Mathematical Preliminaries
1.1.1 (a) Ifu n < A/n p the integral test showsP n u n converges forp >1. (b) Ifu n > A/n,P n u n diverges because the harmonic series diverges.
1.1.2 This is valid because a multiplicative constant does not affect the conver- gence or divergence of a series.
1.1.3 (a) The Raabe testP can be written 1 +(n+ 1) ln(1 +n −1 ) lnn
This expression approaches 1 in the limit of large n But, applying the
Cauchy integral test, Z dx xlnx = ln lnx, indicating divergence.
(b) Here the Raabe testP can be written
1 + 1 n ả +ln 2 (1 +n −1 ) ln 2 n , which also approaches 1 as a large-nlimit But the Cauchy integral test yields Z dx xln 2 x=− 1 lnx, indicating convergence.
1.1.5 (a) Divergent, comparison with harmonic series.
(b) Divergent, by Cauchy ratio test.
(e) Divergent, comparison with 1 2 (n+ 1) −1 or by Maclaurin integral test. 1.1.6 (a) Convergent, comparison withζ(2).
(b) Divergent, by Maclaurin integral test.
(c) Convergent, by Cauchy ratio test.
1.1.7 The solution is given in the text.
1.1.8 The solution is given in the text.
1.1.10 In the limit of largen,u n+1 /u n = 1 + 1 n+O(n −2 ).
Applying Gauss’ test, this indicates divergence.
1.1.11 Lets n be the absolute value of thenth term of the series.
The series converges because the sequence lnn increases at a slower rate than n, resulting in s(n+1) being less than s(n) and the limit of s(n) approaching 0 as n approaches infinity However, since the terms of this series are greater than those of the harmonic series, it is not absolutely convergent.
This new series is created by combining adjacent terms of the same sign from the original series, resulting in an alternating series of decreasing terms that converge to zero as a limit.
2n+ 4, this series converges With all signs positive, this series is the harmonic series, so it is not aboslutely convergent.
(c) Combining adjacent terms of the same sign, the terms of the new series satisfy
The general form of these relations is
An upper limit to the left-hand side member of this inequality is 2/(n−1).
The terms of the new series decrease towards zero, indicating that the original series converges Since all terms are positive, the original series resembles the harmonic series, which is not absolutely convergent.
1.1.12 The solution is given in the text.
To derive the nth term of ζ(2) - c₁α₁ - c₂α₂, we need to select values for c₁ and c₂ such that the numerator remains independent of n when expressed over the common denominator n²(n + 1)(n + 2) The suitable values for c₁ and c₂ that fulfill this requirement are c₁ = c₂ = 1, leading to the conclusion that ζ(2) can be expressed as ζ(2) = α₁ + α₂.
Keeping terms throughn= 10, this formula yieldsζ(2)≈1.6445; to this precision the exact value isζ(2) = 1.6449.
1 (2n) 3 =ζ(3) and that the second term on the left-hand side isζ(3)/8) Our summation therefore has the value 7ζ(3)/8.
1.1.15 (a) Writeζ(n)−1 asP ∞ p=2 p −n , so our summation is
The summation overnis a geometric series which evaluates to p −2
Summing now overp, we get
(b) Proceed in a fashion similar to part (a), but now the geometric series has sum 1/(p 2 +p), and the sum overpis now lacking the initial term of α 1, so
The convergence of the series is optimized if we set B =−1, leading to the final result ζ(3) = 29
(c) Number of terms required for error less than 5×10 −7 : ζ(3) alone, 999; combined as in part (a), 27; combined as in part (b), 11.
1.2.1 (a) Applying Leibniz’ test the series converges uniformly for ε≤x 0 is.
(b) The Weierstrass M and the integral tests give uniform convergence for
1 +ε≤x 0 is chosen.
1.2.2 The solution is given in the text.
1.2.4 From |cosnx| ≤1,|sinnx| ≤1 absolute and uniform convergence follow for−s < x < sfor anys >0.
1.2.6 The solution is given in the text.
1.2.7 The solution is given in the text.
1.2.8 (a) Forn= 0,1,2, we find d 4n+1 sinx dx 4n+1 ¯¯ ¯¯
Taylor’s theorem gives the absolutely convergent series sinx X ∞ n=0
(b) Similar derivatives for cosxgive the absolutely convergent series cosx X ∞ n=0
To check this we substitute this into the first relation, giving η 0+ 1 η 0 −1+ 1 η 0+ 1 η 0 −1−1
0 does not exist, there is no Maclaurin expan- sion.
(b)|x−x 0 |< x 0 because the origin must be excluded.
= lim x→x 0 f 0 (x) g 0 (x), where the intermediate formal expression f(x+ (x 0 −x)) g(x+ (x 0 −x) may be dropped.
1.2.14 The solution is given in the text.
1.2.15 The solutions are given in the text.
The solution provided in the text is valid for the range 0 ≤ x < ∞ While the upper limit x does not need to be small, a larger value may lead to slow convergence, rendering the expansion less effective.
1.3.5 The expansion of the integral has the form
1.3.6 Form= 1,2, the binomial expansion gives (1+x) −m/2 X ∞ n=0 à−m/2 n ả x n
By mathematical induction we show that à−m/2 n ả
1.3.12 The solution is given in the text.
1.3.13 The two series have different, nonoverlapping convergence intervals. 1.3.14 (a) Differentiating the geometric series
(b) Writingx= tany asix= e 2iy + 1 e 2iy −1 we extracty=−i
1.3.16 Start by obtaining the first few terms of the power-series expansion of the expression within the square brackets Write
Inserting this into the complete expression forf(ε), the limit is seen to be 4/3.
Comparing these expansions, we note agreement throughx 2 , and the x 3 terms differ by (2/9)x 3 , or 2/9A 3
1.3.18 (a) Insert the power-series expansion of arctant and carry out the inte- gration The series forβ(2) is obtained.
(b) Integrate by parts, converting lnxinto 1/xand 1/(1+x 2 ) into arctanx.
The integrated terms vanish, and the new integral is the negative of that already treated in part (a).
1.4.1 Use mathematical induction First evaluate the claimed expression for the sum for bothn−1 andn:
Next verify that S n =S n−1+n 4 Complete the proof by verifying that
1.4.2 Use mathematical induction First, differentiate the Leibniz formula for n−1, getting the two terms n−1X j=0 àn−1 j ả "à d dx ả j+1 f(x)
Now change the index of the first summation to (j−1), with j ranging from 1 to n; the index can be extended to j = 0 because the binomial coefficient¡ n−1
−1 ¢vanishes The terms then combine to yield
The sum of two binomial coefficients has the value¡ n j ¢, thereby confirming that if the Leibnize formula is correct for n−1, it is also correct for n.
The binomial coefficient sum can be verified by understanding it as the number of ways to choose j objects from n total objects This can occur in two scenarios: either by selecting j-1 objects from the first n-1, with the nth object being the jth choice, or by making all j selections from the first n-1 objects while leaving the nth object unselected The proof is concluded by recognizing that the Leibniz formula accurately represents the first derivative.
1.5 Operations on Series Expansions of Functions
1.5.1 The partial fraction expansion is
The upper and lower limits give the same result, canceling the factor 1/2.
1.5.2 Start by writing the partial-fraction expansion forp+ 1 using the assumed form of that forpmultiplied by an additional factor 1/(n+p+ 1) Thus, we want to see if we can simplify
X p j=0 àp j ả(−1) j n+j à 1 n+p+ 1 ả to get the expected formula Our first step is to expand the two factors containingninto partial fractions:
Replacing the 1/(n+j) term of our original expansion using this result and adding a new 1/(n+p+ 1) term which is the summation of the above result for allj, we reach
Using the first formula supplied in the Hint, we replace each square bracket by the quantity
, thereby identifying the first summation as all but the last term of the partial-fraction expansion for p+ 1 The second summation can now be written
Using the second formula supplied in the Hint, we now identify the quan- tity within square brackets as
= 1 + (−1) p+1 −1 = (−1) p+1 , so the second summation reduces to
1 n+p+ 1, as required Our proof by mathematical induction is now completed by observing that the partial-fraction formula is correct for the casep= 0.
1.5.3 The formula for u n (p) follows directly by inserting the partial fraction decomposition If this formula is summed for n from 1 to infinity, all terms cancel except that containingu 1, giving the result
The proof is then completed by inserting the value ofu 1(p−1).
By substituting Eq (1.88) into Eq (1.87) and changing the summation variable from n to p (where p = n - j), we can redefine the ranges for both j and p to extend from zero to infinity This allows us to factor the p summation out and isolate the quantities that are independent of j, ultimately leading to the expression for f(x) as the summation from p = 0 to infinity.
