Physicspupil – Page 42

February 21, 2019January 19, 2023

201902210102 Exercise 3.1.1

$\mathbf{A}=\begin{bmatrix} A'^t\\ A'^x\\ A'^y\\A'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} A^t \\ A^x \\ A^y \\ A^z \end{bmatrix}$

$\mathbf{B}=\begin{bmatrix} B'^t\\ B'^x\\ B'^y\\B'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} B^t \\ B^x \\ B^y \\ B^z \end{bmatrix}$

are two four-vectors.

$\begin{aligned} A'^tB'^t & =(\gamma A^t-\gamma\beta A^x)(\gamma B^t-\gamma\beta B^x)\\ & = \gamma^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2\beta^2A^xB^x \end{aligned}$

$\begin{aligned} A'^xB'^x & = (-\gamma\beta A^t+\gamma A^x)(-\gamma\beta B^t+\gamma B^x)\\ & = \gamma^2\beta^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2A^xB^x\\ \end{aligned}$

$\begin{aligned} A'^yB'^y & = A^yB^y \end{aligned}$

$\begin{aligned} A'^zB'^z & = A^zB^z \end{aligned}$

$\begin{aligned} &\quad\enspace -A'^tB'^t+A'^xB'^x+A'^yB'^y+A'^zB'^z \\ & = (-\gamma^2+\gamma^2\beta^2)A^tB^t+(-\gamma^2\beta^2+\gamma^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\gamma^2(1-\beta^2)A^tB^t+\gamma^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^tB^t+\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -A^tB^t+A^xB^x+A^yB^y+A^zB^z \end{aligned}$

Scalar product of any two four-vectors is frame-independent.

February 21, 2019January 19, 2023

201902210054 Exercise 17 Chapter 2

It sometimes occurs that the generalized coordinates appear separately in the kinetic energy and the potential energy in such a manner that $T$ and $V$ may be written in the form

$T=\displaystyle{\sum_i}f_i(q_i)\dot{q_i}^2$ and $V=\displaystyle{\sum_iV_i(q_i)}$ .

Show that Lagrange’s equations then separate, and that the problem can always be reduced to quadratures.

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

From the fact that

$\begin{aligned} \mathcal{L} & =T-V \\ & =\displaystyle{\sum_i f_i(q_i)\dot{q_i}^2-\sum_iV_i(q_i)}\\ & =\sum_i\bigg( f_i(q_i)\dot{q_i}^2-V_i(q_i)\bigg)\\ & =\sum_i(T_i-V_i)\\ & =\sum_iL_i\\\end{aligned}$ ,

the Lagrange’s equation can be separated into $i$ Lagrange’s equations.

Remark.

The solution is incomplete. It remains to be shown how the problem can always be reduced to quadratures.

February 20, 2019January 19, 2023

201902200847 Exercise 11 Chapter 2

When two billiard balls collide, the instantaneous forces between them are very large but act only in an infinitesimal time $\Delta t$ , in such a manner that the quantity

$\displaystyle{\int_{\Delta t}}F\,\mathrm{d}t$

remains finite. Such forces are described as impulsive forces, and the integral over $\Delta t$ is known as the impulse of the force. Show that if impulsive forces are present Lagrange’s equations may be transformed into

$\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{q}_j}} \bigg)_f-\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{q}_j}} \bigg)_i=S_j$ ,

where the subscripts $i$ and $f$ refer to the state of the system before and after the impulse, $S_j$ is the impulse of the generalized impulsive force corresponding to $q_j$ , and $\mathcal{L}$ is the Lagrangian including all the non-impulsive forces.

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

For billiard-balls collision, the Euler-Lagrange (E-L) equation is

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)-\bigg( \frac{\partial L}{\partial q_j}\bigg)=Q_j$ ,

where $Q_j$ is the generalised impulsive force corresponding to $q_j$ and not derivable from the potential.

Taking integral over $\Delta t$ on both sides,

LHS becomes

$\displaystyle{\int_{\Delta t}\frac{\displaystyle{\mathrm{d}\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)}}{\mathrm{d}t}\,\mathrm{d}t}-\int_{\Delta t}\bigg( \frac{\partial L}{\partial q_j}\bigg) \,\mathrm{d}t$ .

The second term upon integration is zero,

because $\displaystyle{\bigg( \frac{\partial L}{\partial q_j}}\bigg) \Delta t=0$ for infinitesimal time $\Delta t\to 0$ .

