Math Methods for Physicist solutions Webber & Arfken

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Physics 451, 452, 725: Mathematical Methods

Russell Bloomer

1

University of Virginia

Note: There is no guarantee that these are correct, and they should not be copied

1

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The cross product of a and b will be one of three cases, which are perpendicular up, down and 0. The same is true for the vectors c and d. We can define vectors e = a × b and f = c × d. Since e and f are either up, down or 0, they are either 0◦ or 180◦to each other. By Eq. 1.442_{, the cross product is 0, due to a sine function. In the case of 0,}

the cross product is 0, because the 0 vector has no magnitude. X

1.3

Mathematical Methods for Physicists 1.4.16 Find the magnetic induction B given the following information :

F = q(v × B) v = ˆx,F q = 2ˆz − 4ˆy v = ˆy,F q = 4ˆx − ˆz v = ˆz,F q = ˆy − 2ˆx

The induction can be found using two cross products from the given information. ˆ x yˆ ˆz 1 0 0 Bx By Bz = 2ˆz − 4ˆy

From this matrix, we find By= 2 and Bz = 4.

Now we use the next relationship to find the last term. ˆ x yˆ ˆz 0 1 0 Bx By Bz = 4ˆx − ˆz

From this matrix, we find Bx= 1.

Combining these three yields B :

B = ˆx + 2ˆy + 4ˆz X

1.4

Mathematical Methods for Physicists 1.5.7 Show that :

a × (b × c) + b × (c × a) + c × (a × b) = 0 From the text, we know that :

a × (b × c) = b(a · c) − c(a · b)

We can infer the other two relationship will be the same. Make these substitutions, we find: b(a · c) − c(a · b) + c(b · a) − a(b · c) + a(c · b) − b(c · a) Because the dot product is associative:

b(a · c) − c(a · b) + c(a · b) − a(b · c) + a(b · c) − b(a · c) = 0 X

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1.5

Mathematical Methods for Physicists 1.5.8

For a vector A, which can be decomposed into a radial component Arand tangential component Atshow that :

(a) Ar= ˆr(A · ˆr)

We know from the definition of the vector that Ar = A · ˆr and that the radial magnitude multiplied by the radial

unit vector will yield the radial vector.

Ar= (Ar)ˆr X (b) At= −ˆr × (ˆr × A) ˆ r × A = ˆ r ˆt nˆ 1 0 0 Ar At 0 = Atnˆ (1.5) ˆ r × Atn =ˆ ˆ r ˆt nˆ 1 0 0 0 0 At = Atˆt = At X (1.6)

1.6

Mathematical Methods for Physicists 1.5.9 Show that if (A), B and C are coplanar, then

A · B × C = 0

We know from the text that A · B × C = AxByCz− AxBzCy+ AyBzCx− AyBxCz+ AzByCx− AzBxCy. If the vectors

are coplanar then one of the components can be set to 0 under a transportation. If one of the components is 0 and all the terms have one of each of the components, then the triple value is 0. Thus it is a necessary condition. Lets assume A · B × C = 0, now show that it is coplanar.

One way to write the dot product is with a cosine, and one way to write the cross product is with a sine. Since we know that it equals 0, then either sine or cosine or both terms are 0. If the cosine term is 0, then the cross product is perpendicular to A, which would place it in the same plane as B and C. Thus they are coplanar. Now if the sine term is 0, then (B) and C form a line. A third vector would then form a plane with that line. Thus it is sufficient for this condition.

A · B × C = 0 X

1.7

Mathematical Methods for Physicists 1.5.12 Show that :

(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C) First we will expand the left side.

(A × B) · (C × D) = ((AyBz− AzBy)ˆx + (AzBx− AxBz)ˆy + (AxBy− AyBx)ˆz) ·((CyDz− CzDy)ˆx + (CzDx− CxDz)ˆy + (CxDy− CyDx)ˆz) = (AyBz− AzBy)(CyDz− CzDy) + (AzBx− AxBz)(CzDx− CxDz) + (AxBy− AyBx)(CxDy− CyDx) = AyBzCyDz− AyBzCzDy− AzByCyDz+ AzByCzDy+ AzBxCzDx− AzBxCxDz− AxBzCzDx +AxBzCxDz+ AxByCxDy− AxByCyDx− AyBxCxDy+ AyBxCyDx

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= (AxCx+ AyCy+ AzCz)(BxDx+ ByDy+ BzDz) − (AxDx+ AyDy+ AzDz)(BxCx+ ByCy+ BzCz)

= (A · C)(B · D) − (A · D)(B · C) This is exactly the same as the right side of the equation. X

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Chapter 2 451: Problem Set 5

2.1

Problem 1. Show that :

Tij= Aij+ Sij+ cδijTkk

Where Aij= −Aji,antisymmetric, and Sij= Sji,symmetric, and Tkk,trace.

We can represent the antisymmetric and symmetric parts with two matrices.

Aij= 0 A12 A13 −A12 0 A23 −A13 −A23 0 (2.1) Sij= 0 S12 S13 S12 0 S23 S13 S23 0 (2.2)

Now combining any ij-term, one finds that as long as i 6= j then Tij= Aij+ Sij. Now we have to find for when i = j.