Using now Eq (1.71), we identify the binomial coefficient in the above equation as à p+j j ả
, so thej summation reduces to
Insertion of this expression leads to the recovery of Eq (1.86).
1.5.5 Applying Eq (1.88) to the coefficients in the power-series expansion of arctan(x), the first 18a n (a 0 througha 17) are:
Using the series expansion in Eq (1.87) with x = 1, we find that the sum of terms up to the 17th provides an approximate value of arctan(1) ≈ 0.785286, which is quite close to the exact value of 0.785398 Notably, the 18th nonzero term in this power series is -1/35, indicating that the series for x = 1 can be effectively truncated after this point.
18 terms would barely give a result good to two significant figures The 18-term Euler expansion yields arctan(1/√
3)≈0.523598, while the exact value at this precision is 0.523599.
1.7.2 The triangle sides are given by
AB=B−A, BC=C−B, CA=A−C withAB+BC+CA= (B−A) + (C−B) + (A−C) = 0.
1.7.3 The solution is given in the text.
1.7.4 Ifv 0 i =v i −v1 ,r 0 i =r i −r1 ,are the velocities and distances, respectively, from the galaxy atr1 , thenv 0 i =H 0(r i −r1) =H 0r 0 i holds, i.e., the same Hubble law.
1.7.5 With one corner of the cube at the origin, the space diagonals of length√
The face diagonals of length√
1.7.6 (a) The surface is a plane passing through the tip ofa and perpendicular toa.
(b) The surface is a sphere havinga as a diameter:
1.7.7 The solution is given in the text.
1.7.8 The path of the rocket is the straight line r=r1+tv, or in Cartesian coordinates x(t) = 1 +t, y(t) = 1 + 2t, z(t) = 1 + 3t.
We now minimize the distance|r−r0 |of the observer at the point r0 (2,1,3) fromr(t), or equivalently (r−r0) 2 =min Differentiating the rocket path with respect totyields ˙r= ( ˙x,y,˙ z) =˙ v Setting d dt(r−r0) 2 = 0 we obtain the condition
The condition ˙r = v indicates that the tangent vector of the line is equal to the velocity of the rocket, implying that the shortest distance vector from point r0 is perpendicular to the trajectory By solving for t, we derive the ratio of scalar products as t = −(r1 − r0) · v / v² = −(−1, 0, −2) · (1, 2, 3).
Substituting this parameter value into the rocket path gives the point r s = (3/2,2,5/2) on the line that is closest to r0 The shortest distance isd=|r0 −r s |=|(−1/2,1,−1/2)|=p
1.7.9 Consider each corner of the triangle to have a unit of mass and be located ata i from the origin where, for example, a1= (2,0,0), a2= (4,1,1), a3= (3,3,2).
Then the center of mass of the triangle is
The three midpoints are located at the point of the vectors
We start from each corner and end up in the center as follows
1.7.10 A 2 =A 2 = (B−C) 2 =B 2 +C 2 −2BCcosθwithθthe angle betweenBˆ andC.ˆ
1.7.11 PandQare antiparallel;Ris perpendicular to bothPandQ.
(x+iy) −1 =e −iθ r =1 r(cosθ−isinθ) = x−iy r 2 = x−iy x 2 +y 2
1.8.2 Ifz=re iθ ,√ z=√ re iθ/2 =√ r(cosθ/2 +isinθ/2).In particular,
1.8.3 e inθ = cosnθ+isinnθ= (e iθ ) n = (cosθ+isinθ) n = P n ν=0 ¡ n ν ¢cos n−ν θ(isinθ) ν
Separating real and imaginary parts we have cosnθ [n/2]X ν=0
1−e ix = e iN x/2 e ix/2 e iN x/2 −e −iN x/2 e ix/2 −e −ix/2
Now take real and imaginary parts to get the result.
All other identities are shown similarly.
= cosz 1cosz 2 −sinz 1sinz 2+i(sinz 1cosz 2+ sinz 2cosz 1).
Separating this into real and imaginary parts for real z 1 , z 2 proves the addition theorems for real arguments Analytic continuation extends them to the complex plane.
1.8.6 (a) Using cosiy= coshy,siniy =i sinhy,etc and the addition theorem we obtain sin(x+iy) = sinxcoshy+icosxsinhy,etc.
(b)|sinz| 2 = sin(x+iy) sin(x−iy) = sin 2 xcosh 2 y+ cos 2 xsinh 2 y
= sin 2 x(cosh 2 y−sinh 2 y) + sinh 2 y= sin 2 x+ sinh 2 y,etc.
1.8.7 (a) Using cosiy= coshy,siniy =i sinhy,etc and the addition theorem we obtain sinh(x+iy) = sinhxcosy+icoshxsiny,etc.
(b)|cosh(x+iy)| 2 = cosh(x+iy) cosh(x−iy) = cosh 2 xcos 2 y+sinh 2 xsin 2 y
1.8.8 (a) Using Exercise 1.8.7(a) and rationalizing we get tanh(x+iy) =sinhxcosy+icoshxsiny coshxcosy+isinhxsiny
2sinh 2x(cos 2 y+ sin 2 y) + 2 i sin 2y(cosh 2 x−sinh 2 x) cosh 2 xcos 2 y+ sinh 2 xsin 2 y
2 sinh 2x+isin 2y cos 2 y+ sinh 2 x = sinh 2x+ sin 2y cos 2y+ cosh 2x.
(b) Starting from cosh(x+iy) sinh(x+iy) this is similarly proved.
1.8.9 The expansions relevant to this exercise are tan −1 x=x−x 3
The desired identity follows directly by comparing the expansion of tan −1 x withi/2 times the difference of the other two expansions.
1.8.10 (a) The cube roots of −1 are −1, e πi/3 = 1/2 +i√
(b) Writeias e πi/2 ; its 1/4 power has values e (πi/2+2nπ)/4 for all integer n; there are four distinct values: e iπ/8 = cosπ/8 +isinπ/8, e 5iπ/8 cos 5π/8 +isin 5π/8,e 9iπ/8 =−e iπ/8 , ande 13iπ/8 =−e 5iπ/8
(c)e iπ/4 has the unique value cosπ/4 +isinπ/4 = (1 +i)/√
1.8.11 (a) (1 +i) 3 has a unique value Since 1 +ihas magnitude√
2 and is at an angle of 45 ◦ =π/4, (1 +i) 3 will have magnitude 2 3/2 and argument 3π/4, so its polar form is 2 3/2 e 3iπ/4
(b) Since−1 =e πi , its 1/5 power will have values e (2n+1)πi for all integer n There will be five distinct values: e kπi/5 withk= 1, 3, 5, 7, and 9.
To begin, we expand the function as a power series in terms of x while maintaining y at its actual value Next, we take each term from the x expansion and further expand it as a power series in y, treating x as a constant The nth term of the x expansion can be expressed as \( \frac{x^n}{n!} \).
Themth term in they expansion of thex n term is therefore x n n! y m m! à ∂
The coefficient in the above equation can be written
Using the right-hand side of the above equation and collecting together all terms with the same value ofm+n, we reach the form given in the exercise.
In the context of differentiating functions, the quantities α i are treated as independent of the variables x i during differentiation When the differential operator is raised to the power of n and combined with t n, the resulting expansion yields terms that apply a total of n derivatives to the function f Each term reflects the number of times each variable x i is differentiated, with the coefficient of each unique term representing the various combinations of derivatives in the expansion Specifically, the number of ways a term where each derivative x j is applied n j times can be calculated using the formula n! / (n1! n2! ), ensuring that the sum of n i equals n This approach leads to the same outcome as expanding the function sequentially in x 1, then x 2, and so forth.
1.10.1 Apply an integration by parts to the integral in Table 1.2 defining the gamma function, for integern >0: Γ(n) Z ∞
Rearranging to Γ(n+ 1) =nΓ(n), we apply mathematical induction, not- ing that if Γ(n) = (n−1)!, then also Γ(n+ 1) =n! To complete the proof, we directly evaluate the integral Γ(1) =R ∞
The integral can be evaluated through contour integration, as illustrated in Example 11.8.5 A method inspired by this section involves multiplying the integrand by e^(-αx) and analyzing the integral's value as α approaches 0 By differentiating the integral with respect to the parameter α, we can derive the necessary results.
