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String Theory: Chapter 5 Problem Set Solutions for Barton Zwiebach's "A First Course in String Theory"

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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.

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Quick Calculations

Quick Calculation 5.1: Calculate the Variation $\delta S_{nr}$ for a Galilean Boost

Quick Calc 5.1: Solution

Quick Calculation 5.2: Show that you can Reparameterize $\frac{d p_{\mu}}{d \tau} = 0$ as $\frac{d p_{\mu}}{d \tau '} = 0$ for $\tau ' (\tau)$.

Quick Calc 5.2: Solution

Problems

Problem 5.1: Point Particle Equation of Motion and Reparameterizations

Problem 5.1: Solution

Problem 5.2: Particle Equation of Motion with Arbitrary Parameterization

Problem 5.2: Solution

Problem 5.3: Four-Current of a Charged Point Particle in D Dimensions

Problem 5.3 (a): Solution

Problem 5.3 (b): Solution

Problem 5.4: Hamiltonian for a Nonrelativistic Charged Particle $(\dagger)$

Problem 5.4 (a): Solution

Problem 5.4 (b): Solution

Problem 5.4 (c): Solution

Problem 5.5: Equations of Motion for a Charged Point Particle

Problem 5.5: Solution

Problem 5.6: Electromagnetic Field Dynamics with a Charged Particle (*) $(\dagger)$

Problem 5.6: Solution

Problem 5.7: Point Particle Action in Curved Space (*)

Problem 5.7: Solution

(*) 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

Quick Calculation 5.1:

Problem Statement: Calculate explicitly the variation of the action $S_{nr}$ under a boost.

Equation (5.2): $S_{nr} = \int L_{nr} dt = \int \frac{1}{2} m v^{2}(t) dt$, $v^{2} \equiv \overrightarrow{v} \cdot \overrightarrow{v}$, $\overrightarrow{v} = \frac{d \overrightarrow{x}}{dt}$, $v = \rvert \overrightarrow{v} \rvert$

Solution:

In context with the textbook, the problem means the variation $\delta$ of the non-relativistic action $S_{nr}$, under a Galilean boost $\overrightarrow{v} \rightarrow \overrightarrow{v} + \overrightarrow{v_{0}}$, where $\overrightarrow{v_{0}}$ is a constant:

$\delta S_{nr} = \delta \int L_{nr} dt = \int \frac{1}{2} m v_{new}^{2}(t) dt$
$\delta S_{nr} = \delta \int \frac{1}{2} m \overrightarrow{v} \cdot \overrightarrow{v_{0}} dt$
$\delta S_{nr} = m \int \overrightarrow{v} \cdot \overrightarrow{v_{0}} dt$

The fraction $\frac{1}{2}$ is cancelled by the symmetry of $\overrightarrow{v}, \overrightarrow{v_{0}}$ varying by the same amount. We replace $v^{2}$ with $\overrightarrow{v} \cdot \overrightarrow{v_{0}}$, the magnitudes of each velocity vector modified by the difference in their directions. Since $\overrightarrow{v_{0}}$ is constant, it will not vary over the interval. Therefore that term can be pulled outside the integral:

$\delta S_{nr} = m \overrightarrow{v_{0}} \cdot \int \overrightarrow{v} dt$
$\delta S_{nr} = m \overrightarrow{v_{0}} \cdot \int \frac{d \overrightarrow{x}}{dt} dt$
$\delta S_{nr} = m \overrightarrow{v_{0}} \cdot \int d \overrightarrow{x}$

Therefore, $\delta S_{nr} = 0$, because it is zero at the end points $x_{i}, x_{f}$. The variation of the action is thus zero under a Galilean boost. (Q.E.D.)

Quick Calculation 5.2:

Problem Statement: Show that $\frac{d p_{\mu}}{d \tau} = 0$ implies $\frac{d p_{\mu}}{d \tau '} = 0$ for an arbitrary parameter $\tau ' (\tau)$. What should be true about $\frac{d \tau '}{d \tau}$ for $\tau '$ to be a good parameter when $\tau$ is?

Solution:

This is just a straight-forward consequence of the chain rule, which implicitly is a change of basis or coordinates:

$\frac{d p_{\mu}}{d \tau} = \frac{d p_{\mu}}{d \tau '} \frac{d \tau ' }{d \tau} = 0$
$\Rightarrow \frac{d p_{\mu} }{d \tau '} = 0$

This holds for arbitrary $\tau ' (\tau)$ provided $\frac{d \tau ' }{d \tau}$ is continuous, such that it obeys the Inverse Function Theorem.

Problems

Problem 5.1

Problem Statement: Point particle equation of motion and reparameterizations.