So the last term must vanish when i¬j, we know that the Kronecker delta function has this behavior. This yields : Tij= Aij+ Sij+ δij

Well this is not all of the value, because the delta function only has a value of one. There must be some constant to yield the proper result. The scalar value must have the property :

n

X

i,j=1

cδij= Tkk

Now combine this last term to the previous value.

Tij= Aij+ Sij+ cδijTkk

So what is important to note is that on any average n-dimensional term c = 1

n, but for an exact value there is a need

for an indency.

2.2

Problem 2. a.

Show for the epsilon tensor that :

εijkεilm= δjlδkm− δjmδkl

The definition for the Levi-Civita Symbol given in Arfken

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for a nonzero value textbf ej× ekmust be in the ei direction. Because of this then

εijkεilm= (ei· ej× ek) (ei· el× em)

= (ej× ek) · (el× em)

From Arfken (pg 33)

(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C) We apply this property to find

(ej× ek) · (el× em) = (ej· el) (ek· em) − (ej· em) (ek· el)

From definition Kronecker delta, eu· ev= δuv with this then

(ej· el) (ek· em) − (ej· em) (ek· el) = δjlδkm− δjmδkl

b.

Evaluate : εijkεijl

Using part a.

εijkεijl= δjjδkl− δjlδjk

Using inspection the last term is one when j=k=l, then the first term is also one therefore it is 0. Now sum over all of the terms. 3 X j=1 3 X k=l6=j=1 δjjδkl− δjlδjk= 3 X k=l=1 2δkl= 6

2.3

Problem 3. Prove that Ai;j− Aj;i= ∂Ai ∂xj − ∂Aj ∂xi

By the definition of a covariant derivative

Ai;j=

∂Ai

∂xj − AkΓ k

ij (2.4)

With switching of the indencies

Aj;i=

∂Aj

∂xi − AkΓ k ji

From Arfken, we know that Γkji= Γkij.

Ai;j− Aj;i= ∂Ai ∂xj − AkΓ k ij− ∂Aj ∂xi + AkΓ k ji = ∂Ai ∂xj − ∂Aj ∂xi + AkΓ k ji− AkΓkij = ∂Ai ∂xj − ∂Aj ∂xi + AkΓ k ji− AkΓkji = ∂Ai ∂xj − ∂Aj ∂xi + 0 = ∂Ai ∂xj − ∂Aj ∂xi

2.4

Problem 4. Not required to do.

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2.5

Problem5.

From the transformation properties of gij and the definition of the Christoffel Symbol Γijkobtain the transformation

law for Γijkgiven in class.

From Arfken (pg 154) Γsij= 1 2g ks ∂gik ∂xj + ∂g_jk ∂xi − ∂g_ij ∂xk (2.5) so then Γ0sij= 1 2g 0ks ∂g0ik ∂x0j + ∂g0jk ∂x0i − ∂g0ij ∂x0k

We also know that

g0kl= gij ∂xi ∂x0k ∂xj ∂x0l (2.6) g0kl= ∂x 0k ∂xi ∂x0l ∂xjg ij (2.7) let us examine on of part of the metric transformation and then we can use this to apply to the other parts.

∂ ∂x0kg 0 ij= ∂ ∂x0k gpm ∂xp ∂x0i ∂xm ∂x0j =∂gpm ∂xl ∂xl ∂x0k ∂xp ∂x0i ∂xm ∂x0j + gpm ∂2xp ∂x0k_∂x0i ∂xm ∂x0j + gpm ∂2xp ∂x0k_∂x0j ∂xm ∂x0i

The other two parts of the metric transformation are just permutations of the pervious one. So ∂ ∂x0ig 0 kj+ ∂ ∂x0jg 0 ki− ∂ ∂x0kg 0 ik= ∂xl ∂x0k ∂xp ∂x0i ∂xm ∂x0j ∂gml ∂xp + ∂gpl ∂xm − ∂gpm ∂xl + 2gpm ∂2xp ∂x0i_∂x0j ∂xm ∂x0k

With the definition of the Christoffel Symbol we find that Γ0sij= ∂x∂s xp ∂xl ∂x0i ∂xm ∂x0jΓ p lm+ ∂x0s ∂xp ∂2_xp ∂x0i_∂x0j (2.8)

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Chapter 3 451: Problem Set 6

3.1

Mathematical Methods for Physicists 3.2.4 a. For a + ib ↔ a b −b a

Show that addition (i) and multiplication (ii) are valid (i)

Let us define another matrix

c + id ↔

c d

−d c

Now add the two complex numbers and convert to matrix form. a + ib + c + id = (a + c) + i(b + d) ↔

a + c b + d −b − d b + c

Now add the two matrices directly. a b −b a + c d −d c = a + c b + d −b − d b + c

This does hold for addition. X

(ii)

First for a real scalar

α(a + ib) = aα + ibα ↔

aα bα

−bα aα

Now the scalar directly to the matrix α a b −b a = aα bα −bα aα

Next for two matrices.

a b −b a c d −d c = ac − bd ad + bc −bc − ad −bd + ac

For two complex numbers multiplied

(a + ib)(c + id) = (ac − bd) + i(ad + cb) ↔

ac − bd ad + bc −ad − bc ac − bd

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b.

