467

Appendix: Physical Constants andMathematical Relations

c©2010 by Harvey Gould and Jan Tobochnik5 February 2010

A.1 Physical Constants and Conversion Factors

constant symbol magnitudeAvogadro’s number NA 6.022 × 1023

Boltzmann’s constant k 1.381 × 10−23 J/Kuniversal gas constant R 8.314 J/(mol K)Planck’s constant h 6.626 × 10−34 J s

~ 1.055 × 10−34 J sspeed of light c 2.998 × 108 m/selectron charge e 1.602 × 10−19 Celectron mass me 9.109 × 10−31 kgproton mass mp 1.672 × 10−27 kg

Table A.1: Useful physical constants.

newton 1 N ≡ 1 kgm/s2

joule 1 J ≡ 1Nmwatt 1 W ≡ 1 J/spascal 1 Pa ≡ 1N/m2

Table A.2: SI derived units. For information about SI units see <physics.nist.gov/cuu/Units/>.

1 atm = 1.013bar= 1.013 × 105 Pa= 760 mmHg

1 cal = 4.186J1 eV = 1.602 × 10−19 J

Table A.3: Conversion factors.

468

A.2 Hyperbolic Functions

coshx =1

2[ex + e−x]. (A.1)

sinhx =1

2[ex − e−x]. (A.2)

tanhx =sinhx

coshx=

ex − e−x

ex + e−x. (A.3)

A.3 Approximations

f(x) = f(a) + (x − a)f ′(a) +1

2!(x − a)2f ′′(a) +

1

3!(x − a)3f ′′′(a) + · · · . (A.4)

ex =

∞∑

n=0

xn

n!≈ 1 + x +

x2

2!+ · · · . (A.5)

sin x =

∞∑

n=0

(−1)n x2n+1

(2n + 1)!≈ x − x3

3!+

x5

5!+ · · · . (A.6)

cosx =

∞∑

n=0

(−1)n x2n

(2n)!≈ 1 − x2

2!+

x4

4!+ · · · . (A.7)

(a + x)p = ap(

1 +x

a

)p

= ap∞∑

n=0

(

p

n

)

(x

a

)n

(A.8)

≈ ap[

1 + p(x

a

)

+p(p − 1)

2!

(x

a

)2

+ · · ·]

. (A.9)

1

1 − x=

∞∑

n=1

xn−1 ≈ 1 + x + x2 + · · · . (A.10)

ln(1 + x) =∞∑

n=1

(−1)n+1 xn

n≈ x − x2

2+

1

3x3 + · · · . (A.11)

tanhx =

∞∑

n=1

22n(22n − 1)

(2n)!B2n x2n−1 ≈ x − x3

3+

2x5

15+ · · · , (A.12)

where Bn are the Bernoulli numbers (see Section A.7).

A.4 Euler-Maclaurin Formula

∞∑

i=0

f(i) =

∫

∞

0

f(x) dx +1

2f(0) − 1

12f ′(0) +

1

720f ′′′(0) + · · · . (A.13)

469

A.5 Gaussian Integrals

In ≡∫

∞

−∞

xne−ax2

dx (n ≥ 0, a > 0). (A.14)

I0 =(π

a

)1/2

. (A.15)

I1 = 0. (A.16)

I2 =1

2

( π

a3

)1/2

. (A.17)

Derivation:

I0 =

∫ +∞

−∞

e−ax2

dx. (A.18)

Because x in the integrand of (A.18) is a dummy variable, we may equally well write I0 as

I0 =

∫ +∞

−∞

e−ay2

dy. (A.19)

To convert the integrand to a form we can integrate, we multiply I0 by itself and write

I20 =

∫ +∞

−∞

e−ax2

dx

∫ +∞

−∞

e−ay2

dy =

∫ +∞

−∞

∫ +∞

−∞

e−a(x2+y2) dx dy. (A.20)

The double integral in (A.20) extends over the entire x-y plane. We introduce the polar coordinatesr and θ, where r2 = x2 + y2. The element of area in polar coordinates is rdr dθ. Hence, I2

0 can berewritten in the form

I20 =

∫ 2π

0

∫

∞

0

e−ar2

rdr dθ = 2π

∫

∞

0

e−ar2

rdr. (A.21)

We let z = ar2, dz = 2ardr, and write

I20 =

π

a

∫

∞

0

e−z dz =π

a

[

− e−z]

∞

0=

π

a. (A.22)

Hence, we obtain the desired result

I0 =

∫ +∞

−∞

e−ax2

dx =(π

a

)1/2

. (A.23)

The values of In for n odd in (A.15) are zero by symmetry. For odd values of n, we define In

as

In =

∫

∞

0

xne−ax2

dx (n odd). (A.24)