0 e −αx sinx dx=− 1 α 2 + 1, where the integral forI 0 is ientified as having the value found in Example 1.10.4 We now integrate the expression forI 0 , writing it as the indefinite integral
The value ofCis now determined from the value ofI(∞), which from the form ofImust be zero Thus,C= tan −1 ∞=π/2, and, since tan −1 0 = 0, we findI(0) =π/2.
Now integrate term by term; each integrand is a simple exponential The result is
The series in parentheses is that discussed in Exercise 1.3.2, with value π/4 Our integral therefore has valueπ/2.
1.10.4 Expand the integrand as a power series in e −ax and integrate term by term:
0 Ăe −ax −e −2ax +e −3ax − ã ã ãÂ
After factoring out (1/a), the series that remains is that identified in
Eq (1.53) as ln 2, so our integral has value ln(2)/a.
1.10.5 Integrate by parts, to raise the power ofxin the integrand:
Note that the integrated terms vanish The integral can now be recognized (see Table 1.2) as−Ci(π).
1.10.6 This is a case of the integralI(α) defined in the solution of Exercise 1.10.2, withα= 1 We therefore have
1.10.7 Write erf as an integral and interchange the order of integration We get
1.10.8 WriteE 1as an integral and interchange the order of integration Now the outer (u) integration must be broken into two pieces:
1.10.9 Change the variable of integration toy=x+ 1, leading to
1.10.10 After the integration by parts suggested in the text, with [tan −1 x] 2 dif- ferentiated anddx/x 2 integrated, the result isI(1), where
We now differentiateI(a) with respect to the parametera, reaching after a partial-fraction decomposition
Integrating with respect toa, we getI(a) =πln(1 +a) +C, withC set to zero to obtain the correct resultI(0) = 0 Then, settinga= 1, we find
1.10.11 Integrating over one quadrant and multiplying by four, the range ofxis
(0, a) and, for givenx, the range ofyis from 0 to the positivey satisfying the equation for the ellipse Thus,
1.10.12 Draw the dividing line aty= 1/2 Then the contribution to the area for eachybetween 1/2 and 1 is 2p
A simple explanation of these two terms is that π/3 is the area of the sector that includes the piece in question, while √
3/4 is the area of the triangle that is the part of the sector not included in that piece.
1.11.1 The mean value theorem gives n→∞ lim
The left-hand side of the equation is nonzero only near the points x = x1 and x = x2, both of which correspond to Exercise 1.11.4 In these instances, the relevant quantity is represented by |x1 - x2|.
1.11.7 Integrating by parts we find
By substituting the specified form for δ n (x) and adjusting the variable of integration to n x, we derive a result that remains constant regardless of n The indefinite integral of 1/cosh²(x) is tanh(x), which approaches +1 as x approaches +∞.
As x approaches negative infinity, the normalization for δn is confirmed The function tanh(x) indicates that for large n and negative x, the right-hand side of the equation tends toward zero, while for large n and positive x, it approaches +1.
Determinants and Matrices
2.1.2 The determinant of the coefficients is equal to 2 Therefore no nontrivial solution exists.
2.1.3 Given the pair of equations x+ 2y= 3, 2x+ 4y= 6.
(a) Since the coefficients of the second equation differ from those of the first one just by a factor 2, the determinant of (lhs) coefficients is zero.
(b) Since the inhomogeneous terms on the right-hand side differ by the same factor 2, both numerator determinants also vanish.
(c) It suffices to solvex+ 2y = 3 Given x, y = (3−x)/2 This is the general solution for arbitrary values ofx.
2.1.4 (a) C ij is the quantity that multiplies a ij in the expansion of the deter- minant The sum overicollects the quantities that multiply all thea ij in columnj of the determinant.
(b) These summations form determinants in which the same column (or row) appears twice; the determinant is therefore zero,
2.1.5 The solution is given in the text.
If a set of forms is linearly dependent, at least one form can be expressed as a linear combination of the others To demonstrate this, create a determinant from their coefficients, ensuring each row corresponds to one form By subtracting a linear combination of the other rows from one row, you can reduce it to zero without altering the determinant's value Consequently, the determinant will equal zero, confirming linear dependence The process of Gauss elimination further supports this conclusion.
2.1.8 (a) δ ii = 1 (not summed) for eachi= 1,2,3.
(b) δ ij ε ijk = 0 becauseδ ij is symmetric ini, jwhileε ijk is antisymmetric ini, j.
(c) For eachεinε ipq ε jpq to be non-zero, leaves only one value foriand j, so thati =j Interchanging pand q gives two terms, hence the factor 2.
(d) There are 6 permutationsi, j, k of 1,2,3 inε ijk ε ijk = 6.
2.1.9 Givenk impliesp6=qforε pqk 6= 0 Forε ijk 6= 0 requires eitheri=pand soj =q, or i=qand thenj=p Henceε ijk ε pqk =δ ip δ jp −δ iq δ jp
2.2.1 Writing the product matrices in term of their elements,
AB= (X m a im b mk ), BC= (X n b in c nk ),
(AB)C ÃX n ÃX m a im b mn
=A(BC) ÃX m a im ÃX n b mn c nk
, because products of real and complex numbers are associative the paren- theses can be dropped for all matrix elements.
, i.e., the correspondence holds for addition and subtraction.
Similarly, it holds for multiplication because first
(a 1+ib 1)(a 2+ib 2) = (a 1 a 2 −b 1 b 2) +i(a 1 b 2+a 2 b 1) and matrix multiplication yields à a 1 b 1
2.2.4 A factor (−1) can be pulled out of each row giving the (−1) n overall.
2.2.5 (a) First we check that à ab b 2
Second, to find the constraints we write the general matrix as à A B
= 0 givingD=−A, D 2 =−BC=A 2 This implies, if we setB=b 2 , C=−a 2 without loss of generality, thatA=ab=−D.
2.2.7 Expanding the commutators we find
[B,[A,C]] C−BCA−ACB+CAB, [C,[A,B]] −CBA−ABC+BAC, and subtracting the last double commutator from the second yields the first one, since theBACandCABterms cancel.
2.2.8 By direct multiplication of the matrices we find [A,B] =C,BA= 0, etc.
2.2.9 These results can all be verified by carrying out the indicated matrix multiplications.
2.2.10 Ifa ik = 0 =b ik fori > k, then also X m a im b mk = X i≤m≤k a im b mk = 0, as the sum is empty fori > k.
2.2.11 By direct matrix multiplications and additions.
2.2.12 By direct matrix multiplication we verify all claims.
2.2.13 By direct matrix multiplication we verify all claims.
2.2.14 Fori6=kanda ii 6=a kk we get for the product elements
(AB) ik = (X n a in b nk ) = (a ii b ik ) = (BA) ik = (X n b in a nk ) = (b ik a kk ).
2.2.15 X m a im b mk =a ii b ii δ ik =X m b im a mk
Since trace ABC equals trace BCA, select one of the previous instances where two commuting matrices are positioned next to each other and swap their order If necessary, perform a cyclic permutation to achieve the arrangement CBA.
2.2.17 Taking the trace, we find from [M i ,M j ] =iM k that itrace(M k ) = trace(M i M j −M j M i ) = trace(M i M j )−trace(M i M j ) = 0.
2.2.18 Taking the trace of A(BA) = −A 2 B=−B yields −tr(B) = tr(A(BA)) tr(A 2 B) = tr(B).
2.2.19 (a) Starting from AB =−BA, multiply on the left byB −1 and take the trace After simplification, we get traceB=−trace B, so traceB= 0. 2.2.20 This is proved in the text.
2.2.21 (a) A unit matrix except thatM ii =k,
(b) A unit matrix except thatM im =−K,
(c) A unit matrix except thatM ii =M mm = 0 andM mi −M im = 1. 2.2.22 Same answers as Exercise 2.2.21.
2.2.24 (a) The equation of part (a) states thatT moves people from areaj but does not change their total number.
(b) Write the component equation P j T ij P j =Q i and sum overi This summation replacesT ij by unity, leaving that the sum overP j equals the sum overQ i , hence conserving people.
2.2.25 The answer is given in the text.
2.2.27 Taking the determinant of ˜AA= 1 and using the product theorem yields det(˜A) det(A) = 1 = det 2 (A) implying det(A) =±1.
2.2.28 If ˜A=−A,˜S=S, then trace(SA) = trace(SA) = trace(˜f AS) =˜ −trace(AS).
2.2.29 From ˜A=A −1 and det(A) = 1 we have
This gives det(A) =a 2 11 +a 2 12 = 1, hencea 11 = cosθ=a 22 , a 12 = sinθ −a 21 ,the standard 2×2 rotation matrix.