Find the matrix for (a + ib)−1

(a + ib)−1↔

a b

−b a −1 From equation 3.50 in Arfken.

a(−1)ij = Cji |A| (3.1) So Cji= a −b b a |A| = a2+ b2 (a + ib)−1= a a2+b2 −b a2+b2 b a2+b2 a a2+b2 X

3.2

Mathematical Methods for Physicists 3.2.5 If A is an n × n matrix, show that

det(−A) = (−1)ndetA Let us look at the case of when the matrix is multiplied by a real scalar,α

αA = α1nA

detαA = det(α1n)detA

The determinate of a diagonal matrix is just the multiplication of the diagonal terms. In this case all the diagonal terms have a value of α.

det(α1n) = n

Y

i=1

αi= αn

Now substitute this term into the previous equation.

detαA = det(α1n)detA = αndetA

Then set α = −1

det(−A) = (−1)ndetAX

3.3

Mathematical Methods for Physicists 3.2.6 a.

Show that the square of the following matrix is 0 A = ab b2 −a2 _−ab Then A2 ab b2 −a2 _−ab ab b2 −a2 _−ab = a2b2− a2 b2 ab3− ab3 −a3_{b + a}3_b _−a2_b2_{+ a}2_b2 = 0 b.

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Show a numerical case for if C=A+B, in general det C 6= det A+det B Let A = 3 1 −9 −3 B = 2 1 −4 −2 Then C = 5 2 −13 −5

Now for the determinants

detA = 0 detB = 0 detC = 1

Therefore, det C 6= det A+det B X

3.4

Mathematical Methods for Physicists 3.2.15 a.

Show that [Mx, My] = iMz and so on.

Lets start with defining all the products of two M-matrices

MxMy= 1 √ 2   0 1 0 1 0 1 0 1 0   1 √ 2   0 −i 0 i 0 −i 0 i 0  = i   1 2 0 − 1 2 0 0 0 1 2 0 − 1 2   (3.2) MyMx= 1 √ 2   0 −i 0 i 0 −i 0 i 0   1 √ 2   0 1 0 1 0 1 0 1 0  = i   −1 2 0 − 1 2 0 0 0 1 2 0 1 2   (3.3) MxMz= 1 √ 2   0 1 0 1 0 1 0 1 0     1 0 0 1 0 1 0 1 0  = i   0 0 0 −_√i 2 0 i √ 2 0 0 0   (3.4) MzMx=   1 0 0 0 0 0 0 0 −1   1 √ 2   0 1 0 1 0 1 0 1 0  = i   0 −_√i 2 0 0 0 0 0 _√i 2 0   (3.5) MyMz= 1 √ 2   0 −i 0 i 0 −i 0 i 0     1 0 0 0 0 0 0 0 −1  = i   0 0 0 1 √ 2 0 − 1 √ 2 0 0 0   (3.6) MzMy=   1 0 0 0 0 0 0 0 −1   1 √ 2   0 −i 0 i 0 −i 0 i 0  = i   0 _√1 2 0 0 0 0 0 −_√1 2 0   (3.7)

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Now for the commutation relations [Mx, My] = MxMy− MyMx= i   1 0 0 0 0 0 0 0 −1  = iMz (3.8) [My, Mx] = MyMx− MxMy= i   −1 0 0 0 0 0 0 0 1  = −iMz (3.9) [Mz, Mx] = MzMx− MzMx= i    0 −_√i 2 0 i √ 2 0 − i √ 2 0 _√i 2 0   = iMy (3.10) [Mx, Mz] = MxMz− MzMx= i    0 √i 2 0 −_√i 2 0 i √ 2 0 −_√i 2 0   = −iMy (3.11) [My, Mz] = MyMz− MzMy= i    0 _√1 2 0 1 √ 2 0 1 √ 2 0 _√1 2 0   = iMx (3.12) [Mz, My] = MzMy− MyMz= i    0 −_√1 2 0 −_√1 2 0 − 1 √ 2 0 −_√1 2 0   = −iMx (3.13) This is equivalent as [Mi, Mj] = iεijkMkX b. Show that M2≡ M2 x+ M 2 y+ M 2 z = 213 Examining term M2x= 1 √ 2   0 1 0 1 0 1 0 1 0   1 √ 2   0 1 0 1 0 1 0 1 0  = 1 2   1 0 1 0 2 0 1 0 1   (3.14) M2y= 1 √ 2   0 −i 0 i 0 −i 0 i 0   1 √ 2   0 −i 0 i 0 −i 0 i 0  = 1 2   1 0 −1 0 2 0 −1 0 1   (3.15) M2z=   1 0 0 0 0 0 0 0 −1     1 0 0 0 0 0 0 0 −1  =   1 0 0 0 0 0 0 0 1   (3.16)

