String Theory: Chapter 3 Problem Set Solutions for Barton Zwiebach's "A First Course in String Theory"
Warning: These are my own solutions for the given problems. If you are a student, use these at your own risk. They have not been subjected
to grading in a university setting. Your professor may require more explicit statements, or other details worked out in the calculations.
Quick Calculations
Problems
(*) Asterisk indicates solution may be fuzzy or lacking sufficient rigor.
(**) Double asterisk indicates solution may be significantly flawed.
\( (\dagger) \) Dagger indicates the problem is marked in the textbook as referenced again in later chapters.
(Unmarked indicates solution should be correct other than minor quibbles.)
Quick Calculations
Problem Statement: Verify that \( \overrightarrow{E} \), given Equation 3.8, is invariant under the gauge transformations given by expressions 3.10
Step (1): Write equations 3.8 and 3.10
Equation 3.8: Electric field in terms of scalar potential and vector potential
\( \overrightarrow{E} = - \frac{1}{c} \frac{\partial \overrightarrow{A}}{\partial t} - \nabla \Phi \)
Equations 3.10: Gauge transformations
\( \Phi \rightarrow \Phi ' = \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \)
\( \overrightarrow{A} \rightarrow \overrightarrow{A'} = \overrightarrow{A} + \nabla \epsilon \)
Step (2): Write expression for \( \overrightarrow{E'} \) and substitute in the gauge transformation equations 3.10
\( \overrightarrow{E'} = - \frac{1}{c} \frac{\partial \overrightarrow{A'}}{\partial t} - \nabla \Phi ' \)
\( \Rightarrow \overrightarrow{E'} = - \frac{1}{c} \frac{\partial}{\partial t}( \overrightarrow{A} + \nabla \epsilon ) - \nabla ( \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} ) \)
But then when you distribute these through, you have products of \( \frac{\partial}{ \partial t}\) and \( \nabla \), which cancel out to zero:
\( \overrightarrow{E'} = - \frac{1}{c} ( \frac{\partial}{\partial t} \overrightarrow{A} + \frac{\partial}{\partial t} \nabla \epsilon ) - ( \nabla \Phi - \frac{1}{c} \nabla \frac{\partial \epsilon}{\partial t} ) \)
\( \overrightarrow{E'} = - \frac{1}{c} ( \frac{\partial}{\partial t} \overrightarrow{A} + 0 ) - ( \nabla \Phi - 0 ) \)
This leaves an expression for \( \overrightarrow{E} \) that is equal to Equation 3.8, which means it is invariant under the gauge transformations 3.10:
\( \overrightarrow{E'} = - \frac{1}{c} \frac{\partial \overrightarrow{A}}{\partial t} - \nabla \Phi \)
\( \Rightarrow \overrightarrow{E'} = \overrightarrow{E} \) Q.E.D.
Problem Statement: Verify that the gauge transformations 3.10 are correctly summarized by expression 3.21
Step (1): Write equations 3.10 and 3.21
Equations 3.10: Gauge transformations
\( \Phi \rightarrow \Phi ' = \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \)
\( \overrightarrow{A} \rightarrow \overrightarrow{A'} = \overrightarrow{A} + \nabla \epsilon \)
Equation 3.21: Gauge transformations in terms of index notation for vector potential \( \overrightarrow{A} \).
\( A_{\mu} \rightarrow A'_{\mu} = A_{\mu} + \partial_{\mu} \epsilon \)
Step (2): Substitute \( \Phi ' \) and A' values into components \( A_{\mu} \). This is taking \( A_{\mu} \rightarrow A'_{\mu} \).
Equation 3.14: \( A_{\mu} = ( - \Phi , A^{1} , A^{2} , A^{3} ) \)
\( \Rightarrow A'_{\mu} = [ - ( \Phi - \frac{1}{c} \frac{\partial}{\partial t} \epsilon ) , (A^{1} + \frac{\partial}{\partial x^{1}} \epsilon) , (A^{2} + \frac{\partial}{\partial x^{2}} \epsilon) , (A^{3} + \frac{\partial}{\partial x^{3}} \epsilon) ] \)
Now, the \( - \Phi \) term is just \( A_{0} \), and \( A^{\mu} = A_{\mu} \) for the spatial indices. This implies:
\( A'_{\mu} = ( A_{0} + \frac{1}{c} \frac{\partial \epsilon}{\partial t} ) + ( \overrightarrow{A} + \nabla \epsilon ) \)
Rearranging these A components and grouping the \( \epsilon \) terms:
\( A'_{\mu} = ( A_{0} + A_{1} + A_{2} + A_{3} ) + ( \frac{1}{c} \frac{\partial}{\partial t} \epsilon + \nabla \epsilon ) \)
The factor of \( \frac{1}{c} \) folds into converting \( \frac{\partial}{\partial t} \) into \( \frac{\partial}{\partial x^{0}} \), as \( \gamma ds = c dt \).
\( A'_{\mu} = ( A_{0} + A_{1} + A_{2} + A_{3} ) + ( \frac{\partial \epsilon }{ \partial x^{0}} + \frac{\partial \epsilon }{ \partial x^{1}} + \frac{\partial \epsilon }{ \partial x^{2}} + \frac{\partial \epsilon }{ \partial x^{3}}) \)
\( \Rightarrow A'_{\mu} = A_{\mu} + \partial_{\mu} \epsilon \) Q.E.D.
Problem Statement: Verify the equations in 3.26, \( T_{\lambda \mu \nu} = - T_{\mu \lambda \nu} \) and \( T_{\lambda \mu \nu} = - T_{\lambda \nu \mu} \).
Step (1): Write down the definition of \( T_{\lambda \mu \nu} \) from Equations 3.23.
Equation 3.23: Definition of field strength combinations \( T_{\lambda \mu \nu} \)
\( T_{\lambda \mu \nu} \equiv \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda} + \partial_{\nu} F_{\lambda \mu} \)
Step (2): Rearrange equations in 3.26 so that they equal zero, then substitute in Equation 3.23 to verify.
\( T_{\lambda \mu \nu} + T_{\mu \lambda \nu} = 0 \)
\( \Rightarrow \partial_{\lambda} (F_{\mu \nu} + F_{\nu \mu}) + \partial_{\mu} (F_{\nu \lambda} + F_{\lambda \nu}) + \partial_{\nu} (F_{\lambda \mu} + F_{\mu \lambda}) = 0 \)
\( T_{\lambda \mu \nu} + T_{\lambda \nu \mu} = 0 \)
\( \Rightarrow \partial_{\lambda} (F_{\mu \nu} + F_{\nu \mu}) + \partial_{\mu} (F_{\nu \lambda} + F_{\lambda \nu}) + \partial_{\nu} (F_{\lambda \mu} + F_{\mu \lambda}) = 0 \)
We know that the field strength tensor \( F_{\mu \nu} \) is antisymmetrical, therefore that the terms in those equations cancel to zero:
\( F_{\mu \nu} = - F_{\nu \mu} \)
\( F_{\lambda \nu} = - F_{\nu \lambda} \)
\( F_{\mu \lambda} = - F_{\lambda \mu} \)
\( \Rightarrow T_{\lambda \mu \nu} + T_{\mu \lambda \nu} = ( 0 + 0 + 0 ) = 0\)
and \( T_{\lambda \mu \nu} + T_{\lambda \nu \mu} = ( 0 + 0 + 0 ) = 0 \)
Which verifies Equations 3.26, that \( T_{\lambda \mu \nu} = - T_{\mu \lambda \nu} \) and \( T_{\lambda \mu \nu} = - T_{\lambda \nu \mu} \). Q.E.D.
Problem Statement: Show that \( F^{\mu \nu} = - F^{\nu \mu} \), \( F^{0 i} = - F_{0 i} \), and \( F^{i j} = F_{i j} \).
Step (1): Write down the definition of \( F^{ \mu \nu } \) from Equations 3.29.
Equation 3.29: Relation of \( F^{ \mu \nu } \) with lower indices \( F_{ \alpha \beta } \), dummy variables \( \alpha , \beta \)
\( F^{\mu \nu} = \eta^{\mu \alpha} \eta^{\nu \beta} F_{\alpha \beta} \)
Step (2): Rearrange them so they sum to zero. Substitute in Equation 3.29 and verify that they sum to zero.
Quick Calculation 3.4.1: \( F^{\mu \nu} = - F^{\nu \mu} \), anti-symmetric upper index field strengths.
\( F^{\mu \nu} + F^{\nu \mu} = 0 \)
\( \Rightarrow \eta^{\mu \alpha} \eta^{\nu \beta} F_{\alpha \beta} + \eta^{\nu \alpha} \eta^{\mu \beta} F_{\alpha \beta} = 0 \)
Index swap the second term so the \( \eta \) parts can be factored out:
\( \eta^{\nu \alpha} \eta^{\mu \beta} F_{\alpha \beta} = \eta^{\mu \alpha} \eta^{\nu \beta} F_{\beta \alpha} \)
This leaves a lower index antisymmetry relation, which is already known from Equation 3.16:
\( \eta^{\mu \alpha} \eta^{\nu \beta} ( F_{\alpha \beta} + F_{ \beta \alpha } ) = 0\)
Equation 3.16: \( F_{\alpha \beta} = - F_{\beta \alpha} \Rightarrow ( F_{\alpha \beta} + F_{ \beta \alpha } ) = 0 \)
\( \Rightarrow F^{\mu \nu} + F^{\nu \mu} = 0 \)
\( \Rightarrow F^{\mu \nu} = - F^{\nu \mu} \) Q.E.D.
Quick Calculation 3.4.2: \( F^{0 i} = - F_{0 i} \), anti-symmetric upper/lower index field strengths for the \( \mu = 0 \) case.
\( F^{0 i} + F_{0 i} = 0\)
Equation 3.29 re-expresses the first term into lower index terms, using dummy variables j & k since "i" is a Latin script:
\( F^{0 i} + F_{0 i} = ( \eta^{0 j} \eta^{i k} F_{ j k } ) + F_{0 i} \)
We need j = 0, but \( \eta^{0 0} = -1 \). Then we will need i = k, so we replace \( \eta \) with the Kronecker \( \delta^{i k} \):
\( F^{0 i} + F_{0 i} = - \delta^{i k} F_{0 k} + F_{0 i} \)
\( \Rightarrow F^{0 i} + F_{0 i} = - F_{0 i} + F_{0 i} = 0 \)
\( \Rightarrow F^{0 i} = - F_{0 i} \) Q.E.D.
Quick Calculation 3.4.3: \( F^{i j} = F_{i j} \), upper/lower index field strengths are equal except for the case in Quick Calculation 3.4.2.
\( F^{i j} - F_{i j} = 0 \)
\( \Rightarrow ( \eta^{i k} \eta^{j l} F_{k l} ) - F_{i j} = 0 \)
We need i = k and j = l, so the \( \eta \) terms are replaced Kronecker delta terms:
\( F^{i j} - F_{i j} = \delta^{i k} \delta^{j l} F_{k l} - F_{i j} \)
\( \Rightarrow F^{i j} - F_{i j} = F_{i j} - F_{i j} = 0 \)
\( \Rightarrow F^{i j} = F_{i j} \) Q.E.D.
Problem Statement: Show that \( vol( B^{d} ) = \frac{ \pi^{\frac{d}{2}} } { \Gamma( 1 + \frac{d}{2} ) } \).
Ball \( B^{d} \) is the space enclosed by the sphere \( vol( S^{d - 1} ) \), so it follows that this sphere is the derivative of the ball with respect to radius R:
\( vol( S^{d - 1} ) = \frac{d}{dR} B(R) \)
Step (1): Express \( vol(S^{d - 1} ) \) from Equation 3.52 explicitly in terms of R in dimensions d - 1:
Equation 3.52: \( vol(S^{d - 1} ) = \frac{ 2 \pi^{\frac{d}{2}} } { \Gamma ( \frac{d}{2} )} \)
\( \Rightarrow vol(S^{d - 1} ) = \frac{ 2 \pi^{\frac{d}{2}} } { \Gamma ( \frac{d}{2} )} R^{d - 1} \)
Step (2): Integrate \( vol(S^{d - 1}) \) with respect to R, to get the equation for \( vol(B^{d}) \)
\( vol(B^{d}) = \int \frac{ 2 \pi^{\frac{d}{2}} } { \Gamma ( \frac{d}{2} )} R^{d - 1} dR \)
\( \Rightarrow vol(B^{d}) = \frac{ 2 \pi^{\frac{d}{2}} } { \Gamma ( \frac{d}{2} )} \int R^{d - 1} dR \)
This yields an expression that is not quite the same as what we want:
\( vol(B^{d}) = \frac{ 2 \pi^{\frac{d}{2}} } { \Gamma ( \frac{d}{2} )} \frac{R^{d}}{d} \)
The form of the equation can be re-expressed by using an identity property of the special function \( \Gamma \):
\( \Gamma(x + 1) = x \Gamma(x) \)
Notice that the denominator can be expressed in the form of the right-hand side of this identity if \( x = \frac{d}{2} \):
\( vol(B^{d}) = \frac{ \pi^{\frac{d}{2}} } { \frac{d}{2} \Gamma ( \frac{d}{2} )} R^{d} \)
\( \Rightarrow vol(B^{d}) = \frac{ \pi^{\frac{d}{2}} } { \Gamma ( 1 + \frac{d}{2} )} R^{d} \)
Dropping the explicit statement of \( R^{d} \) gives back the form of \( vol(B^{d}) \) being sought:
\( vol(B^{d}) = \frac{ \pi^{\frac{d}{2}} } { \Gamma ( 1 + \frac{d}{2} )} \) Q.E.D.
