**Quiz 1.**

**Homework Statement:**

Let u:R3×R→R3 be flow velocity of incompressible fluid. Let fluid be subject to potential force −∇χ. To prove

ddt∫V12ρ⟨u,u⟩dV+∫∂VH⟨u,n⟩dA=0where H:=12ρ⟨u,u⟩+p+χ, and notation ⟨x,y⟩ denote standard inner product on R3.

**Relevant Equations:**

fluid dynamic, euler’s equation of the motion

DuDt=−∇pρ−∇χI re-write the Euler equation for incompressible fluid using suffix notation

∂ui∂t+uj∂ui∂xj+∂∂xi(pρ+χ)=0what theorems applies to the problem?

**Quiz 2.**

A rectangular coil with a height a, width b, of copper wire with a diameter of d with a number of N-½ threads (or turns or loops, not sure which is the proper word) (the last thread/loop is only half) is mounted on the vertical conductive and elastic fiber (Pic. A − 4).

Pic A-4 (I’m attaching the picture)

The coil can deflect/deviates about a vertical axis given by the fiber. The coil is located in a homogeneous magnetic field with induction B in the horizontal direction. The coil is connected to a power supply with a voltage U. The current I of the coil is given by the resistance of the R resistor.

When switch S is switched off –( at rest), the plane of the coil with the direction of the magnetic field forms a zero angle α0 (α = 0). If we move the coil from the equilibrium position by a small angle α and release it, it will oscillate around the equilibrium position with the frequency f0.

After switching on the switch, the current I1 starts to go through the electrical circuit.

a) After stabilization of the coil oscillations, the new equilibrium position of the coil is given by the angle α1 of deflection from the initial position. Write the equation for the angle α1

b) After a small deflection from the new equilibrium position and release, the coil starts to oscillate with frequency f1. Determine the frequency f1 as a function of the angle α1.

c) For a given current I1, the angle α1 = fa (B) is a function of the induction B and the frequency f1 = fb (α1) is a function of the angle α1. Plot graphs of both functions for given values.

d) This way can be measured the horizontal component Bh of magnetic field induction. Determine the induction of Bh using the constructed graphs from the measured values I1 = 2.0A, f0 = 10Hz and f1 = 300Hz and the given coil parameters.

Solve the problem in general and then for values: N = 5, a = 10 cm, b = 10 cm, d = 0.20 mm, I1 = 2.0 A

**Relevant Equations:**

Fm=B*I*l*sinα

f=1/T

v=s/T –>T=s/v

–> f=v/s

**Quiz 3.**

**Homework Statement:**

please see below

**Relevant Equations:**

please see below

For one-dimensional binding potential, a unique energy corresponds to a unique quantum state of the bound particle. In contrast, a particle of unique energy bound in a three-dimensional potential may be in one of several different quantum states. For example, suppose that the three-dimensional box has edges of equal lengths a=b=c, so that it is a cube. Then the energy states are given by

E=h28ma2(nx2+ny2+nz2),nx,ny,nz=1,2,3,4…

(a) what is the lowest possible value for the energy E in the cubical box? show that there is only one quantum state corresponding to this energy (that is, only one choice of the set nx,ny,nz).

(b) What is the second-lowest energy E? Show that this value of energy is shared by three distinct quantum states. Given the probability function |ψ|2 for each of these states, could they be distinguuished from one another?

(c) let n2=nx2+ny2+nz2 be an integer proportional to a given permitted energy. List the degeneracies for energies corresponding to n2=3,6,9,11,12. Can you find a value of n2 for which the energy level is sixfold degenerate?

(a) the lowest possible value for E is

E=h28ma2(12+12+12)=3h28ma2

where the quantum state corresponding to this energy is (nx,ny,nz)=(1,1,1).

(b)the second lowest energy E is

E=h28ma2(22+12+12)=3h24ma2

shared by three distinct quantum states (nx,ny,nz)=(2,1,1) or (1,2,1) or (1,1,2).

They cannot be distinguished from one another because|ψ|2=sin2(nxπxa)sin2(nyπxb)sin2(nzπxc)where a=b=c, and the three quantum states have identical probabilities.