Now combine these terms 1 2   1 0 1 0 2 0 1 0 1  + 1 2   1 0 −1 0 2 0 −1 0 1  +   1 0 0 0 0 0 0 0 1  =   2 0 0 0 2 0 0 0 2  = 2   1 0 0 0 1 0 0 0 1  = 213X c. Show that M2 , Mi = 0 (3.17) Mz, L+ = L+ (3.18) L+ , L− = 2Mz (3.19)

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where

L+≡ Mx+ iMy

L−≡ Mx− iMy

Let’s begin withM2

, Mi = 0. We know that M2= 213, so M2 , Mi = 2 [13, Mi] = 2 (13Mi− Mi13) = 2 (Mi− Mi) = 2 (0) = 0X Now we have to find L+

L+≡ Mx+ iMy= 1 √ 2   0 1 0 1 0 1 0 1 0  + i √ 2   0 −i 0 i 0 −i 0 i 0   =√1 2   0 1 0 1 0 1 0 1 0  + 1 √ 2   0 1 0 −1 0 1 0 −1 0   =√1 2   0 2 0 0 0 2 0 0 0   NowMz, L+ = L+ Mz, L+ = MzL+− L+, Mz =   1 0 0 0 0 0 0 0 −1   1 √ 2   0 2 0 0 0 2 0 0 0  − 1 √ 2   0 2 0 0 0 2 0 0 0     1 0 0 0 0 0 0 0 −1   = √1 2   0 2 0 0 0 0 0 0 0  − 1 √ 2   0 0 0 0 0 −2 0 0 0   = √1 2   0 2 0 0 0 2 0 0 0   = L+X Lastly forL+ , L− = 2Mz. We first need L− L−≡ Mx− iMy= 1 √ 2   0 1 0 1 0 1 0 1 0  − i √ 2   0 −i 0 i 0 −i 0 i 0   =√1 2   0 1 0 1 0 1 0 1 0  − 1 √ 2   0 1 0 −1 0 1 0 −1 0   =√1 2   0 0 0 2 0 0 0 2 0  

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Now forL+ , L− = 2Mz L+ , L− = L+ L−− L−L+ = √1 2   0 2 0 0 0 2 0 0 0   1 √ 2   0 0 0 2 0 0 0 2 0  − 1 √ 2   0 0 0 2 0 0 0 2 0   1 √ 2   0 2 0 0 0 2 0 0 0   = 1 2   4 0 0 0 4 0 0 0 0  − 1 2   0 0 0 0 4 0 0 0 4   = 2   1 0 0 0 0 0 0 0 −1   = 2MzX

3.5

Mathematical Methods for Physicists 3.2.24 If A and B are diagonal, show that A and B commute. For two matrices to commute [A, B] = AB − BA = 0

If A and B are diagonal then they are must be of same dimension n We can define A = n X i,j=1 aijδij (3.20) B = n X i,j=1 bijδij (3.21) Let C = AB and D = BA C = n X i,j cij= n X i,j=1 n X k=1 aikbkjδikδkj D = n X i,j dij= n X i,j=1 n X k=1 bikakjδikδkj

We can reduce this, because of the double δs

n X i,j cij= n X i,j=1 aiibjjδij n X i,j dij= n X i,j=1 biiajjδij n X i,j cijδij= n X i=1 aiibii n X i,j dijδij= n X i=1 biiaii

So C and D are both diagonal matrices, also. Because each term of A and B are scalars, then aiibii = biiaii.

Therefore C = n X i,j cijδij= n X i=1 aiibii= n X i=1 biiaii= n X i,j dijδij= D C = AB = D = BA Therefore [A, B] = AB − BA = 0X

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3.6

Mathematical Methods for Physicists 3.2.33 Show that

detA−1= (detA)−1

Let’s begin with the hint that we are given to use the product theorem of section 3.2 in Arfken, which is det(AB) = detAdetB

so let A = A and B = A−1, then

detAA−1= detAdetA−1 det(13) = detAdetA

−1

By the definition of determinate det(13) = 1, and that the determinate is a scalar value.

det(13) = detAdetA−1

1 = detAdetA−1 (detA)−1= detA−1X

3.7

Mathematical Methods for Physicists 3.2.36 Find the inverse of

A =   3 2 1 2 2 1 1 1 4   (3.22) We begin with   3 2 1 2 2 1 1 1 4 1 0 0 0 1 0 0 0 1  

Now reorder the rows by inspection.