The first term is

$\displaystyle{\int_{\Delta t}\mathrm{d}\bigg( \frac{\partial L}{\partial \dot{q_j}} \bigg)}=\bigg[ \frac{\partial L}{\partial \dot{q_j}}\bigg]^{t+\Delta t}_{t}$ .

Rename $t+\Delta t$ the (final) state of system $f$ after the impulse and $t$ the (initial) state of system $i$ before the impulse.

LHS reads

$\displaystyle{\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_f}-\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_i$

whereas RHS reads

$\displaystyle{\int_{\Delta t}Q_j \,\mathrm{d}t=S_j}$ ,

i.e., the impulse of generalised impulsive force.

The transformed E-L equation in the presence of impulsive forces is

$\displaystyle{\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_f}-\bigg( \frac{\partial L}{\partial \dot{q_j}}\bigg)_i=S_j$ ,

as desired.

February 20, 2019January 19, 2023

201902200839 Exercise 5 Chapter 2

A particle is subjected to the potential $V(x)=-Fx$ , where $F$ is a constant. The particle travels from $x=0$ to $x=a$ in a time interval $t_0$ . Assume the motion of the particle can be expressed in the form $x(t)=A+Bt+Ct^2$ . Find the values of $A$ , $B$ , and $C$ such that the action is a minimum.

Solution.

The solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

1D-case:

$\mathcal{L}=T-V=\displaystyle{\frac{1}{2}}m\dot{x}^2+Fx$ .

Euler-Lagrange (E-L) equation:

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{x}}}\bigg) =\frac{\partial \mathcal{L}}{\partial x}$

gives the path over which the action is stationary.

That is,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}( m\dot{x}) & = F \\ \Rightarrow \ddot{x} & =\displaystyle{\frac{F}{m}}\\ \end{aligned}$ .

On $x(t)=A+Bt+Ct^2$ taking derivative twice,

$\ddot{x}(t)=2C$ .

Equate them,

$C=\displaystyle{\frac{F}{2m}}$ .

The event $(x=0,t=0)$ gives

$A=0$ .

And the event $(x=a,t=t_0)$ gives

$a=x(t_0)=Bt_0+\displaystyle{\frac{F}{2m}}t_0^2$ .

Express it as

$B=\displaystyle{\frac{a-\frac{F}{2m}t_0^2}{t_0}}$ .

Thus,

$x(t)=(0)+\Bigg( \displaystyle{\frac{a-\frac{F}{2m}t_0^2}{t_0}}\Bigg) t+\bigg( \displaystyle{\frac{F}{2m}}\bigg) t^2$

is recovered.

February 20, 2019January 17, 2023

201902200829 Derivation 2.3

Prove that the shortest distance between two points in space is a straight line.

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

Assume the path (of any curve $C$ ) connecting two points $(a,y(a))$ and $(b,y(b))$ is given by a function $C(x)=(x,y(x))$ , with $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}C(x)=(1,y')$ being the first derivative of the curve.

To minimise the path distance

$\displaystyle{\mathcal{L}=\int \| C'\| \,\mathrm{d}x=\int_a^b\sqrt{1+y'^2}\,\mathrm{d}x}$ ,

define now

$f(x,y,y')=\sqrt{1+y'^2}$ ,

having $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y}}=0$ and $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}y'}}=\frac{y'}{\sqrt{1+y'^2}}$ .

From Euler-Lagrange (E-L) equation it follows that

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}}\bigg( \displaystyle{\frac{y'}{\sqrt{1+y'^2}}}\bigg) =0$ ,

i.e., $y'=\mathrm{Const.}$

In conclusion, the shortest distance between two points in space is a straight line.

Lemma. (Fundamental lemma of the calculus of variations)

If $\displaystyle{\int_{x_1}^{x_2}M(x)\eta (x)\,\mathrm{d}x=0}$ for any $\eta (x)$ continuous through second derivative, then $M(x)$ must identically vanish in the interval $x_1,x_2$ .

_{Text on pg.38, Goldstein}

February 20, 2019January 17, 2023

201902200820 Derivation 1.2

Prove that the magnitude $R$ of the position vector for the center of mass from an arbitrary origin is given by the equation

$M^2R^2=M \displaystyle{\sum_i}m_ir_i^2-\displaystyle{\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2}$ .

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

$\displaystyle{M\mathbf{R}}=\sum m_i \mathbf{r}_i$ .

Taking squares on both sides,

$\displaystyle{M^2\mathbf{R}^2}=\sum_{i,j}m_im_j\mathbf{r}_i\cdot \mathbf{r}_j$ .

