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Chapter 5: Q26P (page 239)

Use the method of Lagrange multipliers to find the rectangle of largest area, with sides parallel to the axes that can be inscribed in the ellipse(xa)2+(yb)2=1. What is the maximum area?

Short Answer

Expert verified

The maximum area of an inscribed triangle can have is=2ab

Step by step solution

01

Define area of triangle

The area of a triangle is defined as the total space occupied by a triangle's three sides in a two-dimensional plane. The basic formula for calculating the area of a triangle is half the product of its base and height.

02

Calculating the area of inscribed rectangle

We want area of the inscribed rectangle to be maximum. If its sides are equal to 2x and 2y, its area is: a(x,y)=4xy.

G(x,y,λ)=4xy+λ(xa)2+(yb)2-1∂G∂x=0=4y+2λxa2y=-λx2a2∂G∂y=0=4x+2λyb2⇒4x=-λy2b2=-λ2xa2b2

λ1=0orλ2=±2ab

03

Calculating the maximum area

Because side of a rectangular must be positive

y=-λx2a2=±bxa=bxa∂G∂λ=0=(xa)2+(yb)2-1x2a2+b2x2a2b2=1x2=a22x=a2,y=a2

So maximum area of an inscribed triangle can have is A=4xy=2ab

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Most popular questions from this chapter

Typically, the interaction potential depends only on the vectorr=r1-r2between

the two particles. In that case the Schrodinger equation seperates, if we change variables from r1,r2andR=(m1r1+m2r2)I(m1+m2)(the center of mass).

(a)Show that

localid="1655976113066" r1=R+(μ/m1)r,r2=R-(μ/m)r,and∇1=(μ/m)∇R+∇r,∇2=(μ/m1)∇R-∇r,where

localid="1655976264171" μ=m1m2m1+m2(5.15).

is the reduced mass of the system

(b) Show that the (time-independent) Schrödinger equation (5.7) becomes

-h22m1∇12ψ-h22m2∇22ψ+Vψ=Eψ-h22(m1+m2)∇R2ψ-h22μ∇r2ψ+V(r)ψ=Eψ.(5.7).

(c) Separate the variables, letting ψ(R,r)=ψR(R)ψr(r)Note that ψRsatisfies the

one-particle Schrödinger equation, with the total mass (m1+m2)in place of m, potential zero, and energy ERwhile ψrsatisfies the one-particle Schrödinger equation with the reduced mass in place of m, potential V(r) , and energy localid="1655977092786" Er. The total energy is the sum: E=ER+Er. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics; it reduces the two-body problem to an equivalent one-body problem.

Find the average energy per free electron (Etot/Nd), as a fraction of the

Fermi energy. Answer:(3/5)EF

We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, E=p2/2m,,with the relativistic formula, E=p2c2+m2c4-mc2. Momentum is related to the wave vector in the usual way: p=hk. In particular, in the extreme relativistic limit, E≈pc=hck.

(a) Replace h2k2n Equation 5.55 by the ultra-relativistic expression, hck, and calculateEtotin this regime.

dE=h2k22mVπ2k2dk (5.55).

(b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc , such that gravitational collapse occurs for N>N_{C}is called the Chandrasekhar limit.

(c) At extremely high density, inverse beta decaye-+p+→n+v,converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star non relativistic ally?

(a) Find the chemical potential and the total energy for distinguishable particles in the three dimensional harmonic oscillator potential (Problem 4.38). Hint: The sums in Equations5.785.78and5.795.79can be evaluated exactly, in this case−−no need to use an integral approximation, as we did for the infinite square well. Note that by differentiating the geometric series,

11-x=∑n=0∞xn

You can get

ddx(x1-x)=∑n=1∞(n+1)xn

and similar results for higher derivatives.

(b)Discuss the limiting caserole="math" localid="1658400905376" kBT≪hω.
(c) Discuss the classical limit,role="math" localid="1658400915894" kBT≫hω, in the light of the equipartition theorem. How many degrees of freedom does a particle in the three dimensional harmonic oscillator possess?

(a) Calculate<1/r1-r2>for the stateψ0(Equation 5.30). Hint: Dod3r2integral

first, using spherical coordinates, and setting the polar axis alongr1, so

that

ψ0r1,r2=ψ100r1ψ100r2=8πa3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2cosθ2.

Theθ2integral is easy, but be careful to take the positive root. You’ll have to

break ther2integral into two pieces, one ranging from 0 tor1,the other fromr1to∞

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it toE0(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

E0=8-13.6eV=-109eV(5.31).
See all solutions

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