  1 1 4 3 2 1 2 2 1 0 0 1 1 0 0 0 1 0  

Now cancel all but the diagonal terms.   1 1 4 3 2 1 2 2 1 0 0 1 1 0 0 0 1 0  →   1 1 4 0 −1 −11 0 0 −7 0 0 1 1 0 −3 0 1 −2     1 1 4 0 1 11 0 0 −7 0 0 1 −1 0 3 0 1 −2  →   1 0 −7 0 1 11 0 0 −7 1 0 −2 −1 0 3 0 1 −2     1 0 −7 0 1 11 0 0 1 1 0 −2 −1 0 3 0 −1 7 2 7  →   1 0 0 0 1 0 0 0 1 1 −1 0 −1 11 7 − 1 7 0 −1 7 2 7   So A−1=   1 −1 0 −1 11 7 − 1 7 0 −1 7 2 7   (3.23)

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Now let’s check that this is indeed the inverse. AA−1=   3 2 1 2 2 1 1 1 4     1 −1 0 −1 11 7 − 1 7 0 −1 7 2 7  =   1 0 0 0 1 0 0 0 1   AA−1=   1 −1 0 −1 11 7 − 1 7 0 −1 7 2 7     3 2 1 2 2 1 1 1 4  =   1 0 0 0 1 0 0 0 1  

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Chapter 4 725: Problem Set 4

4.1

Problem 1

Use only known functions (and series) to improve the convergence ofP∞ n=1(−1)

n−1 zn

n so that the final series

con-verges absolutely at z = ±1. Where is the singularity and what happens to it? This summation is equivalent to ln (1 + z) =P∞

n=1(−1) n−1 zn

n. If both sides is multiplied by (1 + a1z) to find

(1 + a1z) ln (1 + z) = (1 + a1z) ∞ X n=1 (−1)n−1z n n = ∞ X n=1 (−1)n−1z n n + a1 ∞ X n=1 (−1)n−1z n+1 n Now making everything to the same power

(1 + a1z) ln (1 + z) = z + ∞ X n=2 (−1)n−1z n n + a1 ∞ X n=2 (−1)n−1 z n n − 1 = z + ∞ X n=2 (−1)n−1 1 n− a1 n − 1 zn n = z + ∞ X n=2 (−1)n−1 n (1 − a1) − 1 n(n − 1) zn Now set a1= 1. So check when z = −1

(z = −1) + ∞ X n=2 (−1)n−1 −1 n(n − 1) (z = −1)n= − 1 + ∞ X n=2 (−1)n−1 −1 n(n − 1) (−1)n = −1 + ∞ X n=2 (−1)2n−1 −1 n(n − 1) = −1 + ∞ X n=2 (−1)2n−2 1 n(n − 1)

Shifting the summation from n = 2 to n = 1, also (−1)2n−2_{= 1}

(z = −1) + ∞ X n=2 (−1)n−1 ₋₁ n(n − 1) (z = −1)n n = −1 + ∞ X n=1 1 n(n + 1) = −1 + 1 = 0

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The summation is known to be 1.1 This means the series converges absolutely at z = −1. Now for z = 1 (z = 1) + ∞ X n=2 (−1)n−1 −1 n(n − 1) (z = 1)n = 1 + ∞ X n=2 (−1)n−1 ₋₁ n(n − 1) (1)n

Now the absolute value of the summation, because if this converges absolutely, so does the actual summation. 1 + ∞ X n=2 1 n(n − 1) Shifting the summation from n = 2 to n = 1

1 + ∞ X n=2 1 n(n − 1) = 1 + ∞ X n=1 1 n(n + 1)

As earlier the summation is absolutely convergent. Therefore the function is absolutely convergent at z = 1. In the original summation there was a singularity at z = −1. This is removed by multiplying by a function, which at −1

equals 0. This suppresses the singularity. X

4.2

Problem 2

What is the threshold energy necessary to produce a ∆ at 1232 MeV in electron scattering according to e+p → e0+∆? The energy squared in the center of mass frame for the system is

s = (me+ mp)2

The threshold energy is the energy of the electron, which in four space Ee= pe(pp+ pe) √ s = pe· pp+ pe· pe √ s

Because of the small size of electrons mass pe· pe = m2e≈ 0. Also, pe· pp= 1₂ (me0+ m∆)2− m2e+ m2p. Again,

all m2e and m2e0 terms are set to 0. The energy becomes

Ee≈ m2∆+ 2m∆me0− m2_p 2√s = m 2 ∆+ 2m∆me0− m2_p 2 (m∆+ me0)

Now inserting the values for the masses Ee≈ 12322+ 2(.511)(1232) − 938.32 2(.511 + 938.3) = 340.1MeV X

4.3

Problem 3 Let A = ∂_∂x2f2, B = ∂2_f ∂x∂y, and C = ∂2_f

∂y2. We have a saddle point if B

2_{− AC > 0. For f (z) = u(x, y) + iv(x, y) apply}

the Cauchy-Riemann conditions and show that u and v do not have both a maximum or minimum in a finite region on the complex plane.

