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1 Physics 2102B - 1st Assignment Your solution to problems 6, 7, and 8 has to be handed in, in class, on Monday, February 3, 2013. 1. When considering the Maxwell-Boltzmann distribution it was assumed to have the following form f v () = α e mv x 2 +v y 2 +v z 2 ( ) 2 kT , (1.1) where α is some normalization constant. a) Consider the Gaussian integral I = e β x 2 dx −∞ , (1.2) with β another constant. Now take the square of this integral (i.e., consider I 2 ), change from Cartesian to polar coordinates, solve the integral for I 2 , and then show that I = π β . (1.3) b) Use your result from a) to show in equation (1.1) α = N m 2π kT 32 , (1.4) if we also require that N = f v () dv x −∞ dv y −∞ dv z −∞ , (1.5) with N the total number of particles in a system that verifies the Maxwell-Boltzmann distribution. c) Show that v x 2 e mv x 2 2 kT dv x −∞ = 1 2π 2π kT m 32 . (1.6) You may want to consider integrating by parts to solve this integral.

1st assignment solution - Western Universityhoude/courses/s/physics2102/1st... · 1 Physics 2102B - 1st Assignment Your solution to problems 6, 7, and 8 has to be handed in, in class,

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Physics 2102B - 1st Assignment

Your solution to problems 6, 7, and 8 has to be handed in, in class, on Monday, February 3, 2013.

1. When considering the Maxwell-Boltzmann distribution it was assumed to have the following form f v( ) =αe−m vx

2+vy2+vz

2( ) 2kT , (1.1) where α is some normalization constant. a) Consider the Gaussian integral I = e−βx

2

dx−∞

∫ , (1.2)

with β another constant. Now take the square of this integral (i.e., consider I 2 ), change from Cartesian to polar coordinates, solve the integral for I 2 , and then show that

I = πβ. (1.3)

b) Use your result from a) to show in equation (1.1)

α = N m2π kT

⎛⎝⎜

⎞⎠⎟3 2

, (1.4)

if we also require that N = f v( )dvx−∞

∫ dvy−∞

∫ dvz−∞

∫ , (1.5)

with N the total number of particles in a system that verifies the Maxwell-Boltzmann distribution. c) Show that

vx2e−mvx

2 2kT dvx−∞

∫ = 12π

2π kTm

⎛⎝⎜

⎞⎠⎟3 2

. (1.6)

You may want to consider integrating by parts to solve this integral.

2

Solution. a) From equation (1.2)we can write

I 2 = e−βx

2

dx−∞

∫( ) e−βy2

dy−∞

∫( )= e−β x2+y2( ) dx

−∞

∫ dy−∞

∫ . (1.7)

We now make the change of variables

x = r cos θ( )y = r sin θ( )

(1.8)

which implies that dxdy = rdrdθ . Equation (1.7) then becomes

I 2 = dθ0

∫ ⋅ e−βr2

r dr0

= 2π ⋅ − 12β

e−βr2⎛

⎝⎜⎞⎠⎟0

= πβ,

(1.9)

which implies that

I = πβ. (1.10)

b) It becomes apparent from equations (1.1) to (1.3) that

N = αe−m vx2+vy

2+vz2( ) 2kT dvx−∞

∫ dvy−∞

∫ dvz−∞

∫=α I β = m 2kT( )⎡⎣ ⎤⎦

3

=α 2π kTm

⎛⎝⎜

⎞⎠⎟3 2

,

(1.11)

and

α = N m2π kT

⎛⎝⎜

⎞⎠⎟3 2

. (1.12)

3

c) We define

I1 = vx2e−mvx

2 2kT dvx−∞

∫= vx ⋅ − kT

me−mvx

2 2kT⎛⎝⎜

⎞⎠⎟

−∞

+ kTm

e−mvx2 2kT dvx−∞

= 0 + kTm

⎛⎝⎜

⎞⎠⎟ I β = m 2kT( )

= 12π

2π kTm

⎛⎝⎜

⎞⎠⎟3 2

,

(1.13)

where we integrated by parts using I1 = uv −∞

∞ − vdu−∞

∫ with

u = vx , du = dvx

dv = vxe−mvx

2 2kT dvx , v = − kTme−mvx

2 2kT (1.14)

and we used equation (1.10).