2.2.30 Becauseεis real, det(A ∗ ) =X i k ε i 1 i 2 i n a ∗ 1i 1 a ∗ 2i 2 ã ã ãa ∗ ni n ÃX i k ε i 1 i 2 i n a 1i 1 a 2i 2 ã ã ãa ni n
Because, for anyA, det(A) = det(˜A), det(A ∗ ) = det(A † ).
2.2.31 IfJ x andJ y are real, so also must be their commutator, so the commuta- tion rule requires thatJ z be pure imaginary.
2.2.33 AsC jk =P n S nj ∗ S nk , trace (C) =P nj |S nj | 2 2.2.34 IfA † =A,B † =B, then
2.2.36 −iC † = (AB−BA) † =B † A † −A † B † −AB=−iC.
2.2.37 (AB) † =B † A † yields [A,B] = 0 as the condition, that is, the answer in the text.
2.2.40 Start by noting the relationships σ i σ j +σ j σ i = 0 ifi6=j, and σ i 2 =1 2 ; see Eq (2.59); for proof add Eqs (2.29) and (2.30) Then,
, and noting fron Eq (2.57) that if C = A⊗B and C 0 = A 0 ⊗B 0 then
It is obvious from the second line of the above equation set that γ 0 γ i + γ i γ 0 = 0; from the third line of the equation set we find γ i γ j +γ j γ i is zero ifj6=ibecause then σ j σ i =−σ i σ j
2.2.42 The anticommutation can be demonstrated by matrix multiplication.
2.2.43 These results can be confirmed by carrying out the indicated matrix op- erations.
2.2.48 (a) Written as 2×2 blocks, the matricesα i and the wave function Ψ are α i à 0 σ i σ i 0
In block form, Eq (2.73) becomes
The solution is completed by moving the right-hand side of the above equation to the left, written in the form à −E 0
! and combining all the terms by matrix addition.
2.2.49 The requirements the gamma matrices must satisfy are Eqs (2.74) and
(2.75) Use the same process that was illustrated in the solution to Exer- cise 2.2.41, but now withγ 0 =σ 1 ⊗12.
2.2.50 In the Weyl representation, the matrices α i and the wave function Ψ, written as 2×2 blocks, take the forms α i à −σ i 0
Then proceed as in the solution to Exercise 2.2.48, obtaining the matrix equation
When m is negligible, the matrix equation simplifies into two independent equations for Ψ1 and Ψ2 In this scenario, one solution set results in Ψ2 being zero while Ψ1 satisfies the equation −σãpΨ1 = EΨ1 Conversely, another solution set features Ψ1 as zero, with Ψ2 mirroring the previously determined Ψ1 values, but with opposite sign values for E.
(b) If for allr,r 0† r=r † U † Ur, then we must haveU † U=1.
Vector Analysis
3.2.2 (AìB) 2 =A 2 B 2 sin 2 θ=A 2 B 2 (1−cos 2 θ) =A 2 B 2 −(AãB) 2 withθthe angle betweenAˆ andB.ˆ
The vector P is oriented at an angle θ from the positive x-axis, while vector Q is positioned at an angle of -ϕ Consequently, the angle between these two vectors is θ + ϕ Both vectors have a unit length, leading to the relationship P·Q = cos(θ + ϕ), and the z-component of the cross product Q × P is given by sin(θ + ϕ).
3.2.5 Ifaandbboth lie in thexy-plane their cross product is in thez-direction.
The same is valid forc×d∼ˆz.The cross product of two parallel vectors is zero Hence (a×b)×(c×d) = 0.
3.2.6 CrossA−B−C= 0 intoAto get−A×C=A×B, orCsinβ=Bsinγ, etc.
3.2.8 (a) AãBìC = 0, A is the plane of B and C The parallelpiped has zero height above theBC plane and therefore zero volume.
3.2.9 Applying the BAC-CAB rule we obtain
[aãcb−aãbc] + [bãac−bãca] + [cãba−cãab] = 0.
3.2.11 The scalar triple productAãBìCis the volume spanned by the vectors.
3.2.13 (AìB)ã(CìD) = [(AìB)ìC]ãD= [(AãC)B−(BãC)A]ãD
3.2.14 Using the BAC-CAB rule withA×Bas the first vector we obtain
(AìB)ì(CìD) = (AìB)ãDC−(AìB)ãCD.
3.2.15 The answer is given in the text.
3.3.1 The trigonometric identities follow from the rotation matrix identity
cosϕ 1cosϕ 2 −sinϕ 1sinϕ 2 sinϕ 1cosϕ 2+ cosϕ 1sinϕ 2
−cosϕ 1sinϕ 2 −sinϕ 1cosϕ 2 −sinϕ 1sinϕ 2+ cosϕ 1cosϕ 2
To align the reflecting surfaces with the xy, xz, and yz planes, it is essential to understand how incoming rays interact with these planes When a ray strikes the xy plane, the z component of its direction is reversed, while a strike on the xz plane reverses the y component, and a strike on the yz plane reverses the x component These properties hold true for any angle of incidence, effectively reversing the propagation direction to the opposite of its initial orientation.
3.3.3 BecauseSis orthogonal, its transpose is also its inverse Therefore
S is identified as an improper rotation, indicating that S(a × b) has components with opposite signs to a₀ × b₀, which confirms that a × b is a pseudovector Additionally, the differing signs between (aìb)ãc and (a₀ ìb₀)ãc₀ demonstrate that (aìb)ãc is a pseudoscalar Furthermore, the equality of the vectors S(aì(bìc)) and a₀ × (b₀ × c₀) establishes that a × (b × c) is indeed a vector.
The Euler rotations discussed here differ from those in the original text due to the inclination of the polar axis, which is now aligned with the x₀₁ axis instead of the x₀₂ axis To achieve the same polar orientation, we need to reposition the x₀₁ axis to the location of the x₀₂ axis as described in the text, necessitating an initial rotation of π/2 After this adjustment, the rotational position is π/2 greater in a counterclockwise direction compared to the original text rotation, requiring the third Euler angle to be reduced by π/2 from its initial value.
(b) The answer is in the text.
The angle changes result in transformations where cosα becomes -cosα and sinα becomes -sinα, while cosβ remains unchanged and sinβ transforms to -sinβ Similarly, for angle γ, sinγ changes to -sinγ and cosγ becomes -cosγ These transformations confirm that each matrix element in Equation (3.37) remains consistent.
3.4.4 (a) Each of the three Euler rotations is an orthogonal matrix, so their matrix product must also be orthogonal Therefore its transpose, ˜S, must equal its inverse,S −1
(b) This equation simply carries out the three Euler rotations in reverse order, each in the opposite direction.
The projection on the rotation axis remains unchanged during rotation, represented as (rãn)ˆˆ n The component of vector r that is perpendicular to the rotation axis can be expressed as r - (rãn)ˆˆ n When this vector undergoes a rotation of angle Φ, it can be decomposed into two components: one in its original direction, (r - (rãn)ˆˆ n) cos Φ, and another perpendicular to both the original direction and the axis, given by (r - (rãn)ˆˆ n) sin Φìn This simplifies to ˆ rìnˆsin Φ, leading to the final result.
(b) If ˆn= ˆe z , the formula yields r 0 =xcos Φˆe x +ycos Φˆe y +zcos Φˆe z +ysin Φˆe x −xsin Φˆe y +z(1−cos Φ)ˆe z
Simplifying, this reduces to r 0 = (xcos Φ +ysin Φ)ˆe x + (−xsin Φ +ycos Φ)ˆe y +zˆe z
This corresponds to the rotational transformation given in Eq (3.35).
(c) Expandingr 0 2 , recognizing that the second term ofr 0 is orthogonal to the first and third terms, r 0 2 =r 2 cos 2 Φ + (rìn)ˆ ã(rìn) sinˆ 2 Φ
+ (ˆnãr) 2 (1−cos Φ) 2 + 2(ˆnãr) 2 cos Φ(1−cos Φ).
Using an identity to make the simplification
(rìn)ˆ ã(rìn) = (rˆ ãr)(ˆnãn)−(rãn)ˆ 2 =r 2 −(rãn)ˆ 2 , we get r 0 2 =r 2 + (rãn)ˆ 2 (−sin 2 Φ + 1 + cos 2 Φ−2 cos 2 Φ) =r 2
3.5.2 The solution is given in the text.
3.5.4 dF=F(r+dr, t+dt)−F(r,t) =F(r+dr, t+dt)−F(r,t+dt)
3.5.5 ∇(uv) =v∇u+u∇v follows from the product rule of differentiation.