If the path of a point particle is parameterized by proper time, the equation of motion (5.29). Consider now a new parameter $\tau = f(x)$. Find the most general function f which (5.29) implies: $\frac{d^{2} x^{\mu}}{d \tau^{2}} = 0$.

Solution:

The "most general" function f would be f(s), where f is a function of the spacetime interval s, rather than a parameter such as $\tau '$. This would follow from the textbook saying the action can be summed up by ds itself without using parameters. But by "general" the problem seems to mean for some arbitrary x:

$\tau = f(x)$

This notation is awkward. Remember $\mu$ is an index, not an exponent, and this is not $\partial_{\mu}$. Then by the chain rule:

$\frac{d x^{\mu}}{d x} = \frac{d x^{\mu}}{d \tau} \frac{d \tau}{d x}$
$\frac{d x^{\mu}}{d x} = \frac{d x^{\mu}}{d \tau} \frac{d f}{d x}$

Then we take the derivative with product rule:

$\frac{d^{2} x^{\mu}}{d x^{2}} = \frac{d}{dx}( \frac{d x^{\mu}}{d \tau} \frac{d f}{d x} )$
$\frac{d^{2} x^{\mu}}{d x^{2}} = \frac{d^{2} x^{\mu}}{d \tau^{2}} ( \frac{d f}{d x} ) ( \frac{d f}{d x} ) + \frac{d x^{\mu} }{d \tau} \frac{d^{2} f}{d x^{2}} = 0$

The extra $\frac{d f}{d x}$ comes from having to do another chain rule to do the x derivative on $\frac{d x^{\mu}}{d \tau}$. The problem gives us $\frac{d^{2} x^{\mu}}{d \tau^{2}} = 0$ as a condition:

$\Rightarrow 0 + \frac{d x^{\mu} }{d \tau} \frac{d^{2} f}{d x^{2}} = 0$

$\frac{d x^{\mu} }{d \tau} \neq 0$ in general. Therefore it follows that: $\frac{d^{2} f}{d x^{2}} = 0$, which integrates out to the function:

$f = ax + b$; a, b are arbitrary constants. (Q.E.D.)

Problem 5.2

Problem Statement: Particle equation of motion with arbitrary parameterization.

Vary the point particle action (5.15) to find a manifestly reparameterization invariant form of the free particle equation of motion.

Solution:

Equation (5.15): $S = -mc \int^{t_{f}}_{t_{i}} \sqrt{ - \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu}}{d \tau} } d \tau$

For the sake of context, "manifestly" reparameterization invariant in the textbook, Section 5.2, refers to the action S having the same form after changing the parameter $\tau$ to $\tau '$ using the chain rule. In Section 5.3 the action is varied $\delta S = -mc \int_{\mathcal{P}} \delta (ds)$, which ultimately leads to the general condition: $\frac{d^{2} x^{\mu}}{ds^{2}} = 0$. Therefore we can begin at this point.

You cannot generally replace ds with $d \tau$ in this equation, $\frac{d^{2} x^{\mu}}{d \tau^{2}} \neq 0$ for any arbitrary parameter $\tau$. The equivalent expression must be expanded in terms of the parameter using the chain rule, and then manipulated into a form that shows it is manifestly reparameterization invariant:

$\frac{d x^{\lambda}}{ds} = \frac{d x^{\lambda}}{d \tau} \frac{d \tau}{d s}$

$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d}{ds}( \frac{d x^{\lambda}}{d \tau} \frac{d \tau}{d s} )$
$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \frac{d \tau}{d s} ) ( \frac{d \tau}{d s} ) + \frac{d x^{\lambda} }{d \tau} \frac{d^{2} \tau}{d s^{2}} = 0$
$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \frac{d \tau}{d s} )^{2} + \frac{d x^{\lambda} }{d \tau} \frac{d^{2} \tau}{d s^{2}} = 0$

Then we also know that ds is parameterized along the worldline with increments $d \tau$:

$ds = \sqrt{ - \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} } d \tau$
$(ds)^{2} = - \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ( d \tau )^{2}$

We can change the last term to get it all in terms of $( \frac{d \tau}{d s} )^{2}$, making use of the chain rule on the last term:

$\frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \frac{d \tau}{d s} )^{2} + \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d s} ( \frac{d \tau}{d s} )^{2} = 0$

We can then substitute in the definition of $\frac{d \tau}{d s}$ from the above relations, but find it is inconveniently reciprocal:

$- \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} )^{-1} + \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d s} ( - \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} )^{- 1 } = 0$
$\Rightarrow \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} )^{-1} + \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d s} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} )^{- 1 } = 0$

This can be re-expressed in terms of positive exponents, by switching to the inverted $( \frac{d \tau}{ds} )^{-1} = \frac{ds}{d \tau}$, and getting all the derivatives in terms of the parameter $\tau$:

