Month: March 2019

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201903310628 Solution to 1980-AL-PHY-I-23

Faraday’s law:

\varepsilon = -N \displaystyle{\frac{\mathrm{d}\Phi}{\mathrm{d}t}}

where \varepsilon is the e.m.f. induced, N the number of turns in the coil, \Phi the magnetic flux, also (\Phi =BA) the product of magnetic flux density B and area A, t the time, and the negative sign due to Lenz’s law.

The magnitude of the e.m.f. induced in the coil is therefore

\varepsilon = \bigg| -N\displaystyle{\frac{\Delta \Phi}{\Delta t}} \bigg| = \bigg| -N \displaystyle{\frac{\Phi -0}{t}} \bigg| = N\Phi /t.

And the answer is C.

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201903080359 Solution to 2005-AL-PHY-IIA-12

We assume that the charges on a small sphere can be treated as one point charge.

At first, P, Q, and R are of charge \pm q, \mp q, and 0.

Then, P and R are put in contact and share their charges evenly. They are now of charge \pm \displaystyle{\frac{q}{2}}. We keep them separated afterwards.

Later on, R and Q are put in contact and share their charges evenly. Each of them is hence of charge:

\displaystyle{\frac{\bigg( \pm \displaystyle{\frac{q}{2}}\bigg) +(\mp q)}{2}}=\mp\displaystyle{\frac{q}{4}}.

We keep them separated afterwards.

The initial magnitude  F of electrostatic force between P and Q is given by:

F=\bigg| \displaystyle{k\frac{(\pm q)(\mp q)}{r^2}} \bigg|=k \displaystyle{\frac{q^2}{r^2}}

The final magnitude F' of electrostatic force between them is related to the initial F by:

F'=\bigg| \displaystyle{k\frac{ ( \pm \frac{q}{2}) (\mp \frac{q}{4})}{r^2}} \bigg|=\displaystyle{\frac{1}{8}}\bigg( k \displaystyle{\frac{q^2}{r^2}} \bigg) = \displaystyle{\frac{1}{8}}F

\begin{aligned} \quad & \quad & \textrm{Initially} & \quad & \rightarrow & \quad & P\textrm{ and }R\textrm{ in touch} &\quad & \rightarrow & \quad & R\textrm{ and }Q\textrm{ in touch}\\ P & & \pm q & & & & \pm \displaystyle{\frac{q}{2}} & & & & \boxed{\pm \displaystyle{\frac{q}{2}}} \\ Q & & \mp q & & & & \boxed{\mp q} & & & & \mp \displaystyle{\frac{q}{4}}\\ R & & 0 & & & & \pm \displaystyle{\frac{q}{2}} & & & & \mp \displaystyle{\frac{q}{4}} \end{aligned}

And the answer is B.

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201903050215 Solution to 2008-AL-PHY-IIA-14

Let there be an x-axis passing through points A and B. Let x=0 at point A and x=d at point B.

Now consider an arbitrary point x=r on the x-axis.

The magnitude E_{rA} of electric field strength due to the point charge +Q at point A is

E_{rA} = \bigg|\displaystyle{k\frac{Q}{r^2}}\bigg| = \displaystyle{k\frac{Q}{r^2}},

whereas the magnitude E_{rB} of electric field strength due to the point charge -2Q at point B is

E_{rB}=\bigg| \displaystyle{k\frac{-2Q}{(r-d)^2}}\bigg|=\displaystyle{2k\frac{Q}{(r-d)^2}}.

The point(s) at which E_{rA}=E_{rB} is found by solving for r:

\begin{aligned} \displaystyle{k\frac{Q}{r^2}} & = \displaystyle{2k\frac{Q}{(r-d)^2}}\\ (r-d)^2 & = 2r^2 \\ r^2-2rd+d^2 & = 2r^2 \\ 0 & = r^2 +2rd -d^2 \\ r & = \displaystyle{\frac{-2d\pm \sqrt{(2d)^2-4(1)(-d^2)}}{2(1)}} \\ & = -d\pm \sqrt{2}d \end{aligned}

There are two points x=-d+\sqrt{2}d and x=-d-\sqrt{2}d at which E_A=E_B.