Problem Statement: Verify that for d = 3, equation (3.74) coincides with equation (3.67) .
Step (1): Write out equations 3.67 and 3.74.
Equation 3.67: Electric field as a function of radius r in three dimension
\( E(r) = \frac{q}{4 \pi r^{2}} \)
Equation 3.74: Electric field as a function of radius r in arbitrary dimensions d
\( E(r) = \frac{ \Gamma (\frac{d}{2} )}{2 \pi^{\frac{d}{2}} } \frac{q}{r^{d - 1} } \)
Step (2): Set dimension d = 3 in Equation 3.74 and show that the result reduces to the form of Equation 3.67.
\( E(r) = \frac{ \Gamma (\frac{3}{2} )}{2 \pi^{\frac{3}{2}} } \frac{q}{r^{3 - 1} } \)
Notice the gamma function identity used in Quick Calculation 3.5, and replace \( \Gamma \) with its value for d = 3:
\( \Gamma (\frac{3}{2}) = \Gamma(1 + \frac{1}{2}) = \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{1}{2} \sqrt{\pi} \)
Insert this into the expression above for E(r) in dimensions d = 3:
\( E(r) = \frac{ ( \frac{1}{2} ) \sqrt{\pi} }{2 \pi^{\frac{3}{2}} } \frac{q}{r^{2} } \)
\( \Rightarrow E(r) = \frac { \pi^{ \frac{1}{2} - \frac{3}{2} } }{ 4 } \frac{q}{r^{2} } \)
\( \Rightarrow E(r) = \frac{q}{4 \pi r^{2}} \) Q.E.D.
Problem Statement: The force \( \overrightarrow{F} \) on a test charge q in an electric field \( \overrightarrow{E} \) is \( \overrightarrow{F} = q \overrightarrow{E} \). What are the units of charge [Q] in various dimensions d?
We know from equation 3.74 that E has units of charge [Q] divided by length [L] raised to the d - 1 power:
\( [Q] = [E] [L]^{d - 1} \)
We know that \( \overrightarrow{F} = q \overrightarrow{E} \), so substituting that into the expression above:
\( [Q] = \frac{[F]}{[Q]} [L]^{d - 1} \)
\( \Rightarrow [Q]^{2} = [F] [L]^{d - 1} \)
Substitute in the dimensions for force [F] in terms of [M], [L], and [T]:
\( [Q]^{2} = \frac{[M][L]}{[T]^{2}} [L]^{d - 1} = \frac{[M][L]^{d}}{[T]^{2}} = [M] [L]^{d} [T]^{-2} \)
Therefore take the square root of \( [Q]^{2} \) to get the units of [Q] in dimensions d:
\( [Q] = [M]^{\frac{1}{2}} [L]^{\frac{d}{2}} [T]^{-1} \) Q.E.D.
Problem Statement: Show that \( l_{P} = (G)^{\alpha} (c)^{\beta} (\hbar)^{\gamma} \) having units of length uniquely fixes \( \alpha = \gamma = \frac{1}{2} \) and \( \beta = - \frac{3}{2} \), and that this results in \( l_{P} = \sqrt{\frac{G \hbar}{c^{3}}} = 1.616 \times 10^{-33} cm \).
Step (1): Write out the dimension of [G], [c], \( [\hbar] \) in terms of [M], [L], [T]. These can be derived from equations or found by inspection from their values in a given system of units.
\( G = 6.675 \times 10^{-11} \frac{m^{3}}{kg \cdot s^{2}} \)
\( c = 2.998 \times 10^{8} \frac{m}{s} \)
\( \hbar = 1.055 \times 10^{-34} \frac{kg \cdot m^{2}}{s} \)
\( \Rightarrow \)
\( [G] = M^{-1} L^{3} T^{-2} \)
\( [c] = L T^{-1} \)
\( [\hbar] = M L^{2} T^{-1} \)
Since \( l_{P} = (G)^{\alpha} (c)^{\beta} (\hbar)^{\gamma} \) is supposed to have units of length, that means the product \( [G][c][\hbar] = [L] \):
\( [G][c][\hbar] = [L] \)
\( \Rightarrow ( M^{-1} L^{3} T^{-2} )^{\alpha} ( L T^{-1} )^{\beta} ( M L^{2} T^{-1} )^{\gamma} = [L] \)
This means the power of [L] is 1, which implicitly means the product has powers of 0 for [M] and [T].
\( ( M^{-1} L^{3} T^{-2} )^{\alpha} ( M^{0} L^{1} T^{-1} )^{\beta} ( M^{1} L^{2} T^{-1} )^{\gamma} = [M]^{0} [L]^{1} [T]^{0} \)
This imposes a set of constraint equations on \( \alpha, \beta, \gamma \):
\( [M] \rightarrow -\alpha + 0 + \gamma = 0 \)
\( [L] \rightarrow 3\alpha + \beta + 2\gamma = 1 \)
\( [T] \rightarrow -2\alpha - \beta - \gamma = 0 \)
This is three equations and three unknowns. Substitutions in terms of each other yields:
\( \alpha = \gamma = \frac{1}{2} \)
\( \beta = \frac{-3}{2} \) Q.E.D.
This constraint means those values can be uniquely plugged into \( l_{P} = (G)^{\alpha} (c)^{\beta} (\hbar)^{\gamma} \):
\( l_{P} = (G)^{\alpha} (c)^{\beta} (\hbar)^{\gamma} = (G)^{\frac{1}{2}} (c)^{\frac{-3}{2}} (\hbar)^{\frac{1}{2}} \)
\( \Rightarrow l_{P} = ( (G)^{\frac{1}{2}} (\hbar)^{\frac{1}{2}} ) (c)^{\frac{-3}{2}} \)
\( \Rightarrow l_{P} = \frac{ \sqrt{G \hbar} }{ \sqrt{c^{3}} } \)
\( \Rightarrow l_{P} = \sqrt{\frac{G \hbar}{c^{3}}} \) Q.E.D.
Expressed in terms of centimeters instead of meters, this equals:
\( l_{P} = \sqrt{\frac{G \hbar}{c^{3}}} = \sqrt{\frac{6.675 \times 10^{-11} \frac{m^{3}}{kg \cdot s^{2}} 1.055 \times 10^{-34} \frac{kg \cdot m^{2}}{s}}{(2.998 \times 10^{8} \frac{m}{s})^{3}}} = \sqrt{\frac{6.675 \times 10^{-5} \frac{cm^{3}}{kg \cdot s^{2}} 1.055 \times 10^{-30} \frac{kg \cdot cm^{2}}{s}}{(2.998 \times 10^{10} \frac{cm}{s})^{3}}} = \sqrt{2.6134 \times 10^{-66} cm^{2}} = 1.616 \times 10^{-33} cm\) Q.E.D.
Problem Statement: Show that the energy equivalent of the Planck mass is \( m_{P} c^{2} = 1.221 \times 10^{19} GeV (1 GeV = 10^{9} eV) \). This energy is called the Planck energy.
Given the mass of the election and its energy equivalence:
\( m_{e} = 0.9109 \times 10^{-27} g \)
\( m_{e}c^{2} = 0.5110 MeV \)
This means that as a short hand, we can define \( c^{2} \) in terms of those two values, so it is already expressed in terms of electron-volts:
\( c^{2} = \frac{0.5110 MeV}{0.9109 \times 10^{-27} g} \)
Then we simply multiply this by the Planck mass, \( m_{P} = 2.176 \times 10^{-5} g \), to find the Planck energy:
\( E_{P} = m_{P} c^{2} \)
\( E_{P} = (2.176 \times 10^{-5} g) \cdot ( \frac{0.5110 MeV}{0.9109 \times 10^{-27} g} ) \)
\( E_{P} = 1.221 \times 10^{22} MeV \)
\( \Rightarrow E_{P} = 1.221 \times 10^{19} GeV \) Q.E.D.
Problem Statement: Prove that if you move a particle along a closed loop in a static gravitational field, the net work that you do against the gravitational field is zero.
Step (1): Write equation 3.96 which is the gravitational vector field in terms of the gradient of the gravitational potential.
Equation 3.96:
\( \overrightarrow{g} = - \overrightarrow{\nabla} V_{g} \)
This means that the gravitational force is a static gradient of a potential \( V_{g} \).
Step (2): Use Stoke's Theorem to relate a closed loop through the gravitational field to the work exerted against the field.
\( \oint \overrightarrow{g} \cdot d\overrightarrow{l} = \iint ( \overrightarrow{\nabla} \times \overrightarrow{g} ) \cdot d\overrightarrow{S} \)
\( \Rightarrow \oint \overrightarrow{g} \cdot d\overrightarrow{l} = - \iint ( \overrightarrow{\nabla} \times \overrightarrow{\nabla}V_{g} ) \cdot d\overrightarrow{S} \)
The curl of the gradient of any twice differentiable scalar field is always zero: \( \overrightarrow{\nabla} \times \overrightarrow{\nabla}V_{g} = 0 \)
\( \Rightarrow \oint \overrightarrow{g} \cdot d\overrightarrow{l} = 0 \) Q.E.D.
Therefore a particle moving in a closed loop in a static gravitational field does zero net work against the field.
Problem Statement: Show that Equations (3.106) and (3.107) are replaced by \( ( l_{P}^{(D)} )^{D-2} = \frac{ \hbar G^{(D)} }{ c^{3} } = (l_{P})^{2} \frac{G^{(D)}}{G} \) .
Step (1): Write equations 3.106 and 3.107.
Equation 3.106:
\( ( l_{P}^{(5)} )^{3} = \frac{ \hbar G^{(5)} }{ c^{3} } \)
Equation 3.107:
\( ( l_{P}^{(5)} )^{3} = \frac{ \hbar G^{(5)} }{ c^{3} } \frac{G^{(5)}}{G} \rightarrow (l_{P})^{2} \frac{G^{(5)}}{G} \)
Step (2): Follow the same derivation argument as the textbook, except for general spacetime dimension D instead of D = 5.
From equation 3.102: \( \nabla^{2} V_{g}^{(D)} = 4 \pi G^{(D)} \rho_{m} \).
\( \nabla^{2} \) always divides by length squared, and \( V_{g} \) has the same units regardless of dimensions. It follow from this that the product \(G^{(D)} \rho_{m} \) has the same units in all dimensions.
Therefore the units for the product of G and the mass density in spacetime dimension D is equal to their product for D = 4:
\( [G^{(D)}] \frac{[M]}{[L]^{D - 1}} = [ G ] \frac{[M]}{[L]^{3}} \)
Substitute in the units \( [G] = \frac{[c]^{3} [L]^{2}}{[\hbar]} \) and simplify:
\( [G^{(D)}] \frac{[M]}{[L]^{D - 1}} = \frac{[c]^{3} [L]^{2}}{[\hbar]} \frac{[M]}{[L]^{3}} = \frac{[c]^{3}}{\hbar} \frac{[M]}{[L]} \)
\( \Rightarrow [G^{(D)}] \frac{[M]}{[L]^{D - 1}} = \frac{[c]^{3}}{[\hbar]} \frac{[M]}{[L]} \)
\( \Rightarrow \frac{[G^{(D)}]}{[L]^{D - 1}} = \frac{[c]^{3}}{[\hbar] [L]} \)
Now, we can replace L with the Planck length in spacetime dimension D, \( l_{P}^{(D)}\):
\( \frac{G^{(D)}}{(l_{P}^{D})^{D-1}} = \frac{c^{3}}{\hbar (l_{P}^{(D)})}\)
\( \Rightarrow \frac{G^{(D)} \hbar}{c^{3}} = \frac{(l_{P}^{D})^{D-1}}{(l_{P}^{(D)})} \)
\( \Rightarrow \frac{G^{(D)} \hbar}{c^{3}} = (l_{P}^{D})^{D-2} \) Q.E.D.
That is the left hand side of the equations we are looking for. To get the right hand side term, we just pull out a factor of G:
\( (l_{P}^{D})^{D-2} = \frac{\hbar G^{(D)} }{c^{3}} \)
\( (l_{P}^{D})^{D-2} = (\frac{\hbar G}{c^{3}} ) (\frac{G^{(D)}}{G}) \)
Since \( l_{P} = \sqrt{\frac{\hbar G}{c^{3}}} \), we can substitute in \( (l_{P})^{2} \):
\( (l_{P}^{D})^{D-2} = \frac{\hbar G^{(D)}}{c^{3}} = (l_{P})^{2} (\frac{G^{(D)}}{G}) \) Q.E.D.
