text:
## Motivation

In a discord thread on the bivector group (a geometric algebra group chat), MoneyKills posts about trouble he has calculating the correct expression for the angular momentum bivector or it’s dual.

This blog post is a more long winded answer than my bivector response and includes this calculation using both cylindrical and spherical coordinates.

## Cylindrical coordinates.

The position vector for any point on a plane can be expressed as
\label{eqn:amomentum:20}
\Br = r \rcap,

where $$\rcap = \rcap(\phi)$$ encodes all the angular dependence of the position vector, and $$r$$ is the length along that direction to our point, as illustrated in fig. 1.

fig. 1. Cylindrical coordinates position vector.

The radial unit vector has a compact GA representation
\label{eqn:amomentum:40}
\rcap = \Be_1 e^{i\phi},

where $$i = \Be_1 \Be_2$$.

The velocity (or momentum) will have both $$\rcap$$ and $$\phicap$$ dependence. By chain rule, that velocity is
\label{eqn:amomentum:60}
\Bv = \dot{r} \rcap + r \dot{\rcap},

where
\label{eqn:amomentum:80}
\begin{aligned}
\dot{\rcap}
&= \Be_1 i e^{i\phi} \dot{\phi} \\
&= \Be_2 e^{i\phi} \dot{\phi} \\
&= \phicap \dot{\phi}.
\end{aligned}

It is left to the reader to show that the vector designated $$\phicap$$, is a unit vector and perpendicular to $$\rcap$$ (Hint: compute the grade-0 selection of the product of the two to show that they are perpendicular.)

We can now compute the momentum, which is
\label{eqn:amomentum:100}
\Bp = m \Bv = m \lr{ \dot{r} \rcap + r \dot{\phi} \phicap },

and the angular momentum bivector
\label{eqn:amomentum:120}
\begin{aligned}
L
&= \Br \wedge \Bp \\
&= m \lr{ r \rcap } \wedge \lr{ \dot{r} \rcap + r \dot{\phi} \phicap } \\
&= m r^2 \dot{\phi} \rcap \phicap.
\end{aligned}

This has the $$m r^2 \dot{\phi}$$ magnitude that the OP was seeking.

## Spherical coordinates.

In spherical coordinates, our position vector is
\label{eqn:amomentum:140}
\Br = r \lr{ \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta },

as sketched in fig. 2.

fig. 2. Spherical coordinates.

We can factor this into a more compact representation
\label{eqn:amomentum:160}
\begin{aligned}
\Br
&= r \lr{ \sin\theta \Be_1 (\cos\phi + \Be_{12} \sin\phi ) + \Be_3 \cos\theta } \\
&= r \lr{ \sin\theta \Be_1 e^{\Be_{12} \phi } + \Be_3 \cos\theta } \\
&= r \Be_3 \lr{ \cos\theta + \sin\theta \Be_3 \Be_1 e^{\Be_{12} \phi } }.
\end{aligned}

It is useful to name two of the bivector terms above, first, we write $$i$$ for the azimuthal plane bivector sketched in fig. 3.

Spherical coordinates, azimuthal plane.

\label{eqn:amomentum:180}
i = \Be_{12},

and introduce a bivector $$j$$ that encodes the $$\Be_3, \rcap$$ plane as sketched in fig. 4.

Spherical coordinates, “j-plane”.

\label{eqn:amomentum:200}
j = \Be_{31} e^{i \phi}.

Having done so, we now have a compact representation for our position vector
\label{eqn:amomentum:220}
\begin{aligned}
\Br
&= r \Be_3 \lr{ \cos\theta + j \sin\theta } \\
&= r \Be_3 e^{j \theta}.
\end{aligned}

This provides us with a nice compact representation of the radial unit vector
\label{eqn:amomentum:240}
\rcap = \Be_3 e^{j \theta}.

Just as was the case in cylindrical coordinates, our azimuthal plane unit vector is
\label{eqn:amomentum:280}
\phicap = \Be_2 e^{i\phi}.

Now we want to compute the velocity vector. As was the case in cylindrical coordinates, we have
\label{eqn:amomentum:300}
\Bv = \dot{r} \rcap + r \dot{\rcap},

but now we need the spherical representation for the $$\rcap$$ derivative, which is
\label{eqn:amomentum:320}
\begin{aligned}
\dot{\rcap}
&=
\PD{\theta}{\rcap} \dot{\theta} + \PD{\phi}{\rcap} \dot{\phi} \\
&=
\Be_3 e^{j\theta} j \dot{\theta} + \Be_3 \sin\theta \PD{\phi}{j} \dot{\phi} \\
&=
\rcap j \dot{\theta} + \Be_3 \sin\theta j i \dot{\phi}.
\end{aligned}

We can reduce the second multivector term without too much work
\label{eqn:amomentum:340}
\begin{aligned}
\Be_3 j i
&=
\Be_3 \Be_{31} e^{i\phi} i \\
&=
\Be_3 \Be_{31} i e^{i\phi} \\
&=
\Be_{33112} e^{i\phi} \\
&=
\Be_{2} e^{i\phi} \\
&= \phicap,
\end{aligned}

so we have
\label{eqn:amomentum:360}
\dot{\rcap}
=
\rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi}.

The velocity is
\label{eqn:amomentum:380}
\Bv = \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} }.

Now we can finally compute the angular momentum bivector, which is
\label{eqn:amomentum:400}
\begin{aligned}
L &=
\Br \wedge \Bp \\
&=
m r \rcap \wedge \lr{ \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } } \\
&=
m r^2 \rcap \wedge \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } \\
&=
m r^2 \gpgradetwo{ \rcap \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } },
\end{aligned}

which is just
\label{eqn:amomentum:420}
L =
m r^2 \lr{ j \dot{\theta} + \sin\theta \rcap \phicap \dot{\phi} }.

I was slightly surprised by this result, as I naively expected the cylindrical coordinate result. We have a $$m r^2 \rcap \phicap \dot{\phi}$$ term, as was the case in cylindrical coordinates, but scaled down with a $$\sin\theta$$ factor. However, this result does make sense. Consider for example, some fixed circular motion with $$\theta = \mathrm{constant}$$, as sketched in fig. 5.

fig. 5. Circular motion for constant theta

The radius of this circle is actually $$r \sin\theta$$, so the total angular momentum for that motion is scaled down to $$m r^2 \sin\theta \dot{\phi}$$, smaller than the maximum circular angular momentum of $$m r^2 \dot{\phi}$$ which occurs in the $$\theta = \pi/2$$ azimuthal plane. Similarly, if we have circular motion in the “j-plane”, sketched in fig. 6.

fig. 6. Circular motion for constant phi.

where $$\phi = \mathrm{constant}$$, then our angular momentum is $$L = m r^2 j \dot{\theta}$$.

## PHY2403H Quantum Field Theory. Lecture 9: Unbroken and spontaneously broken symmetries, Higgs Lagrangian, scale invariance, Lorentz invariance, angular momentum quantization. Taught by Prof. Erich Poppitz

[Click here for a PDF of this post with nicer formatting (and a Mathematica listing that I didn’t include in this blog post’s latex export)]

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Last time

We followed a sequence of operations

1. Noether’s theorem
2. $$\rightarrow$$ conserved currents
3. $$\rightarrow$$ charges (classical)
4. $$\rightarrow$$ “correspondence principle”
5. $$\rightarrow \hatQ$$
• Hermitian operators
• “generators of symmetry”
\label{eqn:qftLecture9:20}
\hatU(\alpha) = e^{i \alpha \hatQ}

We found
\label{eqn:qftLecture9:40}
\hatU(\alpha) \phihat \hatU^\dagger(\alpha) = \phihat + i \alpha \antisymmetric{\hatQ}{\phihat} + \cdots

### Example: internal symmetries:

(non-spacetime), such as $$O(N)$$ or $$U(1)$$.

In QFT internal symmetries can have different “\underline{modes of realization}”.

[I]

1. “Wigner mode”. These are also called “unbroken symmetries”.
\label{eqn:qftLecture9:60}
\hatQ \ket{0} = 0

i.e. $$\hatU(\alpha) \ket{0} = 0$$.
Ground state invariant. Formally $$:\hatQ:$$ annihilates $$\ket{0}$$.
$$\antisymmetric{\hatQ}{\hatH} = 0$$ implies that all eigenstates are eigenstates of $$\hatQ$$ in $$U(1)$$. Example from HW 1
\label{eqn:qftLecture9:80}
\hatQ = \text{“charge” under $$U(1)$$}.

All states have definite charge, just live in QU.
2. “Nambu-Goldstone mode” (Landau-ginsburg). This is also called a “spontaneously broken symmetry”\footnote{
First encounter example (HWII, $$SU(2) \times SU(2) \rightarrow SU(2)$$). Here a $$U(1)$$ spontaneous broken symmetry.}.
$$H$$ or $$L$$ is invariant under symmetry, but ground state is not.

fig. 1. Mexican hat potential.

fig. 2. Degenerate Mexican hat potential ( v = 0)

Example:
\label{eqn:qftLecture9:100}
\LL = \partial_\mu \phi^\conj \partial^\mu \phi – V(\Abs{\phi}),

where
\label{eqn:qftLecture9:120}
V(\Abs{\phi}) = m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi }^2.

When $$m^2 > 0$$ we have a Wigner mode, but when $$m^2 < 0$$ we have an issue: $$\phi = 0$$ is not a minimum of potential.
When $$m^2 < 0$$ we write
\label{eqn:qftLecture9:140}
\begin{aligned}
V(\phi)
&= – m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi}^2 \\
&= \frac{\lambda}{4} \lr{
\lr{ \phi^\conj \phi}^2 – \frac{4}{\lambda} m^2 } \\
&= \frac{\lambda}{4} \lr{
\phi^\conj \phi – \frac{2}{\lambda} m^2 }^2 – \frac{4 m^4}{\lambda^2},
\end{aligned}

or simply
\label{eqn:qftLecture9:780}
V(\phi)
=
\frac{\lambda}{4} \lr{ \phi^\conj \phi – v^2 }^2 + \text{const}.

