Announcements - web.mit.eduweb.mit.edu/8.07/www/slides/sl09-f19.pdfpip e (sides a and b)is grounded, but one end, at x = 0, is main tained at a sp eci ed p oten tial V 0 (y; z), as

8.07 Lecture Slides 9

October 2, 2019

ELECTRIC POTENTIAL:

SEPARATION OF VARIABLES

(CARTESIAN AND SPHERICAL)

Announcements

Quiz 1 is this coming Monday, October 7.

All material referenced in Problem Sets 1, 2, and 3 will be afair subject for problems on the quiz.

Calculators will not be needed and will not be allowed.

I have posted 10 Practice Problems for Quiz 1, with solutions,on the Exams tab of the class website.

I have also posted a Formula Sheet for Quiz 1 on the Examstab. It is a very complete formula sheet, intended to be atool for review. A copy of the formula sheet will be includedwith the quiz, so of course you need not memorize any ofthe formulas. But I recommend that you go over the formulasheet carefully. If you understand the meaning of each of theformulas, you are very well-prepared for the quiz.

Alan Guth

Massachusetts Institute of Technology

8.07 Lecture Slides 9, October 2, 2019 {1{

One of the problems on the quiz will be taken verbatim (orat least almost verbatim) from either the problem sets (extracredit problems are possible), or from the practice problems.

Yitian Sun will run a Review Session for the quiz on Sunday(Oct 6) from 3:00 to 5:00 pm, in Room 2-131.

Recitations this Thursday, with Marin Solja�ci�c, will focus onreviewing for the Quiz.

Upcoming oÆce hours:

Today, 5:00 - 6:00 pm: Me, Room 6-322. Please come!

Thurs, Yitian, 5:30 - 6:30 pm, Room 8-320

Fri, Yitian, 1:30 - 3:30 pm, Room 8-320

Good luck on the quiz.Alan Guth



Announcements:Special Lecture by Marin Soljacic

One week from today, on Wednesday, October 9, Marin will givea special lecture on modern developments in electricity andmagnetism.

The lecture is to broaden your horizons | the material will notbe included in problem sets or quizzes.

The lecture will not be video-recorded, so if you want to see it,you have to come to the lecture.

Alan Guth



Review of Lecture 8

Separation of Variables in Cartesian Coordinates

How to solve r2V = 0?

In Cartesian coordinates,

@2V

@x2+@2V

@y2+@2V

@z2= 0 :

Try a solution of the form

V (x; y; z) = X(x)Y (y)Z(z) :

(Even if this is not general enough, sums of solutions of this form might work.)

Alan Guth



Review of Lecture 8

Laplace's equation implies

Y Zd2X

dx2+XZ

d2Y

dy2+XY

d2Z

dz2= 0 :

Now divide by V = XY Z:

1

X

d2X

dx2| {z }C1

+1

Y

d2Y

dy2| {z }C2

+1

Z

d2Z

dz2| {z }C3

= 0 :

Since the 1st term depends only on x, the 2nd depends only on y, and the 3rddepends only on z, each term must be a constant. So

C1 + C2 + C3 = 0 :

Alan Guth



Review of Lecture 8

General Form for X(x)

We look �rst at X(x). The other two functions Y (y) and Z(z) are analogous.We have

d2X

dx2= C1X :

There are three possibilities:

1) If C1 > 0, write C1 � �2. Then the general solution can be writ-

ten as X(x) = Ae�x +Be��x , or X(x) = A sinh(�x) +B cosh(�x) ,

where A and B are constants.

2) If C1 < 0, write C1 = ��2. The general solution can then be written as

X(x) = Aei�x +Be�i�x , or X(x) = A sin(�x) +B cos(�x) , where

A and B are constants.

3) If C1 = 0, then the general solution can be written as

X(x) = A+Bx .

{6{

Review of Lecture 8

Separation of Variables Example:Infinitely Long Rectangular Pipe

This is GriÆths's Example 3.5, pp. 134{136.

An in�nitely long rectangular metal pipe (sides a and b) is grounded, but oneend, at x = 0, is maintained at a speci�ed potential V0(y; z), as indicatedin Fig. 3.22. Find the potential inside the pipe.

Alan Guth



Review of Lecture 8

Choosing for each variable the only solutions to the di�erential equation withthe right boundary conditions, the full solution becomes

V (x; y; z) = sinn�y

asin

m�z

be� x :

Try a sum of solutions of this form, with as yet unspeci�ed coeÆcients:

V (x; y; z) =1Xn=1

1Xm=1

Vnm sinn�y

asin

m�z

bexp

"�

r�n�a

�2+�m�

b

�2x

#:

Alan Guth



Review of Lecture 8

Summary: Blackboard DiscussionContinuing the Calculation

For any choice of the Vnm's, the solution

1) satis�es r2V = 0;

2) satis�es boundary conditions on the sides of the pipe, at y = 0; a andz = 0; b;

3) satis�es the boundary condition V ! 0 as x!1.