$\frac{d^{2} x^{\lambda}}{d \tau^{2}} ( (\frac{d \tau}{d s})^{-1} )^{2} - \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d \tau} ( ( \frac{d \tau}{d s} )^{-1} )^{2} = 0$
$\Rightarrow \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ) - \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d \tau} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ) = 0$

It is manifestly invariant with the same form under a reparameter change $( \frac{d \tau }{d \tau '} )^{2}$:

$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ) - \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d \tau} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ) = 0$
$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d^{2} x^{\lambda}}{d \tau^{2}} ( \frac{d \tau }{d \tau '} )^{2} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ( \frac{d \tau }{d \tau '} )^{2} ) - \frac{1}{2} \frac{d x^{\lambda} }{d \tau} \frac{d}{d \tau} ( \frac{d \tau }{d \tau '} )^{2} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu} }{d \tau} ( \frac{d \tau }{d \tau '} )^{2} ) = 0$
$\frac{d^{2} x^{\lambda}}{d s^{2}} = \frac{d^{2} x^{\lambda}}{d \tau ' ^{2}} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau '} \frac{d x^{\nu} }{d \tau '} ) - \frac{1}{2} \frac{d x^{\lambda} }{d \tau '} \frac{d}{d \tau '} ( \eta_{\mu \nu} \frac{d x^{\mu}}{d \tau '} \frac{d x^{\nu} }{d \tau '} ) = 0$ (Q.E.D.)

Problem 5.3

Problem Statement: Current of a charged point particle.

Consider a point particle with charge q whose motion in a D = d + 1-dimensional spacetime is described by functions $x^{\mu}(\tau) = \{ x^{0} (\tau), \overrightarrow{x} \}$ , where $\tau$ is a parameter. The moving particle generates an electromagnetic current $j^{\mu} = ( c \rho , \overrightarrow{j} ) .$

(a) Use delta functions to write expressions for the current components $j^{0} (\overrightarrow{x} , t)$ and $j^{i} ( \overrightarrow{x}, t)$.

(b) Show that your answer in (a) arise from the integral representation

$j^{\mu} (t, \overrightarrow{x} ) = qc \int d \tau \delta^{D} ( x - x( \tau ) ) \frac{d x^{ \mu} (\tau )}{d \tau}$.

Here $\delta^{D} (x) \equiv \delta ( x^{0} ) \delta ( x^{1} ) ... \delta ( x^{d} )$.

Solution:

Problem 5.3 (a): Use delta functions to write expressions for the current components $j^{0} (\overrightarrow{x} , t)$ and $j^{i} ( \overrightarrow{x}, t)$.

In non-relativistic classical electromagnetism, the charge moves along path $\overrightarrow{x} (t)$ relative to a coordinate system $\overrightarrow{x}$, and the individual points along that path can be represented as a delta function. This is because the charge is concentrated at that position at that time t, such that the charge density is:

$\rho = q \delta ( \overrightarrow{x} - \overrightarrow{x} (t) )$

Which means the current density $j = \rho \overrightarrow{v}$ is:

$\rho \overrightarrow{v} = q \overrightarrow{v} \delta ( \overrightarrow{x} - \overrightarrow{x} (t) )$

In relativistic spacetime this path is instead the worldline, and t is replaced by the parameter $\tau$ for the proper time, where being stationary to the charges is the charge density and moving relative to the charges is the current density. The time component of the four-current $j^{\mu} = ( c \rho , \overrightarrow{j} )$ is then the charge density, and the spatial components $\overrightarrow{j}$ are the current density:

$j^{0} = c \rho = c q \int \delta ( t - ct(\tau) ) d \tau$
$j^{i} = \overrightarrow{j} = q \int \frac{d x^{i} }{d \tau} \delta^{i} ( x - x(\tau) ) d \tau$

The nature of delta functions is that, integrated over infinity, they are distributions concentrated at one point: $\int^{+\infty}_{-\infty} f(t) \delta (t - T) = f(T)$.

Problem 5.3 (b): Show that your answer in (a) arise from the integral representation $j^{\mu} (t, \overrightarrow{x} ) = qc \int d \tau \delta^{D} ( x - x( \tau ) ) \frac{d x^{ \mu} (\tau )}{d \tau}$.