The potential (called the Mexican hat potential) is illustrated in fig. 1 for non-zero $$v$$, and in
fig. 2 for $$v = 0$$.
We choose to expand around some point on the minimum ring (it doesn’t matter which one).
When there is no potential, we call the field massless (i.e. if we are in the minimum ring).
We expand as
\label{eqn:qftLecture9:160}
\phi(x) = v \lr{ 1 + \frac{\rho(x)}{v} } e^{i \alpha(x)/v },

so
\label{eqn:qftLecture9:180}
\begin{aligned}
\frac{\lambda}{4}
\lr{\phi^\conj \phi – v^2}^2
&=
\lr{
v^2 \lr{ 1 + \frac{\rho(x)}{v} }^2
– v^2
}^2 \\
&=
\frac{\lambda}{4}
v^4 \lr{ \lr{ 1 + \frac{\rho(x)}{v} }^2 – 1 } \\
&=
\frac{\lambda}{4}
v^4
\lr{
\frac{2 \rho}{v} + \frac{\rho^2}{v^2}
}^2.
\end{aligned}

\label{eqn:qftLecture9:200}
\partial_\mu \phi =
\lr{
v \lr{ 1 + \frac{\rho(x)}{v} } \frac{i}{v} \partial_\mu \alpha
+ \partial_\mu \rho
} e^{i \alpha}

so
\label{eqn:qftLecture9:220}
\begin{aligned}
\LL
&= \Abs{\partial \phi^\conj}^2 – \frac{\lambda}{4} \lr{ \Abs{\phi^\conj}^2 – v^2 }^2 \\
&=
\partial_\mu \rho \partial^\mu \rho + \partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }

\frac{\lambda v^4}{4} \frac{ 4\rho^2}{v^2} + O(\rho^3) \\
&=
\partial_\mu \rho \partial^\mu \rho
– \lambda v^2\rho^2
+
\partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }.
\end{aligned}

We have two fields, $$\rho$$ : a massive scalar field, the “Higgs”, and a massless field $$\alpha$$ (the Goldstone Boson).

$$U(1)$$ symmetry acts on $$\phi(x) \rightarrow e^{i \omega } \phi(x)$$ i.t.o $$\alpha(x) \rightarrow \alpha(x) + v \omega$$.
$$U(1)$$ global symmetry (broken) acts on the Goldstone field $$\alpha(x)$$ by a constant shift. ($$U(1)$$ is still a symmetry of the Lagrangian.)

The current of the $$U(1)$$ symmetry is:
\label{eqn:qftLecture9:240}
j_\mu = \partial_\mu \alpha \lr{ 1 + \text{higher dimensional $$\rho$$ terms} }.

When we quantize
\label{eqn:qftLecture9:260}
\alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp

\label{eqn:qftLecture9:280}
j^\mu(x) = \partial^\mu \alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ i \omega_\Bp – i \Bp } e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ -i \omega_\Bp + i \Bp } e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp.

\label{eqn:qftLecture9:300}
j^\mu(x) \ket{0} \ne 0,

instead it creates a single particle state.

## Examples of symmetries

In particle physics, examples of Wigner vs Nambu-Goldstone, ignoring gravity the only exact internal symmetry in the standard module is
$$(B\# – L\#)$$, believed to be a $$U(1)$$ symmetry in Wigner mode.

Here $$B\#$$ is the Baryon number, and $$L\#$$ is the Lepton number. Examples:

• $$B(p) = 1$$, proton.
• $$B(q) = 1/3$$, quark
• $$B(e) = 1$$, electron
• $$B(n) = 1$$, neutron.
• $$L(p) = 1$$, proton.
• $$L(q) = 0$$, quark.
• $$L(e) = 0$$, electron.

The major use of global internal symmetries in the standard model is as “approximate” ones. They become symmetries when one neglects some effect( “terms in $$\LL$$”).
There are other approximate symmetries (use of group theory to find the Balmer series).

### Example from HW2:

QCD in limit
\label{eqn:qftLecture9:320}
m_u = m_d = 0.

$$m_u m_d \ll m_p$$ (the products of the up-quark mass and the down-quark mass are much less than a composite one (name?)).
$$SU(2)_L \times SU(2)_R \rightarrow SU(2)_V$$

### EWSB (Electro-Weak-Symmetry-Breaking) sector

When the couplings $$g_2, g_1 = 0$$. ($$g_2 \in SU(2), g_1 \in U(1)$$).

## Scale invariance

\label{eqn:qftLecture9:340}
\begin{aligned}
x &\rightarrow e^{\lambda} x \\
\phi &\rightarrow e^{-\lambda} \phi \\
A_\mu &\rightarrow e^{-\lambda} A_\mu
\end{aligned}

Any unitary theory which is scale invariant is also \underline{conformal} invariant. Conformal invariance means that angles are preserved.
The point here is that there is more than scale invariance.

We have classical internal global continuous symmetries.
These can be either

1. “unbroken” (Wigner mode)
\label{eqn:qftLecture9:360}
\hatQ\ket{0} = 0.
2. “spontaneously broken”
\label{eqn:qftLecture9:380}
j^\mu(x) \ket{0} \ne 0

(creates Goldstone modes).
3. “anomalous”. Classical symmetries are not a symmetry of QFT.
Examples:

• Scale symmetry (to be studied in QFT II), although this is not truly internal.
• In QCD again when $$\omega_\Bq = 0$$, a $$U(1$$ symmetry (chiral symmetry) becomes exact, and cannot be preserved in QFT.
• In the standard model (E.W sector), the Baryon number and Lepton numbers are not symmetries, but their difference $$B\# – L\#$$ is a symmetry.

## Lorentz invariance.

We’d like to study the action of Lorentz symmetries on quantum states. We are going to “go by the book”, finding symmetries, currents, quantize, find generators, and so forth.

Under a Lorentz transformation
\label{eqn:qftLecture9:400}
x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu,

We are going to consider infinitesimal Lorentz transformations
\label{eqn:qftLecture9:420}
{\Lambda^\mu}_\nu \approx
{\delta^\mu}_\nu + {\omega^\mu}_\nu
,

where $${\omega^\mu}_\nu$$ is small.
A Lorentz transformation $$\Lambda$$ must satisfy $$\Lambda^\T G \Lambda = G$$, or
\label{eqn:qftLecture9:800}
g_{\mu\nu} = {{\Lambda}^\alpha}_\mu g_{\alpha \beta} {{\Lambda}^\beta}_\nu,

into which we insert the infinitesimal transformation representation
\label{eqn:qftLecture9:820}
\begin{aligned}
0
&=
– g_{\mu\nu} +
\lr{ {\delta^\alpha}_\mu + {\omega^\alpha}_\mu }
g_{\alpha \beta}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
\lr{
g_{\mu \beta}
+
\omega_{\beta\mu}
}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
g_{\mu \nu}
+
\omega_{\nu\mu}
+
\omega_{\mu\nu}
+
\omega_{\beta\mu}
{\omega^\beta}_\nu.
\end{aligned}

The quadratic term can be ignored, leaving just
\label{eqn:qftLecture9:840}
0 =
\omega_{\nu\mu}
+
\omega_{\mu\nu},

or
\label{eqn:qftLecture9:860}
\omega_{\nu\mu} = – \omega_{\mu\nu}.

Note that $$\omega$$ is a completely antisymmetric tensor, and like $$F_{\mu\nu}$$ this has only 6 elements.
This means that the
infinitesimal transformation of the coordinates is
\label{eqn:qftLecture9:440}
x^\mu \rightarrow {x’}^\mu \approx x^\mu + \omega^{\mu\nu} x_\nu,

the field transforms as
\label{eqn:qftLecture9:460}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

or
\label{eqn:qftLecture9:760}
\phi'(x^\mu + \omega^{\mu\nu} x_\nu) =
\phi'(x) + \omega^{\mu\nu} x_\nu \partial_\mu\phi(x) = \phi(x),

so
\label{eqn:qftLecture9:480}
\delta \phi = \phi'(x) – \phi(x) =
-\omega^{\mu\nu} x_\nu \partial_\mu \phi.

Since $$\LL$$ is a scalar
\label{eqn:qftLecture9:500}
\begin{aligned}
\delta \LL
&=
-\omega^{\mu\nu} x_\nu \partial_\mu \LL \\
&=

\partial_\mu \lr{
\omega^{\mu\nu} x_\nu \LL
}
+
(\partial_\mu x_\nu) \omega^{\mu\nu} \LL \\
&=
\partial_\mu \lr{

\omega^{\mu\nu} x_\nu \LL
},
\end{aligned}

since $$\partial_\nu x_\mu = g_{\nu\mu}$$ is symmetric, and $$\omega$$ is antisymmetric.
Our current is
\label{eqn:qftLecture9:520}
J^\mu_\omega
=

\omega^{\mu\nu} x_\mu \LL
.