We must still arrange to satisfy the boundary condition at x = 0,

V (0; y; z) = V0(y; z) :

So we need to choose the Vnm's so that

V (0; y; z) =1Xn=1

1Xm=1

Vnm sinn�y

asin

m�z

b= V0(y; z) :

{9{

Review of Lecture 8

So we need to choose the Vnm's so that

V (0; y; z) =1Xn=1

1Xm=1

Vnm sinn�y

asin

m�z

b= V0(y; z) :

Can we do this? YES!

This is a Fourier series. Fourier's theorem (often called Dirichlet's theorem in itsmost rigorous formulation) says that this series is complete. Any piecewisecontinuous bounded function, satisfying the boundary conditions that itvanishes when y = 0 or y = a or z = 0 or z = b, can be expanded this way.

Alan Guth



Review of Lecture 8

We will not prove completeness, which is not an easy thing to prove | so wewill leave it to math classes.

But once one knows that V0(y; z) can be written as such a series, it is easy to�nd the Vnm's that do it. We use the fact that

Z a

0

sinn0�y

asin

n�y

ady =

1

2a Æn0n :

For n0 = n, the integral is evaluated by recognizing that the average valueof sin2 w, when averaged over any number of half periods of sinw, is equalto 1/2. For n0 6= n, trig identities can be used to convert the integral intointegrations of cosw over integral numbers of periods, which vanish.

Alan Guth



Review of Lecture 8

So we integrate both sides,

Z a

0

sinn0�y

a

Z b

0

sinm0�z

b

1Xn=1

1Xm=1

Vnm sinn�y

asin

m�z

bdz dy

=

Z a

0

sinn0�y

a

Z b

0

sinm0�z

bV0(y; z) dz dy :

Using Z a

0

sinn0�y

asin

n�y

ady =

1

2a Æn0n

and the analogous formula for z, we �nd

1

4ab Vn0m0 =

Z a

0

sinn0�y

a

Z b

0

sinm0�z

bV0(y; z) dz dy :

Alan Guth



Review of Lecture 8

1

4ab Vn0m0 =

Z a

0

sinn0�y

a

Z b

0

sinm0�z

bV0(y; z) dz dy :

So we can write the �nal solution as

V (x; y; z) =1Xn=1

1Xm=1

Vnm sinn�y

asin

m�z

bexp

"�

r�n�a

�2+�m�

b

�2x

#;

where

Vnm =4

ab

Z a

0

sinn�y

a

Z b

0

sinm�z

bV0(y; z) dz dy :

We have dropped the primes in the formula above, which we can do becausethe equation no longer contains any instances of the original indices m and n.

Alan Guth



Separation of Variablesfor Spherical Coordinates:

Traceless Symmetric Tensor Approach

Laplace's Equation in Spherical Coordinates:

r2'(r; �; �) =

1

r2@

@r

�r2@'

@r

�+

1

r2r

2ang ' = 0 ;

where

r2ang ' �

1

sin �

@

@�

�sin �

@'

@�

�+

1

sin2 �

@2'

@�2:

Alan Guth



Radial part can be written 2 alternative ways:

1

r2@

@r

�r2@'

@r

�=

1

r

@2

@r2(r') :

Seek solution of form:

'(r; �; �) = R(r)F (�; �) :

Write Laplace's equation as

0 =r2

RFr2' =

1

R

d

dr

�r2dR

dr

�+

1

Fr2ang F :

Since 1st term depends only on r, and 2nd only on � and �, each must be aconstant:

1

Fr2ang F = Cang ;

1

R

d

dr

�r2dR

dr

�= �Cang :

Alan Guth



Expansion of F (�; �)

We wish to �nd the most general solution to

r2ang F = Cang F :

Vocabulary: F is an eigenfunction of the operator r2ang, with

eigenvalue Cang.

Instead of � and �, we will indicate directions by the unit vectorn:

n = sin � cos� e1 + sin � sin� e2 + cos � e3 ;

where ei is the unit vector in the i-direction (where i = 1; 2;or 3).

Alan Guth



Write F (�; �) as a power series in the components of n. (This is equivalent tothe standard spherical harmonic expansion, known to converge for square-integrable piece-wise continuous functions.)