First notice that the $\delta^{D}$ term is composed of $\delta x^{\mu}$ elements, including the time component:

$\delta^{D} (x) \equiv \delta ( x^{0} ) \delta ( x^{1} ) ... \delta ( x^{d} )$

This is really a set of D integrals over the delta functions along the worldline, where the c is absorbed into the time component $x^{0}$:

$j^{\mu} (t, \overrightarrow{x} ) = q [ \int d \tau \delta ( x^{0} - x^{0}( \tau ) ) \frac{d x^{ 0} (\tau )}{d \tau} ] [ \int d \tau \delta ( x^{1} - x^{1}( \tau ) ) \frac{d x^{ 1} (\tau )}{d \tau} ] ... [ \int d \tau \delta ( x^{d} - x^{d}( \tau ) ) \frac{d x^{ d} (\tau )}{d \tau} ]$

Integrating over the parameter $\tau$ gives the charge and current densities at each time point on the worldline. The first integral contains our expression for the charge density at a given time, and the remaining d integrals are our current density definition at each time point:

$j^{0} = q [ \int d \tau \delta ( x^{0} - x^{0}( \tau ) ) \frac{d x^{ 0} (\tau )}{d \tau} d \tau ] = q [ c \int \delta ( t - ct( \tau ) ) d \tau ]$

$j^{i} = q \int d \tau \delta ( x^{i} - x^{i}( \tau ) ) \frac{d x^{ i} (\tau )}{d \tau} d \tau = q \int \frac{d x^{i} }{d \tau} \delta^{i} ( x - x(\tau) ) d \tau$

Thus, those terms arise from the D dimension integral. (Q.E.D.)

Problem 5.4

Problem Statement: Hamiltonian for a nonrelativistic charged particle.

The action for a nonrelativistic particle of mass m and charge q coupled to an electromagnetic field is obtained by replacing the first term in (5.34) by the nonrelativistic action for a free point particle:

$S = \int \frac{1}{2} m v^{2} dt + \frac{q}{c} \int A_{\mu} (x) \frac{d x^{\mu}}{dt} dt$

We have also chosen to use time to parameterize the second integral.

(a) Rewrite the action S in terms of the potentials $( \Phi , \overrightarrow{A} )$ and the ordinary velocity $\overrightarrow{v}$. What is the Lagrangian?

(b) Calculate the canonical momentum $\overrightarrow{p}$ conjugate to the position of the particle and show that it is given by $\overrightarrow{p} = m \overrightarrow{v} + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v}$

(c) Show that the Hamiltonian for the charged particle is: $H = \frac{1}{2m} ( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} + q \Phi$.

Solution:

Equation (5.34): $S = -mc \int_{\mathcal{P}} ds + \frac{q}{c} \int_{\mathcal{P}} A_{\mu} (x) d x^{\mu}$

Problem 5.4 (a): Rewrite the action S in terms of the potentials $( \Phi , \overrightarrow{A} )$ and the ordinary velocity $\overrightarrow{v}$. What is the Lagrangian?

Recall that: $A_{\mu} = ( A_{0} , A_{1} , A_{2} , ... , A_{n} ) = ( - \Phi , \overrightarrow{A} )$ and $\frac{d x^{\mu}}{dt} = ( \frac{d x^{0}}{dt} , \frac{d x^{1}}{dt} , \frac{d x^{2}}{dt} , \frac{d x^{3}}{dt} ) = ( c , \overrightarrow{v} )$, where $A_{0} = - \Phi$ and $x^{0} = ct$.

Then we can treat $\int A_{\mu} (x) \frac{d x^{\mu}}{dt} dt$ as an inner product, since the textbook in Section 5.4 says this is the dot product at each time point:

$S = \int \frac{1}{2} m v^{2} dt + \frac{q}{c} \int A_{\mu} (x) \frac{d x^{\mu}}{dt} dt$
$S = \int \frac{1}{2} m v^{2} dt + \frac{q}{c} \int ( - \Phi c + \overrightarrow{A} \cdot \overrightarrow{v} ) dt$

Since $S = \int L dt$ it follows that the Lagrangian of the nonrelativistic charged particle is defined by the action as:

$L = \frac{1}{2} m v^{2} - q \Phi + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v}$ (Q.E.D.)

Problem 5.4 (b): Calculate the canonical momentum $\overrightarrow{p}$ conjugate to the position of the particle and show that it is given by $\overrightarrow{p} = m \overrightarrow{v} + \frac{q}{c} \overrightarrow{A}$

The canonical momentum $\overrightarrow{p} = \frac{\partial}{\partial \dot{q}}$ conjugate to the position of the nonrelativistic charged particle is just the velocity derivative of the Lagrangian:

$\overrightarrow{p} = \frac{\partial L }{\partial v}$
$\overrightarrow{p} = mv + \frac{q}{c} \frac{\partial }{\partial v} \overrightarrow{A} \cdot \overrightarrow{v}$

However, this last term is an inner product depending on velocity (i.e. it is a scalar whose value changes linearly with $\overrightarrow{v}$ ), so it is a constant of the partial differentiation:

$\overrightarrow{p} = mv + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v}$ (Q.E.D.)

