Category: The Physics Language

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202105241204 Homework 1 (Q3)

Prove

(a) \nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A})-A\times (\nabla f)

(b) \nabla \times (\mathbf{A}\times \mathbf{B})=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}

Attempts. (brute force)

(a)

\begin{aligned} & \quad \nabla\times (f\mathbf{A}) \\ & = \nabla \times (fA_x,fA_y,fA_z) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ fA_x & fA_y & fA_z \end{vmatrix} \\ & = \bigg( \frac{\partial}{\partial y}(fA_z)-\frac{\partial}{\partial z}(fA_y),\, \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z),\, \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x) \bigg) \\ & = \Bigg( \bigg( f\frac{\partial A_z}{\partial y} + \frac{\partial f}{\partial y}A_z - f\frac{\partial A_y}{\partial z} - \frac{\partial f}{\partial z}A_y \bigg) , \\ & \quad \qquad \bigg( f\frac{\partial A_x}{\partial z} + \frac{\partial f}{\partial z}A_x - f\frac{\partial A_z}{\partial x} - \frac{\partial f}{\partial x}A_z \bigg) , \\ & \qquad \qquad \bigg( f\frac{\partial A_y}{\partial x}-\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}A_x - f\frac{\partial A_x}{\partial y} \bigg) \Bigg) \\ & = f\Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg) ,\, \bigg( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial z} \bigg),\, \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\bigg) \Bigg) \\ & \quad \qquad + \bigg( \frac{\partial f}{\partial y}A_z - \frac{\partial f}{\partial z}A_y,\, \frac{\partial f}{\partial z}A_x - \frac{\partial f}{\partial x}A_z,\, \frac{\partial f}{\partial x}A_y - \frac{\partial f}{\partial y}A_x \bigg) \\ & = f(\nabla \times \mathbf{A}) + \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = f(\nabla \times \mathbf{A}) + (\nabla f)\times\mathbf{A} \\ & = f(\nabla \times \mathbf{A}) - \mathbf{A}\times (\nabla f) \\ \end{aligned}

(b)

\begin{aligned} \textrm{LHS}\enspace & = \nabla \times (\mathbf{A}\times \mathbf{B}) \\ & = \nabla \times (A_yB_z-A_zB_y,\, -A_xB_z+A_zB_x,\, A_xB_y-A_yB_x) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_yB_z-A_zB_y & -A_xB_z + A_zB_x & A_xB_y - A_yB_x \end{vmatrix} \\ & = \Bigg( \bigg( \frac{\partial}{\partial y}(A_xB_y) - \frac{\partial}{\partial y}(A_yB_x) - \frac{\partial}{\partial z}(A_xB_z) + \frac{\partial}{\partial z}(A_zB_x) \bigg) ,\, \\ & \quad \qquad \bigg( -\frac{\partial}{\partial x}(A_xB_y) + \frac{\partial}{\partial x}(A_yB_x) + \frac{\partial}{\partial z}(A_yB_z) - \frac{\partial}{\partial z}(A_zB_y) \bigg) ,\, \\ & \qquad \qquad \bigg( \frac{\partial}{\partial x}(-A_xB_z) - \frac{\partial}{\partial x}(A_zB_x) - \frac{\partial}{\partial y}(A_yB_z) + \frac{\partial}{\partial y}(A_zB_y) \bigg) \Bigg) \\ & = \Bigg( \bigg( A_x\frac{\partial B_y}{\partial y} + \frac{\partial A_x}{\partial y}B_y - A_y\frac{\partial B_x}{\partial y} - \frac{\partial A_y}{\partial y}B_x - A_x\frac{\partial B_z}{\partial z} - \frac{\partial A_x}{\partial z}B_z + A_z\frac{\partial B_x}{\partial z} + \frac{\partial A_z}{\partial z}B_x \bigg) ,\, \\ & \quad \qquad \bigg( -A_x\frac{\partial B_y}{\partial x} - \frac{\partial A_x}{\partial x}B_y + A_y\frac{\partial B_x}{\partial x}+\frac{\partial A_y}{\partial x}B_x + A_y\frac{\partial B_z}{\partial z} + \frac{\partial A_y}{\partial z}B_z - A_z\frac{\partial B_y}{\partial z} - \frac{\partial A_z}{\partial z}B_y \bigg) ,\, \\ & \qquad \qquad \bigg( A_x\frac{\partial B_z}{\partial x} + \frac{\partial A_x}{\partial x}B_z - A_z\frac{\partial B_x}{\partial x} - \frac{\partial A_z}{\partial x}B_x - A_y\frac{\partial B_z}{\partial y} - \frac{\partial A_y}{\partial y}B_z + \frac{\partial A_z}{\partial y}B_y + A_z\frac{\partial B_y}{\partial y} \bigg) \Bigg) \\ \end{aligned}