Our Noether current is
\label{eqn:qftLecture9:540}
\begin{aligned}
j^\nu_{\omega^{\mu\rho}}
&= \PD{\phi_{,\nu}}{\LL} \delta \phi – J^\mu_\omega \\
&=
\partial^\nu \phi\lr{ – \omega^{\mu\rho} x_\rho \partial_\mu \phi } + \omega^{\nu \rho} x_\rho \LL \\
&=
\omega^{\mu\rho}
\lr{
\partial^\nu \phi\lr{ – x_\rho \partial_\mu \phi } + {\delta^{\nu}}_\mu x_\rho \LL
} \\
&=
\omega^{\mu\rho} x_\rho
\lr{
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL
}
\end{aligned}

We identify
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

so the current is
\label{eqn:qftLecture9:580}
\begin{aligned}
j^\nu_{\omega_{\mu\rho}}
&=
-\omega^{\mu\rho} x_\rho
{T^\nu}_\mu \\
&=
-\omega_{\mu\rho} x^\rho
T^{\nu\mu}
.
\end{aligned}

Define
\label{eqn:qftLecture9:600}
j^{\nu\mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} },

which retains the antisymmetry in $$\mu \rho$$ yet still drops the parameter $$\omega^{\mu\rho}$$.
To check that this makes sense, we can contract
$$j^{\nu\mu\rho}$$ with $$\omega_{\rho\mu}$$
\label{eqn:qftLecture9:880}
\begin{aligned}
j^{\nu\mu\rho} \omega_{\rho\mu}
&= -\inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} }
\omega_{\mu\rho} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\mu} T^{\nu\rho}
\omega_{\rho\mu} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\rho} T^{\nu\mu}
\omega_{\mu\rho} \\
&=
– x^{\rho} T^{\nu\mu}
\omega_{\mu\rho},
\end{aligned}

which matches \ref{eqn:qftLecture9:580} as desired.

### Example. Rotations $$\mu\rho = ij$$

\label{eqn:qftLecture9:620}
\begin{aligned}
J^{0 i j} \epsilon_{ijk}
&=
\inv{2} \lr{ x^i T^{0j} – x^{j} T^{0i} } \epsilon_{ijk} \\
&=
x^i T^{0j} \epsilon_{ijk}.
\end{aligned}

Observe that this has the structure of $$(\Bx \cross \Bp)_k$$, where $$\Bp$$ is the momentum density of the field.
Let
\label{eqn:qftLecture9:640}
L_k \equiv Q_k = \int d^3 x J^{0ij} \epsilon_{ijk}.

We can now quantize and build a generator
\label{eqn:qftLecture9:660}
\begin{aligned}
\hatU(\Balpha)
&= e^{i \Balpha \cdot \hat{\BL}} \\
&= \exp\lr{i \alpha_k
\int d^3 x x^i \hat{T}^{0j} \epsilon_{ijk}
}
\end{aligned}

From \ref{eqn:qftLecture9:560} we can quantize with $$T^{0j} = \partial^0 \phi \partial^j \phi \rightarrow \hat{\pi} \lr{\spacegrad \phihat}_j$$, or
\label{eqn:qftLecture9:900}
\begin{aligned}
\hatU(\Balpha)
&=
\exp\lr{i \alpha_k
\int d^3 x x^i \hat{\pi} (\spacegrad \phihat)_j \epsilon_{ijk}
} \\
&=
\exp\lr{i \Balpha \cdot
\int d^3 x \hat{\pi} \spacegrad \phihat \cross \Bx
}
\end{aligned}

\label{eqn:qftLecture9:680}
\begin{aligned}
\phihat(\By) \rightarrow \hatU(\alpha) \phihat(\By) \hatU^\dagger(\alpha)
&\approx
\phihat(\By) +
i \Balpha \cdot
\antisymmetric{
\int d^3 x \hat{\pi}(\Bx) \spacegrad \phihat(\Bx) \cross \Bx
}
{
\phihat(\By)
} \\
&=
\phihat(\By) +
i \Balpha \cdot
\int d^3 x
(-i) \delta^3(\Bx – \By)
&=
\phihat(\By) +
\Balpha \cdot \lr{ \spacegrad \phihat(\By ) \cross \By}
\end{aligned}

Explicitly, in coordinates, this is
\label{eqn:qftLecture9:700}
\begin{aligned}
\phihat(\By)
&\rightarrow
\phihat(\By) +
\alpha^i
\lr{
\partial^j \phihat(\By) y^k \epsilon_{jki}
} \\
&=
\phihat(\By) –
\epsilon_{ikj} \alpha^i y^k \partial^j \phihat \\
&=
\phihat( y^j – \epsilon^{ikj} \alpha^i y^k ).
\end{aligned}

This is a rotation. To illustrate, pick $$\Balpha = (0, 0, \alpha)$$, so $$y^j \rightarrow y^j – \epsilon^{ikj} \alpha y^k \delta_{i3} = y^j – \epsilon^{3kj} \alpha y^k$$, or
\label{eqn:qftLecture9:n}
\begin{aligned}
y^1 &\rightarrow y^1 – \epsilon^{3k1} \alpha y^k = y^1 + \alpha y^2 \\
y^2 &\rightarrow y^2 – \epsilon^{3k2} \alpha y^k = y^2 – \alpha y^1 \\
y^3 &\rightarrow y^3 – \epsilon^{3k3} \alpha y^k = y^3,
\end{aligned}

or in matrix form
\label{eqn:qftLecture9:720}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha & 0 \\
-\alpha & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}.

## Totally asymmetric potential

December 16, 2015 phy1520 , , ,

### Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.

(b) If the wave function of such a state is expanded as

\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),

what are the phase restrictions on $$F_{lm}$$?

### A: part (a)

For a time reversal invariant Hamiltonian $$H$$ we have

\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.

If $$\ket{\psi}$$ is an energy eigenstate with eigenvalue $$E$$, we have

\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}

so $$\Theta \ket{\psi}$$ is also an eigenvalue of $$H$$, so can only differ from $$\ket{\psi}$$ by a phase factor. That is

\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}

Now consider the expectation of $$\BL$$ with respect to a time reversed state

\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}

however, we also have

\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}

so we have $$\bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi}$$ which is only possible if $$\expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0$$.

### A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}

and then consider the same state expanded in the position basis

\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}

This demonstrates a relationship between the wave function and its complex conjugate

\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.

Now expand the wave function in the spherical harmonic basis

\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}

so the $$F_{lm}$$ functions are constrained by

\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Angular momentum expectation values

December 14, 2015 phy1520 ,

### Q: [1] pr 3.18

Compute the expectation values for the first and second powers of the angular momentum operators with respect to states $$\ket{lm}$$.

### A:

We can write the expectation values for the $$L_z$$ powers immediately

\label{eqn:angularMomentumExpectation:20}
\expectation{L_z}
= m \Hbar,

and

\label{eqn:angularMomentumExpectation:40}
\expectation{L_z^2} = (m \Hbar)^2.

For the x and y components first express the operators in terms of the ladder operators.

\label{eqn:angularMomentumExpectation:60}
\begin{aligned}
L_{+} &= L_x + i L_y \\
L_{-} &= L_x – i L_y.
\end{aligned}

Rearranging gives

\label{eqn:angularMomentumExpectation:80}
\begin{aligned}
L_x &= \inv{2} \lr{ L_{+} + L_{-} } \\
L_y &= \inv{2i} \lr{ L_{+} – L_{-} }.
\end{aligned}

The first order expectations $$\expectation{L_x}, \expectation{L_y}$$ are both zero since $$\expectation{L_{+}} = \expectation{L_{-}}$$. For the second order expectation values we have

\label{eqn:angularMomentumExpectation:100}
\begin{aligned}
L_x^2
&= \inv{4} \lr{ L_{+} + L_{-} } \lr{ L_{+} + L_{-} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + L_{+} L_{-} + L_{-} L_{+} } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (L_x^2 + L_y^2) } \\
&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (\BL^2 – L_z^2) },
\end{aligned}

and
\label{eqn:angularMomentumExpectation:120}
\begin{aligned}
L_y^2
&= -\inv{4} \lr{ L_{+} – L_{-} } \lr{ L_{+} – L_{-} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – L_{+} L_{-} – L_{-} L_{+} } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (L_x^2 + L_y^2) } \\
&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (\BL^2 – L_z^2) }.
\end{aligned}

Any expectation value $$\bra{lm} L_{+} L_{+} \ket{lm}$$ or $$\bra{lm} L_{-} L_{-} \ket{lm}$$ will be zero, leaving

\label{eqn:angularMomentumExpectation:140}
\begin{aligned}
\expectation{L_x^2}
&=
\expectation{L_y^2} \\
&=
\inv{4} \expectation{2 (\BL^2 – L_z^2) } \\
&=
\inv{2} \lr{ \Hbar^2 l(l+1) – (\Hbar m)^2 }.
\end{aligned}

Observe that we have
\label{eqn:angularMomentumExpectation:160}
\expectation{L_x^2}
+
\expectation{L_y^2}
+
\expectation{L_z^2}
=
\Hbar^2 l(l+1)
=
\expectation{\BL^2},

which is the quantum mechanical analogue of the classical scalar equation $$\BL^2 = L_x^2 + L_y^2 + L_z^2$$.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## L_z and L^2 eigenvalues and probabilities for a given wave function

### Q: [1] 3.17

Given a wave function

\label{eqn:LsquaredLzProblem:20}
\psi(r,\theta, \phi) = f(r) \lr{ x + y + 3 z },

• (a) Determine if this wave function is an eigenfunction of $$\BL^2$$, and the value of $$l$$ if it is an eigenfunction.

• (b) Determine the probabilities for the particle to be found in any given $$\ket{l, m}$$ state,
• (c) If it is known that $$\psi$$ is an energy eigenfunction with energy $$E$$ indicate how we can find $$V(r)$$.

### A: (a)

Using
\label{eqn:LsquaredLzProblem:40}
\BL^2
=
-\Hbar^2 \lr{ \inv{\sin^2\theta} \partial_{\phi\phi} + \inv{\sin\theta} \partial_\theta \lr{ \sin\theta \partial_\theta} },

and

\label{eqn:LsquaredLzProblem:60}
\begin{aligned}
x &= r \sin\theta \cos\phi \\
y &= r \sin\theta \sin\phi \\
z &= r \cos\theta
\end{aligned}

it’s a quick computation to show that

\label{eqn:LsquaredLzProblem:80}
\BL^2 \psi = 2 \Hbar^2 \psi = 1(1 + 1) \Hbar^2 \psi,

so this function is an eigenket of $$\BL^2$$ with an eigenvalue of $$2 \Hbar^2$$, which corresponds to $$l = 1$$, a p-orbital state.