Problems
Problem Statement: Lorentz covariance for motion in an electromagnetic field.
(a) Check explicitly the relativistic equation reproduces Equation 3.5 when \( \mu \) is the spatial index.
(b) What does the relativistic equation give when \( \mu = 0 \)? Does it make sense?
(c) Is the relativistic equation gauge invariant?
Solution Heuristic:
(a) Check each spatial index, plugging in the values of \( F_{\mu \nu} \), and sum them up to show they are equal to the cross product Equation 3.5
(b) Same procedure except only for the \( \mu = 0 \) case.
(c) By inspection.
Solution:
Equation 3.5: \( \frac{d \overrightarrow{p}}{dt} = q ( \overrightarrow{E} + \frac{\overrightarrow{v}}{c} \times \overrightarrow{B} )\)
Relativistic: \( \frac{d p_{\mu}}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^{\nu}}{ds} \)
Problem 3.1 (a): Check explicitly the relativistic equation reproduces Equation 3.5 when \( \mu \) is the spatial index.
We know that the field strength tensor \( F_{\mu \nu} \) is antisymmteric and has values:
$$
F_{\mu \nu}
=
\begin{bmatrix}
0 & -E_{x} & -E_{y} & -E_{z} \\
E_{x} & 0 & B_{z} & -B_{y} \\
E_{y} & -B_{z} & 0 & B_{x} \\
E_{z} & B_{y} & -B_{x} & 0
\end{bmatrix}
$$
We also know the definition of the cross product, using determinants and cofactor expansion, applied to the case \( \overrightarrow{v} \times \overrightarrow{B} \):
\( \overrightarrow{v} \times \overrightarrow{B} = (B_{z} v_{y} - B_{y} v_{z} ) \overrightarrow{x} + ( B_{x} v_{z} - B_{z} v_{x} ) \overrightarrow{y} + ( B_{x} v_{y} - B_{y} v_{x} ) \overrightarrow{z} \)
Case \( \mu = 1 \):
\( F_{1 0} = E_{x} \)
\( F_{1 1} = 0 \)
\( F_{1 2} = B_{z} \)
\( F_{1 3} = -B_{y} \)
Run the \( \nu \) index through the relativistic equation. Note that \( ds = \frac{c}{\gamma} dt \Rightarrow \gamma ds = c dt \Rightarrow \frac{dx^{0}}{dt} = \frac{c dt}{dt} = c \)
\( \frac{d p_{x}}{d t} = \frac{q}{c} [ F_{1 0} \frac{d x^{0}}{dt} + F_{1 1} \frac{d x^{1}}{dt} + F_{1 2} \frac{d x^{2}}{dt} + F_{1 3} \frac{d x^{3}}{dt}] \)
\( \frac{d p_{x}}{d t} = \frac{q}{c} [ c E_{x} + 0 + B_{z} \frac{dy}{dt} - B_{y} \frac{d z}{dt}] \)
\( \frac{d p_{x}}{d t} = q [ E_{x} + \frac{1}{c} (B_{z} v_{y} - B_{y} v_{z})] \)
Case \( \mu = 2 \):
\( F_{2 0} = E_{y} \)
\( F_{2 1} = -B_{z} \)
\( F_{2 2} = 0 \)
\( F_{2 3} = B_{x} \)
Run the \( \nu \) index through the relativistic equation. Note that \( ds = \frac{c}{\gamma} dt \Rightarrow \gamma ds = c dt \Rightarrow \frac{dx^{0}}{dt} = \frac{c dt}{dt} = c \)
\( \frac{d p_{y}}{d t} = \frac{q}{c} [ F_{2 0} \frac{d x^{0}}{dt} + F_{2 1} \frac{d x^{1}}{dt} + F_{2 2} \frac{d x^{2}}{dt} + F_{2 3} \frac{d x^{3}}{dt}] \)
\( \frac{d p_{y}}{d t} = \frac{q}{c} [ c E_{y} - B_{z} \frac{dx}{dt} + 0 + B_{x} \frac{d z}{dt}] \)
\( \frac{d p_{y}}{d t} = q [ E_{y} + \frac{1}{c} ( B_{x} v_{z} - B_{z} v_{x} )] \)
Case \( \mu = 3 \):
\( F_{3 0} = E_{z} \)
\( F_{3 1} = B_{y} \)
\( F_{3 2} = -B_{x} \)
\( F_{3 3} = 0 \)
Run the \( \nu \) index through the relativistic equation. Note that \( ds = \frac{c}{\gamma} dt \Rightarrow \gamma ds = c dt \Rightarrow \frac{dx^{0}}{dt} = \frac{c dt}{dt} = c \)
\( \frac{d p_{z}}{d t} = \frac{q}{c} [ F_{3 0} \frac{d x^{0}}{dt} + F_{3 1} \frac{d x^{1}}{dt} + F_{3 2} \frac{d x^{2}}{dt} + F_{3 3} \frac{d x^{3}}{dt}] \)
\( \frac{d p_{z}}{d t} = \frac{q}{c} [ c E_{z} + B_{y} \frac{dx}{dt} - B_{x} \frac{d y}{dt} + 0] \)
\( \frac{d p_{z}}{d t} = q [ E_{z} + \frac{1}{c} (B_{y} v_{x} - B_{x} v_{y})] \)
Summing these up and noting the definition of the cross product, we see that the relativistic equation for the spatial indices yields Equation 3.5:
\( \frac{d \overrightarrow{p}}{d t} = \frac{d \overrightarrow{p_{x}}}{d t} + \frac{d \overrightarrow{p_{y}}}{d t} + \frac{d \overrightarrow{p_{z}}}{d t}\)
\( \frac{d \overrightarrow{p}}{d t} = \frac{q}{c} ( ( E_{x} + E_{y} + E_{z} ) + \frac{1}{c} [ (B_{z} v_{y} - B_{y} v_{z} ) + ( B_{x} v_{z} - B_{z} v_{x} ) + ( B_{x} v_{y} - B_{y} v_{x} ) ] ) \)
\( \Rightarrow \frac{d \overrightarrow{p}}{d t} = \frac{q}{c} [ \overrightarrow{E} + \frac{1}{c} ( \overrightarrow{v} \times \overrightarrow{B} ) ] \) Q.E.D.
Problem 3.1 (b): What does the relativistic equation give when \( \mu = 0 \)? Does it make sense?
Case \( \mu = 0 \):
\( F_{0 0} = 0 \)
\( F_{0 1} = -E_{x} \)
\( F_{0 2} = -E_{y} \)
\( F_{0 3} = -E_{z} \)
\( p_{0} = - \frac{E}{c} \)
Note, however, that the E in the \( p_{0} \) expression is energy, while the other E terms are electric fields.
\( \frac{d p_{0}}{d t} = - \frac{1}{c} \frac{d E}{dt} \)
\( - \frac{1}{c} \frac{d E}{dt} = - \frac{q}{c} ( 0 + E_{x} v_{x} + E_{y} v_{y} + E_{z} v_{z} ) \)
\( \Rightarrow \frac{d E}{dt} = q ( E_{x} v_{x} + E_{y} v_{y} + E_{z} v_{z} ) \)
This is just the inner product \( \overrightarrow{E} \cdot \overrightarrow{v} \), but note that \( \overrightarrow{F} = q \overrightarrow{E} \):
\( \frac{d E}{dt} = q \overrightarrow{E} \cdot \overrightarrow{v} \)
\( \Rightarrow \frac{d E}{dt} = \overrightarrow{F} \cdot \overrightarrow{v} \)
This is just the definition of power, the time rate of change of energy, which is equal to force times velocity.
Problem 3.1 (c): Is the relativistic equation gauge invariant?
Yes. \( F_{\mu \nu} \) itself is invariant under gauge transformations, while the other terms \( p_{\mu} \) and \(x^{\nu} \) have no dependence on the vector potential terms \( A_{\mu} \), which means the whole expression is gauge invariant.
Problem Statement: Maxwell equations in four dimensions.
(a) Show explicitly that the source-free Maxwell equations emerge from \( T_{\mu \lambda \nu } = 0 \).
(b) Show explicitly that the Maxwell equations with sources emerge from equation (3.34).
Solution:
The source-free Maxwell equations are Equations 3.1 and 3.2:
Equation 3.1: \( \nabla \times \overrightarrow{E} = - \frac{1}{c} \frac{\partial \overrightarrow{B}}{\partial t} \)
Equation 3.2: \( \nabla \cdot \overrightarrow{B} = 0 \)
The Maxwell equations with sources are Equations 3.3 and 3.4:
Equation 3.3: \( \nabla \cdot \overrightarrow{E} = \rho \)
Equation 3.4: \( \nabla \times \overrightarrow{B} = \frac{1}{c} \overrightarrow{j} + \frac{1}{c} \frac{\partial \overrightarrow{E}}{\partial t} \)
We know that the field strength tensor \( F_{\mu \nu} \) is antisymmteric and by equation 3.20 has values:
$$
F_{\mu \nu}
=
\begin{bmatrix}
0 & -E_{x} & -E_{y} & -E_{z} \\
E_{x} & 0 & B_{z} & -B_{y} \\
E_{y} & -B_{z} & 0 & B_{x} \\
E_{z} & B_{y} & -B_{x} & 0
\end{bmatrix}
$$
But the upper index tensor \( F^{\mu \nu} \), Equation 3.33, has flipped signs in the zeroth index values:
$$
F^{\mu \nu}
=
\begin{bmatrix}
0 & E_{x} & E_{y} & E_{z} \\
-E_{x} & 0 & B_{z} & -B_{y} \\
-E_{y} & -B_{z} & 0 & B_{x} \\
-E_{z} & B_{y} & -B_{x} & 0
\end{bmatrix}
$$
Problem 3.2 (a): Show explicitly that the source-free Maxwell equations emerge from \( T_{\mu \lambda \nu } = 0 \).
By equation 3.23, we know the definition of the field strength combination \( T_{\mu \lambda \nu } \):
\( T_{\mu \lambda \nu } = \partial_{\mu} F_{\lambda \nu} + \partial_{\lambda} F_{\nu \mu} + \partial_{\nu} F_{\mu \lambda} \)
We also know from Equation 3.15 that the field strengths are defined as \( F_{\mu \nu} \equiv \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \). This implies \( T_{\mu \lambda \nu } = 0 \) because of the crossing partial derivatives:
\( T_{\mu \lambda \nu } = \partial_{\mu} (\partial_{\lambda} A_{\nu} - \partial_{\nu} A_{\lambda} ) + \partial_{\lambda} (\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu} ) + \partial_{\nu} (\partial_{\mu} A_{\lambda} - \partial_{\lambda} A_{\mu} ) = 0 \)
More close grained, this can happen in different ways, as some index value combinations vanish trivially. If the value is \( \mu = \lambda = \nu \), the inner terms cancel as zero, as these are on the diagonal and \( F_{\mu \nu} \) is antisymmetric, and thus \( T_{\mu \lambda \nu } = 0 + 0 + 0 = 0 \). If two of the indices are equal, \( T_{\mu \lambda \nu } \) also vanishes trivially:
\( T_{\mu \mu \nu } = \partial_{\mu} (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} ) + \partial_{\mu} (\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu} ) + \partial_{\nu} (\partial_{\mu} A_{\mu} - \partial_{\mu} A_{\mu} ) = 0 \)
\( T_{\mu \mu \nu } = \partial_{\mu} \partial_{\mu} A_{\nu} + (- \partial_{\mu} \partial_{\nu} A_{\mu} + \partial_{\mu} \partial_{\nu} A_{\mu} ) - \partial_{\mu} \partial_{\mu} A_{\nu} + 0 = 0 \)
\( T_{\mu \mu \nu } = 0 + 0 + 0 = 0 \)
This leaves the case of each index having its own separate value, which does not vanish trivially, but which still equals zero because of the partial derivatives. These can be found by iterating through each partial derivative index value, left to right, and not using the index value for the missing \( \partial \), yielding four equations:
\( \partial_{0} F_{1 2} + \partial_{1} F_{2 0} + \partial_{2} F_{0 1} = 0 \)
\( \partial_{0} F_{1 3} + \partial_{1} F_{3 0} + \partial_{3} F_{0 1} = 0 \)
\( \partial_{0} F_{2 3} + \partial_{2} F_{3 0} + \partial_{3} F_{0 2} = 0 \)
\( \partial_{1} F_{2 3} + \partial_{2} F_{3 1} + \partial_{3} F_{1 2} = 0 \)
\( \partial_{0} = \frac{1}{c}\frac{\partial}{\partial t} , \partial_{1} = \frac{\partial}{\partial x} , \partial_{2} = \frac{\partial}{\partial y} \partial_{3} = \frac{\partial}{\partial z} \)
From the matrix 3.20 for \( F_{\mu \nu} \) above, we know the components equal:
\( F_{1 2} = B_{z} \)
\( F_{1 3} = - B_{y} \)
\( F_{2 3} = B_{x} \)
\( F_{2 0} = E_{y} \)
\( F_{3 0} = E_{z} \)
\( F_{3 1} = B_{y} \)
\( F_{0 1} = - E_{x} \)
\( F_{0 2} = - E_{y} \)
Therefore you plug these into the set of equations above:
\( \frac{1}{c} \frac{\partial}{\partial t} B_{z} + \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} = 0 \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{y} + \frac{\partial}{\partial x} E_{z} - \frac{\partial}{\partial z} E_{x} = 0 \)
\( \frac{1}{c} \frac{\partial}{\partial t} B_{x} + \frac{\partial}{\partial y} E_{z} - \frac{\partial}{\partial z} E_{y}= 0 \)
\( \frac{\partial}{\partial x} B_{x} + \frac{\partial}{\partial y} B_{y} + \frac{\partial}{\partial z} B_{z} = 0 \)
The first three equations rearrange into a form that equals Equation 3.1: \( \nabla \times \overrightarrow{E} = - \frac{1}{c} \frac{\partial \overrightarrow{B}}{\partial t} \).
