calculation puzzle - I want an exact value of Pi

Use 1, 2, 2, 2 (Each number used exactly once) to form π.

I want an exact value of π, not any approximation.

Available operators:

+ - * / ( ) ^ !
You can use any one of them for any times.

Hint:

Think about a special feature of the "!".

superconductivity - quark momentum and quark mass in unpairing and pairing phase

In this Review paper in p.1464, top left: Rev.Mod.Phys.80:1455-1515,2008 -- Color superconductivity in dense quark matter

It shows a figure of quark momentum of quark $u,d,s$, in unpairing and pairing phase, such that

$$p_d >p_u>p_s$$

however, we know: $$ m_u

so why isnt that $$p_u >p_d>p_s$$ if momentum is inverse proportional to the quark mass? What is going on here for these quark momentum?

optics - Advantage of using a polygonal mirror with larger number of faces in Michelson method of measuring the speed of light and its value

The following image is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Speed of Light", page 447, topic "Michelson Method":

^{For higher image resolution click here.}

The following text is from the "Science Hero" article - Michelson’s Method for Determining Velocity of Light, under the topic "Disadvantages of Michelson’s method":

At high speeds [angular speed of the rotating mirror], the rotating mirror may break. But speed can be reduced by increasing the number of faces of the mirror.

I can understand that when the number of faces in the rotating mirror is increased, the clear image of the source could be seen at lower angular speeds since the angle by which the mirror needs to rotate for the next face to take the position of the adjacent face is decreased.

The speed of light as measured by this method is given by

$$c=\frac{D\omega N}{2\pi}$$

where $D$ is the distance travelled by the light between reflections from the polygonal mirror, $\omega$ is the minimum possible angular speed of rotation of the mirror when the image becomes steady and $N$ is the number of faces in the polygonal mirror.

As $c$ is a constant, the product $\omega N$ is also constant. So, it can be seen that when we increase the number of faces the rotating mirror, the clear image could be obtained at lesser angular speeds. Now as $N$ gradually approaches infinity, i.e., the polygonal mirror becomes a cylindrical mirror, the angular speed $\omega$ tends to zero. So I think there must be a highest possible value for $N$ which gives the most benefit. What is its value, and what is the reason for this choice? Are there any other advantages of using a larger number of faces in the rotating mirror besides the one discussed in the question?

---

^{Related question asked by me: Number of reflecting surfaces in the rotating mirror in the Michelson method of determination of speed of light}

^{I think Michelson method of determining the speed of light is different from the Michelson Morley experiment. So, I had to use the query michelson speed of light -morley as my initial results were populated with the second experiment which has a similar name.}

^{This method of determination of speed of light is briefly discussed here and here.}

Answer

The purpose of having a relatively large number N of mirrors on the polygon is to increase the switching rate for a given rotational speed. This allows the distance to the retroreflecting mirror to be short enough to be practical. An important factor is that the beam size needs to be large enough to ensure that a significant fraction of the beam will reach the distant retroreflector. As you probably know, a light beam spreads faster when its waist diameter is small. This means the mirrors need to be relatively large, depending on how far away the retroreflector is. In turn, this means that N must be relatively small for a fixed-size wheel.

acoustics - Directionality of sound depending on emitter size and wavelength

I'm a biologist who has been having trouble with the above concept; I searched this forum and found some info from the speaker/audio perspective but I'm interested in animal emitters so I'm not sure how similar the concepts are.

First, based on Jakobsen et al. (2013) who quote older work I haven't gotten copies of yet (which the authors say is based on piston/baffle type models, so potentially the same assumptions which go into speaker design), the directionality of a sound beam will increase with sound frequency and with emitter size (larger bats assumed to have larger emitters; wavelength info at bottom). This means that to maintain a beam of given directionality (in black), a smaller bat will have to use higher frequencies.

I should clarify that by 'increased directionality' I mean a narrower sound beam. I have seen on speaker acoustics sites that the same term often means 'a wider beam' (reaching more of the room/audience).

Anyway, no explanation was provided as to why this is true. At first it seemed counterintuitive that a larger emitter would be more directional, but after looking at diffraction animations I got from similar questions at this site, I think it's because the size of the emitter needs to be considered relative to the wavelenth of the emitted frequency:

So, here's what I'm thinking: if a sound's wavelength is equal to or smaller than the emitter/opening, it will pass through with minimal diffraction and therefore be directional, whereas if the sound's wavelength is larger than the emitter/opening, it will diffract and spread out more (becomes less directional).

Is this a correct interpretation, or have I missed something? Any clarifications very much appreciated.

EDIT: I know about Huygen's principle, but admittedly have difficulty with it. If I imagine many points in a slit all emitting circularly, the emitters at the ends of the slit still aren't summing with waves from other point emitters at the end-side, and this is the case regardless of the size of the slit. But perhaps, since there are more point emitters in a larger slit, that means more energy is moving forward in the summed straight wave, and therefore the increased directionality is also due to greater power?