For u(x, y) A = ∂ 2 u ∂x2, B = ∂2u ∂x∂y, C = ∂2u ∂y2 1_{Arfken and Weber. Mathematical Methods for Physicists 5}th_{Ed. pg 316}

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Then B2− AC > 0 becomes ∂2u ∂x∂y 2 − ∂ 2 u ∂x2 ∂2 u ∂y2 > 0 Applying the Cauchy-Riemann Conditions2

∂2u ∂y2 → − ∂2v ∂x∂y → − ∂2u ∂x2 Then B2− AC > 0 simplifies to ∂2_u ∂x∂y 2 + ∂ 2_u ∂x2 2 > 0

This is always greater than 0, because u is a real value function. Therefore there are no maximum nor minimum for u. For v(x, y) A = ∂ 2 v ∂x2, B = ∂2v ∂x∂y, C = ∂2v ∂y2 Then B2_{− AC > 0 becomes} ∂2v ∂x∂y 2 − ∂ 2 v ∂x2 ∂2 v ∂y2 > 0 Applying the Cauchy-Riemann Conditions

∂2_v ∂x2 → − ∂2_u ∂x∂y → − ∂2_v ∂y2 Then B2_{− AC > 0 simplifies to} ∂2v ∂x∂y 2 + ∂ 2 v ∂y2 2 > 0

This is always greater than 0, because v is a real value function, because the imaginary term comes from the i in

front of v. Therefore there are no maximum nor minimum for v. X

4.4

Problem 4

Find the analytic function f (z) = u(x, y) + iv(x, y) a

For u = x3− 3xy

From the Cauchy-Riemann Conditions

∂u ∂x = 3x

2

− 3y2= ∂v ∂y From this, integrate with respect to y to find v

Z ∂v =

Z

3x2− 3y2_{∂y → v = 3x}2

y − y3+ c0 where c0 is a constant of integration. So

w(x, y) = u(x, y) + iv(x, y) + ic0 = x3− 3xy2+ 3ix2y − iy3+ c

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where c = ic0. Now in the form of f (z), where z is the form z = x + iy w(x, y) = x3− 3xy2+ 3ix2y − iy3+ c = (x + iy) x2+ 2ixy − y2 + c

= (x + iy) (x + iy)2+ c → f (z) = z3+ c X b

For v = e−ysin x

Beginning with the Cauchy-Riemann Conditions ∂v ∂y = −e

−y

sin x =∂u ∂x Integrating this with respect to x to find u

Z ∂u =

Z

−eysin x∂x → u = e−ycos x + k where k is a constant of integration. So

w(x, y) = u(x, y) + iv(x, y) = e−ycos x + ie−ysin x + k Now in the form of f (z)

w(x, y) = e−ycos x + ie−ysin x + k = e−y(cos x + i sin x) + k

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Chapter 5 725: Problem Set 5

5.1

Problem 1

Starting with the Euler integral, analytically continue the Gamma function into the left-hand complex plane and determine its poles with their residues.

The Euler integral is

Γ(z) = Z∞

0

e−ttz−1dt (5.1)

To extend into the the left hand side of the complex plane. Consider Γ(z + 1) = Z ∞ 0 e−ttzdt = −ett−z ∞ 0 + Z ∞ 0 e−tztz−1dt = zΓ(z) So consider z = −z Γ(−z + 1) = Z∞ 0 e−tt−zdt = −ett−z ∞ 0 + Z ∞ 0 e−t(−z)t−z−1dt = −zΓ(−z)

This can be continued indefinitely to the left. The poles can be found know that a 0 has to be in the denominator. This occurs for z = 0, −1, −2, . . .. The value at these poles can be found from equation 6.47

f(n)(zo) = n! 2πi I _{f (z)dz} (z − zo)n+1 (5.2)

For the Gamma function from the Euler integral form, f (z) = e−t. The value of f(n)_{(0) = ±1 depending if n is even}

or odd. This becomes f(n)_{(0) = (−1)}n_{. Therefore the residues are} (−1)n n!

5.2

Problem 2

What kind of analytic functions are sin z, 1/ sin z and what are their singularities? Let sin z = sin(x + iy). This can be expanded to

sin(x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y

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At large values for y, the function approaches ∞, so the behavior at large z must be examined. So replace z =_w1, in the summation representation

sin z = ∞ X n=0 (−1)nz2n+1 (2n + 1)! = ∞ X n=0 (−1)n (2n + 1)!w2n+1

So as z → ∞, w → 0 means there are “poles” at z = ∞. The problem is that the largest “poles” has power of ∞. This means that sin z has an essential singularity at z = ∞ X

Now for _{sin z}1 . This function has poles at z = ±nπ, where n is an integer. The pole expansion can be found using the method form equation 7.15.

f (z) = f (0) + ∞ X n=1 bn(z − an) −1 + a−1n (5.3)

For this f (z), the above expansion must include all negative n. For f (z), bn = 1/ cos z|zO. This means bn= ±1.

When n is even b2n= 1 and -1 for odd.

1 sin z = 1 z+ ∞ X n=1 (−1)n 1 z − nπ + 1 nπ + 1 z + nπ+ 1 −nπ =1 z + ∞ X n=1 (−1)n 1 z − nπ+ 1 z + nπ This reduces to 1 sin z = 1 z + ∞ X n=1 (−1)n 1 z − nπ+ 1 z + nπ =1 z + ∞ X n=1 (−1)n 2z z2_{− (nπ)}2 X

5.3

Problem 3 Evaluate the seriesP∞

n=−∞ 1

n2+2−2 at least approximately and exactly if possible.

An approximation of the sum is with an integral.