(c)

n2=3⇒(1,1,1)

n2=6⇒(2,1,1),(1,2,1),(1,1,2)

n2=9⇒(2,2,1),(2,1,2),(1,2,2)

n2=11⇒(3,1,1),(1,3,1),(1,1,3)

n2=12⇒(2,2,2)

When nx≠ny≠nz, the number of degeneracies are given by the number of permutations of nx,ny,nz.

**Quiz 4.**

A proton beam of kinetic energy 20 MeV enters a dipole magnet 2 m in length.

How strong must the field be to deflect the beam by 10 degrees?

**Relevant Equations:**

F = qvB

I haven’t taken a physics courses in some time and I’m having trouble getting started with this textbook question. I know that there will be relativistic effects present, but I can deal with that. The problem is how I can approach the problem. I initially thought of a geometric way to set up the problem where I simply assume magnetic field will exert a force uniformly:

But I’m not sure if this will work out since the magnetic field will technically induce a circular motion. Any guidance would be greatly appreciated!

**Quiz 5.**

###### Calculate the Energy Levels of an Electron in a Finite Potential Well

**Quiz 6.**

**Homework Statement:**

psb

**Relevant Equations:**

psb

The equationℏ22md2udr2−Ze2ru=Eugives the schrodinger equation for the spherically symmetric functions u=rψ for a hydrogen-like atom.

In this equation, substitute an assumed solution of the form u(r)=(Ar+Br2)e−br and hence find the values of b and the ratio B/A for which this form of solution satisfies the equation. Verify that it corresponds to the second energy level, with E=Z2/4 times the groud-state energy of hydrogen, and with B/A=−Z/2a0 where a0 is the Bohr radius for hydrogen. What is the value of the coefficient b in terems of a0?

effort:

given

E1=mZ2e42ℏ2

and

a0=4πℏ2Zme2⇒B/A=−2Z2πℏ2me2

ℏ22m[(A+2Br)e−br−b(Ar+Br2)e−br]−Ze2r(Ar+Br2)e−br=mZ4e48ℏ2(Ar+Br2)e−br

[(ℏ22mr−bℏ22m−Ze2r)Ar+(ℏ22mr−bℏ22m−Ze2r)Br2]e−br

(ℏ22mr−bℏ22m−Ze2r)(Ar+Br2)e−br

b=−m2rZ2e4+h4−2mZe2h2rh4

how do you proceed?

The book says the answer isb=Z2a0=2Z2πℏ2me2

many thanks edited with progress

**Quiz 7.**

**Homework Statement:**

Illustrate the concept that “the physical state of system A is correlated with the state of system B” by considering the momenta of cars on the M25 at rush-hour and the road over the Nullarbor Plain in the dead of night.

**Quiz 8.**

###### Prove the reflection and transmission coefficients always sum to 1

**Quiz 9.**

**Homework Statement:**

Average Energy density and the Poynting vector of an EM wave

**Relevant Equations:**

energy density of the EM fields, Poynting vector

Hi,

In Problem 9.12 of Griffiths Introduction to Electrodynamics, 4th edition (Problem 9.11 3rd edition), in the problem, he says that one can calculate the average energy density and Poynting vector as

using the formula

I don’t really understand how to do this. I show my attempt below, but I don’t think it is good. Can someone explain if it is really all there is to it?

**Quiz 10.**

**Homework Statement:**

Given the Friedmann–Lemaitre equation for a universe containing only matter and radiation)

(1)aa˙2=Cr−ka2

Where a is the dimensionless cosmic scale factor, Cr is simplify a constant and the initial condition (I.C.) is a(0)=0.

a) Solve (1) for a flat universe in terms of the conformal time η. Then show that the conformal time at matter–radiation equality is given by:

(*)ηeq=2(2−1)H0−1(ΩM)0−1/2aeq1/2

b) Use this result to show that the horizon at matter–radiation equality corresponds today

to a scale of 16((ΩM)0h2)−1 Mpc, where h:=H0/(100km s−1Mpc−1).

