## Archive for September, 2010

### Delay…

September 30, 2010

Unfortunately, due to some outside factors, there will be no Physics Friday or Monday Math this weekend. Expect them to return October 8.

### “The War to end all wars”…

September 30, 2010

…officially ends this Sunday:

The final £60 million instalment is part of a £22 billion debt imposed for starting one of the bloodiest conflicts in history will be cleared on what will also be the 20th anniversary of German reunification.

The Allied victors – primarily Britain, France and America set the reparations in 1919’s Treaty of Versailles – a peace agreement – as both compensation and punishment for waging the four year war, which left 10 million soldiers dead, and European towns and cities devastated.

Germany’s Federal Office for Central Services and Unresolved Property Issues said that the bond issued to pay remaining debts stemming from ‘The War To End All Wars’ will be written off on 3 October.

### Monday Math 137

September 27, 2010

Let n be the product of three consecutive positive integers. Show that there is an integer s, greater than the smallest of the consecutive integers, such that n is divisible by s2-1.
Solution:

### Physics Friday 137

September 24, 2010

Part 17: Momentum Transport and Radiation Pressure

Let us consider a linearly-polarized plane wave propagating in the z direction, with plane of polarization in the x direction; so we then have
electric and magnetic fields

,
with .
Let us now compute the Maxwell stress tensor . As you may recall from here, the components of this symmetric (Cartesian) tensor are given by
.
First, we recognise that . Secondly, , and . Thus, for the electric parts, we see  except for when i=j=1, so the electrical component only gives diagonal terms; we have
,
,
and
;
Next, for magnetic components, we have . Secondly, , and ,
so again, we only have diagonal terms,


,
so our tensor’s off-diagonal terms are zero. As for the diagonal terms, we note that for T11,
,
where we used the facts that , and that .
Similarly,
.
Lastly,
.
Recalling that the flux of the ith component of momentum in the j direction is given by , we see that the only momentum flow is of the z component of momentum, pz, flowing in the z direction; the wave carries momentum, aligned with its direction, in the direction of its motion. If we time-average over the cycle of the wave, we obtain .
(This last statement, in terms of the root-mean-square average, is valid for elliptically-polarized light as well).

Now, suppose we have a flat surface, perpendicular to our wave (and thus it is normal to the z axis in our coordinates); and that this surface is a perfect absorber. As the wave arrives at this surface, we see that the average linear momentum per unit area flowing into the surface in unit time is the vector p with components
;
as previously established, this is zero except for the z component:
.
Now, since the wave is being absorbed, this much momentum is lost from the field per unit time per unit of area of the surface. By conservation of the momentum, the surface must be gaining this momentum per unit time, and since change in linear momentum with time is a force (Newton’s second law); we see that the above gives the force on the surface per unit area; that is to say, a pressure equal to
; this is radiation pressure.
Note also that we demonstrated here that the average energy density in the wave is
;
thus, we see that the radiation pressure on a perfect absorber due to a plane electromagnetic wave incident normally to the surface is equal to the energy density of that wave.

### Monday Math 136

September 20, 2010

Here is a very clever geometry proof I once saw. I do not recall where it was I encountered it, but It sticks out in my memory for its creative simplicity; I take no credit for it.

Statement: suppose we have a box (right rectangular prism), which is divided completely into a finite number of smaller boxes, not necessarily the same size or shape. If for each of these smaller boxes, at least one of the edges (dimensions) is of integer length, then at least one edge of the overall box must be of integer length.
Proof:

### This is going to hurt

September 17, 2010

Kevin J. Williamson at NRO‘s “Exchequer” blog: “The Entitlement Bubble: The Bust Is Going to Be a Nightmare

### Blogroll

September 17, 2010

I’ve just added two more blogs to my blogroll. Go check out Secular Right and The Unreligious Right.

### Physics Friday 136

September 17, 2010

Part 16: Energy in Electromagnetic Waves

As we saw here, the flow of energy in electromagnetic fields is given by the Poynting vector , the power flowing per unit of cross-sectional area. Now, let us consider electromagnetic plane waves in vacuum. As we showed here, they are transverse waves with the electric and magnetic fields both perpendicular to the direction of propagation, and that the electric and magnetic fields are perpendicular to each other as well, and proportional, with . Since the fields are perpendicular, the magnitude of their cross product is simply the product of their magnitudes:
, and we can see by the right-hand rule that S will be in the direction of propagation , so electromagnetic waves transport energy in their direction of propagation; an obvious result. We also see that since S is proportional to the square of a field magnitude, we recover the important optical property that the intensity of an eletromagnetic wave is proportional to the square of its amplitude.
More specifically, for a linearly-polarized wave, we have sinusoidally varying fields. With z the direction of propagation and polarization in the x direction, we then have

,
and so
.
The cosine-squared term oscillates between zero and one, and has a time average of 1/2. Taking the time average, then, we have
,
where Erms is the root-mean-square amplitude of the electric field, and which is equal to the peak amplitude over the square root of two.