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{z} = \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{y} = \frac{\partial}{\partial z} E_{x} - \frac{\partial}{\partial x} E_{z} \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{x} = \frac{\partial}{\partial z} E_{y} - \frac{\partial}{\partial y} E_{z} \)
Where the right-hand sides are the curl of \(\overrightarrow{E}\): \( - \frac{1}{c} \frac{\partial}{\partial t}( B_{x} + B_{y} + B_{z} ) = ( \frac{\partial}{\partial z} E_{y} - \frac{\partial}{\partial y} E_{z} ) + (\frac{\partial}{\partial z} E_{x} - \frac{\partial}{\partial x} E_{z} ) + ( \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} )\)
\( \Rightarrow - \frac{1}{c} \frac{\partial \overrightarrow{B}}{\partial t} = \nabla \times \overrightarrow{E} \) Q.E.D.
The fourth equation is the inner product of the Laplacian with the magnetic field B, or the divergence of B, which is Equation 3.2: \( \nabla \cdot \overrightarrow{B} = 0 \).
Problem 3.2 (b): Show explicitly that the Maxwell equations with sources emerge from equation (3.34).
Combining the matrix 3.20 for \( F^{\mu \nu} \) with the definition of current four-vector \( j^{\mu} = ( c\rho, j^{1}, j^{2}, j^{3} ) \):
Equation 3.34: \( \frac{ \partial F^{\mu \nu}}{\partial x^{\nu}} = \frac{1}{c} j^{\mu} \)
Iterating out the \( \nu \) index for each value of \( \mu \) will produce four equations with sources if j is non-zero:
\( \frac{ \partial F^{0 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{0 0}}{\partial x^{0}} + \frac{ \partial F^{0 1}}{\partial x^{1}} + \frac{ \partial F^{0 2}}{\partial x^{2}} + \frac{ \partial F^{0 3}}{\partial x^{3}} = \frac{1}{c} j^{0} \)
\( \frac{ \partial F^{1 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{1 0}}{\partial x^{0}} + \frac{ \partial F^{1 1}}{\partial x^{1}} + \frac{ \partial F^{1 2}}{\partial x^{2}} + \frac{ \partial F^{1 3}}{\partial x^{3}} = \frac{1}{c} j^{1} \)
\( \frac{ \partial F^{2 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{2 0}}{\partial x^{0}} + \frac{ \partial F^{2 1}}{\partial x^{1}} + \frac{ \partial F^{2 2}}{\partial x^{2}} + \frac{ \partial F^{2 3}}{\partial x^{3}} = \frac{1}{c} j^{2} \)
\( \frac{ \partial F^{3 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{3 0}}{\partial x^{0}} + \frac{ \partial F^{3 1}}{\partial x^{1}} + \frac{ \partial F^{3 2}}{\partial x^{2}} + \frac{ \partial F^{3 3}}{\partial x^{3}} = \frac{1}{c} j^{3} \)
These values of F were written out explicitly for lower index in Problem 3.2 (a), except for:
\( F^{1 0} = E_{x} , F^{2 1} = - B_{z} , F^{3 2} = B_{y} \)
For the spatial indices \( F^{i j} = F_{i j} \) per Quick Calculation 3.4.3, but \( F^{0 i} = - F_{0 i} \) is a sign flip compared to matrix 3.20 per Quick Calculation 3.4.2, which are the electric field terms.
So plugging all of these into the equations above, and expressing in terms of (t,x,y,z):
\( \frac{ \partial F^{0 \nu}}{\partial x^{\nu}} = 0 + \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} + \frac{ \partial E_{z}}{\partial z} = \frac{1}{c} j^{0} = \rho \)
\( \frac{ \partial F^{1 \nu}}{\partial x^{\nu}} = - \frac{1}{c} \frac{ \partial E_{x}}{\partial t} + 0 + \frac{ \partial B_{z}}{\partial y} - \frac{ \partial B_{y}}{\partial z} = \frac{1}{c} j^{1} \)
\( \frac{ \partial F^{2 \nu}}{\partial x^{\nu}} = - \frac{1}{c} \frac{ \partial E_{y}}{\partial t} - \frac{ \partial B_{z}}{\partial x} + 0 + \frac{ \partial B_{x}}{\partial z} = \frac{1}{c} j^{2} \)
\( \frac{ \partial F^{3 \nu}}{\partial x^{\nu}} = - \frac{1}{c} \frac{ \partial E_{z}}{\partial t} + \frac{ \partial B_{y}}{\partial x} - \frac{ \partial B_{x}}{\partial y} + 0 = \frac{1}{c} j^{3} \)
The first equation is Equation 3.3, the divergence of the electric field or Gauss's Law:
\( \nabla \cdot \overrightarrow{E} = \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} + \frac{ \partial E_{z}}{\partial z} = \rho \) (Q.E.D.)
Moving the \( \frac{1}{c} \frac{ \partial E_{i}}{\partial t} \) terms to the right-hand side, we see that the left-hand side is the curl of the magnetic field B:
\( \frac{ \partial B_{z}}{\partial y} - \frac{ \partial B_{y}}{\partial z} = \frac{1}{c} j^{1} + \frac{1}{c} \frac{ \partial E_{x}}{\partial t} \)
\( - \frac{ \partial B_{z}}{\partial x} + \frac{ \partial B_{x}}{\partial z} = \frac{1}{c} j^{2} + \frac{1}{c} \frac{ \partial E_{y}}{\partial t} \)
\( + \frac{ \partial B_{y}}{\partial x} - \frac{ \partial B_{x}}{\partial y} = \frac{1}{c} j^{3} + \frac{1}{c} \frac{ \partial E_{z}}{\partial t} \)
\( \Rightarrow \nabla \times \overrightarrow{B} = (\frac{ \partial B_{z}}{\partial y} - \frac{ \partial B_{y}}{\partial z}) + ( \frac{ \partial B_{x}}{\partial z} - \frac{ \partial B_{z}}{\partial x} ) + ( \frac{ \partial B_{y}}{\partial x} - \frac{ \partial B_{x}}{\partial y} ) = \frac{1}{c} ( j^{1} + j^{2} + j^{3} ) + \frac{1}{c} ( \frac{ \partial E_{x}}{\partial t} + \frac{ \partial E_{y}}{\partial t} + \frac{ \partial E_{z}}{\partial t} ) \)
\( \nabla \times \overrightarrow{B} = \frac{1}{c} ( j^{1} + j^{2} + j^{3} ) + \frac{1}{c} ( \frac{ \partial E_{x}}{\partial t} + \frac{ \partial E_{y}}{\partial t} + \frac{ \partial E_{z}}{\partial t} ) \)
This is just Equation 3.4:
\( \nabla \times \overrightarrow{B} = \frac{1}{c} \overrightarrow{j} + \frac{1}{c} \frac{\partial \overrightarrow{E}}{\partial t} \) (Q.E.D.)
Problem Statement: Electromagnetism in three dimensions.
(a) Find the reduced Maxwell's equations in three dimensions by starting with Maxwell's equations and the force law in four dimensions, using the ansatz 3.11, and assuming that no field can depend on the z-direction.
(b) Repeat the analysis of three-dimensional electromagnetism starting with the Lorentz covariant formulation. Take \( A^{\mu} = ( \Phi, A^{1}, A^{2} ) \), examine \( F_{\mu \nu} \), the Maxwell equations (3.34), and the relativistic form of the force law derived in Problem 3.1
Solution:
Problem 3.3 (a): Find the reduced Maxwell's equations in three dimensions by starting with Maxwell's equations and the force law in four dimensions, using the ansatz 3.11, and assuming that no field can depend on the z-direction.
Following the argument in the textbook under Chapter 3.2, the laws of electromagnetism in only three dimensions is counter-intuitive, because the perpendicular aspect of the fields requires three spatial indices. However, this is not a problem mathematically, it is normal to an x,y plane of two spatial dimensions. This allows us to eliminate a row/column by making the electric field in the z-direction zero, the magnetic field along the x,y-plane zero, and not allowing z-dependence in the differential equations:
Ansatz 3.11: \( E_{z} = B_{x} = B_{y} = 0 \)
(1) First use this constraint to reduce Equation 3.1 (Maxwell-Faraday Law): \( \nabla \times \overrightarrow{E} = - \frac{1}{c} \frac{\partial \overrightarrow{B}}{\partial t} \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{z} = \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{y} = \frac{\partial}{\partial z} E_{x} - \frac{\partial}{\partial x} E_{z} \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{x} = \frac{\partial}{\partial z} E_{y} - \frac{\partial}{\partial y} E_{z} \)
\( \Rightarrow \)
\( - \frac{1}{c} \frac{\partial}{\partial t} B_{z} = \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} \)
\( 0 = \frac{\partial}{\partial z} E_{x} - 0 \)
\( 0 = \frac{\partial}{\partial z} E_{y} - 0 \)
Therefore in three dimensions, Maxwell equation 3.1 becomes: \( - \frac{1}{c} \frac{\partial}{\partial t} B_{z} = \frac{\partial}{\partial x} E_{y} - \frac{\partial}{\partial y} E_{x} \) Q.E.D.
(2) Second use this constraint to reduce Equation 3.3 (Gauss's Law): \( \nabla \cdot \overrightarrow{E} = \rho \)
\( \nabla \cdot \overrightarrow{E} = \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} + \frac{ \partial E_{z}}{\partial z} = \rho \)
\( \nabla \cdot \overrightarrow{E} = \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} + 0 = \rho \)
Therefore in three dimensions, Maxwell equation 3.3 becomes: \( \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} = \rho \) Q.E.D.
(3) Third use this constraint to reduce Equation 3.4 (Maxwell-Ampere Law): \( \nabla \times \overrightarrow{B} = \frac{1}{c} \overrightarrow{j} + \frac{1}{c} \frac{\partial \overrightarrow{E}}{\partial t} \)
\( \frac{ \partial B_{z}}{\partial y} - \frac{ \partial B_{y}}{\partial z} = \frac{1}{c} j^{1} + \frac{1}{c} \frac{ \partial E_{x}}{\partial t} \)
\( - \frac{ \partial B_{z}}{\partial x} + \frac{ \partial B_{x}}{\partial z} = \frac{1}{c} j^{2} + \frac{1}{c} \frac{ \partial E_{y}}{\partial t} \)
\( + \frac{ \partial B_{y}}{\partial x} - \frac{ \partial B_{x}}{\partial y} = \frac{1}{c} j^{3} + \frac{1}{c} \frac{ \partial E_{z}}{\partial t} \)
\( \Rightarrow \)
\( \frac{ \partial B_{z}}{\partial y} - 0 = \frac{1}{c} j^{1} + \frac{1}{c} \frac{ \partial E_{x}}{\partial t} \)
\( - \frac{ \partial B_{z}}{\partial x} + 0 = \frac{1}{c} j^{2} + \frac{1}{c} \frac{ \partial E_{y}}{\partial t} \)
\( + 0 - 0 = 0 + 0 \)
Therefore in three dimensions, Maxwell equation 3.4 becomes the two equations:
\( \frac{ \partial B_{z}}{\partial y} = \frac{1}{c} j^{1} + \frac{1}{c} \frac{ \partial E_{x}}{\partial t} \)
\( - \frac{ \partial B_{z}}{\partial x} = \frac{1}{c} j^{2} + \frac{1}{c} \frac{ \partial E_{y}}{\partial t} \) Q.E.D.