**Relevant Equations:**

N/A

a) For a flat universe (k=0), so (1) simplifies to a˙2=Cra2. The solution to this first order, separable ODE (given the I.C. a(0)=0) is

(2)a(t)=(4Cr)1/4t1/2

We switch to conformal time by means of the change of variables

dη=dta

So (2) takes the form

(3)a(η)=(Cr)1/2η

But how to go from (3) to (∗)?

Before discussing b) I need to understand (∗)

Main sources:

– Chapter 38 in Matthias Blau’s GR notes (attached).

– Chapter 1 in Daniel Baumann’s notes.

**Quiz 11.**

**Homework Statement:**

Calculate the wavelength for an Ex polarized wave

**Relevant Equations:**

Unsure

Calculate the wavelength for an Ex polarized wave travelling through an anisotropic material with ϵ¯¯=ϵ0diag(0.5,2,1) and μ¯¯=2μ0 in:

a. the y direction

b. the z direction

Leave answers in terms of the free space wavelength.

All I’ve gotten so far is:

E(y,t)=E0cos(kyy−wt)

λ=2πky

I don’t know how to determine ky or k¯. I’m basically totally stumped on this problem. Of course it needs to be in terms of:

λ0=2πcω0

**Quiz 12.**

###### Lagrangian of a mass between two springs with a pendulum hanging down

###### Homework Statement:

A particle of mass m1 hangs from a rod of negligible mass and length l, whose support point consists of another particle of mass m2 that moves horizontally subject to two springs of constant k each one. Find the equations of motion for this system.

###### Relevant Equations:

L=T−V

What I first did was setting the reference system on the left corner. Then, I said that the position of the mass m2 is x2. I also supposed that the pendulum makes an angle θ with respect to the vertical axis y. So the generalized coordinates of the system would be x2 and θ. Thus, the coordinates of m1 are:

- x1=x2+lsinθ
- y1=−lcosθ

Then I took the time derivative of both x1 and y1:

- x˙1=x˙2+lθ˙cosθ
- y˙1=lθ˙sinθ

The Kinetic energy T would be the sum of the kinetic energy of each mass T=T1+T2. Since T1=12m2\dotx22 and T2=12m1(x˙12+y˙12, I got:

- T=12m2x˙22+12m1(x˙22+l2θ˙2+2x˙2θ˙l)

Now, I have trouble with the potential energy. The potential energy of m1 is easy, it would be just V1=m1gh:

- V1=−m1glcosθ

The potential energy of m2 would be the sum of the potential energy that each spring has on the mass. What I did was saying that that the mass m2 moves a distance x2, thus the lhs spring would have a displacement of x2 and the rhs spring would have a contraction of x2 likewise. So:

- V2=12kx22+12k(−x2)2=kx12

But, according to my teacher, if the first spring moves a distance x2, then the second would contract a distance of a−x2, where a is the distance between the walls, that is constant. I don’t understand why this is the case, can you please explain it better for me to understand? Thanks.

**Quiz 13.**

**Homework Statement:**

A particle of mass m moving with velocity v1, leaves a half-space in which the potential is a constant U1 and enters the other half-space, where the potential energy is a different constant U2.

**Relevant Equations:**

Determine the change in the direction of motion of the particle.

**Quiz 14.**

**Homework Statement:**

Suppose an observer O measures a particle of mass m moving in the x direction to have speed v, energy E, and momentum p. Observer O’, moving at speed u in the x direction; measures v’, E’, and p’ for the same object. (a) Use the Lorentz velocity transformation to find E’ and p’ in terms of m, u, and v. (b) Reduce (E’)^2 – (p’c)^2 to its simplest form.

**Relevant Equations:**

Lorentz velocity transformation:

v′=(v−u)/(1−(uv/c2))

Relationship between energy-momentum:

E2=(pc)2+(mc2)2

I try to use relativistic energy equation:

E′=γmc2

But, I use

γ=1(1−(v′c)2

then I use Lorentz velocity transformation.

v′=v−u1−uvc2

At the end, I end up with messy equation for E’ but still have light speed c in the terms. How should I do to get E’ just in terms of m, u, and v?