### (b)

Recall that the angular representation of $$L_z$$ is

\label{eqn:LsquaredLzProblem:100}
L_z = -i \Hbar \PD{\phi},

so we have

\label{eqn:LsquaredLzProblem:120}
\begin{aligned}
L_z x &= i \Hbar y \\
L_z y &= – i \Hbar x \\
L_z z &= 0,
\end{aligned}

The $$L_z$$ action on $$\psi$$ is

\label{eqn:LsquaredLzProblem:140}
L_z \psi = -i \Hbar r f(r) \lr{ – y + x }.

This wave function is not an eigenket of $$L_z$$. Expressed in terms of the $$L_z$$ basis states $$e^{i m \phi}$$, this wave function is

\label{eqn:LsquaredLzProblem:160}
\begin{aligned}
\psi
&= r f(r) \lr{ \sin\theta \lr{ \cos\phi + \sin\phi} + \cos\theta } \\
&= r f(r) \lr{ \frac{\sin\theta}{2} \lr{ e^{i \phi} \lr{ 1 + \inv{i}} + e^{-i\phi} \lr{ 1 – \inv{i} } } + \cos\theta } \\
&= r f(r) \lr{
\frac{(1-i)\sin\theta}{2} e^{1 i \phi}
+
\frac{(1+i)\sin\theta}{2} e^{- 1 i \phi}
+ \cos\theta e^{0 i \phi}
}
\end{aligned}

Assuming that $$\psi$$ is normalized, the probabilities for measuring $$m = 1,-1,0$$ respectively are

\label{eqn:LsquaredLzProblem:180}
\begin{aligned}
P_{\pm 1}
&= 2 \pi \rho \Abs{\frac{1\mp i}{2}}^2 \int_0^\pi \sin\theta d\theta \sin^2 \theta \\
&= -2 \pi \rho \int_1^{-1} du (1-u^2) \\
&= 2 \pi \rho \evalrange{ \lr{ u – \frac{u^3}{3} } }{-1}{1} \\
&= 2 \pi \rho \lr{ 2 – \frac{2}{3}} \\
&= \frac{ 8 \pi \rho}{3},
\end{aligned}

and

\label{eqn:LsquaredLzProblem:200}
P_{0} = 2 \pi \rho \int_0^\pi \sin\theta \cos\theta = 0,

where

\label{eqn:LsquaredLzProblem:220}
\rho = \int_0^\infty r^4 \Abs{f(r)}^2 dr.

Because the probabilities must sum to 1, this means the $$m = \pm 1$$ states are equiprobable with $$P_{\pm} = 1/2$$, fixing $$\rho = 3/16\pi$$, even without knowing $$f(r)$$.

### (c)

The operator $$r^2 \Bp^2$$ can be decomposed into a $$\BL^2$$ component and some other portions, from which we can write

\label{eqn:LsquaredLzProblem:240}
\begin{aligned}
H \psi
&= \lr{ \frac{\Bp^2}{2m} + V(r) } \psi \\
&=
\lr{
– \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \inv{\Hbar^2 r^2} \BL^2 } + V(r) } \psi.
\end{aligned}

(See: [1] eq. 6.21)

In this case where $$\BL^2 \psi = 2 \Hbar^2 \psi$$ we can rearrange for $$V(r)$$

\label{eqn:LsquaredLzProblem:260}
\begin{aligned}
V(r)
&= E + \inv{\psi} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } \psi \\
&= E + \inv{f(r)} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } f(r).
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