For circularly-polarized waves, the magnitudes of the electric and magnetic fields are constant, so the Poynting vector is a constant; with propagation again in the z direction,
.
For general elliptical polarization, we can again express in terms of the root-mean-square amplitude (with mean over time) as
,
which reduces to the linear or circular case in each of those limits.

Also, since the energy density for electric fields is , and the energy density for magnetic fields is . Applying these to our wave, and taking the time average over a cycle, the average energy density is

And dividing the energy flux by the energy density confirms that the speed of the energy flow is
,
the speed of light in vacuum, as expected.

### Monday Math 135

September 13, 2010

Suppose we have a three-dimensional scalar field a(r) and three-dimensional solenoidal (divergence-free) vector field V(r), both going to zero (sufficiently fast) as |r|→∞. Can we then construct a vector field F(r) such that $\mathbf{\nabla}\cdot\mathbf{F}=a$ and $\mathbf{\nabla}\times\mathbf{F}=\mathbf{V}$; that is to say, can we construct a field with a specified divergence and curl?

As noted in final remarks here, the answer is yes, we can, and given our boundary conditions, it is unique. More specifically, Helmholtz’s theorem, also known as the “fundamental theorem of vector calculus,” detailed here, tells us our field F may be decomposed into irrotational (zero curl) and solenoidal (zero divergence) components. We further noted that the irrotational component may be expressed as the gradient of a scalar field, and the solenoidal term as the curl of a vector field; we thus expressed the decomposition as
$\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}$
Since $\mathbf{\nabla}\times\mathbf{A}$ is the solenoidal term, the divergence of F is just that of the irrotational term, and so our divergence condition is
$\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{F}&=&a\\\mathbf{\nabla}\cdot\left(-\mathbf{\nabla}\phi\right)&=&a\\\nabla^2\phi&=&-a\end{array}$,
Poisson’s equation.
Next, we noted here, that in three-dimensions, the Green’s function for the Laplacian with our boundary conditions is $G(\mathbf{r},\mathbf{r}')=-\frac{1}{4\pi\left|\mathbf{r}-\mathbf{r}'\right|}$. So, we see that
$\phi(\mathbf{r})=\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}\,d^3r'=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$,
since
$\begin{array}{rcl}\nabla_r^2\iiint{G(\mathbf{r},\mathbf{r}')a(\mathbf{r}')}d^3r'&=&\iiint\nabla_r^2\left(G(\mathbf{r},\mathbf{r}')\right)a(\mathbf{r}')d^3r'\\&=&\iiint\delta(\mathbf{r}-\mathbf{r}')a(\mathbf{r}')d^3r'\\&=&a(\mathbf{r})\end{array}$.

Now, let us consider the curl of our field. As the first term is irrotational, we then get curl equation
$\begin{array}{rcl}\mathbf{\nabla}\times\mathbf{F}&=&\mathbf{V}\\\mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)&=&\mathbf{V}\\\mathbf{\nabla}\left(\mathbf{\nabla}\cdot\mathbf{A}\right)-\nabla^2\mathbf{A}&=&\mathbf{V}\end{array}$.
Next, let us consider if our solution A is solenoidal; then the above would reduce to
$\nabla^2\mathbf{A}=-\mathbf{V}$,
the vector Poisson’s equation, whose x, y, and z components are each (scalar) Poisson’s equations analogous to the one for φ. So, we then try solutions like in our previous part:
$A_x(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_x(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$
$A_y(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_y(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$
$A_z(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{V_z(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$,
or combining into a vector equation,
$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$.
Now, we are just left to confirm that this is solenoidal.
First, we note that $\mathbf{\nabla}\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)=-\mathbf{\nabla}'\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)$, where $\mathbf{\nabla}'$ is the del operator with respect to the primed coordinates.
Then, we see
$\begin{array}{rcl}\mathbf{\nabla}\cdot\mathbf{A}&=&\mathbf{\nabla}\cdot\left(\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\right)\\&=&\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\\&=&-\frac{1}{4\pi}\iiint\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'\end{array}$.