Problem 5.4 (c): Show that the Hamiltonian for the charged particle is: $H = \frac{1}{2m} ( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} + q \Phi$

The Hamiltonian in classical mechanics is the Legendre transformation of the Lagrangian:

$H = \sum q^{i}p_{i} - L$

Therefore we can simply write:

$H = \overrightarrow{p} \cdot \overrightarrow{v} - L$
$H = (m \overrightarrow{v} + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v}) \cdot \overrightarrow{v} - ( \frac{1}{2} m v^{2} - q \Phi + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v} )$

Since $\overrightarrow{A} \cdot \overrightarrow{v}$ is a scalar, it is skipped over by the dot product:

$H = (m v^{2} + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v}) - ( \frac{1}{2} m v^{2} - q \Phi + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v} )$
$H = \frac{1}{2} m v^{2} + q \Phi$

Rather than try to derive $H = \frac{1}{2m} ( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} + q \Phi$, we will work backwards. At first glance, this is a problem:

$( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} = ( m\overrightarrow{v} + \frac{q}{c} \overrightarrow{A} \cdot \overrightarrow{v} - \frac{q}{c} \overrightarrow{A} )^{2}$

Notice that $( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2}$ if taken naively is the wrong dimensions. It means subtracting out the $\overrightarrow{A}$ dependence of $\overrightarrow{p}$, the squaring process implicitly involves a dot product of $\overrightarrow{v}$:

$( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} = ( ( m + \frac{q}{c} \overrightarrow{A} - \frac{q}{c} \overrightarrow{A} ) \cdot\overrightarrow{v} )^{2} = ( m^{2} \overrightarrow{v} \cdot \overrightarrow{v} - 0 ) = m^{2} v^{2}$

Therefore it follows that this is an equivalent representation of the Hamiltonian H:

$H = \frac{1}{2m} ( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} + q \Phi = \frac{1}{2} m v^{2} + q \Phi$
$\Rightarrow H = \frac{1}{2m} ( \overrightarrow{p} - \frac{q}{c} \overrightarrow{A} )^{2} + q \Phi$ (Q.E.D.)

Problem 5.5

Problem Statement: Equations of motion for a charged point particle.

Consider the variation of the action (5.34) under a variation $\delta x^{\mu}(x)$ of the particle trajectory. The variation of the first term in the action was obtained in Section 5.3. Vary the second term (written more explicitly in (5.33)) and show that the equation of motion is (5.32). Begin your calculation by explaining why

$\delta A_{\mu} ( x(\tau) ) = \frac{\partial A_{\mu}}{\partial x^{\nu}} ( x( \tau ) ) \delta x^{\nu} (x)$

Solution:

Equation (5.32): $\frac{d p_{\mu}}{d \tau} = \frac{q}{c} F_{\mu \nu} \frac{d x^{\nu}}{d \tau}$

Equation (5.33): $\frac{q}{c} \int_{\mathcal{P}} A_{\mu} ( x ( \tau ) ) \frac{d x^{\mu}}{d \tau}$

Equation (5.34): $S = -mc \int_{\mathcal{P}} ds + \frac{q}{c} \int_{\mathcal{P}} A_{\mu} (x) d x^{\mu}$

The second term in Equation (5.34) is just a more compact way of writing Equation (5.33):

$S = -mc \int_{\mathcal{P}} ds + \frac{q}{c} \int_{\mathcal{P}} A_{\mu} (x) d x^{\mu}$
$S = -mc \int_{\mathcal{P}} ds + \frac{q}{c} \int_{\mathcal{P}} A_{\mu} ( x ( \tau ) ) \frac{d x^{\mu}}{d \tau}$

For the first term, which is Equation (5.5), recall the variation $\delta S$ leads from Equation (5.18) to (5.25):

Equation 5.18: $\delta S = -mc \int \delta (ds)$
Equation 5.25: $\delta S = \int^{t_{f}}_{t_{i}} d \tau \frac{d}{d \tau} ( \delta x^{\mu} p_{\mu} ) - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} ( \tau ) \frac{d p_{\mu} }{d \tau}$

The first term of Equation 5.25 vanishes when evaluated at the end points of the world-line. The second term gives the equation of motion when $\delta S = 0$:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} ( \tau ) \frac{d p_{\mu} }{d \tau}$

Therefore taking the variation of Equation (5.34) yields:

$\delta S = -mc \int_{\mathcal{P}} \delta ( ds ) + \frac{q}{c} \int_{\mathcal{P}} \delta ( A_{\mu} (x) d x^{\mu} )$
$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} ( \tau ) \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ ( \delta A_{\mu} (x) ) \frac{d x^{\mu}}{d \tau} + A_{\mu} (x) ( \delta ( \frac{d x^{\mu}}{d \tau} ) ) ]$