∂v∇v= 0,∇uand∇vare parallel so that (∇u)×(∇v) = 0, and vice versa.
(b) If (∇u)×(∇v) = 0, the two-dimensional volume spanned by∇uand
∇v, also given by the Jacobian
3.5.6 (a) From ˙r = ωr(−ˆxsinωt+ˆycosωt), we get r×˙r =ˆzωr 2 (cos 2 ωt+ sin 2 ωt) =ˆzωr 2
(b) Differentiating ˙rabove we get¨r=−ω 2 r(ˆxcosωt+yˆsinωt) =−ω 2 r.
3.5.7 The time derivative commutes with the transformation because the coef- ficientsa ij are constants ThereforedV j /dtsatisfies the same transforma- tion law asV j
3.5.8 The product rule directly implies (a) and (b).
3.5.9 The product rule of differentiation in conjunction with (aìb)ãc= aã
∇ã(aìb) =bã(∇ìa)−aã(∇ìb).
3.5.10 IfL=−ir×∇, then the determinant form of the cross product gives
When performing the specified operations, it's essential to remember that derivatives affect all elements to their right in the current expression, as well as the function to which the operator is applied.
3.5.12 [aãL,bãL] =a j [L j , L k ]b k =iε jkl a j b k L l =i(aìb)ãL.
3.5.13 The stream lines ofbare solutions of the differential equation dy dx = b y b x = x
The differential equation can be expressed as \(xdx + ydy = 0\), which integrates to \(x^2/2 + y^2/2 = \text{constant}\), representing the equation of a family of circles centered at the origin, \(x^2 + y^2 = C^2\) To analyze the direction of the streamlines, consider the point (+1, 0) on a circle, where \(b_x = 0\) and \(b_y = +1\), indicating a counterclockwise motion.
3.6 Differential Vector Operators: Further Properties
3.6.1 By definition, uìv is solenoidal if ∇ã(uìv) = 0 But we have the identity
∇ã(uìv) =vã(∇ìu)−uã(∇ìv).
If a vectorwis irrotational,∇×w= 0, so ifuandvare both irrotational, the right-hand side of the above equation is zero, proving that u×v is solenoidal.
3.6.2 If∇ìA= 0, then∇ã(Aìr) =rã∇ìA−Aã(∇ìr) = 0−0 = 0. 3.6.3 Fromv=ωìrwe get∇ã(ωìr) =−ωã(∇ìr) = 0.
3.6.4 Forming the scalar product off with the identity
∇×(gf) =g∇×f+ (∇g)×f ≡0 we obtain the result, because the second term of the identity is perpen- dicular tof.
Applying the BAC-CAB rule naively results in the expression (∇ãB)A−(∇ãA)B, where the operator ∇ continues to act on both A and B The product rule of differentiation produces two terms for each variable, ensuring that ∇ acts solely on the subsequent component This leads to the transformation (∇ãB)A into A(∇ãB) + (Bã∇)A, with a similar process applied to the second term, ultimately yielding four distinct terms.
3.6.6 Write the x components of all the terms on the right-hand side of this equation We get
All terms cancel except those corresponding to the x component of the left-hand side of the equation.
3.6.7 Apply the BAC-CAB rule to get
The factor 1/2 occurs because∇ operates only on oneA.
3.6.8 ∇(AãBìr) =∇(rãAìB) = ˆe x (AìB) x + ˆe y (AìB) y + ˆe z (AìB) z
3.6.9 It suffices to check one Cartesian component; we takex Thexcomponent of the left-hand side of Eq (3.70) is
Thexcomponent of the right-hand side is
After canceling the two right-hand-side occurrences of ∂ 2 V x /∂x 2 these two expressions contain identical terms.
3.6.11 (a) IfF or Gcontain an additive constant, it will vanish on application of any component of∇.
(b) If either vector contains a term∇f, it will not affect the curl because
3.6.12 Use the identityvì(∇ìv) =∇(vãv)−(vã∇)v.Taking the curl and noting that the first term on the right-hand side then vanishes, we obtain the desired relation.
3.6.15 From Eq (3.70),∇ì(∇ìA) =−∇ 2 Aif∇ãA= 0.
, which satisfies the heat conduction equation because∇ 2 Φ = 0.
3.6.17 Start by forming the matrixMã∇ We obtain
Apply this matrix to the vectorψ The result (after multiplication byc) is cMã∇ψ
Equating to zero the real and imaginary parts of all components of the above vector, we recover two Maxwell equations.
3.6.18 By direct matrix multiplication we verify this equation.
3.7.1 A triangle ABC has area 1 2 |B−A| |C−A|sinθ, where θ is the angle betweenB−AandC−A This area can be written|(B−A)×(C−A)|/2.
Applying this formula toOAB, we get just|A×B|/2 Continuing to the other three faces, the total area is
To parameterize the circle C, we use the equations x = cosϕ and y = sinϕ, where ϕ represents the polar angle This leads to the derivatives dx = -sinϕ dϕ and dy = cosϕ dϕ Consequently, the force can be expressed as F = -ˆx sinϕ + ˆy cosϕ, which allows us to calculate the work done in this context.
In this analysis, we observe energy expenditure by integrating counterclockwise from ϕ = 0 to π, resulting in a value of -π, indicating that we are moving with the force This path-dependent work aligns with the physical interpretation of the process.
Fãdr ∼ xdy−ydx = L z is proportional to the z-component of orbital angular momentum (involving circulation, as discussed in Section 3.5).
If we integrate along the square through the points (±1,0),(0,−1) sur- rounding the circle we find for the clockwise lower half square path
4 =π, which is consistent with the circular path.
The total work done depends on the chosen path One straightforward option is to first move horizontally from the point (1,1) to (3,1), followed by a vertical movement from (3,1) to (3,3) The work can be calculated by integrating the force along this path For the initial segment of the journey, the work is represented by the integral of the force over the distance traveled.
F x dx; for the second segment it isR
F y dy These correspond to the specific integrals w 1 Z 3
In the first term, the factor x remains constant, making it outside the integral, with a value of 0 on one face of the cube and 1 on the opposite face The same applies to the factors y and z in the other two terms, which contribute equally to the overall result.
3.8.1 For a constant vector a, its divergence is zero Using Gauss’ theorem we have
S dσ, whereS is the closed surface of the finite volumeV Asa6= 0 is arbitrary,
3.8.2 From∇ãr= 3 in Gauss’ theorem we have
S rãdσ, whereV is the volume enclosed by the closed surfaceS.
3.8.3 Cover the closed surface by small (in general curved) adjacent rectangles
S i whose circumference are formed by four lines L i each Then Stokes’ theorem gives
Aãdl= 0 because all line integrals cancel each other.
3.8.4 Apply Gauss’ theorem to∇ã(ϕE) =∇ϕãE+ϕ∇ãE=−E 2 +ε −1 0 ϕρ, whereR
3.8.5 First, show thatJ i =∇ã(xJ) by writing
SinceJis zero on the boundary, so isxJ, so by Gauss’ theorem we have
3.8.6 By direct calculation we can find that ∇×t = 2e z Then, by Stokes’ theorem, the line integral has the value 2A.
3.8.7 (a) Asr×dr/2 is the area of the infinitesimal triangle, H r×dris twice the area of the loop.
(b) Fromdr= (−ˆxasinθ+ˆybcosθ)dθandˆx×yˆ=ˆzwe obtainr×dr ˆ zab(cos 2 θ+ sin 2 θ) andH r×dr=ˆzabR 2π
3.8.8 We evaluate the surface integral withP=r Note thatdσ= ˆe z dA, and that, evaluating components, dσ×∇ ã
Then form (dσ×∇)×r The x and y components of this expression vanish; thez component is à
The surface integral then has the value−2A, whereA is the area of the loop Note that the alternate form of Stokes’ theorem equates this surface integral to−H r×dr.
3.8.9 This follows from integration by parts shifting∇ fromv to u The inte- grated term cancels for a closed loop.
3.8.10 Use the identity of Exercise 3.8.9, i.e.
∇(uv)ãdλ= 0, and apply Stokes’ theorem to
3.8.11 Starting with Gauss’ theorem written as
By substituting B with a×P, where a represents a constant vector and P is arbitrary, the left-hand integrand transforms into (a·P)∇dσ = (P·dσ)∇a Meanwhile, the right-hand integrand expands to P∇(a) - a∇(P), with the first term disappearing since a is constant Consequently, our equation based on Gauss' theorem simplifies to a∇.
= 0, we note that because the constant direction ofais arbitrary the quantity in square brackets must vanish; its vanishing is equivalent to the relation to be proved.