2. Consider equation (1.24) of Chapter 1 of the lecture notes for the one-dimensional wave equation for the electric field

∂2Ez

∂x2− 1c2

∂2Ez

∂t 2= 0. (2.1)

a) Let us now study how this wave equation changes under a Galilean transformation

t = ′tx = ′x − v ′t= ′x − vt,

(2.2)

where ′x and ′t are the spatial and temporal coordinates attached to an inertial reference frame ′K moving at a velocity v relative to another inertial frame K to which x and t are attached. First use the chain rule to show that

∂∂t

= ∂∂ ′t

+ v ∂∂ ′x

∂∂x

= ∂∂ ′x

. (2.3)

4

Second, derive corresponding equations for ∂2 ∂t 2 and ∂2 ∂x2 , and then express equation (2.1) in ′K . Is the wave equation covariant (or invariant) under a Galilean transformation? Are Maxwell’s equations covariant under a Galilean transformation? b) Now let us consider a Lorentz transformation

′t = γ t + v

c2x⎛

⎝⎜⎞⎠⎟

′x = γ x + vt( ), (2.4)

where γ = 1− v2 c2( )−1 2 is the Lorentz factor. Derive the needed relations for ∂2 ∂t 2 and

∂2 ∂x2 , and express equation (2.1) in ′K . Is the wave equation covariant (or invariant) under a Galilean transformation? Are Maxwell’s equations covariant under a Galilean transformation? [Note: To a good approximation the electric field is the same in the two referential frames, i.e., Ez ′Ez , for both types of transformation when v is small relative to c (i.e., when γ 1 .] Solution. a) The chain rule states that

∂∂t

= ∂ ′t∂t

∂∂ ′t

+ ∂ ′x∂t

∂∂ ′x

= ∂∂ ′t

+ v ∂∂ ′x

∂∂x

= ∂ ′t∂x

∂∂ ′t

+ ∂ ′x∂x

∂∂ ′x

= ∂∂ ′x

(2.5)

from equations (2.2). We then have

∂2

∂t 2= ∂

∂ ′t+ v ∂

∂ ′x⎛⎝⎜

⎞⎠⎟

∂∂ ′t

+ v ∂∂ ′x

⎛⎝⎜

⎞⎠⎟

= ∂2

∂ ′t 2+ 2v ∂

∂ ′t∂∂ ′x

+ v2 ∂2

∂ ′x 2

(2.6)

and

5

∂2

∂x2= ∂∂ ′x

∂∂ ′x

⎛⎝⎜

⎞⎠⎟

= ∂2

∂ ′x 2 . (2.7)

If we now apply equations (2.6) and (2.7) to (2.1) we find that

∂2

∂x2− 1c2

∂2

∂t 2⎛⎝⎜

⎞⎠⎟Ez =

∂2

∂ ′x 2 −1c2

∂2

∂ ′t 2− 2 v

c2∂∂ ′t

∂∂ ′x

− v2

c2∂2

∂ ′x 2

⎛⎝⎜

⎞⎠⎟Ez

= ∂2

∂ ′x 2 1−v2

c2⎛⎝⎜

⎞⎠⎟− 1c2

∂2

∂ ′t 2− 2 v

c2∂∂ ′t

∂∂ ′x

⎣⎢

⎦⎥Ez

≠ ∂2

∂ ′x 2 −1c2

∂2

∂ ′t 2⎛⎝⎜

⎞⎠⎟Ez .