F (n) = C(0) + C(1)i ni + C

(2)ij ninj + : : :+ C

(`)i1i2:::i`

ni1ni2 : : : ni` + : : : ;

where repeated indices are summed from 1 to 3 (as Cartesian coordinates).

The general term C(`)i1i2:::i`

ni1ni2 : : : ni` is indicated by the label `. It has `indices i1, i2, : : : , i`, which are each summed from 1 to 3.

Vocabulary: The C(`)i1i2:::i`

are called tensors, and the number of indices is called

the rank of the tensor. C(0), C(1)i , and C

(2)ij are special cases of tensors |

they can also be called a scalar, a vector, and a matrix.

Alan Guth



Restrictions on the C(`)i1i2:::i`

Without loss of generality, we can insist:

1) C(`)i1i2:::i`

are symmetric under any reordering of the indices:

C(`)i1i2:::i`

= C(`)j1j2:::j`

;

where fj1; j2; : : : ; j`g is any permutation of fi1; i2; : : : ; i`g.

2) The C(`)i1i2:::i`

are traceless: if any two indices are set equal to each otherand summed, the result is equal to zero. Since they are symmetric, it doesnot matter which indices are summed.

C(`)i1i2:::i`�2jj

= 0 :

Alan Guth



Evaluation of r2F`(n)

De�ne

F`(n) = C(`)i1i2:::i`

ni1 ni2 : : : ni` :

That is, F`(n) is the `'th term of the series.

It is useful to introduce a new variable r, and de�ne

~r � rn � xiei = x1e1 + x2e2 + x3e3 :

We further de�ne

F`(~r ) � C(`)i1i2:::i`

xi1xi2 : : : xi` = r`F`(n) :

Why do we do all this? Because

r2angF`(~r ) = 0 :

{19{

[r2angF`(~r ) = 0 was derived on the blackboard.]

So what does this tell us about r2angF`(n)?

0 = r2F`(~r ) =1

r2@

@r

�r2@F`(~r )

@r

�+

1

r2r2ang F`(~r )

=1

r2d

dr

�r2dr`

dr

�F`(n) +

1

r2r`r2

ang F`(n)

= r`�2h`(`+ 1)F`(n) +r

2ang F`(n)

i;

and therefore

r2ang F`(n) = �`(`+ 1)F`(n) :

Alan Guth



General Solution to Laplace's Equationin Spherical Coordinates

Recall

1

Fr2ang F = Cang ;

1

R

d

dr

�r2dR

dr

�= �Cang :

We now know that Cang must have the form �`(`+ 1), where ` is an integer.

Look at radial equation:

d

dr

�r2dR

dr

�= `(`+ 1)R :

Alan Guth



Look at radial equation:

d

dr

�r2dR

dr

�= `(`+ 1)R :

Try a solution R(r) = rp. We �nd consistency provided that p(p + 1) =`(` + 1). This quadratic equation has two roots, p = ` and p = �(` + 1).Since we found two solutions to a second order linear di�erential equation,we know that any solution can be written as a linear sum of these two.Thus we can write

R`(r) = r` or R`(r) =1

r`+1:

The most general solution to Laplace's equation, in spherical coordinates,can then be written as

�(~r ) =1X`=0

C(`)i1i2:::i`

r` +C0(`)i1i2:::i`

r`+1

!ni1ni2 : : : ni` ;

where C(`)i1i2:::i`

and C0(`)i1i2:::i`

are arbitrary traceless symmetric tensors, and~r = rn.

{22{

Connection to Standard Y`m(�; �)'s

Later we will be precise, but for now we state that the `'th term,

F`(n) � C(`)i1i2:::i`

ni1ni2 : : : ni` is equivalent to the sum over allY`m's for a given `.

Since m runs from �` to `, there are 2` + 1 possible values, sowe expect that it must take 2`+ 1 independent parameters to

specify the general traceless symmetric tensor C(`)i1i2:::i`

. Let'scheck that.

[The derivation was given on the blackboard.]

Alan Guth



Trace Decomposition Theorem

Any symmetric matrix Si1:::i` can be uniquely written in the form

Si1:::i` = S(TS)i1:::i`

+ Symi1:::i`

�Mi1:::i`�2Æi`�1;i`

�;

where S(TS)i1:::i`

is a traceless symmetric tensor, Mi1:::i`�2 is asymmetric tensor, and

Symi1:::i`

[xxx]

means to symmetrize the expression xxx in the indices i1 : : : i`.Alan Guth



Documents

Announcements - web.mit.eduweb.mit.edu/8.07/www/slides/sl09-f19.pdfpip e (sides a and b)is grounded, but one end, at x = 0, is main tained at a sp eci ed p oten tial V 0 (y; z), as