Notice $\mathbf{r}_{ij}=\mathbf{r}_i-\mathbf{r}_j$ , squaring it,

$r_{ij}^2=r_i^2-2\mathbf{r}_i\cdot \mathbf{r_j}+r_j^2$ ,

arranging,

$\displaystyle{\mathbf{r}_i}\cdot \mathbf{r}_j=\frac{1}{2}(r_i^2+r_j^2-r_{ij}^2)$ .

Plugging this,

$\begin{aligned} \displaystyle{M^2R^2} & =\frac{1}{2}\sum_{i,j}m_im_jr_i^2+\frac{1}{2}\sum_{i,j}m_im_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{\frac{1}{2}M\sum_im_ir_i^2}+\frac{1}{2}M\sum_jm_jr_j^2-\frac{1}{2}\sum_{i,j}m_im_jr_{ij}^2\\ & = \displaystyle{M\sum_im_ir_i^2}-\frac{1}{2}\sum_{i\neq j}m_im_jr_{ij}^2 \\ \end{aligned}$

February 20, 2019January 17, 2023

201902200814 Derivation 1.1

Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy:

$\displaystyle{\frac{\mathrm{d}T}{\mathrm{d}t}}=\mathbf{F}\cdot\mathbf{v}$ ,

while if the mass varies with time the corresponding equation is

$\displaystyle{\frac{\mathrm{d}(mT)}{\mathrm{d}t}}=\mathbf{F}\cdot\mathbf{p}$ .

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

Single particle with constant mass:

$\begin{aligned} \frac{\mathrm{d}T}{\mathrm{d}t} & =\frac{\mathrm{d}\big(\frac{1}{2}mv^2\big)}{\mathrm{d}t} \\ & =m\mathbf{v}\cdot \mathbf{\dot{v}} \\ & =m\mathbf{a}\cdot \mathbf{v} \\ & =\mathbf{F}\cdot \mathbf{v} \end{aligned}$

if mass varies with time:

$\begin{aligned} \frac{\mathrm{d}(mT)}{\mathrm{d}t} & =\frac{\mathrm{d}\big( m\times \frac{1}{2}mv^2\big)}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{(mv)^2}{2}\bigg) \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{p^2}{2}\bigg) \\ & =\mathbf{p}\cdot \mathbf{\dot{p}} \\ & =\mathbf{F}\cdot \mathbf{p} \end{aligned}$

Remark.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}v^2 & =\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot \mathbf{v}) \\ & =\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\cdot \mathbf{v}+\mathbf{v} \cdot \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} \\ & =2\mathbf{v}\cdot \mathbf{a} \end{aligned}$

February 19, 2019January 17, 2023

201902190842 Electric field representation

The blogger claims no originality of his question posted here.

Think of a field in some representation other than a diagram of field lines. For instance, suppose any point in a field is assigned a number to its field strength magnitude, but the direction of field strength of each point is not indicated. Now, could we deduce a rough picture of the (vector) E-field from these (scalar) numbers, provided that we know the distribution of these numbers and the position of charges?

Visualizing_the_Efield_by_numbers_distribution

In the figure, $A$ , $B$ and $C$ are charges of unknown charge quantities and sign. Each single-digit number at a point represents the magnitude of E-field strength at that point. For example, $6$ stands for $6\,\mathrm{N/C}$ , $0$ stands for zero E-field, i.e., a neutral point. By deducing from the figure, determine which of the following statements is/are correct.

(1) Charge quantity of $A$ is larger than that of either $B$ or $C$ .
(2) Charge quantity of $B$ equals charge quantity of $C$ .
(3) Charge $A$ is positive. Charge $B$ and charge $C$ are both negative.

A. (1) only
B. (2) only
C. (1) and (2) only
D. (1), (2), and (3)

Answer. D

February 19, 2019January 17, 2023

201902190827 Electric circuit diagrams (Elementary) Q13

The blogger claims no originality of his idea here.

We are given the following diagram:

Circuit0013

We would like to classify the nodes by the electric potential there. We label the nodes with $a$ , $b$ , $c$ , etc. where the potential at $a$ is larger than that at $b$ , the potential at $b$ is larger than that at $c$ , etc. Therefore we have the following diagram:

Circuit0013Sol001

Aligning $a$ , $b$ , and $c$ from left to right would give the main current direction. And we put each resistor back in between two consecutive nodes, (e.g., $a$ and $b$ , $b$ and $c$ , etc.) according to the two labels nearest to its two ends:

Circuit0013Sol002