\textrm{RHS}=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}

Inspect these four terms on the right hand side by expanding one after the other.

The first term being

\begin{aligned} (\mathbf{B}\cdot\nabla )\mathbf{A} & = \bigg( B_x\frac{\partial}{\partial x} + B_y\frac{\partial}{\partial y} + B_z\frac{\partial}{\partial z} \bigg) \mathbf{A} \\ & = \bigg( B_x\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_x}{\partial y} + B_z\frac{\partial A_x}{\partial z},\, \\ & \quad \qquad B_x\frac{\partial A_y}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_y}{\partial z},\, \\ & \qquad \qquad B_x\frac{\partial A_z}{\partial x} + B_y\frac{\partial A_z}{\partial y} + B_z\frac{\partial A_z}{\partial z} \bigg)\end{aligned}

the second term being

\begin{aligned} (\nabla \cdot \mathbf{B})\mathbf{A} & = \bigg( \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\bigg)\mathbf{A} \\ & = \bigg( \frac{\partial B_x}{\partial x}A_x + \frac{\partial B_y}{\partial y}A_x + \frac{\partial B_z}{\partial z}A_x ,\, \\ & \quad \qquad \frac{\partial B_x}{\partial x}A_y + \frac{\partial B_y}{\partial y}A_y + \frac{\partial B_z}{\partial z}A_y , \, \\ & \qquad \qquad \frac{\partial B_x}{\partial x}A_z + \frac{\partial B_y}{\partial y}A_z + \frac{\partial B_z}{\partial z}A_z \bigg) \\ \end{aligned}

the third term being

\begin{aligned} -(\mathbf{A}\cdot\nabla )\mathbf{B} & = - \bigg( A_x\frac{\partial}{\partial x} + A_y\frac{\partial}{\partial y} + A_z\frac{\partial}{\partial z} \bigg) \mathbf{B} \\ & = -\bigg( A_x\frac{\partial B_x}{\partial x} + A_y\frac{\partial B_x}{\partial y} + A_z\frac{\partial B_x}{\partial z},\, \\ & \quad\qquad A_x\frac{\partial B_y}{\partial x} + A_y\frac{\partial B_y}{\partial y} + A_z\frac{\partial B_y}{\partial z},\, \\ & \qquad\qquad A_x\frac{\partial B_z}{\partial x} + A_y\frac{\partial B_z}{\partial y} + A_z\frac{\partial B_z}{\partial z} \bigg) \\\end{aligned}

and the fourth and last term being

\begin{aligned} -(\nabla \cdot \mathbf{A})\mathbf{B} & = -\bigg( \frac{\partial A_x}{\partial x} +\frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \bigg)\mathbf{B} \\ & = - \bigg( B_x\frac{\partial A_x}{\partial x} + B_x\frac{\partial A_y}{\partial y} + B_x\frac{\partial A_z}{\partial z} ,\, \\ & \quad\qquad B_y\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_y\frac{\partial A_z}{\partial z} ,\, \\ & \qquad \qquad B_z\frac{\partial A_x}{\partial x} + B_z\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_z}{\partial z}\bigg) \\ \end{aligned}

One can check that \textrm{LHS}=\textrm{RHS}.