(2.8)

Since we have an inequality for the last equation, it follows that the wave equation changes mathematical form when subjected to a Galilean transformation and is therefore not covariant in that respect. The same conclusion applies to Maxwell’s equations since the wave equation was derived from them. That is, Maxwell’s equations are not covariant under a Galilean transformation. b) For a Lorentz transformation we have

∂∂t

= ∂ ′t∂t

∂∂ ′t

+ ∂ ′x∂t

∂∂ ′x

= γ ∂∂ ′t

+ v ∂∂ ′x

⎛⎝⎜

⎞⎠⎟

∂∂x

= ∂ ′t∂x

∂∂ ′t

+ ∂ ′x∂x

∂∂ ′x

= γ vc2

∂∂ ′t

+ ∂∂ ′x

⎛⎝⎜

⎞⎠⎟ .

(2.9)

It then follows that

∂2

∂t 2= γ 2 ∂

∂ ′t+ v ∂

∂ ′x⎛⎝⎜

⎞⎠⎟

∂∂ ′t

+ v ∂∂ ′x

⎛⎝⎜

⎞⎠⎟

= γ 2 ∂2

∂ ′t 2+ 2v ∂2

∂ ′x ∂ ′t+ v2 ∂2

∂ ′x 2

⎛⎝⎜

⎞⎠⎟

(2.10)

and

6

∂2

∂x2= γ 2 v

c2∂∂ ′t

+ ∂∂ ′x

⎛⎝⎜

⎞⎠⎟

vc2

∂∂ ′t

+ ∂∂ ′x

⎛⎝⎜

⎞⎠⎟

= γ 2 vc2

⎛⎝⎜

⎞⎠⎟2 ∂2

∂ ′t 2+ 2 v

c2∂2

∂ ′x ∂ ′t+ ∂2

∂ ′x 2

⎣⎢

⎦⎥.

(2.11)

Using these last two equations, we can write

∂2

∂x2− 1c2

∂2

∂t 2= γ 2 1− v

2

c2⎛⎝⎜

⎞⎠⎟

∂2

∂ ′x 2 − 1− v2

c2⎛⎝⎜

⎞⎠⎟1c2

∂2

∂ ′t 2⎡

⎣⎢

⎦⎥

= ∂2

∂ ′x 2 −1c2

∂2

∂ ′t 2,

(2.12)

since γ 2 = 1− v2 c2( )−1 . It follows that both the wave equation and Maxwell’s equations are covariant under a Lorentz transformation (remember that Ez = ′Ez ).

3. A schematic of a diffraction grating is shown in Figure 1, where a plane wave is incident on the left side and propagates in the x direction towards a detector located at a very far distance (practically at infinity) on the right. A number of N infinitesimally narrow slits (identified with n = 0, 1, 2,… , N −1) are spaced a distance d apart in the y direction, as shown, with each of them extending infinitely in the z direction. These slits let part of the plane wave through. As can be seen in the figure, when considering propagation to the detector in a direction at angle θ from the x -axis the contributions emanating from the slits travel different distances depending on their position on the grating.

a) Model the incident electric field associated with the plane wave with

Figure 1 – Schematic of a diffraction grating.

7

E = E0e− j kx−ωt( )ez , (3.1)

where k =ω c is the wave number and ω the angular frequency. Show that the amplitude of the electric at the detector will, up to a global phase factor, scale as

Ed ∝ E0sin Nkd sin θ( ) 2⎡⎣ ⎤⎦sin kd sin θ( ) 2⎡⎣ ⎤⎦

. (3.2)

b) Show that the intensity of the detected electric field will peak when d sin θ( ) = mλ, (3.3) where m is an integer and λ the wavelength of the incident wave. Solution. a) If the wave emanating through the first slit located at n = 0 travels a distance L to the detector, then contributions coming from the other slit will travel L + nd sin θ( ) . The diffracted field at the detector will then be proportional to

Ed ∝ E0e− jkL e− jnkd sin θ( )

n=0

N−1

∑ . (3.4)