**Quiz 15.**

**Homework Statement:**

Evaluate χ(t) for the model function

χ~(ω)=Nq2Vmϵ01ω02−ω2−iωγ=Nq2Vmϵ012ω02−γ24(1ω+iγ2+ω02−γ24−1ω+iγ2−ω02−γ24) ,

and interpret the result.

**Relevant Equations:**

Complex polarization: P(t)=χ~ϵ0E~0e−iωt

I have not studied the Fourier transform (FT) in great detail, but came across a problem in electrodynamics in which I assume it is needed. The problem goes as follows:

Evaluate χ(t) for the model function

χ~(ω)=Nq2Vmϵ01ω02−ω2−iωγ=Nq2Vmϵ012ω02−γ24(1ω+iγ2+ω02−γ24−1ω+iγ2−ω02−γ24) ,

and interpret the result.

To find χ(t), one needs to evaluate the integral χ(t)=12π∫−∞∞χ~(ω)e−iωtdω, right? Are there any tricks to simplify the integral?

**Quiz 16.**

**Homework Statement:**

Plane wavefront is incident at angle α on a plane mirror and reflects off at angle β – treat the situation as a 1d Fraunhofer diffraction problem, finding the phase of the diffracted wavefront as a function of distance from the mirror centre and β. Find the intensity distribution.

**Relevant Equations:**

Fraunhofer intensity

**Quiz 17.**

**Homework Statement:**

Derive the Penrose equation for a vacuum spacetime∇λ∇λRμνρσ=2RκμλσRλρκν−2RκνλσRλρκμ−RκλσρRλκμν

**Relevant Equations:**

N/A

Okay so for this one we can consider the Bianchi identity again∇[λRσρ]μν=2∇λRσρμν+2∇ρRλσμν+2∇σRρλμν=0∇λ∇λRσρμν=∇λ∇ρRσλμν+∇λ∇σRλρμν∇λ∇λRσρμν=gϵλ∇ϵ∇ρRσλμν+gϵλ∇ϵ∇σRλρμνHere I wrote it in a form that’s suggestive of another identity(∇a∇b−∇b∇a)Tc1,…,ckd1,…,dl=∑j=1lRabdjeTc1,…,ckd1,…,e,…dl−∑i=1kRabeciTc1,…,e,…,ckd1,…,dlLet’s say we consider T=R, i.e.(∇ϵ∇ρ−∇ρ∇ϵ)Rσλμν=RϵρσeReλμν+RϵρλeRσeμν+RϵρμeRσλeν+RϵρνeRσλμeIs there a way to tidy this up using some other symmetries of the Riemann tensor

**Quiz 18.**

###### Energy-Momentum Tensor for Electromagnetism in curved space

**Homework Statement:**

Consider

L=−g(−14FμνFμν+AμJμ)

a) Find Tμν

b) Adding a new term to the Lagrangian L′=βRμνgρσFμρFνσ. How are Maxwell’s eqs altered? How about Einstein’s Eqs? Is the current still conserved?

**Relevant Equations:**

N/A

a) I’d separated the Lagrangian into:

L=LMax+Lint

in which LMax=−14−gFμνFμν and Lint=−gAμJμ

Thus:

TMaxμν=FμλFνλ−14FαβFαβ….

**Quiz 19.**

###### Describe the motion of yoyos suspended from the ceiling

**Homework Statement:**

Determine the motion of yoyos for $n=1,2,3$

**Relevant Equations:**

J=mr22

Jθ¨=Fr

I have trouble solving this problem any help would be appreciated.**Problem statement**

J=mr22**a)** Determine the motion of yoyos for n=1,2,3

The case for n=1 is simple, however, I am having trouble with n=2 and n=3.

for n=2 I started by drawing all the forces:

and then I wrote out the equations of motions.