Solution. (proof)

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Using Einstein summation (/notation) and the Levi-Civita symbol \varepsilon_{ijk},

(a)

\begin{aligned} & \quad \nabla \times (f\mathbf{A}) \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(f\mathbf{A}_k)\cdot\varepsilon_{ijk}\qquad\qquad\qquad i,j,k\in\{ x,y,z\} \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\bigg(\frac{\partial}{\partial j}f\bigg)\cdot A_k\cdot\varepsilon_{ijk}+\sum_{i,j,k}\hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_k \bigg)\cdot f\cdot \varepsilon_{ijk} \\ & = (\nabla f)\times \mathbf{A} + f\cdot (\nabla\times\mathbf{A}) \\ & = (\nabla f)\times \mathbf{A} - A\times (\nabla f) \end{aligned}

(b)

\begin{aligned} \mathbf{A}\times\mathbf{B} & = \sum_{k,l,m}\hat{\mathbf{e}}_kA_lB_m\varepsilon_{klm}\\ \nabla\times (\mathbf{A}\times\mathbf{B}) & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(\sum_{l,m}A_lB_m\varepsilon_{klm})\varepsilon_{ijk} \\ & = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] \varepsilon_{klm}\varepsilon_{ijk} \\ \textrm{by } & \varepsilon_{klm}\varepsilon_{ijk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \\ \textrm{Thus, }& = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \\ & = \sum_{i,j,k,l,m}\bigg[ \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m (-\delta_{im}\delta_{jl}) \\ & \quad\qquad + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}B_m\bigg) A_l\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i \bigg( \frac{\partial}{\partial j}B_m \bigg) A_l (-\delta_{im}\delta_{jl}) \bigg] \\ & = (\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B} \end{aligned}

QED

and the proof is more concise.

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202104241713 Homework 1 (Q7)

The height of a mountain is given by h(x,y)=3000-2x^2-y^2, where the y-axis points east, the x-axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point (30,\, -20,\, 800), will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude h(x,y) is given by a function of x and y:

h(x,y)=3000-2x^2-y^2.

Now that the climber moves in the southwest direction

\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}

\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}

At point (30,\, -20,\, 800),

\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}

he will ascend southwesterly.

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202104221542 Homework 2 (Q4)

Evaluate the limit, or explain why the limit fails to exist.

(a) \displaystyle{\lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}}};

(b) \displaystyle{\lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4}}

Solution.

(a)

\begin{aligned} & \quad \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(x-y)}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} x(\sqrt{x}+\sqrt{y}) \\ & = (0)\big( \sqrt{(0)} + \sqrt{(0)} \big) \\ & = 0 \end{aligned}

(b)

Roughwork.

\begin{aligned} & \quad \lim_{(x,y)\to (2,0)} \frac{x^2-y^2-4x+4}{x^2+y^2-4x+4} \\ & = \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ \end{aligned}

If we take limits along the path (2,y)\to (2,0),

\begin{aligned} & \quad \lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4} \\ & = \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{\big((2)-2\big)^2-y^2}{\big((2)-2\big)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{-y^2}{y^2} \\ & = -1 \end{aligned}

whereas if we take limits along the path (x,0)\to (2,0),

\begin{aligned} & \quad \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-(0)^2}{(x-2)^2+(0)^2} \\ & = 1 \end{aligned}

The limit fails to exist because the limiting values vary with the paths of taking the limit \lim_{(x,y)\to (2,0)}.

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202104181519 Homework 1 (Q4)

i. Find the infinitesimal small vector \mathbf{dr} in the cylindrical coordinate induced by an infinitesimal small changes of \mathrm{d}\rho, \mathrm{d}\theta, and \mathrm{d}z in terms of \rho, \theta, z, \mathrm{d}\rho, \mathrm{d}\theta, \mathrm{d}z and the corresponding unit vector.

ii. f(u_1, u_2, u_3) is defined in \mathbf{r}=(u_1, u_2, u_3) coordinate. Its gradient is defined

\displaystyle{\lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}}_l}

where \Delta l_i and \Delta f_i are respectively the changes in length and functional value induced purely by the infinitesimal change in u_i. \hat{\mathbf{u}}_l is the unit vector of \mathbf{u}_i. Thus find the gradient of f in cylindrical coordinate.