It is straightforward to show that

I1 =1

2a. (A.25)

The integrals In and In for n > 1 can be found from the integrals I0 or I1 using the recursionrelation

In = −∂In−2

∂a. (A.26)

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A.6 Stirling’s Approximation

We first use the definition of N ! as

N ! = 1 × 2 × 3 × · · · × N (A.27)

to obtain the weaker form of Stirling’s approximation

lnN ! = ln 1 + ln 2 + ln 3 + · · · + lnN (A.28a)

≈∫ N

1

lnxdx =[

x ln x − x]N

1(A.28b)

= N lnN − N + 1. (A.28c)

For N ≫ 1 we have ln N ! ≃ N lnN − N .

A more accurate approximation for N ! can be found from the integral representation [see(A.36)]:

N ! =

∫

∞

0

xNe−x dx. (A.29)

In the integrand, xN is a rapidly increasing function of x for large N , and e−x is a decreasingfunction of x. Hence f(x) = xNe−x exhibits a sharp maximum for some value of x. As usual, it iseasier to consider ln f(x) and find where it has its maximum:

d ln f(x)

dx=

d

dx(N lnx − x) =

N

x− 1. (A.30)

We next do a Taylor expansion of ln f(x) about its maximum at x = N . We write x = N + z, andexpand in powers of z:

N ln(N + z) − (N + z) = N lnN + N ln(

1 +z

N

)

− N − z (A.31a)

≈ N lnN + N( z

N− 1

2

z2

N2

)

− N − z (A.31b)

= N lnN − N − z2

2N. (A.31c)

Hence,

f ≈ NNe−Ne−Nz2/2, (A.32)

and thus

N ! =

∫

∞

0

f(x) dx ≈ NNe−N

∫

∞

−N

e−Nz2/2 dz. (A.33)

Because the integrand is sharply peaked about z = 0, we can extend the lower limit of integrationto −∞. Hence, we obtain

N ! ≈ NNe−N

∫

∞

−∞

e−Nz2/2 dz = NNe−N (2πN)1/2. (A.34)

471

and therefore

lnN ! = N lnN − N +1

2ln(2πN) (Stirling’s approximation). (A.35)

The Gamma function is defined as

Γ(n) =

∫

∞

0

xn−1 e−x dx (Gamma function). (A.36)

It is easy to show by integrating by parts that

Γ(n + 1) = nΓ(n) = n! (positive integer n). (A.37)

Note that Γ(1) = Γ(2) = 1. Hence Γ(n) can be interpreted as a generalization of the factorialfunction.

For half integer arguments, Γ(n/2) has the special form

Γ(n

2

)

=(n − 2)!!

√π

2n−1)/2, (A.38)

where n!! = n × (n − 2) × · · · × 3 × 1 if n is odd and n!! = n × (n − 2) × · · · × 4 × 2 if n is even.We also have −1!! = 0!! = 1, Γ(1

2 ) =√

π, and Γ(32 ) =

√π/2.

A.7 Bernoulli Numbers

The Bernoulli numbers are the coefficients of xn/n! in the expansion of

x

ex − 1=

∑

n=0

Bnxn

n!. (A.39)

All the Bernoulli numbers Bn with odd n are zero except for B1, that is, B2n+1 = 0 for n > 0:

B0 = 1, B1 = −1

2, B2 =

1

6, B4 = − 1

30, B6 =

1

42, B8 = − 1

30. (A.40)

A.8 Probability Distributions

PN (n) =N !

n! (N − n)!pnq(N−n) (binomial distribution). (A.41)

The binomial distribution is specified by the probability p = 1 − q and N .

P (x) =1√

2πσ2e−(x−x)2/2σ2

(Gaussian distribution). (A.42)

The Gaussian distribution is specified by x and σ2 = x2 − x2:

P (n) =λn

n!e−λ (Poisson distribution). (A.43)

The Poisson distribution is specified only by the parameter λ = n = pN .