Equation 3.2 where \( \nabla \cdot \overrightarrow{B} = 0 \) is trivial (Gauss's Magnetism Law), as \( B_{x} = B_{y} = 0 \) leaves only \( \frac{\partial B_{z}}{\partial z} = 0 \) and there is no spatial change possible in the z-direction. Q.E.D.
Problem 3.3 (b): Repeat the analysis of three-dimensional electromagnetism starting with the Lorentz covariant formulation. Take \( A^{\mu} = ( \Phi, A^{1}, A^{2} ) \), examine \( F_{\mu \nu} \), the Maxwell equations (3.34), and the relativistic form of the force law derived in Problem 3.1
Equation 3.20: $$
F_{\mu \nu}
=
\begin{bmatrix}
0 & -E_{x} & -E_{y} \\
E_{x} & 0 & B_{z} \\
E_{y} & -B_{z} & 0
\end{bmatrix}
$$
Equation 3.33: $$
F^{\mu \nu}
=
\begin{bmatrix}
0 & E_{x} & E_{y} \\
-E_{x} & 0 & B_{z} \\
-E_{y} & -B_{z} & 0
\end{bmatrix}
$$
Step (1): Applying Equation 3.25 \( \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda} + \partial_{\nu} F_{\lambda \mu} = 0\) gives the sourceless reduced Maxwell's equations from Problem 3.3 (a), and applying equation 3.34 \( \frac{\partial F^{\mu \nu}}{\partial x^{\nu}} = \frac{1}{c} j^{\mu} \) where \( j^{\mu} = (c \rho, j^{1}, j^{2}) \) gives the reduced Maxwell's equations with sources from Problem 3.3 (a) (the Maxwell-Faraday equations).
In reduced dimensions Equation 3.25 is only run through 0,1,2 for both \( \mu \) and \( \nu \), which eliminates three of the four equations from Problem 3.2 (1):
\( \partial_{0} = \frac{1}{c} \frac{\partial}{\partial t} , \partial_{1} = \frac{\partial}{\partial x} , \partial_{2} = \frac{\partial}{\partial y} \)
\( \partial_{0} F_{1 2} + \partial_{1} F_{2 0} + \partial_{2} F_{0 1} = 0 \)
\( \Rightarrow \frac{1}{c} \frac{\partial}{\partial t} B_{z} + \frac{\partial}{\partial_{x}} E_{y} - \frac{\partial}{\partial y} E_{x} = 0 \)
This produces the single \(\nabla \times \overrightarrow{E} \) Faraday term from Problem 3.3 (a): \( - \frac{1}{c} \frac{\partial}{\partial t} B_{z} = \frac{\partial}{\partial_{x}} E_{y} - \frac{\partial}{\partial y} E_{x} \) Q.E.D.
For the reduced Maxwell's equations with sources, we recapitulate Problem 3.2 (b) while dropping the value 3 from \( \mu \) and \( \nu \):
\( \frac{ \partial F^{0 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{0 0}}{\partial x^{0}} + \frac{ \partial F^{0 1}}{\partial x^{1}} + \frac{ \partial F^{0 2}}{\partial x^{2}} = \frac{1}{c} j^{0} \)
\( \frac{ \partial F^{1 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{1 0}}{\partial x^{0}} + \frac{ \partial F^{1 1}}{\partial x^{1}} + \frac{ \partial F^{1 2}}{\partial x^{2}} = \frac{1}{c} j^{1} \)
\( \frac{ \partial F^{2 \nu}}{\partial x^{\nu}} = \frac{ \partial F^{2 0}}{\partial x^{0}} + \frac{ \partial F^{2 1}}{\partial x^{1}} + \frac{ \partial F^{2 2}}{\partial x^{2}} = \frac{1}{c} j^{2} \)
Which expressed in terms of (t,x,y) becomes:
\( \frac{ \partial F^{0 \nu}}{\partial x^{\nu}} = 0 + \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} = \frac{1}{c} j^{0} = \rho \)
\( \frac{ \partial F^{1 \nu}}{\partial x^{\nu}} = - \frac{1}{c} \frac{ \partial E_{x}}{\partial t} + 0 + \frac{ \partial B_{z}}{\partial y} = \frac{1}{c} j^{1} \)
\( \frac{ \partial F^{2 \nu}}{\partial x^{\nu}} = - \frac{1}{c} \frac{ \partial E_{y}}{\partial t} - \frac{ \partial B_{z}}{\partial x} + 0 = \frac{1}{c} j^{2} \)
This produces the reduced Maxwell's equations with sources, Gauss's law \( \nabla \cdot \overrightarrow{E} \) and the \( \nabla \times \overrightarrow{B} \) Ampere-Maxwell current law:
\( \frac{ \partial E_{x}}{\partial x} + \frac{ \partial E_{y}}{\partial y} = \frac{1}{c} j^{0} = \rho \)
\( \frac{ \partial B_{z}}{\partial y} = \frac{1}{c} j^{1} + \frac{1}{c} \frac{ \partial E_{x}}{\partial t} \)
\( - \frac{ \partial B_{z}}{\partial x} = \frac{1}{c} j^{2} + \frac{1}{c} \frac{ \partial E_{y}}{\partial t} \) Q.E.D.
Step (2): Apply the relativistic force equation from Problem 3.1, using only index values 0,1,2:
\( \frac{d p_{\mu}}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^{\nu}}{ds} \)
Case \( \mu = 1 \):
\( F_{1 0} = E_{x} \)
\( F_{1 1} = 0 \)
\( F_{1 2} = B_{z} \)
Run the \( \nu \) index through the relativistic equation. Note that \( ds = \frac{c}{\gamma} dt \Rightarrow \gamma ds = c dt \Rightarrow \frac{dx^{0}}{dt} = \frac{c dt}{dt} = c \)
\( \frac{d p_{x}}{d t} = \frac{q}{c} [ F_{1 0} \frac{d x^{0}}{dt} + F_{1 1} \frac{d x^{1}}{dt} + F_{1 2} \frac{d x^{2}}{dt} ] \)
\( \frac{d p_{x}}{d t} = \frac{q}{c} [ c E_{x} + 0 + B_{z} \frac{dy}{dt}] \)
\( \frac{d p_{x}}{d t} = q [ E_{x} + \frac{1}{c} (B_{z} v_{y} )] \) Q.E.D.
Case \( \mu = 2 \):
\( F_{2 0} = E_{y} \)
\( F_{2 1} = -B_{z} \)
\( F_{2 2} = 0 \)
Run the \( \nu \) index through the relativistic equation. Note that \( ds = \frac{c}{\gamma} dt \Rightarrow \gamma ds = c dt \Rightarrow \frac{dx^{0}}{dt} = \frac{c dt}{dt} = c \)
\( \frac{d p_{y}}{d t} = \frac{q}{c} [ F_{2 0} \frac{d x^{0}}{dt} + F_{2 1} \frac{d x^{1}}{dt} + F_{2 2} \frac{d x^{2}}{dt} ] \)
\( \frac{d p_{y}}{d t} = \frac{q}{c} [ c E_{y} - B_{z} \frac{dx}{dt} + 0 ] \)
\( \frac{d p_{y}}{d t} = q [ E_{y} - \frac{1}{c} ( B_{z} v_{x} )] \) Q.E.D.
Case \( \mu = 0 \):
\( F_{0 0} = 0 \)
\( F_{0 1} = -E_{x} \)
\( F_{0 2} = -E_{y} \)
\( p_{0} = - \frac{E}{c} \)
Note, however, that the E in the \( p_{0} \) expression is energy, while the other E terms are electric fields.
\( \frac{d p_{0}}{d t} = - \frac{1}{c} \frac{d E}{dt} \)
\( - \frac{1}{c} \frac{d E}{dt} = - \frac{q}{c} ( 0 + E_{x} v_{x} + E_{y} v_{y}) \)
\( \Rightarrow \frac{d E}{dt} = q ( E_{x} v_{x} + E_{y} v_{y} ) \) Q.E.D.
Problem Statement: Electric fields and potentials of point charges.
(a) Show that for time-independent fields, the Maxwell equation \( T_{0 i j} = 0 \) implies that \( \partial_{i} E_{j} - \partial_{j} E_{i} = 0 \). Explain why this condition is satisfied by the ansatz \( \overrightarrow{E} = - \nabla \Phi \).
(b) Show that with d spatial dimensions, the potential \( \Phi \) due to a point charge q is given by \( \Phi (r) = \frac{\Gamma ( \frac{d}{2} - 1 ) }{4 \pi^{\frac{d}{2}}} \frac{q}{r^{d-2}} \)
Solution:
Problem 3.4 (a): Show that for time-independent fields, the Maxwell equation \( T_{0 i j} = 0 \) implies that \( \partial_{i} E_{j} - \partial_{j} E_{i} = 0 \). Explain why this condition is satisfied by the ansatz \( \overrightarrow{E} = - \nabla \Phi \).
This begins with the Maxwell equation definitions in terms of field strengths from Equation 3.23 \( T_{ \lambda \mu \nu } \), except it is time independent, so \( \lambda = 0 \) and the other two indices become the spatial indices \( \mu = i , \nu = j \):
Equation 3.23: \( T_{ \lambda \mu \nu } \equiv \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda} + \partial_{\nu} F_{\lambda \mu} = 0\)
Equaling zero because of the field strength definition:
Equation 3.15: \( F_{\mu \nu} \equiv \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \)
So by substituting into Equation 3.23 the indices \( \lambda = 0, \mu = i , \nu = j \):
\( T_{ 0 i j } \equiv \partial_{0} F_{i j} + \partial_{i} F_{j 0} + \partial_{j} F_{0 i} = 0\)
Since these are time independent fields, we know that \( \partial_{0} F_{i j} = 0 \). Thus, we have the equation:
\( T_{ 0 i j } = \partial_{i} F_{j 0} + \partial_{j} F_{0 i} = 0 \)
We know that \( F_{j 0} \) being the 0 column, and it is lower index F, these are the positive electric field strengths \( E_{j} \), while \( F_{0 i} \) for the 0 row are the negative values \( - E_{i} \). This comes from matrix 3.20:
$$
F_{\mu \nu}
=
\begin{bmatrix}
0 & -E_{x} & -E_{y} & -E_{z} \\
E_{x} & 0 & B_{z} & -B_{y} \\
E_{y} & -B_{z} & 0 & B_{x} \\
E_{z} & B_{y} & -B_{x} & 0
\end{bmatrix}
$$
Therefore this becomes:
\( T_{ 0 i j } = \partial_{i} E_{j} - \partial_{j} E_{i} = 0 \)
As in Problem 3.2 (a) this corresponds to the cross product terms \( \nabla \times \overrightarrow{E} \), except here there are no time varying terms, no values of magnetic field B. The ansatz \( \overrightarrow{E} = - \nabla \Phi \) explains this because the curl of a gradient is zero: \( \nabla \times \overrightarrow{E} = - \nabla \times ( \nabla \Phi ) = 0 \). Q.E.D.
Problem 3.4 (b): Show that with d spatial dimensions, the potential \( \Phi \) due to a point charge q is given by \( \Phi (r) = \frac{\Gamma ( \frac{d}{2} - 1 ) }{4 \pi^{\frac{d}{2}}} \frac{q}{r^{d-2}} \).
Equation 3.75: \( \overrightarrow{E}= - \nabla \Phi \)
Equation 3.52: \( vol(S^{d-1}) = \frac{2 \pi^{\frac{d}{2}}}{\Gamma (\frac{d}{2})} \)
Therefore Equation 3.75 expressed as a radial dependence: \( \overrightarrow{E}= - \frac{d \Phi}{dr} \)
The flux of the electric field through a surface S is \( \iint \overrightarrow{E} \cdot d \overrightarrow{S} = q\), and we know from the divergence theorem argument in Chapter 3.5 that flux is equal to the magnitude of the electric field times the volume of \(S^{d -1}(r)\):
Equation 3.73: \( E(r) vol(S^{d-1}(r)) = q \)
Which combined with Equation 3.52 is Equation 3.74: \( E(r) = \frac{\Gamma(\frac{d}{2})}{2 \pi^{\frac{d}{2}}} \frac{q}{r^{d-1}}\)
Therefore \( \Phi (r) = - \int E(r)dr \):
\( \Phi (r) = - \frac{q \Gamma(\frac{d}{2})}{2 \pi^{\frac{d}{2}}} \int \frac{1}{r^{d-1}} dr \)
\( \Rightarrow \Phi (r) = - \frac{q \Gamma(\frac{d}{2})}{2 \pi^{\frac{d}{2}}} \frac{-1}{(d-2) r^{d-2}} \)
The identity property for the special function \( \Gamma \) can be rearranged into the form:
\( \Gamma(x ) = \frac{\Gamma(x + 1)}{x} \)
Pull out a factor of 2 in the denominator:
\( \Phi (r) = \frac{\Gamma(\frac{d}{2})}{4 \pi^{\frac{d}{2}}} \frac{q}{(\frac{d}{2}-1) r^{d-2}} = \frac{\Gamma(\frac{d}{2})}{4 \pi^{\frac{d}{2}}} \frac{q}{(\frac{d}{2}-1) r^{d-2}} \)
Let \( x = \frac{d}{2} - 1\):
\( \Rightarrow \frac{\Gamma(\frac{d}{2}) }{(\frac{d}{2}-1)} = \frac{\Gamma(x + 1) }{x} = \Gamma(x) = \Gamma(\frac{d}{2} - 1) \)
This simplifies \( \Phi \) to the form desired:
\( \Phi (r) = \frac{\Gamma ( \frac{d}{2} - 1 ) }{4 \pi^{\frac{d}{2}}} \frac{q}{r^{d-2}} \) (Q.E.D.)