We now calculate the following

Ed 1− e− jkd sin θ( )⎡⎣ ⎤⎦ ∝E0 e− jnkd sin θ( ) − e− j n+1( )kd sin θ( )⎡⎣ ⎤⎦

n=0

N−1

∑∝E0 1− e

− jNkd sin θ( )⎡⎣ ⎤⎦∝E0e

− jNkd sin θ( ) 2 ⋅2 j ⋅sin Nkd sin θ( ) 2⎡⎣ ⎤⎦,

(3.5)

and therefore

Ed ∝E0e− j N−1( )kd sin θ( ) 2 sin Nkd sin θ( ) 2⎡⎣ ⎤⎦

sin kd sin θ( ) 2⎡⎣ ⎤⎦. (3.6)

We thus find that, up to a global phase factor,

Ed ∝ E0sin Nkd sin θ( ) 2⎡⎣ ⎤⎦sin kd sin θ( ) 2⎡⎣ ⎤⎦

. (3.7)

8

b) The intensity is proportional to the square of the electric field, and at the detector it will be proportional to the square of equation (3.7). This signal will then be at a maximum when the denominator goes to zero, at which point the numerator is also zero. Using L’Hôpital’s rule we have lim

kd sin θ( ) 2→mπEd ∝ NE0. (3.8)

The condition for this to happen is therefore

12kd sin θ( ) = π d

λsin θ( )

= mπ (3.9)

or d sin θ( ) = mλ, (3.10) with m is an integer called the order number.

4. Start with Planck’s blackbody radiation law expressed as a function of the wavelength

F(λ,T ) = 2πc2h

λ 51

ehc λkT −1 (4.1)

and derive the corresponding law written as a function of the frequency ν = c λ instead. Solution. Since the energy in a given wavelength band dλ must be the same as that contained in the corresponding frequency band

dν = dνdλ

=d c λ( )dλ

= cλ 2 dλ,

(4.2)

we must have F ν ,T( )dν = F λ,T( )dλ . We then write

9

F ν ,T( ) = F λ,T( ) dλdν

= 2πc2h

λ 51

ehc λkT −1⋅ λ

2

c

= 2πhν3

c21

ehν kT −1.

(4.3)

5. (Prob. 51, Ch. 3, in Thornton and Rex.) Derive the relation for the recoil kinetic energy of the electron and its recoil angle φ in Compton scattering. Show that

Ke =

Δλ λ1+ Δλ λ

cot φ( ) = 1+ hνmc2

⎛⎝⎜

⎞⎠⎟ tan

θ2

⎛⎝⎜

⎞⎠⎟ .

(5.1)

Solution. By conservation of energy we know the electron’s recoil energy equals the energy lost by the photon, which with λ λ λ′ = +Δ yields

Ke =hcλ

− hc′λ

= hcΔλ′λ λ

= hcΔλ λλ 1+ Δλ λ( )

= Δλ λ1+ Δλ λ

hν .

(5.2)

Using the last two of equations (2.48) from the lecture notes for the conservation of linear momentum

hλ= h

′λcos θ( ) + pe2 cos φ( )

0 = h′λsin θ( )− pe2 sin φ( ),

(5.3)

we find when dividing the first by the second

10

cot φ( ) = ′λ λ − cos θ( )

sin θ( )

=Δλ λ +1− cos θ( )

sin θ( ) . (5.4)

But using the result for Δλ derived in the lecture notes for the Compton effect (i.e., equation (2.52)

Δλ = hmc

1− cos θ( )⎡⎣ ⎤⎦, (5.5)

and inserting it in the last of equations (5.4) gives us

cot φ( ) = hmcλ

+1⎛⎝⎜

⎞⎠⎟1− cos θ( )sin θ( )

⎣⎢

⎦⎥

= 1+ hvmc2

⎛⎝⎜

⎞⎠⎟

2sin2 θ 2( )2sin θ 2( )cos θ 2( )

⎣⎢

⎦⎥

= 1+ hvmc2

⎛⎝⎜

⎞⎠⎟ tan θ 2( ),

(5.6)

where we used λ = c ν and

sin a( )sin b( ) = 1

2cos a − b( )− cos a + b( )⎡⎣ ⎤⎦

sin a( )cos b( ) = 12sin a − b( ) + sin a + b( )⎡⎣ ⎤⎦.