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A.9 Fourier Transforms

The Fourier transform of the function f(x) is defined as

f(k) =

∫

f(x) e−ikx dx. (A.44)

Similarly, the inverse transform of f(k) is defined as

f(x) =1

2π

∫

f(k)eikx dk. (A.45)

Note the sign of the exponential in (A.44) and the presence of the factor of 1/2π in (A.45). Otherdefinitions of the Fourier transform are common, especially in fields other than physics. In threedimensions we have

f(k) =

∫

f(r)e−ik·r d3r. (A.46)

The three integrals in (A.46) can be reduced to a single integral if f(r) depends only on r = |r|by using spherical coordinates. We write dr = dx dy dz → r2 sin θ dr dθ dφ and k · r → kr cos θ.Hence, we can write

f(k) =

∫

∞

0

∫ π

0

∫ 2π

0

f(r)e−ikr cos θ r2dr sin θdθ dφ. (A.47)

The integral over φ gives 2π. We can do the integral over θ by making the substitution x = cos θand writing dx = − sin θ dθ:

∫ π

0

e−ikr cos θ sin θ dθ → −∫

−1

1

e−ikrx dx =1

ikr

[

e−ikrx]x=−1

x=1(A.48a)

=1

ikr[eikr − e−ikr] =

2

krsin kr, (A.48b)

where [eikr − e−ikr ]/2i = sin kr. Hence (A.46) reduces to

f(k) = 4π

∫

∞

0

f(r) rsin kr

kdr. (A.49)

A.10 The Delta Function

The Kronecker delta δij is defined as

δij =

{

1 if i = j,

0 if i 6= j,(A.50)

where i and j are integers. The Kronecker delta has the property that

∞∑

i=−∞

δijai = aj . (A.51)

473

The Dirac delta function δ(x) can be thought of as a generalization of the Kronecker deltafunction for continuous variables. Loosely speaking, δ(x) is zero everywhere except at x = 0 whereits value is infinitely large such that its integral is 1, that is,

∫

∞

−∞

δ(x) dx = 1. (A.52)

The Dirac delta function has the useful property that

∫

∞

−∞

f(x)δ(x) dx = f(0), (A.53)

or more generally∫

∞

−∞

f(x)δ(x − a) dx = f(a). (A.54)

Note that the property of the Dirac delta in (A.54) is analogous to (A.51). Another useful propertyis that

∫

∞

−∞

δ(λx) dx =1

|λ| . (A.55)

Despite its name, the Dirac delta function is not really a function and was not first introduced byDirac.

The Fourier transform can be used to obtain a useful representation of the Dirac delta function.The Fourier transform of δ(x) is

δ(k) =

∫

∞

−∞

δ(x)e−ikx dx = 1. (A.56)

The inverse transform gives the desired representation:

δ(x) =1

2π

∫

∞

−∞

eikx dk. (A.57)

A.11 Convolution Integrals

A convolution integral expresses the amount of overlap of one function as its argument is shiftedfrom the argument of another function. The form of the convolution integral in one dimension is

C(x) =

∫

∞

−∞

f(x′)g(x − x′) dx′. (A.58)

The integral (A.58) can be expressed as a product of two Fourier transforms. To derive this resultwe use (A.45) to write the right-hand side of (A.58) as

C(x) =1

(2π)2

∫∫∫

f(k)g(k′)eikx′

eik′(x−x′)dk dk′ dx′. (A.59)

474

We first group terms that depend on x′ and do the integral over x′ using (A.57):∫

eix′(k−k′)dx′ = 2πδ(k − k′). (A.60)

We then use this result to write the right-hand side of (A.59) as

C(x) =1

2π

∫∫

f(k)g(k′)eik′xδ(k − k′)dk dk′ (A.61a)

=1

2π

∫

f(k)g(k)eikxdk. (A.61b)

A.12 Fermi and Bose Integrals

The integrals that commonly occur in the context of the ideal Fermi gas have the form∫

∞

0

xn

ex + 1dx =

(

1 − 1

2n

)

Γ(n + 1)ζ(n + 1), (A.62)

where the Riemann zeta function is defined by

ζ(n) =

∞∑

k=0

1

(k + 1)n. (A.63)

The values of ζ(n) that we will use most often are

ζ(3

2

)

≈ 2.612, (A.64a)

ζ(2) =π2

6≈ 1.645, (A.64b)

ζ(5

2

)

≈ 1.341, (A.64c)

ζ(3) ≈ 1.202, (A.64d)

ζ(4) =π4

90≈ 1.082. (A.64e)

The integrals that are needed in the context of the ideal Bose gas have the form

∫

∞

0

dxxn

ex − 1=

∫

∞

0

dxxne−x

1 − e−x=

∫

∞

0

dxxn∞∑

k=0

e−(k+1)x (A.65a)

=

∞∑

k=0

∫

∞

0

dxxne−(k+1)x =

∞∑

k=0

1

(k + 1)n+1

∫

∞

0

dy yne−y. (A.65b)

If we use the definition of the Riemann zeta function in (A.62) and the definition of the Gammafunction in (A.36), we obtain

∫

∞

0

dxxn

ex= ζ(n + 1)Γ(n + 1). (A.66)

475

If n is an integer, than (A.66) reduces to

I(n) = n!ζ(n + 1). (A.67)