Bãdr and substituteB=ϕa, whereais a constant vector andϕis an arbitrary scalar function Because a is constant, the quantity ∇×ϕa reduces to (∇ϕ)×a, and the left-side integrand is manipulated as follows:
The Stokes’ theorem formula can then be written aã
Because a is arbitrary in direction, the integrals on the two sides of this equation must be equal, proving the desired relation.
3.8.13 Starting from Stokes’ theorem as written in the solution to Exercise 3.8.12, setB=a×P This substitution yields
Applying vector identities and remembering that a is a constant vector, the left- and right-side integrands can be manipulated so that this equation becomes
Bringinga outside the integrals and rearranging, we reach aã ãZ
Since the direction ofais arbitrary, the quantity within the square brackets vanishes, thereby confirming the desired relation.
3.9.1 The solution is given in the text.
3.9.3 The gravitational acceleration in the z-direction relative to the Earth’s surface is
3.9.4 The answer is given in the text.
3.9.5 The answer is given in the text.
3.9.7 (a) This is proved in Exercise 3.6.14.
∇Λãdr= Λ| b a = 0 forb=ain a closed loop.
3.9.9 Using Green’s theorem as suggested in the problem and the formula for the Laplacian of 1/r(whereris the distance fromP), the volume integral of Green’s theorem reduces to
The surface integrals, for a sphere of radiusacentered atP, are
Using the equation ∇(1/r) = −ˆe r /r², the second term of the surface integral results in 4π times the average value of ϕ on the sphere According to Gauss' theorem, the first term of the surface integral disappears since ∇ã∇ϕ is zero throughout the sphere Consequently, we arrive at the final equation: 4πϕ₀ = 4πhϕi.
∂t +J=∇ì(∇ìA)/à= (∇∇ãA−∇ 2 A)/à=J so that−∇ 2 A=àJfollows.
3.9.11 Start from Maxwell’s equation for∇×B and substitute for the fieldsB andEin terms of the potentialsAandϕ The relevant equations are
Next manipulate the left-hand side using Eqs (3.70) and (3.109):
Inserting this result for∇×(∇×A), the terms in∂ϕ/∂tcancel and the desired formula is obtained.
The evaluation of (∇ ×A) y is now completed by using the fact that
∇ãB= 0, so we continue to
Tensors and Differential Forms
4.1.1 This is a special case of Exercise 4.1.2 withB ij 0 = 0.
4.1.2 IfA 0 ij =B ij 0 in one frame of reference, then define a coordinate transfor- mation from that frame to an arbitrary one: x i =x i (x 0 j ), so that
4.1.3 Make a boost in the z-direction If A z =A 0 z =A 0 = 0,then A 0 0 = 0 in the boosted frame by the Lorentz transformation, etc.
In the context of isotropy, the relationship between the components of the tensor T can be expressed as T_ik = cosθsinθ T_11 + cos²θ T_12sin²θ T_21 - sinθcosθ T_22 For a rotation by π, we find that T_12 0 = T_12, while for a rotation by π/2, T_12 equals -T_21 Isotropy necessitates that T_21 and T_12 both equal zero, leading to the conclusion that all off-diagonal components must vanish, and the diagonal components must be equal.
The four-dimensional fourth-rank Riemann–Christoffel curvature tensor in general relativity, denoted as R iklm, consists of 256 components However, the antisymmetry of the first and second pairs of indices reduces these to 36 components, which can be organized into a 6×6 matrix Further symmetry from the exchange of index pairs simplifies this matrix to 21 components Additionally, the Bianchi identity introduces a constraint that reduces the number of independent components to 20 Notably, through the application of permutation symmetries, it is possible to set the first index to zero while ensuring that the remaining indices are all distinct.
4.1.6 Each component has at least one repeated index and is therefore zero.
4.1.7 As the gradient transforms like a vector, it is clear that the gradient of a tensor field of ranknis a tensor of rankn+ 1.
4.1.8 The contraction of two indices removes two indices, while the derivative adds one, so (n+ 1)−2 =n−1.
4.1.9 The scalar product of the four-vectors
4.1.10 The double summation K ij A i B j is a scalar That K ij is a second-rank tensor follows from the quotient theorem.
4.1.11 SinceK ij A jk =B ik is a second-rank tensor the quotient theorem tells us thatK ij is a second-rank tensor.
4.2.1 The direct product ε ijk C lm is a tensor of rank 5 Contracting 4 indices leaves a tensor of rank 1, a vector Inverting givesC jk =ε jki C i ,a tensor of rank 2.
4.2.2 The generalization of the totally antisymmetric ε ijk from three to n di- mensions hasnindices Hence the generalized cross productε ijk A i B j is an antisymmetric tensor of rankn−26= 1 forn6= 3.
4.2.3 The solution is given in the text.
4.2.4 (a) As eachδ ij is isotropic, their direct product must be isotropic as well.
This is valid for any order of the indices The last statement implies (b) and (c).
4.2.5 The argument relating to Eq (4.29) holds in two dimensions, too, with δ 0 ij = det(a)a ip a jq δ pq No contradiction arises becauseε ij is antisymmetric whileδ ij is symmetric.
! is a rotation, then à cosϕ sinϕ
4.2.7 IfA k = 1 2 ε ijk B ij withB ij =−B ji , then
2ε mnk A k =ε mnk ε ijk = (δ mi δ nj −δ mj δ ni )B ij =B mn −B nm = 2B mn
4.3.1 The vector ε i is completely specified by its projections onto the three linearly independentε k , i.e., by the requirements thatε i ãε j =δ j i Taking the form given in the exercise, we form ε i ãε i =(ε j ìε k )ãε i
(ε j ìε k )ãε i = 0, the zero occurring because the three vectors in the scalar triple product are not linearly independent The above equations confirm thatε i is the contravariant version ofε i
4.3.2 (a) From the defining formula, Eq (4.40), the orthogonality of theε i im- plies thatg ij = 0 when i6=j.
(b) See the answer to part (c).
(c) From Eq (4.46) with theε i orthogonal, theε i must also be orthog- onal and have magnitudes that are the reciprocals of the ε i Then, from
Eq (4.47), theg ii must be the reciprocals of theg ii
4.3.3 This exercise assumes use of the Einstein summation convention Inserting the definitions of theε i andε i and evaluating the scalar products, we reach
The term of the product arising from the first term of each factor has the form ∂q i
∂q k , where we have noted that thej summation is the chain-rule expansion for
∂x/∂x, which is unity The products arising from the second terms and third terms of both factors have analogous forms, and the sum of these
“diagonal” terms is also a chain-rule expansion:
The remaining terms of the original product expression all reduce to zero; we illustrate with
Here thej summation is the chain-rule expansion of∂y/∂x and therefore vanishes.
4.3.4 Starting from Eq (4.54), Γ m jk =ε m ã(∂ε k /∂q j ), we see that a proof that
∂ε k /∂q j = ∂ε j /∂q k would also demonstrate that Γ m jk = Γ m kj From the definition ofε k , we differentiate with respect toq j , reaching
The differentiability of the coordinates ensures that the equation's right-hand side remains symmetric in j and k, allowing for their interchange without affecting the left-hand side's value.
The covariant metric tensor is diagonal, featuring nonzero elements represented as g_ii = h²_i, where g_ρρ = g_zz = 1 and g_ϕϕ = ρ² Similarly, the contravariant metric tensor is also diagonal, with nonzero elements being the reciprocals of the covariant components, resulting in g_ρρ = g_zz = 1 and g_ϕϕ = 1/ρ².
4.3.6 Lets be the proper time on a geodesic andu à (s) the velocity of a mass in free fall Then the scalar d ds(V ãu) =dV ds ãu+V β d 2 x β ds 2 à
∂ à V α −Γ β αà V β ¢ involves the covariant derivative which is a four-vector by the quotient theorem Note that the use of the geodesic equation ford 2 x β /ds 2 is the key here.
4.3.7 For this exercise we need the identity
∂q j , which can be proved by writingg ik =ε i ãε k and differentiating.
In the above we have used the metric tensor to raise or lower indices and the relationA k B k =A k B k , and have identified Christoffel symbols using the definition in Eq (4.54).
The last line of the above series of equations can be rearranged to the form constituting a solution to the exercise.
4.3.9 All but three of the covariant derivative components of a contravariant vectorVare of the form
The remaining three components are
To determine the condition g ij ;k = 0, we start by taking the covariant derivative of the identity g im g mj = δ i j This leads to the equation 0 = g im;k g mj + g im g mj ;k, which simplifies to g im g ;k mj By multiplying this expression by g ni and applying the relation g ni g im = δ m n, we conclude that g ;k nj = 0.