We notice from the problem statement that $\delta A_{\mu} ( x( \tau ) ) = \frac{\partial A_{\mu}}{\partial x^{\nu}} ( x( \tau ) ) \delta x^{\nu} (x)$, which makes sense because it expresses the variation of $A_{\mu}$ with respect to the variation of different coordinates $x^{\nu}$. Then we can substitute this into the above equation for $\delta S$ and suppress all the parameter parentheses for clarity:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\mu}}{\partial x^{\nu}} \delta x^{\nu} ) \frac{d x^{\mu}}{d \tau} + A_{\mu} ( \delta ( \frac{d x^{\mu}}{d \tau} ) ) ]$

Since we want to get the equation of motion, we have to use the standard trick of the derivative product rule to get the variations all in terms of $\delta x$:

Product Rule:
$\frac{d}{d \tau} ( A_{\mu} \delta x^{\mu} ) = \frac{d A_{\mu}}{d \tau} \delta x^{\mu} + A_{\mu} \delta ( \frac{d x^{\mu}}{d \tau} )$
$\Rightarrow A_{\mu} \delta ( \frac{d x^{\mu}}{d \tau} ) = \frac{d}{d \tau} ( A_{\mu} \delta x^{\mu} ) - \frac{d A_{\mu}}{d \tau} \delta x^{\mu}$

Substituting this in (and recalling that expressions of the form $\int^{t_{f}}_{t_{i}} \frac{d}{d \tau} ( A_{\mu} \delta x^{\mu} ) d \tau = 0$ at the end points of $\delta x^{\mu}$ ):

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\mu}}{\partial x^{\nu}} \delta x^{\nu} ) \frac{d x^{\mu}}{d \tau} + \frac{d}{d \tau} ( A_{\mu} \delta x^{\mu} ) - \frac{d A_{\mu}}{d \tau} \delta x^{\mu} ]$
$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\mu}}{\partial x^{\nu}} \delta x^{\nu} ) \frac{d x^{\mu}}{d \tau} - \frac{d A_{\mu}}{d \tau} \delta x^{\mu} ]$

This is in terms of variations $\delta x^{\nu}$ and $\delta x^{\mu}$. To express it in terms of $A_{\nu}$, we need another $\frac{\partial A_{\mu}}{\partial x^{\nu}}$ term:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\mu}}{\partial x^{\nu}} \delta x^{\nu} ) \frac{d x^{\mu}}{d \tau} - \frac{\partial A_{\mu}}{\partial x^{\nu}} \frac{d A_{\nu}}{d \tau} \delta x^{\mu} ]$

To get it in terms of only $\delta x^{\mu}$, we do a coordinate change on the left term: $(\frac{\partial A_{\mu}}{\partial x^{\nu}} \delta x^{\nu} ) \frac{d x^{\mu}}{d \tau} = \frac{\partial A_{\nu}}{\partial x^{\mu}} \frac{\partial x^{\mu}}{\partial x^{\nu}} \frac{d x^{\mu}}{d \tau} \delta x^{\nu}$

$\Rightarrow \delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\nu}}{\partial x^{\mu}} \delta x^{\mu} ) \frac{d x^{\nu}}{d \tau} - \frac{\partial A_{\mu}}{\partial x^{\nu}} \frac{d A_{\nu}}{d \tau} \delta x^{\mu} ]$

With the $\frac{d A_{\nu}}{d \tau}$ term, we just have A variation along the $x^{\nu}$ coordinates:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\nu}}{\partial x^{\mu}} \delta x^{\mu} ) \frac{d x^{\nu}}{d \tau} - \frac{\partial A_{\mu}}{\partial x^{\nu}} \frac{d x^{\nu}}{d \tau} \delta x^{\mu} ]$
$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ (\frac{\partial A_{\nu}}{\partial x^{\mu}} - \frac{\partial A_{\mu}}{\partial x^{\nu}} ) \frac{d x^{\nu}}{d \tau} \delta x^{\mu} ]$

Therefore since $F_{\mu \nu} = \frac{\partial A_{\nu}}{\partial x^{\mu}} - \frac{\partial A_{\mu}}{\partial x^{\nu}}$:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \frac{d x^{\nu}}{d \tau} \delta x^{\mu} ]$

$\Rightarrow \delta S = \int^{t_{f}}_{t_{i}} d \tau ( - \frac{d p_{\mu} }{d \tau} + \frac{q}{c} F_{\mu \nu} \frac{d x^{\nu}}{d \tau} ) \delta x^{\mu}$

Therefore with $\delta S = 0$ the Euler-Lagrange equation of motion is given by $\frac{d p_{\mu} }{d \tau} = \frac{q}{c} F_{\mu \nu} \frac{d x^{\nu}}{d \tau}$, which is Equation (5.32). (Q.E.D.)