∞ X n=−∞ 1 n2_{+ 2}−2 ≈ Z ∞ −∞ dx x2_{+ 2}−2

This integral can be evaluated in the complex plane on the semi circle of the upper plane. Z ∞ −∞ dx x2_{+ 2}−2 = I dz (z + 1/2i) (z − 1/2i) The only pole in the upper plane is zo= 1/2i. The integral becomes

I

dz

(z + 1/2i) (z − 1/2i)= 2πi X f (z) zo = 2πi 1 1/2i + 1/2i = 2π

This value is a good approximation of the real value, which is 2π coth π/2. The approximation is within 10% of the

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5.4

Problem 4 Evaluate the integrals

Z ∞ 0 dx 1 + 2x2_{+ x}4, Z ∞ 0 dx 1 + 2x + 2x2_{+ x}3 (5.4)

For the first integral, the function is an even function it can be extended to the negative axis. So Z ∞ 0 dx 1 + 2x2_{+ x}4 = 1 2 Z ∞ −∞ dx 1 + 2x2_{+ x}4

So then map it in complex plane to find

1 2 Z ∞ −∞ dz 1 + 2z2_{+ z}4

Now factor into its roots

1 2 Z ∞ −∞ dz 1 + 2z2_{+ z}4 = 1 2 Z ∞ −∞ dz (z − i)2(z + i)2

There are roots of second power. Applying equation 6.461 _{for z = i, because the contour is only of the upper plane.}

1 2 Z ∞ −∞ dz (z − i)2(z + i)2 = 2πi 2 " −2 (zo+ i)3 zo=i # = πi 1 4i =π 4 X

For the second integral, the use of natural logarithm as discussed in lecture will be applied. The integral can be written as Z∞ 0 dx 1 + 2x + 2x2_{+ x}3 = Z ∞ 0 dx (x + 1)x +1+i √ 3 2 x +1−i √ 3 2 From lecture I f (z) ln zdz = − Z ∞ 0 f (x)dx ln xdx + Z ∞ 0 f (x)dx ln (x + 2πi)dx = 2πi Z ∞ 0 f (x)dx Or in terms of the real integral

Z ∞ 0 f (x)dx = 1 2πi I f (z) ln zdz = −XRes f (z) ziln zi

where zi are the poles. The poles are first order, so

Z ∞ 0 dx (x + 1)x +1+i √ 3 2 x +1−i √ 3 2 = 1 2πi Z ∞ 0 ln zdz (z + 1)z +1+i√3 2 z +1−i√3 2 =XRes f (z) ziln zi = ln (−1) −1 + 1+i√3 2 −1 +1−i√3 2 + ln−1−i √ 3 2 −1−i√3 2 + 1 −1−i√3 2 + 1−i√3 2 + ln−1+i √ 3 2 −1+i√3 2 + 1+i√3 2 −1+i√3 2 + 1 = −π √ 3 9 + π√3 9 + iπ 3 + π√3 9 − iπ 3 =π √ 3 9 X

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Chapter 6 725: Problem Set 6

6.1

Problem 1 Solve the ODE (1 + x2_)y0

+ (1 − x)2_{y(x) = xe}−x

with initial condition y(0) = 0 First solve the homogenous equation

(1 + x2)y0+ (1 − x)2y(x) = 0 Writing y0=dy_dx. Placing all the y’s on one side and the x’s the other

dy y = − (1 − x)2 1 + x2 dx → Z dy y = − Z (1 − x)2 1 + x2 dx − ln C1+ ln y = −x + ln 1 + x2→ y = C1(1 + x2)e −x

So the homogenous solution is

yh= C1(1 + x2)e−x (6.1)

Now solve for the particular solution with the guess y = Axe−x+ Be−x (1 + x2)y0+ (1 − x)2y = xe−x→ (1 + x2)×

(−Ax + A − B) e−x+ (1 − x)2(Ax + B) e−x= xe−x

Since e−xappears in every term it can be cancelled. Now group everything in terms of powers of x −Ax3+ Ax3= 0 → Nothing is learned

(A − B)x2− 2Ax2+ Bx2= 0 → A = 0 −Ax + Ax − 2Bx = x → B = −1/2

A − B + B = 0 → Nothing is learned So the particular solution is

yp= −

1 2e

−x

(6.2) Finally solve for the boundary condition

y(0) = 0 → yh(0) + yp(0) = 0 y = C1(1 + x2)e−x− 1 2e −x → C1(1 + 02)e−0− 1 2e −0 = 0 C1− 1 2= 0 → C1= 1 2 Then the solution is

y =1 2(1 + x 2 )e−x−1 2e −x → y =1 2x 2 e−x X

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6.2

Problem 2 Solve the ODE y00−5

6y 0

+ 9

25y(x) = 0 with initial conditions y(1) = 0, y 0

(1) = 2.