4.3.11 To start, note that the contravariantV k are, in the notation of Eq (3.157),
V r ,V θ /r,V ϕ /rsinθand that [det)g)] 1/2 =r 2 sinθ Then the tensor deriva- tive formula, Eq (4.69), evaluates straightforwardly to Eq (3.157). 4.3.12 ∂ à Φ ;ν =∂ ν ∂ à Φ−Γ α àν ∂ α Φ≡∂ ν Φ ;à =∂ à ∂ ν Φ−Γ α νà ∂ α Φ.
4.4.1 ∇(uv) =v∇u+u∇v follows from the product rule of differentiation.
∂v∇v= 0,∇uand∇vare parallel, so that (∇u)×(∇v) = 0.
(b) If (∇u)×(∇v) = 0, the two-dimensional volume spanned by∇uand
∇v, also given by the Jacobian
To compute the partial derivatives ∂(x, y)/∂(u, v), it is essential to express x and y in terms of u and v The explicit formulas for these variables are given by x = uv/(v + 1) and y = u/(v + 1) This formulation simplifies the process of obtaining the required derivatives.
(b) Here we first needJ −1 , computed as follows:
Taking the reciprocal to obtainJ, and rewriting in terms ofuandv (the form usually needed ifJ is to be inserted into an integral overuandv), we get
(v+ 1) 2 , in agreement with the answer to part (a).
In Example 4.5.2, the results for the starred quantities ∗1 and ∗(dt∧dx 1 ∧dx 2 ∧dx 3) were thoroughly analyzed, along with the value of ∗dx 1 The signs of the results for ∗dx i (where i ≠ 1) align due to the cyclic ordering of differentials dx i, dt, dx j, and dx k, which shares the same parity as dx 1, dt, dx 2, and dx 3 For ∗dt, the standard ordering is achieved when dt is followed by the other differentials, confirming that dt, being the only starred differential, has a metric tensor element g tt = +1, thus validating the value assigned to ∗dt.
Example 4.5.2 derived a value for ∗(dt∧dx 1) Corresponding results for
∗(dt∧dx i ) hold because the cyclic ordering of dx i , dx j , dx k causes the permutation to standard order to be the same for alli.
To confirm that ∗(dx j ∧ dx k) = dt ∧ dx i, we observe that the sequence dx j, dx k, dt, dx i represents an even permutation of the standard order Additionally, since both g jj and g kk equal -1, there is no sign change in the outcome.
In examining the expression ∗(dx 1 ∧ dx 2 ∧ dx 3), we observe that the components dx 1, dx 2, dx 3, and dt form an odd permutation of the standard order However, since the starred quantity relates to three negative diagonal elements of the metric tensor, the resulting value is positive, as demonstrated in Equation (4.82).
The final case to examine involves the expression ∗(dt∧dx i ∧dx j ) It is important to note that the elements dt, dx i, dx j, and dx k represent an even permutation of the standard order Additionally, the components dt and dx i, dx j contribute two minus signs from the metric tensor elements Consequently, the overall sign of the expression is positive, as illustrated in Equation (4.82).
In a constant force field, the work done during motion in the x direction can be expressed as a constant multiplied by the differential displacement, represented as a_x dx This concept similarly extends to motions in the y and z directions, leading to the conclusion that the work can be described by the 1-form w + a.
(b)dω 2=dx∧dy−dy∧dx= 2dx∧dy.
(c)d(dx∧dy) =d(dx)∧dy−dx∧d(dy) = 0.
4.6.2 dω 3=y dx∧dz+xdy∧dz+z dx∧dy+xdz∧dy−z dy∧dx−zdy∧dx
= 2y dx∧dz+ 2z dx∧dy. d(dω 3) = 2dy∧dx∧dz+ 2dz∧dx∧dy= 0.
4.6.3 (a) (xdy−ydx)∧(xy dz+xz dy−yz dx) =x 2 y dy∧dz
−xy 2 dx∧dz+x 2 z dy∧dy−xyz dx∧dy−xyz dy∧dx+y 2 z dx∧dx
=x 2 y dy∧dz−xy 2 dx∧dz.
Applyd: d(x 2 y dy∧dz)−d(xy 2 dx∧dz) = 2xy dx∧dy∧dz
+x 2 dy∧dy∧dz−y 2 dx∧dx∧dz−2xy dy∧dx∧dz
(b)dω 2 ∧ω 3 −ω 2 ∧dω 3= 2dx∧dy∧(xy dz+xz dy−yz dx)
−(x dy−y dx)∧(2y dx∧dz+ 2z dx∧dy)
= 2xy dx∧dy∧dz−2xy dy∧dx∧dz
4.7.1 Let dx = a 1 du+a 2 dv +a 3 dw, dy = b 1 du+b 2 dv+b 3 dw, and dz c 1 du+c 2 dv+c 3 dw Then, dx∧dy∧dz= (a 1 du+a 2 dv+a 3 dw)∧(b 1 du+b 2 dv+b 3 dw)∧(c 1 du+c 2 dv+c 3 dw).
By expanding the right-hand side and eliminating terms with duplicate differentials, we can rearrange the wedge products into standard order with the appropriate sign assignments This process leads us to the expression dx∧dy∧dz = (a1b2c3 - a1b3c2 - a2b1c3 + a2b3c1 + a3b1c2 - a3b2c1) du∧dv∧dw.
We recognize the coefficient ofdu∧dv∧dw as the determinant
To complete the idenfication ofJ as a Jacobian, note that a 1 =∂x/∂u, a 2=∂x/∂v, and so on, and thereforeJ =∂(x, y, z)/∂(u, v, w).
4.7.3 ydx+xdy: Closed; exact because it isd(xy). ydx+xdy x 2 +y 2 : Not closed because
[ln(xy) + 1]dx+x y dy: Closed because ∂A/∂y= 1/y=∂B/∂x.
It is exact, beingd(xlnxy).
This exercise closely resembles a prior differential form, differing only in the sign of the dx term It is identified as a closed form and is exact, represented as dtan(y/x) The function f(z)dz can be expressed as f(x+iy)(dx+idy).
∂A/∂y=∂B/∂x=if 0 (z) It is closed; also exact because Aand B can be obtained as derivatives of the indefinite integralR f(z)dz.
Vector Spaces
5.1.1 Using orthogonality the hφ n |fi=a n Z b a w(x)f(x)φ n (x)dxare derived fromf and therefore unique.
(c i −c 0 i )φ i = 0.Letc k −c 0 k 6= 0 be the first non-zero term Then φ k =− 1 c k −c 0 k
(c i −c 0 i )φ i would say that φ k is not linearly independent of theφ i , i > k,which is a contradiction.
This results also from minimizing the mean square error
|hf|gi| 2 ≤ hf 2 ihg 2 ibecause the double integral is nonnegative.
5.1.7 Theϕ j are assumed to be orthonormal ExpandingI, we have
I=hf|fi −X i a ∗ i hϕ i |fi −X i a i hf|ϕ i i+X ij a ∗ i a j hϕ i |ϕ j i ≥0.
Using the relationa i =hϕ i |fiand the orthonormality conditionhϕ i |ϕ j i δ ij ,
5.1.8 The expansion we need is sinπx=X i hϕ i |sinπxi hϕ i |ϕ i i ϕ i (x).
Figure 5.1.8 Red line is approximation throughϕ 3 , black line is exact. sinπx=2/π
This series converges fairly rapidly See Fig 5.1.8.
By integration we finda 0= 1/2,a 1= 1/4,a 2= 1/8,a 3= 1/16 Thus, e −x = 1
This expansion when terminated afterL 3fails badly beyond aboutx= 3. See Fig 5.1.9.
5.1.10 The forms P i |ϕ i ihϕ i | and P j |χ j ihχ j | are resolutions of the identity. Therefore
The coefficients off in theϕbasis area i =hϕ i |fi, so the above equation is equivalent to f =X j b j χ j , with b j =X i hχ j |ϕ i ia i
5.1.11 We assume the unit vectors are orthogonal Then,
This expression is a component decomposition ofa.
Figure 5.1.9 Red line is approximation through L 3, black line is exact.
5.1.12 The scalar productha|aimust be positive for every nonzero vector in the space If we writeha|aiin the form (a 1 −a 2) 2 + (k−1)a 2 2 , this condition will be violated for some nonzeroa unlessk >1.