Now, recall vector integration by parts; specifically, for scalar function φ(r) and vector function v(r), then one has on volume V with bounding surface S:
$\iiint_{V}\mathbf{\nabla}\phi\cdot\mathbf{v}\,dV=\iint_{S}\phi\mathbf{v}\cdot\mathbf{\hat{n}}dS-\iiint_{V}\phi\mathbf{\nabla}\cdot\mathbf{v}\,dV$.
Here, our vector field is V(r‘), and our scalar field is $\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}$, also as a function of r‘. Then, we see that

$\begin{array}{rcl}\iiint_{V'}\mathbf{V}(\mathbf{r}')\mathbf{\nabla}'\cdot\left(\frac1{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\,d^3r'&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'-\iiint_{V'}\frac{\mathbf{\nabla}'\cdot\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'\\&=&\iint_{S'}\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\cdot\mathbf{\hat{n}}'dS'\end{array}$,
as V is solenoidal, so the numerator in the integrand of the right-hand-side volume integral is zero. Now, if V(r) goes to zero faster than |r|-1 as |r|→∞, then the surface integral goes to zero as the volume integral is expanded to all space, and thus, our solution is solenoidal, as we assumed, and so satisfies $\mathbf{\nabla}\times\left(\mathbf{\nabla}\times\mathbf{A}\right)=\mathbf{V}$.
Thus, we have (unique) solution
$\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}$,
where
$\phi(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{a(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$,
and
$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\iiint\frac{\mathbf{V}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\,d^3r'$.

### Physics Friday 135

September 10, 2010

Part 15: Polarization of Electromagnetic Waves

Last week, we began discussing electromagnetic waves in a vacuum, obtaining the fact that light and other electromagnetic waves are transverse waves, with the electric and magnetic fields perpendicular to the direction of propagation. We also established that the fields are perpendicular to each other, and proportional: for waves in vacuum, we have .
So, since the magnetic field is determined by the electric field and the propagation direction, we will focus on the electric field E for the following description, with the magnetic field behaving similarly due to the above.
Given our unit vector in the direction , we may choose a pair of orthogonal unit vectors  and  in the plane with normal , so that the three form an orthonormal basis for space. However, to simplify the discussion, let the wave propagate in the x direction, so that  becomes , while  and  become  and . Now, we can express our electric field in terms of x and y components. Recall that we have electric field , which for our wave propagating in the z direction is
,
or, breaking into components,
,
where Ex and Ey are complex numbers, and we take the real component of the right-hand side above to get the physical field. Specifically, that Ex and Ey are complex means each carries a phase, not necessarily the same.
Suppose, though, that Ex and Ey do have the same phase, so that  is real. Then, taking the real part, we have
,
where φ is the phase. We see then that the electric field vector maintains a constant angle θ with the x axis, , and oscillates sinusoidally in this line. Light in this mode is a linearly polarized wave; and we have retrieved another classic optical phenomenon from Maxwell’s laws.
Now, suppose our phases are different. In particular, let us start with the simple case that the two components have the same magnitude, but differ in phase by π/2 (90°); then , and we have
, or letting Ex set our zero phase,
.
Note that at a particular fixed point in space, then, the electric field vector is of constant magnitude, but rotates in direction with angular frequency ω, sweeping out a circle. Thus, we dub such a wave circularly polarized. Specifically, as we look into the wave, we note that for the upper (+) sign, the vector rotates counter-clockwise, and in the lower (-) sign, the vector rotates clockwise. We say that the former is left circularly polarized, or has positive helicity, while the latter is right circularly polarized, or has negative helicity.
In the case where the phases of Ex and Ey differ by a value other than zero or π/2, or they differ in both phase and amplitude, then the tip of the electric field vector traces an ellipse in the xy-plane, and we call the light elliptically polarized.
Now, note that if we add two circularly polarized waves of equal magnitude and opposite helicity, the result is a linearly polarized wave, with the angle of polarization depending on the phase difference between the circularly polarized waves. More generally, defining the complex-valued vectors ,
we see that left circularly polarized light is given by
,
and right circularly polarized light is given by
,
where E+ and E are complex amplitudes.
Further, any complex vector  may be expressed as a complex linear combination of , and so the general wave may be expressed as
,
and the circular polarizations form a basis: any electromagnetic wave may be decomposed into left and right circularly polarized components. [This is key to the phenomenon of optical rotation.]
For the above basis, let us set , so that r is the ratio of the magnitudes of the components, and φ the phase difference. Then, with a little math, one can show that the ellipse traced by the electric field vector has semi-minor axis to semi-major axis ratio of
, and thus eccentricity of , and that the major and minor axes of this ellipse are rotated from the x and y axes by an angle φ/2.
If r=1, we get, as previously noted, a linearly polarized wave, at an angle θ=φ/2.