Answer

Your argument is correct, though the bat presumably emits sound through an orifice, i.e. a 2D surface, rather than a slit. However the sound diffracts in basically the same was as for a slit.

If we make the usual physicists assumption that everything is spherical, or in this case a circular disk, the diffraction pattern created by a circular orifice is an Airy disk. If we take the first minimum to give the radius of the beam of sound then the half-angle of the beam is given by:

$$ \sin\theta \approx 1.22 \frac{\lambda}{d} $$

where $\lambda$ is the wavelength and $d$ is the diameter of the orifice.

cosmology - Is there any primordial gravitational wave(form) that could explain dark energy?

First, I asked myself the question: Is there any gravitational waveform, however fantastical, that could explain dark energy.

I suppose for a ubiquitous gravitational wave to be an explanation for dark energy, then it would have to have the same effect: Dark energy effectively stretches space itself at every point in the universe. What kind of primordial gravitational 'wave' could effectively produce such an effect in the universe?

Please point out any absurdity in this thought experiment:

********************** Begin Edit June 27 2017 **************************

Imagine a 2-D model of our universe before inflation as an arbitrarily tiny patch of space on that drumhead.

A drumstick strikes anywhere on that drumhead, initiating a wave that stretches the drumhead equally in all directions.

• Now inflationary epoch begins.


• The drumhead grows at least 10^26 bigger in O(10^-33) seconds.


• Inflationary epoch ends

What is the state of the initial drumhit waveform now?

Very much like the CMB, it would effectively still be propagating as a very nearly uniform wave coming from all directions.

Now lets make some big assumptions and work backwards.

Extrapolate the 2-D universe to 3-D, and the drumhit to gravitational waves (GWs).

The drum-hit would need to be a collection of interfering GWs, such that:

• 
  

  GWs stretch our tiny patch of space equally in 3 dimensions. The drumhead analogy falls apart here, and dreaming up a set of cancelling GWs is non-trivial. Is cancellation of gravitational waves possible?

  
  

  * begin edit June 28th It appears that it is not possible for any combination of GWs to stretch space equally in 3 dimensions. As a GW passes a given volume of space, the volume contained cannot change. Any stretching in one dimension must be counterbalanced by a compression in the other dimension(s). end edit June 28th *

  
  
  


• their wavelength(s) have now been stretched to O(Hubble length).


• period has also been stretched to O(Hubble age).


• amplitude is O(lambda Cosmological constant)

If you were to work backwards, could you come up with plausible initial conditions for the GWs before inflation started? Certainly someone has attempted this math in the past? ********************** End Edit June 27 2017 ****************************

Original thought experiment (2016):

• a gravitational wave must propagate at the speed of light


• if the wave was/is produced by the big bang itself, the source of the gravitational wave comes from every direction, and is nearly uniform in every direction.



• I’m thinking along the lines of a single giant, long period gravitational wave, amplitude of up to 10^N meters (the size of the universe), and a wavelength of ~ 8 billion years.


• a wavelength of ~ 8 billion years doesn’t really make sense, unless you say that due to the inflationary epoch, its wavelength is being stretched out.


• gravitational waves are complex 3 dimensional waves... however:
  
  
  
  • because it comes from every direction for any observer, the ‘stretching’ of spacetime happens evenly in every direction radially. I'm thinking of a polar coordinate system.
  
  
  • stretching in the other polar dimensions are cancelled out from any observer’s point of view, again since the wave is coming from every direction radially.
  
  
  • Maybe there is a flaw here... would the wave also cancel out in the radial dimension itself?
  
  
  
  


• Since it was/is produced by the big bang itself, its ’source’ is ~380,000 years beyond the observable universe’s event horizon (i.e. 380,000 years beyond the cosmic background radiation). Again, its wavelength stretched out because of inflation.

electrostatics - Why does flux in closed surface remain constant if exterior charge is altered?

Q. Charges $q_1$ and $q_2$ lie inside and outside respectively of a closed surface $S$. Let $E$ be the field at any point on $S$ and $\Phi$ be the flux of $E$ over $S$.

One of the answer is: if $q_2$ changes, $E$ will change but flux will not change.

According to Gauss law,

$$\Phi = \oint_S E \cdot ds \, . \tag{1}$$

Also,

$$\Phi_\text{external} = q/\epsilon_0 \, . \tag{2}$$

But according to the given answer if the electric field $E$ is changing then the flux should also change since $ds$ is constant.

Also, by formula (1), the flux coming out of the surface should remain constant as $q_1$ inside is constant. In short, electric field changes, $q_2$ changes, $ds$ is constant. Then, by the given formulas how can the flux change?

The two formulas are contradictory. The answer given in the book is surely correct.

Answer

But according to the given answer if the electric field $E$ is changing then the flux should also change since $ds$ is constant.