The necessary and sufficient condition for \mathbf{E}_{A}+\mathbf{E}_{B}=\mathbf{0} is that they are equal in magnitude: E_{A}=E_{B} and opposite in sign: \mathbf{E}_A=-\mathbf{E}_B,

i.e.,

\mathbf{E}_{A}+\mathbf{E}_{B}=\mathbf{0}

\Leftrightarrow

i. E_{A}=E_{B} and ii. the directions of \mathbf{E}_{A} and \mathbf{E}_{B} are opposite to each other.

Suffice it to check whether \mathbf{E}_A and \mathbf{E}_B at the above-mentioned two points are in opposite direction.

The first candidate x=-d+\sqrt{2}d, where 0<x<d, is in between point A and point B. If a positive test charge were placed there, both the electric field strength \mathbf{E}_A due to +Q at A and \mathbf{E}_B due to -2Q at B would point to the right. That is, this point fails to meet our requirement ii.

The second candidate x=-d-\sqrt{2}d, where -2d<x<0, is to the left of point A. If a positive test charge were placed at x, the electric field strength \mathbf{E}_A there due to +Q at A would point to the left. Whereas the electric field strength \mathbf{E}_B there due to -2Q at B would point to the right.

We conclude that the resultant electric field strength is zero at only one point, i.e., the point x=-d-\sqrt{2}d.

And the answer is C.

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201903030616 Exercise 17.2.1

Let \mathbf{A} and \mathbf{B} be arbitrary vector field and covariant vector field.

Following Eq. (17.17), the scalar product A^\mu B_{\mu} upon absolute differentiation is by definition:

\begin{aligned} \nabla_\alpha (A^\mu B_{\mu}) & = (\nabla_\alpha A^\mu )B_\mu + A^\mu (\nabla_\alpha B_\mu ) \\ & = \bigg( \displaystyle{\frac{\partial A^{\mu}}{\partial x^\alpha}} +\Gamma^{\mu}_{\alpha\nu} A^\nu \bigg) B_\mu + A^\mu (\nabla_\alpha B_\mu ) \end{aligned}

On the other hand, following Eq. (17.18),

\begin{aligned} \nabla_\alpha (A^\mu B_\mu ) & =\partial_\alpha (A^\mu B_\mu ) \\ & = \displaystyle{\frac{\partial A^\mu}{\partial x^\alpha}}B_\mu + A^\mu \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} \end{aligned}.

Equating Eq. (17.17) with Eq. (17.18):

\begin{aligned} \bigg( \displaystyle{\frac{\partial A^{\mu}}{\partial x^\alpha}} +\Gamma^{\mu}_{\alpha\nu} A^\nu \bigg) B_\mu + A^\mu (\nabla_\alpha B_\mu ) & = \displaystyle{\frac{\partial A^\mu}{\partial x^\alpha}}B_\mu + A^\mu \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} \\ \Gamma^\mu_{\alpha\nu}A^\nu B_\mu + A^\mu (\nabla_\alpha B_\mu ) & = A^\mu \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} \\ A^\mu (\nabla_\alpha B_\mu ) & = A^\mu \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} - \Gamma^\mu_{\alpha\nu} A^\nu B_\mu \\ A^\mu (\nabla_\alpha B_\mu ) & = A^\mu \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} - \Gamma^\nu_{\alpha\mu} A^\mu B_\nu \\ \nabla_\alpha B_\mu & = \displaystyle{\frac{\partial B_\mu}{\partial x^\alpha}} - \Gamma^\nu_{\alpha\mu}B_\nu \end{aligned}

I have proven Eq. (17.7).

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201903030526 Problem 6.1

The covariant derivative of v^\mu is defined:

v^\mu_{;\alpha}=\partial_\alpha v^\mu +\Gamma^\mu_{\alpha\beta}v^\beta.