Let’s guess that the solution has the form y = erx_{, so the differential equation becomes}

r2erx−5 6re rx + 9 25e rx = 0 r2−5 6r + 9 25= 0 Solving for r r = 5 12± r 25 144− 9 25 = 5 12± i 60 √ 671 So the solution as the form

y = C1e( 5 12+ i 60 √ 671)x + C2e( 5 12− i 60 √ 671)x

For ease of calculation

ω1= 5 12+ i 60 √ 671, ω2= 5 12− i 60 √ 671 Now solve for the boundary conditions.

y(1) = 0 → C1eω1+ C2eω2= 0 → C2= −C1eω1 −ω₂

Now the second condition

y0(1) = 2 → C1ω1eω1+ C2ω2eω2= 2 C1ω1eω1− C1eω1 −ω₂ ω2eω2= 2 C1 ω1eω1− ω2eω1 −ω₂+ω2_{= 2} C1= 2e−ω1 ω1− ω2 Therefore C2= −C1eω1−ω2→ C2= −2e−ω2 ω1− ω2

The final solution is then

y = 2e −ω1 ω1− ω2 eω1x₋ 2e ω2 ω1− ω2 eω2x y = ( −60ie−125− i√671 60 + 5 12+ i√671 60 x √ 671 +60ie −5 12+i √ 671 60 + ₅ 12−i √ 671 60 x √ 671 X

6.3

Problem 3

Solve (x2_{− 2x + 1)y}00_{− 4(x − 1)y}0_{− 14y(x) = x}3_{− 3x}2_{+ 3x − 8 for the general solution. Then adjust it to the initial}

condition y(0) = 9/20, y0(0) = 11/10

This differential equation can be written as

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Define z ≡ (x − 1), then differential becomes

z2y00− 4zy0− 14y = z3_{− 7}

The homogenous equation is just Euler differential equation with y = zp

zp(p(p − 1) − 4p − 14) = 0 → p2− 5p − 14 = 0 The solution for p is p = 7, −2. So the homogenous solution is then

yh= C1z7+ C2z−2 (6.3)

The particular solution is of the form

y = Az3+ Bz2+ Cz + D

Now, because the second derivative is multiplied by z2 _{and the first derivative is multiplied by z, all terms but the}

constant and z3are zero, ie B = C = 0. So

6Az3− 12Az3− 14Az3− 14D = z3− 7 A = −1

20, D = 1 2 Therefore the particular solution is

yp= −1 20z 3 +1 2 (6.4)

Finally for the constants with the boundary conditions y(−1) = 9/20 and y0(−1) = 11/10. For y(−1) = 9/20 −C1+ C2+ 1/20 + 1/2 = 9/20 → C2= C1− 1/10

Now, y0(−1) = 11/10

7C1+ 2 (C1− 1/10) − 3/20 = 11/10 → C1= 29/180

Then

C2= 29/180 − 1/10 → C2= 11/180

The solution becomes when z = x − 1 is substituted back in y = 29 180(x − 1) 7 + 11 180(x − 1) −2 − 1 20(x − 1) 3 +1 2 X (6.5)

6.4

Problem 4

Determine the general solution of the ODE xy00= y(x)y0 that depends on two integration constants. Find another solution that cannot be obtained from the general solution. Explain.

The right hand side can be written as

yy0= d dx 1 2y 2

Integrating both sides

Z xy00= Z d dx 1 2y 2 → Z xy00= 1 2y 2 + C10

The left hand side can be done by partial integration with u = x, du = dx, v = y0 and dv = y00, so Z

xy00= xy0− Z

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Combining the two constants in to one, then xy0− y =1 2y 2 + C1→ dy 1 2y2+ y + C1 = dx x Integrating both sides

Z dy 1 2y2+ y + C1 = Z dx x 2 tan−1 1 + y √ −1 + 2C1 _√ −1 + 2C1= ln x + C2 Solving for y tan−1 1 + y √ −1 + 2C1 =p−1/2 + C1(ln x + C2) 1 + y √ −1 + 2C1 = tanhp−1/2 + C1(ln x + C2) i y = −1 +√−1 + 2C1tan hp −1/2 + C1(ln x + C2) i

The solution that depends on two constant is

y = −1 +√−1 + 2C1tan

hp

−1/2 + C1(ln x + C2)

i

(6.6) The third solution that cannot be obtained form the general solution is y = constant. This solution cause the differential equation to become trivial, because y0 = 0. Therefore, y00= 0. This is a solution because 0 times any

Math Methods for Physicist solutions Webber & Arfken

Physics 451, 452, 725: Mathematical Methods

Russell Bloomer

University of Virginia

Note: There is no guarantee that these are correct, and they should not be copied

Contents

451: Problem Set 2

451: Problem Set 5

451: Problem Set 6

725: Problem Set 4

725: Problem Set 5

725: Problem Set 6

Special Functions

Chapter 1

451: Problem Set 2

1.1

1.2

1.3

1.4

1.5

1.6

1.7

Chapter 2

451: Problem Set 5

2.1

2.2

2.3

2.4

2.5

Chapter 3

451: Problem Set 6

3.1

3.2

3.3

3.4

3.5

3.6

3.7

Chapter 4

725: Problem Set 4

4.1

4.2

4.3

4.4

Chapter 5

725: Problem Set 5

5.1

5.2

5.3

5.4

Chapter 6

725: Problem Set 6

6.1

6.2

6.3

6.4

Appendix A

Special Functions

Index