(5.7)

6. (Prob. 28, Ch. 3, in Thornton and Rex.) We have waves in a one-dimensional box, such that the wave displacement ψ x,t( ) = 0 for x = 0 and x = L , where L is the length of the box, and

∂2ψ∂x2

− 1c2

∂2ψ∂t 2

= 0. (6.1)

a) Show that the solutions for equation (6.1) are of the form

ψ x,t( ) = a t( )sin nπ x

L⎛⎝⎜

⎞⎠⎟ , n = 1, 2, 3,… (6.2)

11

and a t( ) satisfies the (harmonic oscillator) equation

d 2a t( )dt 2

+ω n2a t( ) = 0, (6.3)

where ω n = nπc L is the angular frequency. b) Consider a general solution for equation (6.3) and show that ψ x,t( ) is composed in families of waves travelling in opposite directions. Clearly explain why the waves travel in a given direction. Solution. a) We start by taking spatial and time derivatives of equation (6.2) with

∂2ψ∂x2

= −a t( ) nπL

⎛⎝⎜

⎞⎠⎟2

sin nπ xL

⎛⎝⎜

⎞⎠⎟

∂2ψ∂t 2

=d 2a t( )dt 2

sin nπ xL

⎛⎝⎜

⎞⎠⎟ ,

(6.4)

which when inserted in the wave equation yield

d 2a t( )dt 2

+ω n2a t( ) = 0, (6.5)

with ω n = nπc L . b) Equation (6.5) is a typical linear second-order differential equation, which allows any or all of the following functions as solutions: sin ω nt( ) , cos ω nt( ) , and e± jωnt . Choosing the sine function as an example, we find that

ψ x,t( ) = An sin ω nt( )sin nπ xL

⎛⎝⎜

⎞⎠⎟n

= 12

An cosnπ xL

−ω nt⎛⎝⎜

⎞⎠⎟ − cos

nπ xL

+ω nt⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥n

= 12

An cos knx −ω nt( )− cos knx +ω nt( )⎡⎣ ⎤⎦n∑ ,

(6.6)

where we have introduced the wave number kn =ω n c = nπ L for convenience. We can now define the phases of the two families of cosine functions as

12

φn± x,t( ) = knx ±ω nt. (6.7)

If we concentrate on the spatial location of a specific maximum on the ‘+ ’ wave where φn

+ x,t( ) = 0 (i.e., when cos φn+ x,t( )⎡⎣ ⎤⎦ = 1), we then find that

x = ω n

knt = ct, (6.8)

which implies that this maximum on the wave is travelling at the speed c in the positive x -direction as time passes on. A similar calculation for the ‘− ’ wave reveals that x = −ct , and the waves travels in the negative x -direction at speed c as time passes on.

7. Consider an ideal gas at temperature T composed of molecules of mass m and of number density (i.e., number per volume) n . a) Assume that the gas is contained within a box and show that the number of molecules that strike a surface of area A on a wall in a time interval Δt is approximately given by nA kT m dt 2 , with k the Boltzmann constant. To simplify this calculation, consider that the wall is perpendicular to the x -axis and make the radical assumption that the x components of the velocities of all molecules have the same magnitude. That is, assume that all molecules have the same vx , with half the molecules moving to the right and the other half moving to the left. Consider that the equipartition of energy applies for this gas. b) Abandon the “radical assumption” of part a) and use instead a Maxwell-Boltzmann distribution for vx (i.e., see equation (1.1) above, but simplified to the one-dimension case) to obtain a more exact solution to the problem of part a). c) For molecular oxygen at the so-called standard temperature and pressure (STP: T = 273 K and p = 1 atm = 1.01×105 N/m2 ), use the result of part b) to calculate the average number density of particles that strike a 1 cm2 surface in 1 s? Solution. a) Under the suggested “radical assumption,” with all molecules having the same vx and half of them moving each way, the number density of molecules moving to the right is n 2 . In a time interval Δt , these molecules move a distance vxΔt to the right. Therefore, the molecules that hit the wall are precisely those right-moving molecules in the volume to the right of the dashed line in the accompanying figure, with length vxΔt and area A . The number of such molecules in this volume is nAvxΔt 2 .