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CARAIB is a mechanistic ... [more ▼]As a part of the MACSUR task L2.4, we probabilistically calibrated the CARAIB dynamic vegetation model by Markov chain Monte Carlo (MCMC) simulation with the DREAMZS sampler.. CARAIB is a mechanistic model that calculates the carbon assimilation of the vegetation as a function of the soil and climatic conditions, and can thus be used for simulating grassland production under cutting or grazing management. Bayesian model inversion was performed at 4 grassland sites across Europe: Oensingen, CH; Grillenburg, DE; Laqueuille, FR and Monte-Bodone, IT. Four daily measured variables from these sites: the Gross Primary Productivity (GPP), Net Ecosystem Exchange (NEE), Evapotranspiration (ET) and Soil Water Content (SWC) were used to sample 10 parameters related to rooting depth, stomatal conductance, specific leaf area, carbon-nitrogen ratio and water stresses. The maximized likelihood function therefore involved four objectives, whereas the applied Bayesian framework allowed for assessing the so called parameter posterior probability density function (pdf), which quantifies model parameter uncertainty caused by measurement and model errors. Sampling trials were performed using merged data from all sites (all-sites-sampling) and for each site (site-specific sampling) separately. The derived posterior parameter pdfs from the all-sites sampling and site-specific sampling runs showed differences in relation with the specificities of each site. Analysis of these distributions also revealed model sensitivity to parameters conditioned on the measured data, as well as parameter correlations. [less ▲]Detailed reference viewed: 15 (0 ULg) Can a global dynamic vegetation model be used for both grassland and crop modeling at the local scale?Minet, Julien ; Tychon, Bernard ; Jacquemin, Ingrid et alPoster (2014, February)We report on the use of a dynamic vegetation model, CARAIB, within two modeling exercises in the framework of MACSUR. CARAIB is a physically-based, mechanistic model that calculates the carbon ... [more ▼]We report on the use of a dynamic vegetation model, CARAIB, within two modeling exercises in the framework of MACSUR. CARAIB is a physically-based, mechanistic model that calculates the carbon assimilation of the vegetation as a function of the soil and climatic conditions. Within MACSUR, it was used in the model intercomparison exercises for grassland and crop modeling, in the LiveM 2.4 and CropM WP4 tasks, respectively. For grassland modeling, blind model runs at 11 locations were performed for various time ranges (few years). For crop modeling, a sensitivity analysis for building impact response surfaces (IRS) was performed, based on a bench of model runs at different levels of perturbation in the temperature and precipitation input data over 30 years. For grassland modeling, specific management functions accounting for the cutting or grazing of the grass were added to the model, in the framework of the MACSUR intercomparison. Initially developed for modeling the carbon dynamics of the natural vegetation, CARAIB was already adapted for crop modeling but further modifications regarding the management, i.e., yearly-dependent sowing dates, were introduced. For grassland modeling, simulation results will be further intercompared with other modeling groups, but preliminary results showed that the model could cope with the introduction of the grass cutting module. For crop modeling, building the IRS over 30 years permitted to assess the sensitivity of the model to temperature and precipitation changes. So far, the participation of CARAIB in the intercomparison exercises within MACSUR resulted in further improvements of the model by introducing new functionalities. [less ▲]Detailed reference viewed: 7 (1 ULg) Implementing agricultural land-use in the CARAIB dynamic vegetation modelFrançois, Louis ; Jacquemin, Ingrid ; Fontaine, Corentin et alConference (2014)CARAIB (Dury et al., iForest - Biogeosciences and Forestry, 4:82-99, 2011) is a state-of-the-art dynamic vegetation model with various modules dealing with (i) soil hydrology, (ii) photosynthesis/stomatal ... [more ▼]CARAIB (Dury et al., iForest - Biogeosciences and Forestry, 4:82-99, 2011) is a state-of-the-art dynamic vegetation model with various modules dealing with (i) soil hydrology, (ii) photosynthesis/stomatal regulation, (iii) carbon allocation and biomass growth, (iv) litter/soil carbon dynamics, (v) vegetation cover dynamics, (vi) seed dispersal, and (vii) vegetation fires. Climate and atmospheric CO2 are the primary inputs. The model calculates all major water and CO2/carbon fluxes and pools. It can be run with plant functional types or species (up to 100 different species) at various spatial scales, from the municipality to country or continental levels. Within the VOTES project (Fontaine et al., Journal of Land Use Science, 2013, DOI:10.1080/1747423X.2013.786150), the model has been improved to include crops and meadows, and some modules have been written to translate model outputs into quantitative indicators of ecosystem services (e.g., evaluate crop yield from net primary productivity or calculate soil erosion from runoff, slope, grown species and various soil attributes). The model was run over an area covering four municipalities in central Belgium, where land-use is dominated by crops, meadows, housing and some forests and was introduced in the model at the land parcel level. Simulations were also performed for the future. In these simulations, CARAIB was combined with the Aporia Agent-Based Model, to project land-use changes up to 2050. This approach is currently extended within the MASC project (funded by Belgian Science Policy, BELSPO) to the whole Belgian territory (at 1 km2) and to Western Europe (at 20 km x 20 km). [less ▲]Detailed reference viewed: 9 (0 ULg) CARAIB USER'S GUIDEMinet, Julien ; Jacquemin, Ingrid ; François, Louis Learning material (2013)Detailed reference viewed: 38 (17 ULg) Community gardening in Wallonia and Brussels : proposals for research and actionsMinet, Julien ; Stevenne, Kari; Loicq, Gaël et alScientific conference (2013, May 23)Detailed reference viewed: 9 (0 ULg) POLICY & FUNDING INFORMATION: Environmentally sustainable growth and intensification of agriculture (Belgium)Minet, Julien Poster (2013, April 10)Detailed reference viewed: 7 (0 ULg) MAPPING OF CURRENT RESEARCH: Environmentally sustainable growth and intensification of agriculture (Belgium)Minet, Julien Poster (2013, April 10)Detailed reference viewed: 10 (1 ULg) Validation of ground penetrating radar full-waveform inversion for field scale soil moisture mappingMinet, Julien ; Bogaert, P.; Vanclooster, M. et alin Journal of Hydrology (2012), 424-425Ground penetrating radar (GPR) is an efficient method for soil moisture mapping at the field scale, bridging the scale gap between small-scale invasive sensors and large-scale remote sensing instruments ... [more ▼]Ground penetrating radar (GPR) is an efficient method for soil moisture mapping at the field scale, bridging the scale gap between small-scale invasive sensors and large-scale remote sensing instruments. Nevertheless, commonly-used GPR approaches for soil moisture characterization suffer from several limitations and the determination of the uncertainties in GPR soil moisture sensing has been poorly addressed. Herein, we used an advanced proximal GPR method based on full-waveform inversion of ultra-wideband radar data for mapping soil moisture and uncertainties in the soil moisture maps were evaluated by three different methods. First, GPR-derived soil moisture uncertainties were computed from the GPR data inversion, according to measurements and modeling errors and to the sensitivity of the electromagnetic model to soil moisture. Second, the reproducibility of the soil moisture mapping was evaluated. Third, GPR-derived soil moisture was compared with ground-truth measurements (soil core sampling). The proposed GPR method appeared to be highly precise and accurate, with spatially averaged GPR inversion uncertainty of 0.0039 m3m-3, a repetition uncertainty of 0.0169 m3m-3 and an uncertainty of 0.0233 m3m-3 when compared with ground-truth measurements. These uncertainties were mapped and appeared to be related to some local model inadequacies and to small-scale variability of soil moisture. In a soil moisture mapping framework, the interpolation was found to be the determinant source of the observed uncertainties. The proposed GPR method was proven to be largely reliable in terms of accuracy and precision and appeared to be highly efficient for soil moisture mapping at the field scale. [less ▲]Detailed reference viewed: 35 (1 ULg) Integrating coarse-scale uncertain soil moisture data into a fine-scale hydrological modelling scenarioVernieuwe, H.; De Baets, B.; Minet, Julien et alin Hydrology & Earth System Sciences (2011), 15Detailed reference viewed: 9 (0 ULg) Reconstruction of sub-wavelength fractures and physical properties of masonry media using full-waveform inversion of proximal penetrating radarPatriarca, C.; Lambot, S.; Mahmoudzadeh, M. R. et alin Journal of Applied Geophysics (2011), 74(1)Detailed reference viewed: 17 (0 ULg) Effect of high-resolution spatial soil moisture variability on simulated runoff response using a distributed hydrologic modelMinet, Julien ; Laloy, E.; Lambot, S. et alin Hydrology & Earth System Sciences (2011), 15The importance of the spatial variability of antecedent soil moisture conditions on runoff response is widely acknowledged in hillslope hydrology. Using a distributed hydrologic model, this paper aims at ... [more ▼]The importance of the spatial variability of antecedent soil moisture conditions on runoff response is widely acknowledged in hillslope hydrology. Using a distributed hydrologic model, this paper aims at investigating the effects of soil moisture spatial variability on runoff in various field conditions and at finding the structure of the soil moisture pattern that approaches the measured soil moisture pattern in terms of field scale runoff. High spatial resolution soil moisture was surveyed in ten different field campaigns using a proximal ground penetrating radar (GPR) mounted on a mobile platform. Based on these soil moisture measurements, seven scenarios of spatial structures of antecedent soil moisture were used and linked with a field scale distributed hydrological model to simulate field scale runoff. Accounting for spatial variability of soil moisture resulted in higher predicted field scale runoff as compared to the case where soil moisture was kept constant. The ranges of possible hydrographs were delineated by the extreme scenarios where soil moisture was directly and inversely modelled according to the topographic wetness index (TWI). These behaviours could be explained by the sizes and relative locations of runoff contributing areas, knowing that runoff was generated by infiltration excess over a certain soil moisture threshold. The most efficient scenario for modeling the within field spatial structure of soil moisture appeared to be when soil moisture is directly arranged according to the TWI, especially when measured soil moisture and TWI were correlated. The novelty of this work is to benefit from a large set of high-resolution soil moisture measurements allowing to model effectively the within field distribution of soil moisture and its impact on the field scale hydrograph. These observations contributed to the current knowledge of the impact of antecedent soil moisture spatial variability on the field scale runoff. [less ▲]Detailed reference viewed: 46 (1 ULg) Mapping shallow soil moisture profiles at the field scale using full-waveform inversion of ground penetrating radar dataMinet, Julien ; Wahyudi, A.; Bogaert, P. et alin Geoderma (2011), 161Full-waveform inversions were applied to retrieve surface, two-layered and continuous soil moisture profiles from ground penetrating radar (GPR) data acquired in an 11-ha agricultural field situated in ... [more ▼]Full-waveform inversions were applied to retrieve surface, two-layered and continuous soil moisture profiles from ground penetrating radar (GPR) data acquired in an 11-ha agricultural field situated in the loess belt area in central Belgium. The radar system consisted of a vector network analyzer combined with an off-ground horn antenna operating in the frequency range 200–2000 MHz. The GPR system was computer controlled and synchronized with a differential GPS for real-time data acquisition. Several inversion strategies were also tested using numerical experiments, which in particular demonstrated the potentiality to reconstruct simplified two-layered configurations from more complex, continuous dielectric profiles as prevalent in the environment. The surface soil moisture map obtained assuming a one-layered model showed a global moisture pattern mainly explained by the topography while local moisture patterns indicated a line effect. Two-layered and profile inversions provided consistent estimates with respect to each other and field observations, showing significant moisture increases with depth. However, some discrepancies were ob- served between the measured and modeled GPR data in the higher frequency ranges, mainly due to surface roughness effects which were not accounted for. The proposed GPR method and inversion strategies showed great promise for high-resolution, real-time mapping of soil moisture at the field scale. [less ▲]Detailed reference viewed: 35 (1 ULg) Spatial fields' dispersion as a farmer strategy to reduce agro-climatic risk at the household level in pearl millet-based systems in the Sahel: A modeling perspectiveAkponikpe, Pierre B. I.; Minet, Julien ; Gerard, Bruno et alin Agricultural and Forest Meteorology (2011), 151(2), 215-227Detailed reference viewed: 24 (2 ULg) Soil surface water content estimation by full-waveform GPR signal inversion in the presence of thin layersMinet, Julien ; Lambot, Sébastien; Slob, Evert C. et alin IEEE Transactions on Geoscience and Remote Sensing (2010), 48We analyzed the effect of shallow thin layers on the estimation of soil surface water content using full-waveform inversion of off-ground ground penetrating radar (GPR) data. Strong dielectric contrasts ... [more ▼]We analyzed the effect of shallow thin layers on the estimation of soil surface water content using full-waveform inversion of off-ground ground penetrating radar (GPR) data. Strong dielectric contrasts are expected to occur under fast wetting or drying weather conditions, thereby leading to constructive and destructive interferences with respect to the surface reflection. First, synthetic GPR data were generated and subsequently inverted considering different thin-layer model configurations. The resulting inversion errors when neglecting the thin layer were quantified, and then, the possibility to reconstruct these layers was investigated. Second, laboratory experiments reproducing some of the numerical experiments configurations were conducted to assess the stability of the inverse solution with respect to actual measurement and modeling errors. Results showed that neglecting shallow thin layers may lead to significant errors on the estimation of soil surface water content ($\Delta\theta$ > 0.03 $m^3/m^3$), depending on the contrast. Accounting for these layers in the inversion process strongly improved the results, although some optimization issues were encountered. In the laboratory, the proposed full-waveform method permitted to reconstruct thin layers with a high resolution up to 2 cm and to retrieve the soil surface water content with an rmse less than 0.02 $m^3/m^3$, owing to the full-waveform inverse modeling. These results suggest that the proposed GPR approach is promising for field-scale mapping of soil surface water content of nondispersive soils with low electrical conductivity and for instances when soil layering is encountered. [less ▲]Detailed reference viewed: 18 (0 ULg) A generalized frequency domain reflectometry forward and inverse modeling technique for soil electrical properties determinationMinet, Julien ; Lambot, Sébastien; Delaide, Géraldine et alin Vadose Zone Journal (2010), 9(4)We have developed a generalized frequency domain reflectometry (FDR) technique for soil characterization that is based on an electromagnetic model decoupling the cable and probe head from the ground using ... [more ▼]We have developed a generalized frequency domain reflectometry (FDR) technique for soil characterization that is based on an electromagnetic model decoupling the cable and probe head from the ground using frequency-dependent reflection and transmission transfer functions. The FDR model represents an exact solution of Maxwell’s equations for wave propagation in one-dimensional multilayered media. The benefit of the decoupling is that the FDR probe can be fully described by its characteristic transfer functions, which are determined using only a few measurements. The soil properties are retrieved after removing the probe effects from the raw FDR data by iteratively inverting a global reflection coefficient. The proposed method was validated under laboratory conditions for measurements in water with different salt concentrations and sand with different water contents. For the salt water, inversions of the data led to dielectric permittivity and electrical conductivity values very close to the expected theoretical or measured values. In the frequency range for which the probe is efficient, a good agreement was obtained between measured, inverted and theoretically predicted signals. For the sand, results were consistent with the different water contents and also in close agreement with traditional time domain reflectometry measurements. The proposed method offers great promise for accurate soil electrical characterization because it inherently permits maximization of the information that can be retrieved from the FDR data and shows a high practicability. [less ▲]Detailed reference viewed: 17 (1 ULg) 1


text:
In this paper, a new model is proposed to study the coupled axial–torsional vibration of the drill string. It is assumed that rotary table angular speed is constant and equals to the nominal angular speed of the drill string. In addition, axial displacement of any point on the drill string is considered to be as the sum of rigid-body motion and elastic vibrations. The depth of cut is defined using instantaneous dynamic states instead of using the delayed model as presented in previous researches. A velocity-weakening function is introduced for modeling the behavior of the frictional component of the torque-on-bit (TOB) with respect to the bit angular speed. After discretizing vibration equations, stability analysis of the system is investigated by linearizing the nonlinear system around its steady-state response point. Considering nominal weight-on-bit (WOB) ($W0$) and nominal rotational speed ($Ω$) as the input parameters of the drilling, variation of maximum allowable value of ($W0$) is presented with respect to variation of $Ω$ . It is shown that the maximum allowable value of $W0$ has an increasing–decreasing behavior with respect to $Ω$. The effect of drill string upper and lower part lengths is studied on the stability of the system, and practical results are presented both in the condition that $W0$ is constant and in the condition that the hook upward force is constant. It is shown that by increasing the drill string length, the system is more exposed to instability, and this must be considered in regulating the input parameters of drilling.