Problem 5.6

Problem Statement: Electromagnetic field dynamics with a charged particle.

The action for the dynamics of both a charged point particle and the electromagnetic field is given by

$S' = -mc \int_{\mathcal{P}} ds + \frac{q}{c} \int_{\mathcal{P}} A_{\mu} (x) d x^{\mu} - \frac{1}{4c} \int d^{D} x F_{\mu \nu} F^{\mu \nu}$

Here $d^{D} = dx^{0} dx^{1} ... dx^{d}$. Note that the action S' is a hybrid; the last term is an integral over spacetime, and the first two terms are integrals over the particle world-line. While included for completeness, the first term will play no role here. Obtain the equation of motion for the electromagnetic field in the presence of the charged particle by calculating the variation of S' under a variation $\delta A_{\mu}$ of the gauge potential. The answer should be equation (3.34), where the current is the one calculated in Problem 5.3. [Hint: to vary $A_{\mu} (x)$ in the world-line action it is useful to rewrite this term as a full spacetime integral with the help of delta functions.]

Solution:

(Warning: This derivation may be essentially correct, but the last step eliminating the $\frac{1}{4c}$ term might be an improper handwave.)

Equation (3.34): $\frac{\partial F^{\mu \nu}}{\partial x^{\nu}} = \frac{1}{c} j^{\mu}$

Problem 5.3: $j^{\mu} (t, \overrightarrow{x}) = qc \int d \tau \delta^{D} ( x - x ( \tau ) ) \frac{d x^{\mu} ( \tau )}{d \tau}$; $\delta^{D} (x) \equiv \delta (x^{0}) \delta (x^{1}) ... \delta (x^{d})$

We know from Problem 5.5 that the variation of the charged particle world-line by $\delta A_{\mu}$ yields:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \frac{d x^{\nu}}{d \tau} \delta x^{\mu} ]$

The first term is irrelevant. The second term is the charged particle along its world-line, expressed in terms of $F_{\mu \nu}$ instead of the gauge potential. If we want to convert this into an integral over spacetime in D dimensions, we can use delta functions like in Problem 5.3, which isolate the particle along its position at each time point:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \frac{q}{c} \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \delta^{D} ( x - x ( \tau ) ) \frac{d x^{\nu}}{d \tau} \delta x^{\mu} ]$

Recalling that the c term is swallowed by $x^{0}$, we see that this second integral is really:

$\delta S = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \frac{1}{c} j^{\mu} \delta x^{\mu} ]$

Then we can replace this into the expression for S' and similarly eliminate c, with the factor of 4 coming from the symmetrical contributions of the $\mu , \nu$ variations from $F_{\mu \nu} F^{\mu \nu}$: (*)

$\delta S' = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \frac{1}{c} j^{\mu} \delta x^{\mu} ] - \frac{1}{4c} \int d^{D} x F_{\mu \nu} F^{\mu \nu}$
$\delta S' = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} \frac{1}{c} j^{\mu} \delta x^{\mu} ] - \int^{t_{f}}_{t_{i}} d \tau F_{\mu \nu} F^{\mu \nu} \frac{d x^{\nu}}{d \tau} \delta x^{\mu}$
$\delta S' = - \int^{t_{f}}_{t_{i}} d \tau \delta x^{\mu} \frac{d p_{\mu} }{d \tau} + \int^{t_{f}}_{t_{i}} d \tau [ F_{\mu \nu} ( \frac{1}{c} j^{\mu} - \frac{\partial ^{F^{\mu \nu} }}{\partial x^{\nu}} ) ] \delta x^{\mu}$

The problem is not concerned with the first term. The equation of motion for the electromagnetic field in the presence of the charged particle (when $\delta S'_{field} = 0$ ) is therefore Equation 3.34:

$\Rightarrow \frac{1}{c} j^{\mu} = \frac{\partial ^{F^{\mu \nu} }}{\partial x^{\nu}}$ (Q.E.D.)

Problem 5.7

Problem Statement: Point particle action in curved space.

In Section 3.6 we considered the invariant interval $ds^{2} = - g_{\mu \nu} (x) d x^{\mu} d x^{\nu}$ in a curved space with metric $g_{\mu \nu} (x)$. The motion of a point particle of mass m on a curved space is studied using the action

$S = -mc \int ds$

Show that the equation of motion obtained by variation of the world-line is

$\frac{d}{ds} [ g_{\mu \rho} \frac{d x^{\mu}}{ds} ] = \frac{1}{2} \frac{\partial g_{\mu \nu} }{\partial x^{\rho}} \frac{d x^{\mu}}{ds} \frac{d x^{\nu}}{ds}$

This is called the geodesic equation. When the metric is constant we recover the equation of motion of a free point particle.