As you say, $E$ does change. However, it changes both where it enters $S$ and where it exits $S$. In the integral, the sign of $E \cdot ds$ depends on whether the field is pointing into or out of the surface. To make this more clear, it is better to remember that $E$ and $ds$ are both vectors. We should really write $\vec{E}$ and $d\vec{S}$, thus writing the integral as

$$\Phi = \oint_S \vec{E} \cdot d\vec{S} \, .$$

The vector $d\vec{S}$ has magnitude given by the size of the area element and by convention the direction points outward from the surface. Now it is clear that if the field is pointing into the surface the contribution to the integral is negative, while if the field points out from the surface the contribution is positive. This is illustrated in the diagram.

So you see that if we e.g. double the value of $q_2$ we double the amount of field entering and exiting $S$, so the total contribution is unchanged (and in particular it's still zero).

We could also move $q_2$ around, while keeping it outside of $S$. However, even moving $q_2$ around does not change $\Phi$. The reason for this is quite deep, but can be derived from the simple fact that the electric field of a charge has zero divergence,

$$\vec{\nabla} \cdot \vec{E} = 0 \, .$$

The divergence theorem says that given a vector field $\vec{U}$ and a surface $S$ which encloses a volume $V$,

$$\int_V \vec{\nabla}\cdot \vec{U} dV = \int_S \vec{U} \cdot d\vec{S} \, .$$

Since for the electric field of a charge $\vec{\nabla}\cdot \vec{E}$ is always zero, then the integral is zero no matter whether we move the charge around or change its amplitude.

Now, there is a very important detail I left out. The divergence of $\vec{E}$ is not truly zero everywhere. In fact, it is zero everywhere except right at the charge itself. At the charge, the divergence is actually infinite. The proper way to write the divergence of the electric field caused by a charge $q$ at position $\mathbf{x}_0$ is

$$\vec{\nabla} \cdot \vec{E} = \delta(\mathbf{x} - \mathbf{x}_0) q / \epsilon_0 \, .$$

If you plug this into the divergence theorem you find that the flux through a surface $S$ caused by this charge is zero if the charge is not inside the surface and $q/\epsilon$ if the charge is inside the surface.

homework and exercises - Maxwell equations of motion from $S = frac{-1}{2} int F wedge ast F$

I'm trying to understand the following equation, used in the derivation of the equations of motion. Let $S = \frac{-1}{2} \int F \wedge \ast F$ and $F = dA$. Let $\delta$ denote variation. Then

$$\delta S = - \int d \delta A \wedge \ast F.$$

(In the full derivation we then int. by parts and get $d\ast F = 0$ for the EOM with no currents.)

In tensor notation, $F = F_{\mu\nu}$ and (I think) $(\ast F) = (\ast F)_{\gamma \delta} = \sqrt{|g|} \epsilon^{\alpha\beta}_{\,\,\,\,\,\, \gamma \delta} F_{\alpha \beta}$ (where $|g|$ is the determinant of the metric). The wedge product is antisymmetric, so (I think) $F\wedge \ast F = \epsilon^{\mu\nu \gamma\delta} F_{\mu\nu} (\ast F)_{\gamma \delta}$. After these considerations, I do not see how we end up with the above result; I'd expect two terms from the product rule for variations, for instance, and where did the factor of $\frac12$ go?

Answer

Actually, once being in differential form formalism, the best is to stay in it and thereby benefit from it. In this sense let's evaluate the variation of the EM-action using F=dA:

$$\delta S =-\frac{1}{2}\delta\int_V dA \wedge\star dA = -\frac{1}{2}\int_V(d\delta A\wedge \star dA + dA \wedge \star d \delta A)=-\frac{1}{2}\int_V 2 d\delta A\wedge \star dA$$

Here we used a hodge operator rule for forms $\lambda, \omega \in \Lambda^k$ for $\lambda = d\delta A$ and $\omega=dA$:

$$\lambda \wedge \star \omega = (-1)^q <\lambda,\omega>e = (-1)^q <\omega,\lambda>e =\omega\wedge \star \lambda$$

with $$e: = e_1 \wedge e_2 \wedge \ldots \wedge e_n \in \Lambda^n$$ where $e_i$ form an orthogonal base in $\Lambda$ and the index $q$ is related with the signature of the metric of the used space. The braket pair $$ represents the scalar product on the space $\Lambda^k$.

Second step (product rule):

$$ d(\delta A\wedge\star dA) = d\delta A\wedge\star dA -\delta A\wedge d\star dA$$

It follows:

$$ -d\delta A\wedge\star d A = -d(\delta A\wedge\star dA)-\delta A\wedge d\star dA$$

Therefore we get for the varied action integral:

$$\delta S = \int_V d(\delta A\wedge\star dA) + \int_V \delta A\wedge d\star dA $$

We transform the first volume integral into an integral over the volume's surface according to Stokes theorem and realize that the variations $\delta A$ vanish on the surface: $$0=\delta S = \int_{\partial V}\delta A\wedge\star dA + \int_V \delta A\wedge d\star dA = \int_V \delta A\wedge d\star dA$$

As the integral $$\int_V \delta A\wedge d\star dA=0$$ for all variations $\delta A$ we can conclude that:

$$ d\star dA=0 $$ which corresponds to the second part of the free Maxwell equations.