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Home >> Seminar
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# Senator: The average per capita after tax income for

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Senator: The average per capita after tax income for [#permalink]

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Senator: The average per capita after tax income for residents of Eastbury is $30,105. 2 years ago, it was 11% lower. This can be directly attributed to the comprehensive set of tax cuts that I helped get approved in Congress. Which of the following statements, if true, most strengthens the conclusion drawn by the senator ? A. The average per capita after tax income for the residents of Eastbury who fall below poverty line increased by 16% in the last 5 years. B. The senator has not personally benefited from the tax cuts. C. The number of residents of Eastbury has not substantially changed in the last two years. D. The federal tax rose 6% during the 4 years prior to the implementation of the tax cuts. E. A recent change in the estate laws did not substantially increase the average per capita before tax income of the residents of Eastbury. In my Opinion Answer should be C Conclusion: Tax cuts have resulted in the increase average per capita income, which was 11% lower So the thought process which i took was Average per capita income = Total income / No of residents Avg per capita income could have also increased if no of residents would have decreased so if we ascertain that there was no substantial change in the no of residents/ population then it is strengthening the conclusion. So i chose Option C Please explain why Option C is wrong and Option E is correct, as I find both are correct in this argument, as they both strengthen the argument. [Reveal] Spoiler: OA Manager Joined: 04 Jun 2010 Posts: 113 Concentration: General Management, Technology Schools: Chicago (Booth) - Class of 2013 GMAT 1: 670 Q47 V35 GMAT 2: 730 Q49 V41 Followers: 14 Kudos [?]: 194 [0], given: 43 Re: Senator - Tax cuts [#permalink] ### Show Tags 20 Sep 2010, 12:27 sachinrelan wrote: Senator: The average per capita after tax income for residents of Eastbury is$30,105. 2 years ago, it was 11% lower. This can be directly attributed to the comprehensive set of tax cuts that I helped get approved in Congress.

Which of the following statements, if true, most strengthens the conclusion drawn by the senator ?

A. The average per capita after tax income for the residents of Eastbury who fall below poverty line increased by 16% in the last 5 years.
B. The senator has not personally benefited from the tax cuts.
C. The number of residents of Eastbury has not substantially changed in the last two years.
D. The federal tax rose 6% during the 4 years prior to the implementation of the tax cuts.
E. A recent change in the estate laws did not substantially increase the average per capita before tax income of the residents of Eastbury.

In my Opinion Answer should be C

Conclusion: Tax cuts have resulted in the increase average per capita income, which was 11% lower

So the thought process which i took was

Average per capita income = Total income / No of residents

Avg per capita income could have also increased if no of residents would have decreased so if we ascertain that there was no substantial change in the no of residents/ population then it is strengthening the conclusion. So i chose Option C

Please explain why Option C is wrong and Option E is correct, as I find both are correct in this argument, as they both strengthen the argument.

Well I agree with you that this a somewhat tricky question. IMO C is wrong because the question clearly states that the income 2 years ago and now is per capita meaning that according to this data the number of residents is somewhat irrelevant. The comparison is between the averages. Also note that in order to strengthen the senator statement you should refer to the tax cuts because this is part of his conclusion. If there was no other answer that mentioned the tax cuts and that strengthened his argument, maybe than you could consider C.
Also pay attention to the number of logical steps you had to take in order to come up with the conclusion that C strengthens the argument, this is a clear warning sign. The fewer the steps the better.

Hope that helps
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Re: Senator - Tax cuts [#permalink]

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20 Sep 2010, 12:47
Senator: The average per capita after tax income for residents of Eastbury is $30,105. 2 years ago, it was 11% lower. This can be directly attributed to the comprehensive set of tax cuts that I helped get approved in Congress. Which of the following statements, if true, most strengthens the conclusion drawn by the senator ? Synopsis: After Comprehensive tax cut --- Per capita income after tax increased we need to strengthen it. Premise + Assumption -------> Conclusion 2 years ago per capita income after t was 11% lower + ( Assumption)---> Per capita income after tax increased A)The average per capita after tax income for the residents of Eastbury who fall below poverty line increased by 16% in the last 5 years. - this tells above " people who fall below poverty line ". - Wrong answer. B)The senator has not personally benefited from the tax cuts.- Junck Answer c)The number of residents of Eastbury has not substantially changed in the last two years.- we dont need to consider the no of residents. d)The federal tax rose 6% during the 4 years prior to the implementation of the tax cuts.- Nothing to do with it. e)A recent change in the estate laws did not substantially increase the average per capita before tax income of the residents of Eastbury.- There are no other factor increased the per capita income.So only tax cuts increase the per capita income. Ms. Big Fat Panda Status: Three Down. Joined: 09 Jun 2010 Posts: 1922 Concentration: General Management, Nonprofit Followers: 421 Kudos [?]: 1800 [1] , given: 210 Re: Senator - Tax cuts [#permalink] ### Show Tags 20 Sep 2010, 12:53 1 This post received KUDOS Expert's post The per capita incomes are averages which already take into account the population. So the factor of population isn't really counted here. However, stating that some other law has NOT increased the income directly tells us that this tax cut was an important player. Hence the answer is E. This is a tricky one, though. The kind that easily tricks people. GMAT Tutor Joined: 24 Jun 2008 Posts: 1181 Followers: 368 Kudos [?]: 1208 [0], given: 4 Re: Senator - Tax cuts [#permalink] ### Show Tags 20 Sep 2010, 13:53 Expert's post sachinrelan wrote: Senator: The average per capita after tax income for residents of Eastbury is$30,105. 2 years ago, it was 11% lower. This can be directly attributed to the comprehensive set of tax cuts that I helped get approved in Congress.

Which of the following statements, if true, most strengthens the conclusion drawn by the senator ?

A. The average per capita after tax income for the residents of Eastbury who fall below poverty line increased by 16% in the last 5 years.
B. The senator has not personally benefited from the tax cuts.
C. The number of residents of Eastbury has not substantially changed in the last two years.
D. The federal tax rose 6% during the 4 years prior to the implementation of the tax cuts.
E. A recent change in the estate laws did not substantially increase the average per capita before tax income of the residents of Eastbury.

In my Opinion Answer should be C

Conclusion: Tax cuts have resulted in the increase average per capita income, which was 11% lower

So the thought process which i took was

Average per capita income = Total income / No of residents

Avg per capita income could have also increased if no of residents would have decreased so if we ascertain that there was no substantial change in the no of residents/ population then it is strengthening the conclusion. So i chose Option C

Please explain why Option C is wrong and Option E is correct, as I find both are correct in this argument, as they both strengthen the argument.

I think this is a very bad question - where is it from?

We know that the per capita after-tax income has risen over the last two years. The senator attributes this to his or her tax cuts. This is not a particularly strong argument on the surface. After-tax income can increase because taxes go down, or because pre-tax income goes up. In addition, even if taxes went down, there may have been other tax cuts besides those introduced by the senator that are primarily responsible. Further, the average per capita income could change because the population changed in some way; perhaps there was an influx of high-income earners moving to Eastbury, or an exodus of low-income earners.

By POE we can work out what answer the question designer intends to be correct. A and D talk about time periods irrelevant to the argument (we're only concerned with what has happened over the last two years). B is entirely irrelevant.

C does not appreciably strengthen nor weaken the argument. If the population had increased, say, then not only would the number of people change, but so would the total income earned by Eastbury residents. We'd need to know something about how much these new people earned in order to determine what effect they would have on the per capita income; if they were all earning $1,000,000 per year, then the per capita income would go up, but if they were all unemployed and earning$0 per year, then per capita income would go down. In that C rules out an alternative explanation (change in population) for the increase in per capita income, it strengthens the argument, but only in a very insignificant way.

E is also a bad answer here. While we do want to know that average income has not risen substantially if we want to strengthen the argument, we have no compelling reason to think that estate laws have any significant effect on the average per capita income in Eastbury. There are dozens of reasons why average income could increase (employment rate might increase, or average wage might increase, for example); estate laws are (perhaps) just one of these, and there's nothing in the passage to make us think that estate laws are particularly important to Eastbury residents. So sure, E rules out one of the dozens of possible factors that could have led income to increase, but that alone does almost nothing to strengthen the argument. E would only strengthen the argument if it ruled out a significant alternative explanation for the facts in the stem, and it simply doesn't do that.

It's just not a good question.
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20 Sep 2010, 13:55
Expert's post
I'd add that some might interpret E to say that average income has not increased. That is not what E says. It says that the change to estate laws did not increase average income. Other factors might still have increased average income, however; E in no way rules out that possibility.
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20 Sep 2010, 17:45
E rules out alternate reason for increase and it's correct answer.
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Re: Senator - Tax cuts [#permalink]

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21 Sep 2010, 06:42
IanStewart wrote:
I'd add that some might interpret E to say that average income has not increased. That is not what E says. It says that the change to estate laws did not increase average income. Other factors might still have increased average income, however; E in no way rules out that possibility.