Problem 3.5: Calculating the divergence in higher dimensions.
Problem Statement:
Let \( \overrightarrow{f} = f(r) \hat{r} \) be a vector function in \( \mathbb{R}^{d} \). Here \( \hat{r} \) is a unit radial vector, and r is the radial distance to the origin. Derive a formula for \( \nabla \cdot \overrightarrow{f} \) by applying the divergence theorem to a spherical shell of radius r and width dr. Check that for d = 3 your answer reduces to \( \nabla \cdot \overrightarrow{f} = f'(r) + \frac{2}{r} f(r) \).
The divergence \( \nabla \cdot \overrightarrow{f} = \nabla \cdot f(r) \hat{r} \) involves a scalar function f(r) multiplied by the unit radial vector \( \hat{r} \). This is subject to the generalized product rule in vector calculus for this case:
\( \nabla \cdot ( f(r) \hat{r} ) = (\nabla f(r)) \cdot \hat{r} + f(r) \nabla \cdot \hat{r} \)
\( \nabla \cdot ( f(r) \hat{r} ) = \frac{\partial f(r)}{\partial r} \cdot \hat{r} + f(r) \nabla \cdot \hat{r} \)
\( \Rightarrow \nabla \cdot ( f(r) \hat{r} ) = f'(r) + f(r) \nabla \cdot \hat{r} \)
Recall that the unit radial vector \( \hat{r} = \frac{\overrightarrow{r}}{r} \), so we have to take into account this r dependence in the product rule:
\( \nabla \cdot ( f(r) \hat{r} ) = f'(r) + f(r) \nabla \cdot \frac{\overrightarrow{r}}{r} \)
This involves the same kind of vector calculus product rule, except for the scalar function \( \frac{1}{r} \):
\( \nabla \cdot ( f(r) \hat{r} ) = f'(r) + f(r) [ \nabla \frac{1}{r} \cdot \overrightarrow{r} + \frac{1}{r} \nabla \cdot \overrightarrow{r} ] \)
These are \( \overrightarrow{r} \) in d dimensions. \( \nabla \cdot \overrightarrow{r} = d \), as the inner product with \( \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, ... ) \) makes all the individual vector components \( x \hat{x} + y \hat{y} + ... \) equal 1, summing up as (1 + 1 + ... = d).
\( \frac{1}{r} \) is to the first power even though it has d components, so \( \frac{\partial}{\partial r} r^{-1} \cdot r \hat{r} = ( \frac{-1}{r^{2}} ) r \hat{r} = \frac{-1}{r} \). Combining these gives:
\( \nabla \cdot \overrightarrow{f} = \nabla \cdot ( f(r) \hat{r} ) = f'(r) + f(r) [ \frac{-1}{r} + \frac{d}{r} ] \)
\( \Rightarrow \nabla \cdot \overrightarrow{f} = f'(r) + \frac{d - 1}{r} f(r) \)
In the case of d = 3, this becomes: \( \nabla \cdot \overrightarrow{f} = f'(r) + \frac{2}{r} f(r) \). (Q.E.D.)
Problem 3.6: Analytic continuation for gamma functions.
Problem Statement:
Consider the definition of the gamma function for complex arguments z whose real part is positive:.
\( \Gamma (z) = \int^{\infty}_{0} dt e^{-t} t^{z - 1} , R(z) > 0 \).
Use this equation to show that for R(z) > 0:
\( \Gamma (z) = \int^{1}_{0} dt t^{z-1} ( e^{-t} - \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} ) + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} + \int^{\infty}_{1} dt e^{-t} t^{z - 1} \)
Explain why the above right-hand side is well defined for R(z) > - N - 1. It follows that this right-hand side provides the analytic continuation of \( \Gamma (z) \) for R(z) > -N - 1. Conclude that the gamma function has poles at 0, -1, -2, ..., and give the value of the residue at z = -n (with n a positive integer).
Interpreting this problem to mean that we are showing the second equation is defined for R(z) > 0 and reduces to the first equation under R(z) > - N - 1 , not deriving the equation itself, then we start with the equation and rearrange it to pull out the definition from the first equation:
\( \Gamma (z) = \int^{1}_{0} dt t^{z-1} ( e^{-t} - \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} ) + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} + \int^{\infty}_{1} dt e^{-t} t^{z - 1} \)
\( \Gamma (z) = \int^{1}_{0} dt t^{z-1} e^{-t} - \int^{1}_{0} dt t^{z-1} \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} + \int^{\infty}_{1} dt e^{-t} t^{z - 1} \)
\( \Gamma (z) = ( \int^{1}_{0} dt t^{z-1} e^{-t} + \int^{\infty}_{1} dt e^{-t} t^{z - 1} ) - \int^{1}_{0} dt t^{z-1} \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} \)
\( \Gamma (z) = ( \int^{\infty}_{0} dt e^{-t} t^{z - 1} ) - \int^{1}_{0} dt t^{z-1} \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} \)
\( \Gamma (z) = \Gamma (z) + ( \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \frac{1}{z + n} - \int^{1}_{0} t^{z-1} \sum\limits_{n=0}^N \frac{(-t)^{n}}{n!} dt ) \)
Then we can pull out the series term \( \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} \), notably leaving \( t^{n} \) after removing \( (-1)^{n} \):
\( \Gamma (z) = \Gamma (z) + \sum\limits_{n=0}^N \frac{(-1)^{n}}{n!} ( \frac{1}{z + n} - \int^{1}_{0} t^{n+z-1} dt ) \)
If R(z) > - N - 1 then the integral \( \int^{1}_{0} t^{n+z-1} dt \) always converges (the exponent is -2 for N=0 and a bigger negative number for N > 0), so the equation is well defined, which leads to \( \Gamma (z) = \Gamma (z) \) since the integral evaluates as \( \frac{1}{z + n} \). This is an analytical continuation beyond R(z) > 0. In terms of finding the residues (singular points where gamma becomes undefined), consider a recurrence of the identity property:
\( \Gamma (z) = \frac{\Gamma ( z + 1) }{z} \)
\( \Rightarrow \Gamma (z + 1) = \frac{\Gamma (z + 2) }{z + 1} \Rightarrow z \Gamma (z) = \frac{\Gamma (z + 2) }{z + 1} \)
\( \Rightarrow \Gamma (z) = \frac{\Gamma (z + 2) }{z(z + 1)} \)
Repeating this N times will generalize as:
\( \Gamma (z) = \frac{\Gamma (z + N + 1) }{z(z + 1) \cdot \cdot \cdot (z + N)} \)
This implies that the poles of the \( \Gamma (z) \) are at the values \( z = 0, -1, ..., -N \). We can plug in some value \( z = -n \) were n > 0 into this recurrence relation to find the value of the residue \( Res(\Gamma,-n) \):
\( (z + n) \Gamma (z) = \frac{\Gamma (z + n + 1) }{z(z + 1) \cdot \cdot \cdot (z + n - 1)} \)
\( \displaystyle\lim_{z\to -n} [ (z + n) \Gamma (z) ] = Res(\Gamma,-n) = \frac{\Gamma ( -n + n + 1) }{-n(1 - n) \cdot \cdot \cdot (n - 1 - n)} \)
\( Res(\Gamma,-n) = \frac{\Gamma(1)}{(-1)(n)(-1)(n-1)(-1)(n-2)...(-1)(n-n+1)} = \frac{1}{(-1)^{n} n!}\)
\( \Rightarrow Res(\Gamma,-n) = \frac{(-1)^{n}}{ n!} \) (Q.E.D.)
Problem 3.7: Simple quantum gravity effects are small.
Problem Statement:
(a) What would be the "gravitational" Bohr radius for a hydrogen atom if the attraction binding the electron to the proton was gravitational? The standard Bohr radius is \( a_{0} = \frac{\hbar^{2}}{m e^{2}} = 5.29 \times 10^{-9} cm \).
(b) In "units" where G, c, and \( \hbar \) are set equal to one, the temperature of a black hole is given by \( kT = \frac{1}{8 \pi M} \). Insert back the factors of G, c, and \( \hbar \) into this formula. Evaluate the temperature of a black hole of a million solar masses. What is the mass of black hole whose temperature is room temperature?
Solution:
Problem 3.7 (a): What would be the "gravitational" Bohr radius for a hydrogen atom if the attraction binding the electron to the proton was gravitational? The standard Bohr radius is \( a_{0} = \frac{\hbar^{2}}{m e^{2}} = 5.29 \times 10^{-9} cm \)
The derivation of the Bohr radius starts with the quantization of angular momentum:
\( m v_{n} r_{n} = n \hbar \)
\( \Rightarrow v_{n} = \frac{n \hbar}{ m r_{n} } \)
Then the value for the radius comes from the ground state, after setting centripetal force equal to the electric force, and in our units setting Z = 1 and k = 1:
\( m v_{n}^{2} = \frac{e^{2}}{r_{n}} \)
\( \Rightarrow \frac{n^{2} \hbar^{2}}{ m r_{n}^{2} } = \frac{e^{2}}{r_{n}} \)
\( \Rightarrow r_{n} = \frac{n^{2} \hbar^{2}}{ m e^{2} } \)
\( \Rightarrow a_{0} = r_{1} = \frac{ \hbar^{2}}{ m e^{2} } = 5.29 \times 10^{-9} cm \)
Therefore if we replace the electric force with the gravitational force, we can find the "gravitational" Bohr radius:
\( m_{e} v_{n}^{2} = \frac{G M_{p} }{r_{n}} \)
\( \Rightarrow \frac{n^{2} \hbar^{2}}{ m_{e} r_{n}^{2} } = \frac{G M_{p} m_{e}}{r_{n}} \)
\( \Rightarrow r_{n} = \frac{n^{2} \hbar^{2}}{ G M_{p} m_{e}^{2} } \)
\( \Rightarrow a_{0} = r_{1} = \frac{ \hbar^{2}}{G M_{p} m_{e}^{2} } \)
This evaluates into centimeters as:
\( a_{0} = \frac{ \hbar^{2} }{G M_{p} m_{e}^{2} } = \frac{ ( 1.0546 \times 10^{-34} \frac{ kg m^{2} }{s} )^{2} }{ ( 6.6743 \times 10^{-11} \frac{m^{3}}{kg s^{2}} ) ( 1.6726 \times 10^{-27} kg ) (9.1093 \times 10^{-31} kg )^{2} } = \frac{ ( 1.0546 \times 10^{-30} \frac{ kg cm^{2} }{s} )^{2} }{ ( 6.6743 \times 10^{-5} \frac{cm^{3}}{kg s^{2}} ) ( 1.6726 \times 10^{-27} kg ) (9.1093 \times 10^{-31} kg )^{2} }\)
\( a_{0} = \frac{ ( 1.1122 \times 10^{-60} \frac{ kg^{2} cm^{4} }{s^{2}} ) }{ ( 6.6743 \times 10^{-5} \frac{cm^{3}}{kg s^{2}} ) ( 1.6726 \times 10^{-27} kg ) (82.9793 \times 10^{-62} kg^{2} ) } \)
\( a_{0} = \frac{1.1122}{(6.6743)(1.6726)(82.9793)} \times \frac{ ( 10^{-60} \frac{ kg^{2} cm^{4} }{s^{2}} ) }{ ( 10^{-5} \frac{cm^{3}}{kg s^{2}} ) ( 10^{-27} kg ) ( 10^{-62} kg^{2} ) } \)
\( a_{0} = \frac{1.1122}{926.3340} \times \frac{10^{-60}}{(10^{-5})(10^{-27} ) ( 10^{-62} )} cm \)
\( a_{0} = \frac{1.1122}{926.3340} \times \frac{10^{-60}}{10^{-94} } cm \)
\( a_{0} = 0.0012 \times 10^{34} cm \)
\( \Rightarrow a_{0} = 1.2 \times 10^{31} cm \) (Q.E.D.)
To put this into context, the diameter of the observable universe is 93 billion light-years, which is only \( 8.8 \times 10^{28} cm\).
Problem 3.7 (b): In "units" where G, c, and \( \hbar \) are set equal to one, the temperature of a black hole is given by \( kT = \frac{1}{8 \pi M} \). Insert back the factors of G, c, and \( \hbar \) into this formula. Evaluate the temperature of a black hole of a million solar masses. What is the mass of black hole whose temperature is room temperature?