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

i.

\mathbf{dr}=\mathrm{d}\rho\,\hat{\boldsymbol{\rho}}+\rho\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+\mathrm{d}z\,\hat{\mathbf{z}}

Compare to the figure below.

ii.

\begin{aligned} \nabla f & = \lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}_i} \\ & = \lim_{\Delta\rho\to 0} \frac{\Delta f_\rho}{\Delta \rho}\,\hat{\boldsymbol{\rho}} + \lim_{\Delta\theta\to 0} \frac{\Delta f_\theta}{\rho\Delta\theta}\,\hat{\boldsymbol{\theta}} + \lim_{\Delta z\to 0}\frac{\Delta f_z}{\Delta z}\,\hat{\mathbf{z}} \\ & = \frac{\partial f}{\partial \rho}\,\hat{\boldsymbol{\rho}} + \frac{1}{\rho}\frac{\partial f}{\partial \theta}\,\hat{\boldsymbol{\theta}} + \frac{\partial f}{\partial z}\,\hat{\mathbf{z}} \end{aligned}

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202104162147 Homework 1 (Q2)

The angle a of a triangle ABC is increasing at a rate of 3\,\mathrm{^\circ\, s^{-1}}, the side of AB is increasing at a rate of 1\,\mathrm{cm\, s^{-1}}, and the side of AC is decreasing at a rate of 2\,\mathrm{cm\, s^{-1}}. How fast is the side BC changing when a=30^\circ, AB=10\,\mathrm{cm}, and AC=24\,\mathrm{cm}? Is the length of BC increasing or decreasing?

Solution.

Draw a figure below:

Rephrase the problem.

Given that
\begin{aligned} \frac{\mathrm{d}a}{\mathrm{d}t} & = + 3\,\mathrm{^\circ\, s^{-1}} \\ \frac{\mathrm{d}x}{\mathrm{d}t} & = + 1\,\mathrm{cm\, s^{-1}} \\ \frac{\mathrm{d}y}{\mathrm{d}t} & = -2\,\mathrm{cm\, s^{-1}} \end{aligned}
If a=30^\circ, x=10\,\mathrm{cm}, and y=24\,\mathrm{cm},
then \displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}=\enspace ?}

By cosine law,

z^2=x^2+y^2-2xy\cos a.

Taking ordinary derivatives w.r.t. time t,

\displaystyle{2z\frac{\mathrm{d}z}{\mathrm{d}t} = 2x\frac{\mathrm{d}x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} - 2x\cos a\frac{\mathrm{d}y}{\mathrm{d}t} - 2y\cos a\frac{\mathrm{d}x}{\mathrm{d}t}}

\begin{aligned} z & =\sqrt{x^2+y^2-2xy\cos a} \\ & = \sqrt{(10)^2+(24)^2-2(10)(24)\cos 30^\circ} \\ & = 16.1341\qquad (4\,\mathrm{d.p.}) \end{aligned}

Plugging in the value of each,

\displaystyle{2(\cdot\cdot )\frac{\mathrm{d}z}{\mathrm{d}t} = 2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )(\cdot\cdot )\sin (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big)}

you will know what \displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}} is.

But now, I intend to treat it with partial derivatives.

Let f(x,y,a) = x^2+y^2-2xy\cos a = z^2.

\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial a}\frac{\mathrm{d}a}{\mathrm{d}t} \\ & = (2x-2y\cos a)\frac{\mathrm{d}x}{\mathrm{d}t} + (2y-2x\cos a)\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} \\ \end{aligned}

After \displaystyle{\frac{\mathrm{d}f}{\mathrm{d}t}} is sought, recognise that

\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & =2z\frac{\mathrm{d}z}{\mathrm{d}t} \\ \frac{\mathrm{d}z}{\mathrm{d}t} & = \bigg(\frac{1}{2z}\bigg)\frac{\mathrm{d}f}{\mathrm{d}t} \end{aligned}

you could have it also.

(to be continued)