with the exterior derivative $d$ replaced by the twisted differential $D_a$.

Now, we have : $$\delta\mathcal{F}[a;B]=(D_{a})_{\mu}\delta B^{\mu}= (D_{a})_{\mu}(D_A)^{\mu} \delta U$$

Thus the Faddeev-Popov Laplacian is:

$$M=(D_{a})_{\mu}(D_{A})^{\mu}$$

Please notice that this is different from Witten's result, the second covariant derivative is with respect to the full gauge potential: background $+$fluctuation. We get Witten's result only if we assume that the fluctuations are small and neglect terms higher than the quadratic term. This fact was noticed by Birmingham, Blau, Rakowski and Thompson, where a careful evaluation is given in equations 6.14 and 6.15 on page 195.

Time-dependent time dilation for circular orbits around a Schwarzschild black hole?

Take two clocks, one in a circular orbit around a Schwarzschild black hole and another with a distant observer.

The time dilation factor between the two clocks is said (at https://en.wikipedia.org/wiki/Gravitational_time_dilation ) to be given by $\sqrt{1 - 3GM/rc^2}$ (versions of the proof can be found at Time Dilation in Orbits in the Schwarzschild Metric and Is gravitational time dilation different from other forms of time dilation? ).

What is troubling me is whether this represents some sort of average? Surely a clock in orbit around the black hole is periodically travelling towards/away from the distant observer (unless the orbital plane is at right angles), which will result in a time-dependent time-dilation factor (and indeed a time-dependent doppler shift).

Can someone enlighten me?

Answer

Time-dilation is defined independent of retardation. That is, you are assumed to use some kind of range sensitive detector (perhaps a really big phased array) to collect the light of distant events and the reconstruct their "real" time and position in your frame of reference before computing the dilation.

A result of this is that red- or blue-shift of a light source alone is not sufficient to give you the time-dilation factor in all cases. (Note that if it was then time-dilation would depend on whether the source was approaching or receding.)

So for distant observers at rest with respect to the black hole (or more precisely to the barycenter of the hole and clock), I believe the orientation of the orbit doesn't change the fact that the relative speed of the observer and the orbiting clock is constant, and that therefore the orientation of the orbit doesn't matter.

The situation is complicated if the observer moves with respect to the hole, because then the relative velocity between clock and observe does vary in time (in some orbital orientations).

Understanding degenerate vacua in a Quantum field theory with sponatneously broken symmetry

$\bullet$ Consider the following Hamiltonian (density) $$\mathscr{H}=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)+\mathcal{V}(\phi) \hspace{0.4cm}\text{where}\hspace{0.4cm}\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2\tag{1}$$ is a double-well potential. Expanding the potential $\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2$ about one of the minima (say, $\phi=+v$) and retaining up to quadratic term in the potential, one obtains a perturbative vacuum $|0+\rangle$. This is not the exact vacuum of the full Hamiltonian because higher order terms in the potential have been neglected. Similarly, expanding $V(\phi)$ about $\phi=-v$, and again retaining up to quadratic term one obtains another perturbative vacuum $|0-\rangle$. This too, by the same reason, is not the exact vacuum of the full Hamiltonian.

$\bullet$ Similar procedure in a quantum mechanical problem with the Hamiltonian $$H=\frac{p^2}{2m}+V(x)=\frac{p^2}{2m}+k(x^2-a^2)^2,$$ will lead to two perturbative ground states $|0+\rangle$ and $|0-\rangle$. They are not the exact ground states of $H$. The exact ground state is either given by $|S\rangle$ or $|A\rangle$ which are the symmetric and antisymmetric linear combinations of the perturbative ground states $|0+\rangle$ and $|0-\rangle$.

In this case, one cannot talk about the energy of the states $|0+\rangle$ or $|0-\rangle$ because they are not the eigenstates of H. One can only talk about energies of $|S\rangle$ and $|A\rangle$.

Question 1: If this is true, how is it that in QFT, we talk about the states $|0+\rangle$ and $|0-\rangle$ being degenerate (in energy)? They are just the perturbative vacua and not the exact vacua. Hence, not the eigenstates of $\mathscr{H}$. For a reference, see A Modern Introduction to Quantum Field Theory by Michele Maggiore, page 254-255.

Question 2: What is/are the exact ground states of $\mathscr{H}$ and what is their relation to $|0+\rangle$ and $|0-\rangle$?

homework and exercises - Trajectory of a swimmer trying to reach the opposite bank

I came across the following question 2 days ago:

$Q$. A swimmer is swimming in a river with a speed $u$ with respect to the speed of water which is $v$. It is assumed that the river has straight parallel banks and the speed of water is the same at all points and has a direction parallel to the banks. Assume that the swimmer starts swimming at a point $A$. He always faces a point $B$ on the opposite bank. $B$ is the point that is directly opposite to $A$. Find the trajectory of the swimmer. The width of the river is $w$.