Step (1): Recover the constants G, c, and \( \hbar \) in the right combination using dimensional analysis.
kT has dimensions of energy, therefore \( [E] = M L^{2} T^{-2} \). Inserting the term \( G^{\alpha} c^{\beta} \hbar^{\gamma} \) lets us find the three unknown exponents with three equations, given the dimensions of G, c, and \( \hbar \). From Quick Calculation 3.8:
\( [G] = M^{-1} L^{3} T^{-2} \)
\( [c] = M^{0} L^{1} T^{-1} \)
\( [\hbar] = M^{1} L^{2} T^{-1} \)
Applying the unknown combination of constants to kT: \( kT = ( 8 \pi M )^{-1} G^{\alpha} c^{\beta} \hbar^{\gamma} \).
\( [E] = [G]^{\alpha} [c]^{\beta} [\hbar]^{\gamma} M^{-1} \)
\( M L^{2} T^{-2} = M^{ -\alpha + \gamma - 1} L^{ 3\alpha + \beta + 2 \gamma } T^{-2\alpha - \beta - \gamma } \)
Equating the exponent terms on the left-hand and right-hand sides gives three equations and three unknowns:
\( 1 = -\alpha + \gamma - 1 \)
\( 2 = 3\alpha + \beta + 2 \gamma \)
\(-2 = - 2\alpha - \beta - \gamma \)
Substitutions and rearrangements give:
\( \alpha = -1 , \beta = 3 , \gamma = 1 \)
\( \Rightarrow kT = \frac{1}{ 8 \pi M} \frac{\hbar c^{3}}{G}\) (Q.E.D.)
Step (2): Find the temperature T for a black hole of 1 million solar masses.
Now that we have recovered the constants, \( M_{solar} = 2 \times 10^{30} kg \), and the problem is asking to solve for T given \( M = 10^{6} M_{solar} = 2 \times 10^{36} kg \):
\( T = \frac{1}{ 8 \pi M} \frac{\hbar c^{3}}{k G} \)
\( \Rightarrow T = \frac{1}{50.266 \times 10^{36} kg} \frac{ ( 1.055 \times 10^{-34} \frac{ kg m^{2} }{s} ) ( 2.998 \times 10^{8} \frac{m}{s} )^{3} }{ ( 1.381 \times 10^{-23} \frac{ kg m^{2} }{s^{2} K} ) ( 6.674 \times 10^{-11} \frac{m^{3}}{kg s^{2}} ) } \)
\( \Rightarrow T = \frac{1}{50.266 \times 10^{36} kg} \frac{ ( 1.055 \times 10^{-34} \frac{ kg m^{2} }{s} ) ( 26.946 \times 10^{24} \frac{m^{3}}{s^{3}} ) }{ ( 1.381 \times 10^{-23} \frac{ kg m^{2} }{s^{2} K} ) ( 6.674 \times 10^{-11} \frac{m^{3}}{kg s^{2}} ) } \)
\( \Rightarrow T = \frac{1}{50.266 \times 10^{36} } \frac{ ( 1.055 \times 10^{-34} ) ( 26.946 \times 10^{24} ) }{ ( 1.381 \times 10^{-23} ) ( 6.674 \times 10^{-11} ) } K \)
\( \Rightarrow T = \frac{ ( 1.055 ) ( 26.946 ) }{ (50.266 )( 1.381 ) ( 6.674 ) } \times \frac{ 10^{-34} 10^{24} }{ 10^{36} 10^{-23} 10^{-11} }K \)
\( \Rightarrow T = \frac{ ( 28.428 ) }{ ( 463.291 ) } \times \frac{ 10^{-10} }{ 10^{2} }K \)
\( \Rightarrow T = 0.06136 \times 10^{-12} K \)
\( \Rightarrow T = 6.14 \times 10^{-14} K \) (Q.E.D.)
Step (3): Find the mass of a black hole at room temperature of 300 K.
To find the mass of a black hole at room temperature, which we will take to be 300 Kelvin, you rearrange the equation to swap M and T:
\( M = \frac{1}{ 8 \pi T} \frac{\hbar c^{3}}{k G} \)
As a short cut we have already calculated that \( 8 \pi T M = (6.136 \times 10^{-14} K)(50.266 \times 10^{36} kg) = 306.432 \times 10^{22} kg K = \frac{\hbar c^{3}}{k G} \):
\( \Rightarrow M = \frac{1}{ 8 \pi T} ( 306.432 \times 10^{22} kg K ) \)
\( M = \frac{1}{ 25.133 (300 K)} ( 306.432 \times 10^{22} kg K ) = \frac{1}{ 7540 K} ( 306.432 \times 10^{22} kg K ) \)
\( M = ( 1.326 \times 10^{-4} K^{-1} ) ( 306.432 \times 10^{22} kg K ) \)
\( M = 0.0406 \times 10^{22} kg \)
Which we find to be a black hole with 10 orders of magnitude less mass than 1 solar mass, smaller black holes having higher temperature:
\( M = 4.1 \times 10^{20} kg \) (Q.E.D.)
Problem 3.8: Vacuum energy and an associated length scale.
Problem Statement:
Observations indicate that the expansion of the universe is currently accelerating possibly due to a vacuum energy density. The mass density associated with this energy is approximately \( \rho_{vac} = 7.7 \times 10^{-27} \frac{kg}{m^{3}} \). Some physicists try to understand the acceleration of the universe by introducing modifications to gravity. It is then useful to know what length scales could be important. If one assumes that the only relevant prameters are \( \rho_{vac} , \hbar \), and c, one can construct a length parameter \( l_{vac} \) by multiplying powers: \( l_{vac} = \rho_{vac}^{\alpha} \hbar^{\beta} c^{\gamma} \). What must be the values of \( \alpha, \beta , \) and \( \gamma \) in the above equation? What is the numerical value of \( l_{vac} \)? Express your answer in \( \mu m \), where \( \mu m = 10^{-6} m \).
Consider the dimensions of \( \rho_{vac} , \hbar \), and c:
\( \rho_{vac} = M^{1} L^{-3} T^{0} \)
\( \hbar = M^{1} L^{2} T^{-1} \)
\( c = M^{0} L^{1} T^{-1} \)
Then consider the characteristic length scale:
\( l_{vac} = \rho_{vac}^{\alpha} \hbar^{\beta} c^{\gamma} \)
In dimensional terms this becomes:
\( M^{0}L^{1}T^{0} = M^{(\alpha + \beta)} L^{(-3\alpha + 2\beta + \gamma)} T^{ ( -\beta -\gamma ) } \)
This means the exponents turn into three equations and three unknowns:
\( 0 = \alpha + \beta \)
\( 1 = -3\alpha + 2\beta + \gamma \)
\( 0 = -\beta - \gamma \)
\( \Rightarrow \alpha = -\beta = \gamma \)
\( \alpha = -\frac{1}{4} , \beta = \frac{1}{4} , \gamma = -\frac{1}{4} \)
So the length scale becomes:
\( l_{vac} = ( \frac{\hbar}{\rho_{vac} c} )^{ \frac{1}{4} } \)
The numerical value of this length is:
\( l_{vac} = ( \frac{ 1.055 \times 10^{-34} \frac{ kg m^{2} }{s} }{( .77 \times 10^{-26} \frac{kg}{m^{3}} ) ( 2.998 \times 10^{8} \frac{m}{s} )} )^{ \frac{1}{4} } \)
\( l_{vac} = (\frac{1.055}{.77 \times 2.998})^{ \frac{1}{4} } (\frac{10^{-34}}{ 10^{-26} 10^{8} })^{ \frac{1}{4} } m \)
\( l_{vac} = (\frac{1.055}{2.308})^{ \frac{1}{4} } (\frac{10^{-34}}{ 10^{-26} 10^{8} })^{ \frac{1}{4} } m \)
\( l_{vac} = (0.457)^{ \frac{1}{4} } \times (10^{-16})^{\frac{1}{4}} m \)
\( l_{vac} = 0.822 \times 10^{-4} m \)
\( \Rightarrow l_{vac} = 82.2 \mu m \)
Problem 3.9: Planetary motion in four and higher dimensions.
Problem Statement:
Consider the motion of planets in planar circular orbits around heavy stars in our four-dimensional spacetime and in spacetimes with additional spatial dimensions. We wish to study the stability of these orbits under perturbations that keep them planar. Such a perturbation would arise, for example, if a meteorite moving on the plane of the orbit hits the planet and changes its angular momentum. Show that while planetary circular orbits in our four-dimensional world are stable under such perturbations, they are not so in five or higher dimensions. [Hint: you may find it useful to use the effective potential for motion in a central force field.]
Consider the effective potential for a central gravitational force field, where L is the angular momentum:
\( V_{eff} (r) = V_{g} (r) + \frac{L^{2}}{2mr^{2}} \)
We have to find the equation for \( V_{g} \) in d dimensions first, before doing stability analysis with derivatives. Consider the divergence of gravitational field, analogous to Problem 3.4 (b):
\( \int_{S^{d-1} } \overrightarrow{g} \cdot d\overrightarrow{S} = \int_{B^{d} } \nabla \cdot \overrightarrow{g} = - 4 \pi G^{D} m \)
The argument is exactly the same as Problem 3.4 (b), where: ( \( q \rightarrow - 4 \pi G^{(D)} m \) ), ( \( E(r) \rightarrow g(r) \) ), and ( \( \Phi (r) \rightarrow V_{g} (r) \) ):
\( g(r) vol(S^{d-1}) = 4 \pi G^{D} m \)
\( g(r) = - \frac{4 \pi G^{D} m}{vol(S^{d-1})} \)
\( g(r) = - \frac{\Gamma(\frac{d}{2})}{2 \pi^{\frac{d}{2}}} \frac{4 \pi G^{D} m}{r^{d-1}} \)
\( \Rightarrow g(r) = - \frac{2 \Gamma(\frac{d}{2})}{ \pi^{\frac{d}{2}-1}} \frac{ G^{D} m}{r^{d-1}} \)
Then in analogy of Equation 3.75 with the electric field \( \overrightarrow{E} \) and potential \( \Phi \):
\( \overrightarrow{g} = - \nabla V_{g} = - \frac{d}{dr} V_{g} (r) \)
The integral of g(r) with respect to r gives \( V_{g} \), which simplifies with the \( \Gamma \) identity:
\( V_{g} (r) = - \frac{\Gamma ( \frac{d}{2} - 1 ) }{ \pi^{\frac{d}{2}-1}} \frac{G^{(D)} m}{r^{d-2}} \)
Now, we have the effective potential for d dimension:
\( V_{eff} (r) = - \frac{\Gamma ( \frac{d}{2} - 1 ) }{ \pi^{\frac{d}{2}-1}} \frac{G^{(D)} m}{r^{d-2}} + \frac{L^{2}}{2mr^{2}} \)
This is where it gets a little tricky. We want to find the first two derivatives with respect to r of the effective potential, when the first derivative is equal to zero, while the second derivative gives us the conditions for stability. We already know this for \( V_{g} \), because it is the expression for g(r):
\( \frac{d}{dr} V_{eff} = \frac{d}{dr} V_{g} + \frac{d}{dr} \frac{L^{2}}{2mr^{2}} = 0 \)
For simplicity we will sweep the ugly constant of g(r) under a symbol C:
\( \frac{d}{dr} V_{eff} = C [\frac{1}{r^{d-1}}] - \frac{L^{2}}{mr^{3}} = 0 \)
Since this is the point \( r_{0} \), being circular planetary orbit in a plane, where \( \frac{d}{dr} V_{eff} = 0\):
\( C [\frac{1}{r_{0}^{d-1}}] = \frac{L^{2}}{mr_{0}^{3}} \) (Eqn. 1)
When we do the second derivative, which gives us the stability, we can substitute the right-hand side into it to replace the ugly left-hand side:
\( \frac{d^{2}}{dr^{2}} V_{eff} = C [\frac{(1-d)}{r^{d}}] - \frac{(-3) L^{2}}{mr^{4}} \)
We can substitute into the left-hand side the equation above marked (Eqn. 1) and use the radial distance \( r_{0}\) :
\( [\frac{L^{2}}{m r_{0}^{3}}] \frac{(1-d)}{r_{0}} = C [\frac{(1-d)}{r_{0}^{d}}] \)
Which means the second derivative of the effective potential is:
\( \frac{d^{2}}{dr^{2}} V_{eff} = [\frac{L^{2}}{m r_{0}^{3}}] \frac{(1-d)}{r_{0}} - \frac{(-3) L^{2}}{mr_{0}^{4}} \)
\( \Rightarrow \frac{d^{2}}{dr^{2}} V_{eff} = \frac{(4 - d) L^{2}}{mr_{0}^{4}} \) (Q.E.D.)