13

Finally, since there is one degree of freedom associated with motions along the x -axis equipartition of energy implies that

12mvx

2 = 12kT , (7.1)

which upon insertion in the previous equation gives nA kT m dt 2 for the number of molecules that strike a surface of area A on a wall in a time interval Δt . b) The Maxwell-Boltzmann distribution for this problem is

f vx( ) = N m2πkT

e−mvx2 2kT , (7.2)

for −∞ < vx < ∞ , and the number of molecules having a velocity ranging from vx to vx + dvx being f vx( )dvx (as always, N is the total number of particles in the gas). The number Nw of these that hit the wall on the right within a time Δt is the number in the volume to the right of the dashed line in the figure and is found to be

Nw =AvxΔtV

f vx( )dvx0

∫ , (7.3)

with V the total volume of the box. Realizing that n = N V we write

Nw = nAΔtm

2πkTvxe

−mvx2 2kT dvx0

= nAΔt m2πkT

− kTme−mvx

2 2kT⎛⎝⎜

⎞⎠⎟ 0

= nAΔt kT2πm

.

(7.4)

c) We use the ideal gas law for this calculation, which states that P = nkT . (7.5) Inserting this relation in equation (7.4) we have

14

Nw =pAΔt2πkT

= 1.01×105 N/m2 ⋅10−4 m2 ⋅1 s2π ⋅1.38 ×10−23 J/K ⋅273 K ⋅ 32 ×1.66 ×10−27kg( )

= 2.8 ×1023.

(7.6)

8. (Prob. 60, Ch. 3, in Thornton and Rex.) a) Show that the maximum kinetic energy of the recoil electron in Compton scattering (refer to Figure 8 in Chap. 2 of the Lecture Notes) is given by

Ke,max = hν2hν mc2

1+ 2hν mc2⎛⎝⎜

⎞⎠⎟, (8.1)

where ν is the frequency of the incident photon. At what angles θ and φ does this occur? b) If we detect a scattered electron of 100 keV at φ = 0 , what was the energy of the incident photon? Solution. a) We find from the equation

Δλ = hmc

1− cos θ( )⎡⎣ ⎤⎦ (8.2)

that the maximum change in photon wavelength, which will coincide with a maximum electron recoil, happens when θ = π . Then from the equation derived in Problem 5

cot φ( ) = 1+ hνmc2

⎛⎝⎜

⎞⎠⎟ tan

θ2

⎛⎝⎜

⎞⎠⎟ (8.3)

the corresponding electron recoil angle must be φ = 0 . This is because we then have cot φ( ) = tan θ 2( ) = ∞ , which is a necessary condition for equation (8.3) to remain valid. At any rate, we have in this case

Δλ = 2hmc. (8.4)

15

Using another equation derived in Problem 5 for the kinetic energy of the recoil electron we write

Ke =

Δλ λ1+ Δλ λ

= 2hν mc2

1+ 2hν mc2hν ,

(8.5)

as required. b) We can rearrange equation (8.5) as follows

Ke,max 1+2hνmc2

⎛⎝⎜

⎞⎠⎟ =

2 hν( )mc2

(8.6)

or

2mc2

⎛⎝⎜

⎞⎠⎟ hν( )2 − 2Ke,max

mc2⎛⎝⎜

⎞⎠⎟hν( )− Ke,max = 0. (8.7)

This last equation is quadratic in hν that can be solved with Ke,max = 100 keV to yield

hν =

2Ke,max

mc2⎛⎝⎜

⎞⎠⎟±

2Ke,max

mc2⎛⎝⎜

⎞⎠⎟2

+ 4Ke,max2mc2

⎛⎝⎜

⎞⎠⎟

2 2mc2

⎛⎝⎜

⎞⎠⎟

, (8.8)

for which the only positive solution using the ‘+ ’ sign (a necessary condition for the energy of a photon…) is hν = 217 keV .