The Christoffel symbols are all zeros for Cartesian coordinates in the plane. For the metric tensor of Cartesian coordinates being the 2\times 2 identity matrix, and all its elements constants, the Christoffel symbols of the first kind as well as the second should be zero:

\begin{aligned} V^x_{;x}&=\partial_xV^x\\ V^y_{;x}&=\partial_xV^y\\ V^x_{;y}&=\partial_yV^x\\ V^y_{;y}&=\partial_yV^y \end{aligned}

The metric tensor for polar coordinates in the plane is derived below:

\mathbf{r}=(r\sin\theta ,r\cos\theta)

From

\begin{aligned} \hat{\mathbf{e}}_r&=\frac{\partial}{\partial r}(\mathbf{r})=(\sin\theta ,\cos\theta) \\ \hat{\mathbf{e}}_\theta &=\frac{\partial}{\partial \theta}(\mathbf{r})=(r\cos\theta ,-r\sin\theta ) \end{aligned},

and hence

\begin{aligned} \hat{\mathbf{e}}_r\cdot \hat{\mathbf{e}}_r&=\sin^2\theta +\cos^2\theta =1\\ \hat{\mathbf{e}}_\theta\cdot \hat{\mathbf{e}}_\theta &=r^2\cos^2\theta +(-r)^2\sin^2\theta =r^2 \\ \hat{\mathbf{e}}_r\cdot \hat{\mathbf{e}}_\theta &=\hat{\mathbf{e}}_\theta\cdot \hat{\mathbf{e}}_r=r\sin\theta\cos\theta -r\sin\theta\cos\theta =0 \end{aligned},

it follows that the metric tensor \mathbf{G}=(g_{ij}) is

\begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}.

Since only the (2,2)^{\textrm{th}} element is a non-constant dependent on r, I expect that some of the Christoffel symbols will be zeros.

Two formulae for Christoffel symbols of the second kind:

\begin{aligned} \Gamma^\alpha_{\alpha\beta}&(=\Gamma^\alpha_{\beta\alpha})=\displaystyle{\frac{\partial}{\partial x^\beta}\bigg( \frac{1}{2}\ln |g_{\alpha\alpha}| \bigg)}\\ \Gamma^\alpha_{\beta\beta}&=\displaystyle{-\frac{1}{2g_{\alpha\alpha}}\frac{\partial}{\partial x^\alpha}(g_{\beta\beta})}\qquad\quad (\alpha\neq\beta) \end{aligned}

Because the (1,1)^\textrm{th}, (1,2)^\textrm{th}, and (2,1)^\textrm{th} elements are constants, referring to the preceding formulae I can preclude these terms 0= \Gamma^1_{11}=\Gamma^1_{12}= \Gamma^1_{21}=\Gamma^2_{11}=\Gamma^2_{22} and consider only the terms \Gamma^1_{22}, \Gamma^2_{12}, and \Gamma^2_{21}.

Neatly written,

\begin{aligned} \Gamma^1_{22}&(=\Gamma^r_{\theta\theta})=\displaystyle{-\frac{1}{2(1)}\frac{\partial }{\partial r}(r^2)}=-r\\ \Gamma^2_{12}&(=\Gamma^\theta_{r\theta})=\displaystyle{\frac{\partial}{\partial r}}\bigg( \displaystyle{\frac{1}{2}\ln r^2}\bigg) =\displaystyle{\frac{1}{r}} \\ \Gamma^2_{21}&(=\Gamma^\theta_{\theta r}) =\displaystyle{\frac{1}{r}} \end{aligned}

By the formula for covariant differentiation

v^\mu_{;\alpha}=\partial_\alpha v^\mu +\Gamma^\mu_{\alpha\beta}v^\beta

the following can be obtained:

\begin{aligned} V^r_{;r}&=\partial_rV^r,\\ V^r_{;\theta}&=\partial_\theta V^r-rV^\theta ,\\ V^\theta_{;r}&=\partial_rV^\theta +\frac{1}{r}V^\theta ,\\ V^\theta_{;\theta}&=\partial_\theta V^\theta +\frac{1}{r}V^r \end{aligned}.