Used POE to reach E. No doubt that there can be other reasons for increase in after tax per capita income, however, within the boundaries of the 5 choices, E is the only one that made sense.

Influx of high income earners or departure of low income earners can very well be the reason. However, these are not given in the choices. Hence, as far as the reader is concerned, these are not applicable.
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Re: Senator - Tax cuts [#permalink]

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21 Sep 2010, 07:00
Used POE to reach E. No doubt that there can be other reasons for increase in after tax per capita income, however, within the boundaries of the 5 choices, E is the only one that made sense.

Influx of high income earners or departure of low income earners can very well be the reason. However, these are not given in the choices. Hence, as far as the reader is concerned, these are not applicable.

IMO E is outside the scope of the question because it talks about estate laws and average per capita before tax (not after tax) and does not strenghten the conclusion.
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21 Sep 2010, 07:09
schemer wrote:
Used POE to reach E. No doubt that there can be other reasons for increase in after tax per capita income, however, within the boundaries of the 5 choices, E is the only one that made sense.

Influx of high income earners or departure of low income earners can very well be the reason. However, these are not given in the choices. Hence, as far as the reader is concerned, these are not applicable.

IMO E is outside the scope of the question because it talks about estate laws and average per capita before tax (not after tax) and does not strenghten the conclusion.

Before tax income, in any way, is relevant to the statement.

Agreed that E is not the perfect choice, but by POE, the only one that can makes any sense. Apart from a very stretched assumption that high income earners moved in or low income earners moved out, the only other possibility is increase in before tax income. How does that happen ? Employers increased salaries, either due to good business or a law made them do it (E.g - increase in minimum wage). Since none of the choices deal with the former, we chose the one that deals with the latter. (you can't ask for fish, when they only have chicken on the menu !)

Like Ian said, a bad question.
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21 Sep 2010, 07:57
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Re: Senator: The average per capita after tax income for [#permalink]

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text:
# How do you find the vertex of the parabola y=2x^2-8x+7?

Feb 9, 2015

You can find the coordinates of your vertex either using a mnemonic or the derivative:

1) Mnemonic: the coordinates of the vertex of a parabola in the form:
$y = a {x}^{2} + b x + c$ are:
${x}_{v} = - \frac{b}{2 a}$
${y}_{v} = - \frac{\Delta}{4 a}$ where $\Delta = {b}^{2} - 4 a c$
${x}_{v} = 2$
${y}_{v} = - 1$

2) Derivative: determine the derivative of your function and set it equal to zero:
$y ' = 4 x - 8$
$4 x - 8 = 0$ gived $x = {x}_{v} = 2$
$y = {y}_{v} = 2 \cdot 4 - 8 \cdot 2 + 7 = - 1$

Graphically:
graph{2x^2-8x+7 [-4.385, 4.386, -2.19, 2.193]}

Mar 5, 2016

A slight variant on method

color(blue)("Vertex" ->(x,y)->(2,-1)

#### Explanation:

Given:$\text{ } 2 {x}^{2} - 8 x + 7$

Write as:$\text{ } 2 \left({x}^{2} - \textcolor{red}{\frac{8}{2}} x\right) + 7$

Consider the $\textcolor{red}{- \frac{8}{2}} \text{ from } - \frac{8}{2} x$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- \frac{8}{2}}\right) = + \frac{8}{4} = 2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute $x = \textcolor{m a \ge n t a}{2}$ in the original equation to find ${y}_{\text{vertex}}$

y=2x^2-8x+7" "->" "y_("vertex")=2(color(magenta)(2))^2-8(color(magenta)(2))+7

$\textcolor{b l u e}{{y}_{\text{vertex}} = 8 - 16 + 7 = - 1}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Vertex" ->(x,y)->(2,-1)


text:
# What is the Euler product?

I have two or three questions directly related to this wikipedia article:

wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function

I think it is exactly about my main Q (see my other Q, on hold), and I understand it quite well (sort of), but:

What actually is "the proof of Euler product formula"?

There is, of course the obvious proof under "The case s=1", but it is also obvious what it prooves:

"harmonic series" =

product of all prime numbers,

divided by the

product of (all those prime numbers minus one)

Under way, the euler product series gets inverted (1/n) and reversed (left to right). Poor euler product!

This equation is very very interesting, but what is prooved? Can you not just factor it out?

The article starts off with:

The Euler product formula for the Riemann zeta function...:

Why "...for the R. z. function "?

Then, the article has that animated 120-grid sieve (why 120 ?), and speaks of:

"There is a certain sieving property that we can use to our advantage"

I somehow see what follows, but:

Can not that "property" be explained more clearly? Why that animated sieve at all?

Can not "s" be left out, at first, in the method-prooving part ? (And afterwards expanded ?)

Why Riemann-zeta?, why any s different than 1 ?

Euler was more than 100 years before Riemann, and also Dirichlet. I don't find much/anything what HE actually did with his product formula, basically. (I know he did tons of things -- that is not the problem)

Is this not something between Eratosthenes and Euler?

Please don't blame the article -- I have been searching a bit left and right for a long time, for this "euler product series" (I don't define a name, not even "p"):

$$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} \cdot \frac{16}{17} \cdot \frac{\cdots}{...} \cdot \frac{p-1}{p} \cdot$$

Above is the inverse of the harmonic series.

For me, this is the basic form of the "prime product" (try to google that!). Because it looks nice and simple. This also:

$$\prod_{n = 1} 1 - \frac {1}{p_n }$$

...should be the same. The index $$n$$ is for all consecutive prime numbers $$p_n$$.

Is it only: $$\zeta(1)^{-1}$$

...this? (should also be the same)

Does the Euler product not exist apart from harmonic series and riemann zeta?

Well yes, euler phi totient function: another "loose end".

• – rastafile Nov 23 '19 at 11:29
• the euler_product article itself has centrally: as is the case for the Riemann zeta-function, where a(n)=1 – rastafile Nov 23 '19 at 11:35

Theorema 7 of Euler's "Series Infinitas" goes like this:

The infinitely continued product out of these fractions

$$\frac {2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19}{1\cdot2\cdot4\cdot6\cdot10\cdot12\cdot16\cdot18}\ \small \ {etc.}$$

where the numerators are all the prime numbers, but the denominators are deficient by unity from the numerators,

This product is the same as the sum of this series

$$1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6} + \small{etc.}$$

also infinitely.

Then he factors out the harmonic series and arrives at:

$$\frac {1\cdot2\cdot4\cdot6\cdot10\cdot12\cdot16\cdot18\cdot22\ etc.} {2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\ etc.} \cdot x = 1$$

...which he has to turn upside down because his "Observationes" are all about Goldbach and that (harmonic) series.

The next theorema has s=2, but has nothing to do with Riemann, or prime numbers, directly.

More with $$\pi$$ (the circle one, not the prime function one)

So this is Euler's (algebraic) proof of reciprocity between "his" prime product and the harmonic series.

Acutally, "then he factors out" is a big oversimplification. It looks like the "certain sieving method" is applied between the factoring out steps. And because the first step is one half, the other half has the same size...with the second step, the 1/3, you see that it is the other 2/3 that get substituted.

So Euler sieves out the denominators "one by one": first all even ones, then all with 3 as divisor, then with 5, etc.

$$x=1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6} + \small{etc.}$$

becomes

$$\frac{1}{2}\cdot x = \frac{1}{2}+ \frac{1}{4}+ \frac{1}{6}+ \frac{1}{8}+ {etc.}$$

This is all the even denominators. The other half is the uneven denominators. (Euler only "explains": of this halved(? lat.: demta) series reamains:)

$$\frac{1}{2}\cdot x = 1 + \frac{1}{3}+ \frac{1}{5}+ \frac{1}{7} + {etc.}$$

(I keep the original "1" instead of 1/1 -- either way it is an additional possible confusion)

Then he repaeats factoring out with 1/3:

$$\frac{1}{2}\cdot\frac{1}{3}\cdot x = \frac{1}{3}+ \frac{1}{9}+ \frac{1}{15} + \frac{1}{21}+{etc.}$$

The other two thirds of that half are:

$$\frac{1}{2}\cdot\frac{2}{3}\cdot x = 1 + \frac{1}{5}+ \frac{1}{7}+ \frac{1}{11} + \frac{1}{13}+{etc.}$$

This is the series without the "nec per 2 nec per 3" divisible denominators.

He shows the same trick for "5" i.e. factor out 1/5 and then lists the other 4/5 (of the two thirds of the half of the series), and arrives at above formula.

I think this is very nice, but the subtle (?) problems with infinity etc. can be avoided by leaving out the harmonic series, and instead calculate the unsieved coprimes in the next primorial (2,6,30,210, etc.).


text:
Browse Questions

# True or False: The minimum value of n for which $\tan^{-1}\Large {\frac{n}{\pi}}\;>\Large {\frac{\pi}{4}},\normalsize\;n\in N,$ is valid is 5

$\begin{array}{1 1} \text{False, Minimum Value is 4} \\ \text{Yes Minimum Value if 5} \\ \text{False, Minimum Value is 3} \\ \text{Cannot be Determined }\end{array}$

Toolbox:
• $tan\large\frac{\pi}{4}=1$
• if x>$\large\frac{\pi}{4}$, then tanx >1
• The value of $\pi$ is 3.14 approximately.
Ans - False
we know if x>$\large\frac{\pi}{4}$, then tanx >1 and
$given:tan^{-1}\large\frac{n}{\pi}>\frac{\pi}{4}$

$\Rightarrow\:tan(tan^{-1}\large\frac{n}{\pi})>tan\large\frac{\pi}{4}$
$\Rightarrow\:tan(tan^{-1}\large\frac{n}{\pi})>1$
$\Rightarrow\: \large\frac{n}{\pi}>1 \Rightarrow n > \pi$
$\Rightarrow\:n>3.14$
$\Rightarrow\:$ minimum integer value of n =4

edited Mar 19, 2013


text:
## Stream: maths

### Topic: Bijective Ring Homorphism an Isomorphism

#### Jordan Brown (Jul 29 2020 at 22:50):

Does mathlib have the theorem showing that a bijective ring homomorphism is an isomorphism? It seems to be missing from the data.equiv.ring, is there somewhere else I should look?