1 yoyo:

mx1¨=Fg1+Fs12−Fs0

Jθ¨1=Fs0r

2 yoyo:

mx2¨=Fg2−Fs21

Jθ¨2=Fs21r

What I wanted to do here was also introduce a constraint x2˙=x1˙+θ˙2r, however, I am not sure whether this is correct. It feels right because the speed of the 2nd one is gonna depend on how fast the 2nd yoyo spins and how fast the 1st one moves. From here I continued

θ¨1r=x1¨Jθ¨1=Fs0r⟹Fs0=Jr2x1¨

θ¨2r=x2¨−x1¨⟹θ¨2=x2¨−x1¨r⟹Fs21=J(x2¨−x1¨)r2

Plugging both forces into the equations of motions and using that |Fs21|=|Fs12| I got

mx1¨=Fg1+Fs12−Fs0⟹mx1¨=mg+J(x2¨−x1¨)r2−Jr2x1¨

mx2¨=Fg2−Fs21⟹mx2¨=mg−J(x2¨−x1¨)r2

Plugging in J=mr22

mx1¨=mg+m(x2¨−x1¨)2−m2x1¨

mx2¨=mg−m(x2¨−x1¨)2

diving by m

x1¨=g+(x2¨−x1¨)2−12×1¨⟹2×1¨−12×2¨=g

x2¨=g−(x2¨−x1¨)2⟹32×2¨−12×1¨=g

**Quiz 20.**

**Homework Statement:**

There are two ledges with of U potential energy, and a cavern in between them of 0 potential energy. Such that

U(x) = { x <= 0, U, x > 0, x >= L, U }

Find the probability distribution for all 3 regions

**Relevant Equations:**

(ħ[SUP]2[/SUP]/2m)((d[SUP]2[/SUP]Ψ)/(dx[SUP]2[/SUP]) + U(x)v = EΨ

∫ Ψ * Ψ dx = 1

I’m following Griffith’s Modern Physics 2nd edition chapter 5.

I got to the part where we make Ψ_{I}(0) = Ψ_{II}(0) I get that

αCe^{α(0)} = QAsin(Q(0)) – QBsin(Q(0)) => C = QA/α

But when I try to graph it, the region I distribution doesn’t seem to equal the region II distribution at 0.

The book goes on on to insert C into an equation to solve for G (-αGe^{-α(L)} for the other side of the well), but I don’t understand why it does this.

-αGe^{-α(L)} = (αC/Qs)sin(Q(L)) + C/cos(Q(L))

Why does what happens on one side affect what happens on the other?

###### Electron-positron annihilation, photon emission angle

**Homework Statement:**

In the annihilation process e+e−→γγ, seen in the reference system of the laboratory, in which the electron is stationary and the positron has energy E, it may happen that one of the two photons is emitted at right angles to the direction of incidence of the positron. Calculate, in this case, the energy and the scattering angle of the other photon.

###### Heat Diffusion Equation – Using BCs to model as an orthonormal system

**Homework Statement:**

Consider heat equation

κ∂2ψ∂x2=∂ψ∂t

κ is positive, x⊂[0,a], ψ is real

For t>0: ψ(t,0)=ψ0 at x=0

∂ψ∂x(t,a)=0 at x=a

ψ(0,x)=0

We introduce gk(x)=2asin(qkx)

where qk=πa(k+12)

k=0,1,…

b) Argue that the functions gk form an ortho-normal basis of the space Lb2([0,a]), of square integrable functions f on [0,a] with a Dirichlet boundary condition f(0)=0 at x=0 and a von Neumann condition f′(a)=0 at x=a.

c) Based on the results in (a) (I’ve done this part – it’s d2dx2gk=−qk2gk) and (b) argue that the most general ψ with the correct boundary conditions can be written as ψ(t,x)=ψ0+Σ0infTk(t)gk(x). Find the solutions for the functions Tk.

d) Fix the remaining constants in your solution by imposing the initial condition. Compute the average value ψavg(t) of ψ(x,t) by averaging over x ∈ [0, a] and find an approximate equation for the time as a function ofr:=(ψ0−ψavg(t))/ψ0.

**Relevant Equations:**

Fourier series, Dirichlet, Von Neumann

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