I assumed that point $A$ is the origin i.e., $A\equiv (0,0)$ and $B\equiv (0,w)$. Let's say the swimmer is at point $P\equiv (x,y)$ at some point of time. At that point, he makes an angle $\theta$ with the horizontal with respect to point $B$. Therefore $\cos\theta=\frac x{\sqrt{x^2+(w-y)^2}}$ and $\sin\theta=\frac {w-y}{\sqrt{x^2+(w-y)^2}}$. Now I got the following two equations:-$$\frac {dx}{dt}=v-u\cos\theta=v-\frac {ux}{\sqrt{x^2+(w-y)^2}}\quad and \quad\frac {dy}{dt}=u\sin\theta=\frac {u(w-y)}{\sqrt{x^2+(w-y)^2}}$$Dividing them I got this differential equation:-$$\frac {dy}{dx}=\frac {u(w-y)}{v\sqrt{x^2+(w-y)^2}-ux}$$How do I solve this differential equation? Is there any solution that doesn't use calculus in such type of questions? You may also use the fact that $$\frac d{dt}(\sqrt{x^2+(w-y)^2})=v\cos\theta-u=\frac {vx}{\sqrt{x^2+(w-y)^2}}-u$$

Answer

Let me suppose the swimmer begins at a point $I = (\omega, 0)$ and that your target is the origin $ F = (0,0)$ - I think some of the equations are simpler if the initial position, rather than the target position, is at the origin. In my co-ordinates, the river bank is in the $y$-direction.

The river flows at speed $v$ and the swimmer's speed is $u$. We have that $$ \begin{align} \dot x =& -u\cos\theta\\ \dot y =& -u\sin\theta + v \end{align} $$ We ultimately want $y(x)$ so we can plot the swimmers trajectory. We can write that $$ \frac{dy}{dx} = \frac{\dot y}{\dot x} = \tan\theta - \frac vu \frac1{\cos\theta} = \frac yx - \frac vu \sqrt{1 + \left(\frac yx\right)^2} $$ Now let $p = \frac yx$ such that $\frac{dy}{dx} = p + x\frac{dp}{dx}$. We find $$ \begin{align} p + x\frac{dp}{dx} =& p - \frac vu \sqrt{1 + p^2}\\ - \frac vu \frac{dx}{x} =& \frac{dp}{\sqrt{1 + p^2}} \end{align} $$ You can integrate the RHS with the subsitution $p=\tan \theta$. $$\begin{align}\int {\frac{dp}{\sqrt{1 + p^2}}}&=\int {\sec^2 \theta d\theta \over \sec \theta}\\&=\int{\sec\theta d\theta}\\&=\ln(\sec\theta+\tan\theta)+C\\&=\ln(\sqrt{1+p^2}+p)+C\end{align}$$ Hence $$ - \frac vu \ln x = \ln(\sqrt{1 + p^2} + p) + C $$ which you can solve to find that $$ p = \frac12 \left[ e^C x^{-v/u} - e^{-C} x^{v/u}\right] $$ Finally you can find that $$ y = \frac12 \left[ e^C x^{1-v/u} - e^{-C} x^{1+v/u}\right] $$ Now put it your initial condition that $y(\omega) = 0$, you can solve for $C$ and find that, $$ y = \frac\omega2 \left[ \left(\frac{x}{\omega}\right)^{1-v/u} - \left(\frac{x}{\omega}\right)^{1+v/u}\right] $$

If $v/u < 1$, the swimmer hits the target from a direction parallel to the riverbank, hitting the bank exactly at the target, swimming directly against the current when he hits the target. The bigger $v/u < 1$, the further he travels downstream before making it to his target (though this is bounded at $\omega/2$ - the longest paths approximately hit the river bank at $\omega/2$ then move towards the target). If $v/u > 1$, then $y(0)$ divergences - the current takes you infinitely down the river, asymptotically reaching the shore-line. If $v/u=1$, I think you come to a rest on the riverbank at $\omega/2$ downstream of your target.

I doubt there are any solutions without calculus, other than general reasoning about the swimmer's trajectory in the 3 cases above.

forces - How does centripetal acceleration produce the tangential velocity?

I just got started on my physics course, and already I am confused with this stuff called centripetal acceleration.

One thing I don't understand is what produces the velocity that is tangent to the circular path.

I searched on google, and got from PBS Education that there is a change in the velocity which produces the acceleration: But my textbook tells me that there is no change in the magnitude of the velocity only that its direction is changing.

On another website, there is a proof explaining that there is an inward acceleration because the change-in-velocity vector points inward which indicated that the acceleration vector should also be in the same direction.

If you wish to explain it in calculus, bear in mind that I don't know anything about it.

Answer

A small addition to @Hypnosifl's answer:

Key to understanding the circular motion is the fact that at any moment in time the acceleration (force keeping the horse / object in a circular path) is exactly perpendicular to the direction of motion. We have to do "almost calculus" to explain this... but maybe you will be able to follow:

Imagine we initially have a horizontal velocity, and the acceleration is vertical (downwards). How come the new velocity isn't faster than the old one?