For d = 3 or lower (note: d is spatial dimensions, D is spacetime dimension), the second derivative is positive, which means the orbits are stable if perturbed. For d = 4 or higher, the orbits are unstable. This means that in five spacetime dimensions or higher, circular planetary orbits are unstable under same plane perturbations.
Problem 3.10: Gravitational field of a point mass in a compactified five-dimensional world.
Problem Statement:
Consider a five-dimensional spacetime with space coordinates \( (x,y,z,w) \) not yet compactified. A point mass M is located at the origin \( (x,y,z,w) = (0,0,0,0) \).
(a) Find the gravitational potential \( V_{g}^{(5)}(r) \). Write your answer in term of M, \( G^{(5)} \) , and \( r = (x^{2} + y^{2} + z^{2} + w^{2} )^{\frac{1}{2}} \). [Hint: use \( \nabla^{2} V_{g}^{(5)} = 4 \pi G^{(5)} \) and the divergence theorem.]
Now let w become a circle with radius a while keeping the mass fixed, as shown in Figure 3.3.
(b) Write an exact expression for the gravitational potential \( V_{g}^{(5)} (x,y,z,0) \). This potential is a function of \( R \equiv (x^{2} + y^{2} + z^{2} )^{\frac{1}{2}} \) and can be written as an infinite sum.
(c) Show that for \( R >> a \) the gravitational potential takes the form of a four-dimensional gravitational potential, with Newton's constant \( G^{(4)} \) given in terms of \( G^{(5)} \) as in (3.115). [Hint: turn the infinite sum into an integral.]
These results confirm both the relation between the four- and five-dimensional Newton constants in a compactification and the emergence of a four-dimensional potential at distances large compared to the size of the compact dimension.
Solution:
Problem 3.10 (a): Find the gravitational potential \( V_{g}^{(5)}(r) \). Write your answer in term of M, \( G^{(5)} \) , and \( r = (x^{2} + y^{2} + z^{2} + w^{2} )^{\frac{1}{2}} \). [Hint: use \( \nabla^{2} V_{g}^{(5)} = 4 \pi G^{(5)} \) and the divergence theorem.]
This was already derived above in Problem 3.9 for arbitrary dimensions d,D:
\( V_{g}^{(D)} (r) = - \frac{\Gamma ( \frac{d}{2} - 1 ) }{ \pi^{\frac{d}{2}-1}} \frac{G^{(D)} m}{r^{d-2}} \)
Let D = 5, and therefore d = 4:
\( V_{g}^{(5)} (r) = - \frac{\Gamma ( \frac{4}{2} - 1 ) }{ \pi^{\frac{4}{2}-1}} \frac{G^{(5)} M}{r^{4-2}} \)
\( V_{g}^{(5)} (r) = - \frac{\Gamma ( 2 - 1 ) }{ \pi^{2-1}} \frac{G^{(5)} M}{r^{2}} \)
\( \Rightarrow V_{g}^{(5)} (r) = - \frac{\Gamma ( 1 ) }{ \pi} \frac{G^{(5)} M}{r^{2}} \)
For integers n \( \epsilon \) \( \mathbb{Z}^{+} \):
\( \Gamma (n) = ( n - 1 )! \)
\( \Rightarrow \Gamma (1) = 0! = 1 \)
\( \Rightarrow V_{g}^{(5)} (r) = - \frac{G^{(5)} M}{\pi r^{2}} \) (Q.E.D.)
Problem 3.10 (b): Write an exact expression for the gravitational potential \( V_{g}^{(5)} (x,y,z,0) \). This potential is a function of \( R \equiv (x^{2} + y^{2} + z^{2} )^{\frac{1}{2}} \) and can be written as an infinite sum.
This compactification of one dimension into a circle of radius a means there is a periodicity of length a on the w axis:
\( \mathbb{R}^{1}: w \) ~ \( w + 2 n \pi a \)
That means we can separate out the w in the definition of r in Problem 3.10(a) and substitute in the condition:
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi r^{2}} = - \frac{G^{(5)} M}{\pi ( (x^{2} + y^{2} + z^{2}) + w^{2}) } \)
\( \Rightarrow V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi ( R^{2} + w^{2}) } \)
\( \Rightarrow V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi ( R^{2} + ( 2 n \pi a )^{2}) } \)
The exact expression for all values of n is therefore an infinite sum:
\( V_{g}^{(5)} (x,y,z,0) = - \sum\limits_{n=-\infty}^\infty \frac{G^{(5)} M}{\pi ( R^{2} + ( 2 n \pi a )^{2})} \) (Q.E.D.)
Problem 3.10 (c): Show that for \( R >> a \) the gravitational potential takes the form of a four-dimensional gravitational potential, with Newton's constant \( G^{(4)} \) given in terms of \( G^{(5)} \) as in (3.115). [Hint: turn the infinite sum into an integral.]
R >> a implies that the infinite sum can be turned into an infinite integral instead, as the scale of R dominates over the discreteness of n:
\( V_{g}^{ (5) } (x,y,z,0) = - \int^{+\infty}_{-\infty} \frac{G^{(5)} M}{\pi ( R^{2} + ( 2 \pi a t)^{2})} dt \)
This is taking a continuum limit of \( n \rightarrow t \). This integral is analytically solvable, the limits of arctan and +/- \( \infty \):
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ \pi R^{2}} \int^{+\infty}_{-\infty} \frac{1}{( 1 + ( \frac{2 \pi a}{R})^{2} t^{2})} dt \)
Note that this is the integral identity for arctan(x) involving a constant j:
\( \int \frac{1}{1 + j^{2} x^{2} } = \frac{tan^{-1}(j x )}{j}\)
\( \Rightarrow j = \frac{2 \pi a}{R} \)
Therefore, for the limit of \( t \rightarrow +/- \infty \):
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ \pi R^{2}} [\frac{R tan^{-1}( \frac{2 \pi a}{R} t ) }{( 2 \pi a )}]^{+\infty}_{-\infty} \)
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } [ \frac{tan^{-1}( \frac{2 \pi a}{R} t ) }{( \pi )}]^{+\infty}_{-\infty} \)
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \frac{1}{\pi}[ tan^{-1}( \frac{2 \pi a}{R} t ) ]^{+\infty}_{-\infty} \)
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \frac{1}{\pi} [ [ tan^{-1}( \frac{2 \pi a}{R} t ) ]^{+\infty}_{0} + [ tan^{-1}( \frac{2 \pi a}{R} t ) ]^{0}_{-\infty} ] \)
The limits of arctan at infinity are \( arctan(+ \infty) = \frac{\pi}{2} , arctan (- \infty) = - \frac{\pi}{2} \):
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \frac{1}{\pi} [ [ tan^{-1}( +\infty ) - tan^{-1}( 0 ) ] + [ tan^{-1}( 0 ) - tan^{-1}( -\infty ) ] ] \)
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \frac{1}{\pi} [ [ \frac{\pi}{2} - 0 ] + [ 0 - ( - \frac{\pi}{2} ) ] ] \)
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \frac{\pi}{\pi} \)
\( \Rightarrow V_{g}^{ (5) } (x,y,z,0) = - \frac{G^{(5)} M}{ 2 \pi a R } \)
Since Equation 3.115 says \( \frac{G^{(5)}}{G} = 2 \pi R \equiv l_{C}\), it follows that \( \frac{G^{(5)}}{2 \pi R} = G \):
\( V_{g}^{ (5) } (x,y,z,0) = - \frac{G M}{ R } \) (Q.E.D.)
This shows that compactifying the fourth spatial dimension, with R >> a, makes gravity behave as it does in three spatial dimensions where spacetime D = 4.
Problem 3.11: Exact answer for the gravitational potential.
Problem Statement:
The infinite sum in Problem 3.10 can be evaluated exactly using the identity \( \sum\limits_{n=- \infty}^\infty \frac{1}{1 + (\pi n x)^{2}} = \frac{1}{x} coth( \frac{1}{x})\)
(a) Find an exact closed-form expression for the gravitational potential \( V_{g}^{(5)}(x,y,z,0) \) in the compactified theory.
(b) Expand the answer to calculate the leading correction to the gravitational potential in the limit when R >> a. For what value of R/a is the correction of order 1%?
(c) Use the exact answer in (a) to expand the potential when R << a. Give the first two terms in the expansion. Do you recognize the leading term?
Solution:
Problem 3.11 (a): Find an exact closed-form expression for the gravitational potential \( V_{g}^{(5)}(x,y,z,0) \) in the compactified theory.
From Problem 3.10 (b) we have:
\( V_{g}^{(5)} (x,y,z,0) = - \sum\limits_{n=-\infty}^\infty \frac{G^{(5)} M}{\pi ( R^{2} + ( 2 n \pi a )^{2})} \)
Therefore we re-express this in terms of the identity \( \sum\limits_{n=- \infty}^\infty \frac{1}{1 + (\pi n x)^{2}} = \frac{1}{x} coth( \frac{1}{x})\):
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi R^{2}} \sum\limits_{n=-\infty}^\infty \frac{1}{( 1 + ( \pi n ( \frac{ 2 a }{R} ))^{2})} \)
In this case \( x = \frac{2 a}{R} \), therefore:
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi R^{2}} \frac{R}{2 a} coth( \frac{R}{2 a}) \)
\( \Rightarrow V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{2 \pi a R} coth( \frac{R}{2 a}) \)
As in Problem 3.10 (c), we can use Equation 3.115 ( \( \frac{G^{(5)}}{G} = 2 \pi R \) ) to re-express this in terms of Newton's constant in D = 4, which is just G:
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G M}{R} coth( \frac{R}{2 a}) \)
Problem 3.11 (b): Expand the answer to calculate the leading correction to the gravitational potential in the limit when R >> a. For what value of R/a is the correction of order 1%?
This involves approximating \( coth( \frac{R}{2 a}) \) to first order, ignoring higher order terms because R >> a. We want the form (1 + x), where x is the leading correction, since we're multiplying through \( \frac{G M}{R} \). Re-express coth(x) into the more familiar exponential form:
\( coth ( x ) = \frac{ e^{x} + e^{-x} }{ e^{x} - e^{-x} } \)
\( coth ( x ) = \frac{ e^{-x} }{ e^{x} } \frac{1 + e^{2x}}{1 - e^{-2x}} \)
\( coth ( x ) = e^{-2x} \frac{1 + e^{2x}}{1 - e^{-2x}} \)
\( coth ( x ) = \frac{ e^{-2x} + 1}{1 - e^{-2x}} = \frac{ 1 + e^{-2x}}{1 - e^{-2x}} \)
Having this expression, we can see that \( \gamma = 2x = \frac{R}{a} \):
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G M}{R} \frac{ 1 + e^{-\gamma}}{1 - e^{-\gamma}} \)
This is the form \( \frac{(1 + x)}{(1 - x)} \), which we can simplify with commonly used linear approximations:
\( \frac{(1 + x)}{(1 - x)} = (1 + x) [ \frac{1}{(1 - x)} ] \approx (1 + x) [ (1 + x) ] = (1 + x)^{2} \approx (1 + 2x) \)
\( \Rightarrow V_{g}^{(5)} (x,y,z,0) = - \frac{G M}{R} ( 1 + 2 e^{-\gamma} ) \)
The term \( 2 e^{-\gamma} \) is the leading correction for \( \frac{G M}{R} \), so we find the value of \( \gamma \) for a correction of 1%:
\( 0.01 = 2 e^{-\gamma} \)
\( 0.005 = e^{-\gamma} \)
\( ln(0.005) = -\gamma \)
\( -5.3 = -\gamma \)
\( \Rightarrow \gamma = \frac{R}{a} = 5.3 \) (Q.E.D.)
Problem 3.11 (c): Use the exact answer in (a) to expand the potential when R << a. Give the first two terms in the expansion. Do you recognize the leading term?
Since R << a the approximation in (b) is invalid, and we have to use the Taylor expansion of coth(x) instead:
\( coth(x) = \frac{1}{x} + \frac{x}{3} - \frac{x^{3}}{45} + ... \)
Therefore we substitute this series expansion into the equation from 3.11 (a), where \( x = \frac{R}{2a} \), only for the first two terms:
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{2 \pi a R} coth( \frac{R}{2 a}) \)
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{2 \pi a R} [ \frac{2a}{R} + \frac{R}{6a} ] \)
It asks about recognizing the leading term. If we pull out \( \frac{2a}{R} \) so the first term equals 1, and the second term is a correction:
\( V_{g}^{(5)} (x,y,z,0) = - \frac{G^{(5)} M}{\pi R^{2}} [ 1 + \frac{R^{2}}{12a^{2}} ] \)
The first term is just \( V_{g}^{(5)} = - \frac{G^{(5)} M}{\pi R^{2}} \), the gravitational potential from Problem 3.10 (a). This is the five dimensional spacetime gravitational potential without compactification, and so the expansion terms following it are the corrections adjusting for the compactied w dimension.
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Last updated: 12/15/2020