5.2.1 The solution is given in the text Note thata 10=−1/2, a 20=−1/3, a 21 −1/2, a 30=−1/4, a 31=−9/20, a 32=−1/4.
5.2.2 The solution is given in the text Note thata 10=−1, a 20=−2, a 21= 4.
5.2.3 The solution is given in the text Note that a 10=−2, a 20=−6, a 21 −6√
5.2.4 The solution is given in the text Note thata 10= 0, a 20=−1/2, a 21= 0.
5.2.5 Relying without comment on the integral formulas in Exercise 13.3.2, we compute first hx 0 |x 0 i Z 1
Note that some integrals are zero by symmetry.
The polynomial \( T_0 \) is defined as \( c_0 x_0 \), with \( c_0 \) satisfying the condition \( \langle c_0 x_0 | c_0 x_0 \rangle = |c_0|^2 \langle x_0 | x_0 \rangle = \pi \), leading to \( c_0 = 1 \) and \( T_0 = 1 \) Due to symmetry, the polynomial \( T_1 \) is a linear combination of \( x_0 \) and \( x_1 \) but is constrained to be an odd function dependent solely on \( x_1 \), taking the form \( c_1 x \) This polynomial is inherently orthogonal to \( T_0 \), and \( c_1 \) must fulfill the requirement \( \langle c_1 x | c_1 x \rangle = |c_1|^2 \langle x | x \rangle = \pi \).
Becausehx|xi=π/2, we have c 1= 1 andT 1=x.
The determination of T 1 is a bit less trivial T 2 will be an even function ofx, and will be of the general form
The constantc 2 is now determined from the normalization condition: hT 2 |T 2 i=|c 2 | 2 ¿à x 2 −1 2 ả ¯¯ ¯¯ à x 2 −1 2 ả À
2 , from which we findc 2= 2 and T 2= 2x 2 −1.
5.2.6 From the formula given in the Hint, we have hx 0 |x 0 i Z 1
Taking U 0 = c 0 x 0 , we find |c 0 | 2 hx 0 |x 0 i = π/2, so c 0 = 1 and U 0 = 1. TheU n have even/odd symmetry, so U 1 =c 1 x, and|c 1 | 2 hx|xi=π/2, so
5.2.7 The solution is given in the text Note thata 10=−1/√ π.
5.2.8 Let the orthonormalized vectors be denoted b i First, Make b1 a nor- malized version of c1: b1=c1 /√
3 Then obtainb 2 (denoting b2 before normalization) as b2=c2 −(b1 ãc2)b1
2b3 Collecting our answers, the orthonormal vec- tors are b 1
5.3.1 For arbitraryϕandψwithin our Hilbert space, and an arbitrary operator
Since the first and last expressions in this chain of equations are equal for allA,ψ, and ϕ, we may conclude that (A † ) † =A.
5.3.3 (a) (A1) ij =hx i |A 1 |x j i A corresponding formula holds forA2 Comput- ing for eachi andj, we find
5.3.4 (a) First computeAP n (P n are the normalized polynomials).
Using the above and noting that our basis is theP n , we construct
Note: We built the matrix ofAdirectly from the expansions An alternate and equally valid approach would be to identify the matrix elements as scalar products.
6/5; the coefficients ofP 2 andP 0 vanish becausex 3 is odd From the above data, we get x 3 = (2√
6/5)P 1(x) Thus, the column vector representingx 3 is x 3 ←→
Inserting the explicit forms ofP 1 andP 3, we find
2x) = 3x 3 , in agreement with the directly computed value.
5.4.2 (AB) † =B † A † =BA=ABif and only if [B, A] = 0.
5.4.3 (AB−BA) † = (iC) † =B † A † −A † B † =−iC † =BA−AB=−iC.
5.4.5 (a) For the normalization ofϕ 3=Cz/r=Ccosθ, we need the following integral:
The normalized form of ϕ 3 is thereforep
3/4π(z/r) To check orthogo- nality, we need integrals such as
The integral of the function cos(ϕ)sin(ϕ)sin(2ϕ)/2 vanishes when evaluated over two complete periods, demonstrating its symmetry This symmetry suggests that all other normalization and orthogonality integrals yield similar results.
(b) It is useful to note that ∂(1/r)/∂x=−x/r 3 ; similar expressions are obtained ifxis replaced byy or z Now,
BecauseL z is antisymmetric inxandy, we also have
Combining the above into a matrix representation ofL z ,
Similar processes (or cyclic permutation of x, y, z) lead to the matrix representations
(c) Form the matrix operations corresponding toL x L y −L y −L x :
Carrying out the matrix multiplication and subtraction, the result is i times the matrix ofL z
5.5.1 (a) (1) The column vector representingf(θ, ϕ) isc
5.5.2 (a) Theith column ofU describesϕ i in the new basis Thus,
(b) The transformation is a counterclockwise rotation of the coordinate system about the y axis; this corresponds to the Euler angles α = 0, β = π/2, γ = 0 The above U is reproduced when these angles are substituted into Eq (3.37).
This vector corresponds to f 0 =−x−3y+ 2z, which is consistent with application of the relevant basis transformation tof.
5.5.3 Since the matrixUfor the transformation of Exercise 5.5.2 is unitary, the inverse transformation has matrixU † , which is
The above indicates thatf(x) = (cosθ+3isinθ)χ 1+(−3cos 2 θ+isinθ(cosθ−
Using the above, we find
2 sin 2 θ−3 cosθ+isinθcosθ cosθ+ 2isinθ
OnlyVUf gives the correct result that we found in part (a).
5.5.5 (a) The normalized versions of theP n , denotedP n , areP n =p
(2n+ 1)/2P n The normalized versions of the F n , denoted F n , are F 0 = p
1/8F 2. (b) The transformation matrixU has elementsu ij =hF i |P j i For exam- ple, u 02 Z 1
The complete transformation matrix is
(c)Vhas elementsv ij =hP i |F j i Thus,
(d) By matrix multiplication we can verify that UV = 1, showing that
V = U −1 Since V is also U † , we can also conclude that U and V are unitary.
Lettingcandc 0 be the vectors describing the expansions off(x) respec- tively in theP n and theF n bases, c
5.6.1 (a) The first column of S x shows the result of its operation on α; the second column describes S x β Similar observations apply toS y amd S z
(b) (1) Check thathα+β|α−βi= 0 Expanding, we havehα|αi − hα|βi+ hβ|αi − hβ|βi= 1 + 0 + 0−1 = 0.
(2) A similar expansion shows that hα+β|α+βi = 2, so a proper value ofCis 1/√
2 The same result is obtained forhϕ 0 2 |ϕ 0 2 i.
(3) The matrix elements of the transformation are u ij = hϕ 0 i |ϕ j i.
(c) In the transformed basis, the matrix of an operator S becomes S 0 U S U −1 Noting thatU −1 =U, we compute
Similar operations forS y and S z yield
Therefore the matrix ofL x for this basis is
(b) Form UL x U −1 U is unitary; this can be checked by verifying that
(c) The new basis functions have coefficients (in terms of the original basis) that are the columns ofU † Reading them out, we have ϕ 0 1 =Cxe −r 2 , ϕ 0 2 = C
In this section, we define determinants D1, D2, and D3 based on the overlap matrix elements of the basis functions The overlap matrix elements are represented as s_ij = hχ_i | χ_j i, leading to D1 = S_11 and D2 = S_11 S_22 - S_12 S_21 By substituting these into the formulas derived in Section 5.2, we establish systematic formulas for the basis functions, with the first function defined as ϕ_1 = χ_1.
Comparing with the matrix T as defined in Example 5.6.1, we see that itsjth column consists of the coefficients of theχ i in the formula for ϕ j From the above formulas, we find
5.7.1 Replacexbyx 0 =UxU −1 andp 0 =pbyUpU −1 , so
5.7.2 We need σ 0 1 =Uσ 1U † Ã sin 2θ cos 2θ cos 2θ −sin 2θ
5.7.3 (a) From the equationsL x ϕ 1= 0,L x ϕ 2=iϕ 3,L x ϕ 3=−iϕ 2, we see that
L x applied to any function in the space spanned byϕ 1,ϕ 2,ϕ 3a function that remains within that space The above equations correspond to the action on theϕbasis of the matrix
If this equation is transformed byU, the quantities in it become
, and the transformed matrix equation is
(d) The ϕ 0 are those linear combinations of theϕ with coefficients that are complex conjugates of the corresponding row ofU, and areϕ 0 1 =ϕ 1 x e −r 2 ,ϕ 0 2 = (ϕ 2+iϕ 3)/√
(e) The matrix equation is equivalent to
This is a result that was proved in part (b).