Mental picture:

When the change in velocity due to acceleration $a$ for time $\Delta t$ is added to the initial velocity $v_{init}$, the new velocity has a magnitude

$$|v_{new}| = \sqrt{v_{init}^2 + (a\Delta t)^2}$$

When $a\Delta t$ is small, we can do a Taylor expansion and get

$$|v_{new}| = v_{init}\left(1 + \frac{(a\Delta t)^2}{2 v_{init}}\right)$$

Now if we look at a shorter and shorter interval $\Delta t$ we can see that the change in magnitude in velocity, which goes with the square of the interval, will become zero.

This means that if we keep the acceleration at right angles to the velocity (which is done in circular motion) then the speed will be constant - which is of course what we observe for circular motion of things like satellites, the moon, etc (barring small effects like tides, atmospheric drag etc - which are "imperfections" and not part of the essence of the orbital motion).

quantum field theory - Traceless of stress-energy tensor in $d=2$

This is a question regarding Francesco, section 4.3.3. In this section, he considers the two-point function $$ S_{\mu\nu\rho\sigma}(x) = \left< T_{\mu\nu}(x) T_{\rho\sigma}(0)\right> $$ He then goes on to claim that symmetry of the stress-energy tensor implies $$S_{\mu\nu\rho\sigma}(x) = S_{\nu\mu\rho\sigma}(x)~~~(1)$$ Though he doesn't mention this, I presume this is true only when $x \neq 0$ since the EM tensor is symmetric in a correlation as long as the other fields in the correlator are not evaluated at the same point.

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EDIT: Due to some comments, I'll explain why I think so. If a theory is Poincare invariant, it has conserved currents $T^{\mu\nu}$ for translations and $$ j^{\mu\nu\rho} = T^{\mu\nu} x^\rho - T^{\mu\rho} x^\nu $$ for Lorentz transformations. For completeness, we also note that if the theory has scale invariance the dilation current is $$ j^\mu_D = T^{\mu\nu} x_\nu $$ In a classical theory, conservation of these currents implies symmetry and tracelessness of the stress-energy tensor. In a quantum theory, we have a Ward Identity, which for each of the currents reads \begin{equation} \begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \partial_\mu \left< j^{\mu\nu\rho} X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\rho\frac{\partial}{\partial x_i^\nu} - x_i^\nu\frac{\partial}{\partial x_i^\rho} - i S_i^{\mu\nu} \right) \left< X \right> \\ \partial_\mu \left< j^\mu_D X \right> &= - \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\alpha \frac{\partial}{\partial x_i^\alpha} + \Delta_i \right) \left< X \right> \end{split} \end{equation} where $X = \Phi_1(x_1) \cdots \Phi_n(x_n)$, $S^{\mu\nu}_i$ is the representation of the Lorentz algebra under which $\Phi_i(x_i)$ transforms and $\Delta_i$ is the scaling dimension of $\Phi_i(x_i)$. Now plugging in the exact forms of the currents $j^{\mu\nu\rho}$ and $j^\mu_D$, we find \begin{equation} \begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \left< \left( T^{\mu\nu} - T^{\nu\mu} \right) X \right> &= i \sum\limits_{i=1}^n \delta^d(x-x_i) S_i^{\mu\nu} \left< X \right> \\ \left< T^\mu{}_\mu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \Delta_i \left< X \right> \end{split} \end{equation} Clearly, the EM tensor is not symmetric under correlation functions at the points $x = x_i$.

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Now, using these symmetry properties and certain other properties under parity, he argues that $$ S^\mu{}_\mu{}^\sigma{}_\sigma(x) = \left< T^\mu{}_\mu(x) T^\sigma{}_\sigma(0)\right> = 0 $$ Following the above arguments, this should then only be true at $x \neq 0$. However, Francesco claims that this holds everywhere and therefore concludes that $\left< T^\mu{}_\mu(0)^2 \right> = 0$. How does this makes sense?

quantum electrodynamics - Faddeev-Popov Gauge-Fixing in Electromagnetism

Reading section 9.4 in Peskin, I am wondering about the following:

The functional integral on $A_{\mu}$ diverges for pure-gauge configurations, because for those configurations, the action is zero.

To "fix" this, we recognize that anyway we would not have liked to get contributions from pure-gauge field configurations, because field-configurations in the same gauge-orbit correspond to identical physical field configurations. Ultimately, we would like to make a functional integral which ranges over only distinct gauge-orbits, taking each time only one representative from each gauge orbit.

The way to do this technically is to insert a functional-delta function into the functional integral, where this delta function is always zero unless the field configuration obeys a particular gauge-condition, which is non-zero only once in each gauge-orbit.

So far, so good.

However, then Peskin chooses as the gauge condition the Lorenz gauge condition. I'm wondering: why is that valid? The Lorenz gauge condition does not completely fix the gauge: one can still make further gauge transformations by harmonic functions.

What gives?