#### Matt Earnshaw (Jul 29 2020 at 23:59):

import algebra.category.CommRing.basic
#check ring_equiv.to_Ring_iso

#### Thomas Browning (Jul 30 2020 at 00:05):

But there you're already assuming that you have a ring equivalence (rather than a bijective ring homomorphism)

oops

#### Adam Topaz (Jul 30 2020 at 00:46):

You can combine src#equiv.of_bijective to get this easily.

#### Adam Topaz (Jul 30 2020 at 00:51):

import data.equiv.basic
import data.equiv.ring

variables {A : Type*} [ring A]
variables {B : Type*} [ring B]

noncomputable def ring_equiv_of_bij_hom (f : A →+* B) (h : function.bijective f) : A ≃+* B :=
{ map_mul' := by apply ring_hom.map_mul,
..show A ≃ B, by exact equiv.of_bijective _ h }

There might be a better way. Or mathlib might already have something like this.

#### Johan Commelin (Jul 30 2020 at 03:55):

Can't you golf that to

noncomputable def ring_equiv_of_bij_hom (f : A →+* B) (h : function.bijective f) : A ≃+* B :=
{ .. f, .. equiv.of_bijective _ h }

Last updated: May 19 2021 at 00:40 UTC


text:
# How to make slash (/) paste with correct Unicode codepoint depending on context?

Unicode offers different characters for the slash:

• U+002F (/) for the ordinary slash
• U+2215 (∕) for the mathematical division operator
• U+2044 (⁄) to designate vulgar fractions

How can I have these automatically paste with the right Unicode codepoint? Some starter code is this:

\documentclass{article}
\input glyphtounicode
\pdfgentounicode=1
\usepackage[T1]{fontenc}
\usepackage{nicefrac}

\begin{document}

textual: w/o (intended: U+002F)

in an equation: $$12/4 = 3$$ (intended: U+2215)

in a vulgar fraction (produced by package \texttt{nicefrac}): 1\nicefrac{1}{2} (intended: U+2044)

\end{document}

Note that "-" in math mode already pastes correctly as the "minus sign" (U+2212, "−"), so that is why I suspect that addressing this issue is possible.

-

The three different slashes all use the same glyph name slash in the PDF file. Therefore only one mapping can be set via feature \pdfgentounicode. The PDF format provides another feature ActualText that can be used with package accsupp. The example lets the normal slash untouched, it is mapped to U+002F via \pdfgentounicode/glyphtounicode.tex. The other two are defined as macros. At some stage macros are necessary for accsupp, because it must put some markup in the PDF page contents stream before and after the glyph.

\documentclass{article}

\pdfgentounicode=1
\input{glyphtounicode}
\usepackage{accsupp}
\usepackage{nicefrac}
\usepackage{etoolbox}

% U+2215
\newcommand*{\mathdivide}{%
\ensuremath{%
\BeginAccSupp{method=hex,unicode,ActualText=2215}/\EndAccSupp{}%
}%
}
% U+2044
\newcommand*{\fractionslash}{%
\ensuremath{%
\BeginAccSupp{method=hex,unicode,ActualText=2044}/\EndAccSupp{}%
}%
}

% Patch nice frac macros to insert the right slashes
\makeatletter
\expandafter\patchcmd\csname \@backslashchar @UnitsUglyFrac \endcsname
{/}{\mathdivide}{%
\@latex@info@no@line{Successful patched \string\@UnitsUglyFrac}%
}{%
\@latex@error{Could not patch \string\@UnitsUglyFrac}%
}

\expandafter\patchcmd\csname \@backslashchar @UnitsNiceFrac \endcsname
{/}{\fractionslash}{%
\expandafter\patchcmd\csname \@backslashchar @UnitsNiceFrac \endcsname
{/}{\fractionslash}{%
\@latex@info@no@line{Successful patched \string\@UnitsNiceFrac}%
}{%
\@latex@error{Could not patch \string\@UnitsNiceFrac}%
}%
}{%
\@latex@error{Could not patch \string\@UnitsNiceFrac}%
}
\makeatother

\begin{document}
/ $a\mathdivide b$
\nicefrac{a}{b} $\nicefrac{a}{b}$
\end{document}

The result from copy and paste:

• "/ a∕b a⁄b a⁄b"
• "/ a\u2215b a\u2044b a\u2044b" (escape notation)
-
I know that this is advertising, but with xpatch instead of etoolbox you can more simply say \xpatchcmd{\@UnitsNiceFrac}{...}{...}{...}{...} – egreg Sep 20 '12 at 20:08
Thanks, nice to know. – Heiko Oberdiek Sep 20 '12 at 21:05
Thanks. The patch for nicefrac is great. I was wondering whether there is a way to simply write / instead of defining a new macro \mathdivide using accsupp. The reason I suspect such a solution could be possible is that the hyphen-minus is somehow being treated in two different ways by (La)TeX depending on the mode, leading to two different pastable Unicode characters in the output. So there might be a way to patch/amend TeX internals? – Lover of Structure Sep 21 '12 at 17:55
The hyphen and minus sign are different glyphs usually (the minus sign is longer). – Heiko Oberdiek Sep 21 '12 at 18:24
Yes -- What I mean is that I can input the same character "hyphen-minus" (-) in the source; when it is in text mode it will paste as the "hyphen-minus", but when it is in math mode it will paste as the "minus sign". That means that LaTeX has a mechanism by which it treats the same input character in a way that leads to two different, context-dependent output pastings. The fact that both slashes look the same (assuming that they do) shouldn't be an obstacle in theory: perhaps one can define two characters generated by textmode-/ and mathmode-/ that look the same but paste differently. – Lover of Structure Sep 21 '12 at 22:31


text:
# Best Choice of Estimator. How to compute Variance of Estimator.Basis for it?

I am working on an exercise asking which of the two following estimators; $$X1, X2$$for the population mean of a normal population with parameters $$\mu, \sigma$$ is best and why

$$X1:=\frac {X_1+X_2+....X_{2n}}{2n}$$,

$$X2:=\frac {X_1+X_2+....X_n}{n}$$

I computed the expectation and both are unbiased estimators. I tried to find the variance of the estimators, but the computations became unwieldy:

$$Var(X1)= E[(X1- \mu)^2]=$$

$$\frac {1}{4n^2}E[( \Sigma _{i=1}^{2n}X_i-\mu)^2]=$$

$$E[\frac{1}{4n^2} (X_1+X_2+...+ X_{2n})^2 - 2\mu X1+ \mu^2]$$

and in order to clear this I need to compute $$E[X_i^2]$$ ,

Which I don't see how to do. I tried standardizing but did not see it helping. Without this computation, I cant decide the best estimator for the pop mean based on neither variance nor bias-variance tradeoff.

Any ideas?

Thanks.

It's possible to derive the Expectation of $$X_i^2$$, however there's a much simpler way to calculate your variance:

$$Var(X1) = Var(\frac{1}{2n} \sum_{i=1}^{2n} X_i) \\ = \frac{1}{4n^2} \sum_{i=1}^{2n} Var(X_i) \\ = \frac{1}{4n^2} 2 n \sigma^2 \\ = \frac{\sigma^2}{2n}$$ This comes from the fact that the variance is linear.

• Thank you Erin. I guess we need indepence to use this formula, otherwise we need to use covariance/correlation, right? – MSIS Apr 8 at 23:43
• That's true. But the exercise you are working on is of typical nature where draws of the population are supposedly stochastically independent. If you are interested in getting the expectation of X_i^2 I suggest to look at the theoretical variance by integration and reformulate that. Then you'll see that E[X_i^2] is Var(X) + mu^2. – Erin Sprünken Apr 8 at 23:47
• Erin, I selected your answer. One last question, please : If we have two estimators E1, E2 of the same parameter and they are both unbiased and they have the same variance/s.e, what criteria would we use to select one over the other. Or maybe that should be a separate question? – MSIS Apr 11 at 4:29
• I'd recommend opening another question for this since other people may have a different view or more experience. However, If they're both unbiased and equal variances, then I'd choose the one which is easier to compute (or faster to compute on a PC). Although, what I teach to my students, is that unbiasedness is not the non-plus-ultra. It is really nice to have and sometimes important. However, there are cases where a biased estimator has a lower MSE. Just as an additional information. – Erin Sprünken Apr 11 at 14:29


text:
# What is the distance between the following polar coordinates?:  (2,(pi)/2), (2,(17pi)/12)

Jan 1, 2018

$\overline{A B} \setminus \approx 3.9658$

#### Explanation:

Graph the two points:

A triangle is formed when the two points $A$ and $B$ are connected to the origin $O$. We know the angle $\angle A O B$ is $\frac{17 \pi}{12} - \frac{\pi}{2} = \frac{11 \pi}{12}$, and the two sides $\overline{A O} = \overline{B O} = 2$.

Then, $\overline{A B}$ can be found using the law of cosines:
${\overline{A B}}^{2} = {\overline{A O}}^{2} + {\overline{B O}}^{2} - 2 \cdot \overline{A O} \cdot \overline{B O} \cdot \cos \left(\angle A O B\right)$
${\overline{A B}}^{2} = {2}^{2} + {2}^{2} - 2 \cdot 2 \cdot 2 \cdot \cos \left(\frac{11 \pi}{12}\right)$
${\overline{A B}}^{2} \setminus \approx 15.7274$

$\therefore \overline{A B} \setminus \approx 3.9658$