Solution:

(Warning: This derivation may be essentially correct overall, but the index manipulations may have been done too sloppily in reaching the equation of motion.)

Following the argument in Section 5.3 for Minkowski spacetime, we vary $\delta S$ in terms of $\delta (ds)$:

$\delta S = -mc \int \delta (ds)$

It is more straight forward to instead vary $ds^{2}$, and use proper time s itself instead of a parameter $\tau$:

$\delta (ds^{2}) = \delta ( - g_{\mu \nu} (x) d x^{\mu} d x^{\nu} )$
$2 ds \delta (ds) = - \delta ( g_{\mu \nu} (x) \frac{d x^{\mu}}{d s} ) \frac{d x^{\nu}}{d s} ds^{2}$
$2 ds \delta (ds) = - ( \delta ( g_{\mu \nu} ) \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} + g_{\mu \nu} \delta (\frac{d x^{\mu}}{d s} ) \frac{d x^{\nu}}{d s} ) ds^{2}$
$2 ds \delta (ds) = - ( \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} \delta x^{\rho} + g_{\mu \nu} (\frac{d \delta x^{\mu}}{d s} ) \frac{d x^{\nu}}{d s} ) ds^{2}$

The metric $g_{\mu \nu}$ is dynamical, so it cannot be skipped over when doing the variation. We replace the right term with a product rule:

Product Rule:
$\frac{d}{d s} ( \delta x^{\mu} \frac{d x^{\nu}}{d s} ) = \frac{d \delta x^{\mu} }{d s} \frac{ d x^{\nu} }{d s} + \delta x^{\mu} \frac{d}{d s} \frac{d x^{\nu}}{d s}$
$\frac{d}{d s} ( \delta x^{\mu} \frac{d x^{\nu}}{d s} ) - \delta x^{\mu} \frac{d}{d s} \frac{d x^{\nu}}{d s} = \frac{d \delta d x^{\mu} }{d s} \frac{ d x^{\nu} }{d s}$

Inserting into the equation above gives the following, but the $\frac{d}{d s} ( \delta x^{\mu} \frac{d x^{\nu}}{d s} )$ vanishes at the end points of the world-line:

$2 ds \delta (ds) = - ( \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} \delta x^{\rho} + g_{\mu \nu}[ \frac{d}{d s} ( \delta x^{\mu} \frac{d x^{\nu}}{d s} ) - \delta x^{\mu} \frac{d}{d s} \frac{d x^{\nu}}{d s} ] ) ds^{2}$
$2 ds \delta (ds) = - ( \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} \delta x^{\rho} + g_{\mu \nu} \delta x^{\mu} \frac{d}{d s} \frac{d x^{\nu}}{d s} ) ds^{2}$

Replace these coordinate variations with a differential amount of the spacetime interval instead:

$2 ds \delta (ds) = - ( \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} + g_{\mu \nu} \frac{d^{2} x^{\nu}}{d s^{2}} ) ds^{2} \delta (ds)$

The second term is just the constancy of the momentum along the world-line, like in Equation 5.29 it vanishes, leaving the spacetime curvature term (the problem is specifically asking about the effect of the world-line variation):

$2 ds \delta (ds) = - ( \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} ) ds^{2} \delta (ds)$
$\Rightarrow ds \delta (ds) = (- \frac{1}{2} \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s} ) ds^{2} \delta (ds)$

This is consistent with the problem statement, which implies the form:

$\frac{d}{ds} [ g_{\mu \rho} \frac{d x^{\mu}}{ds} ] = ( \frac{d}{ds} g_{\mu \rho} ) \frac{d x^{\mu}}{ds} + g_{\mu \rho} \frac{d}{ds} ( \frac{d x^{\mu}}{ds} )$
$\frac{d}{ds} [ g_{\mu \rho} \frac{d x^{\mu}}{ds} ] = ( \frac{d}{ds} g_{\mu \rho} ) \frac{d x^{\mu}}{ds} + g_{\mu \rho} ( \frac{d^{2} x^{\mu}}{ds^{2}} )$

Thus, when we consider the condition $\frac{d^{2} x^{\mu}}{ds^{2}} = 0$, leaving only the spacetime curvature term:

$\Rightarrow \frac{d}{ds} [ g_{\mu \rho} \frac{d x^{\mu}}{ds} ] = \frac{1}{2} \frac{ \partial g_{\mu \nu}}{\partial x^{\rho}} \frac{d x^{\mu}}{d s} \frac{d x^{\nu}}{d s}$

Therefore plugging that result into $\delta S = 0$ simply gives this as the equation of motion. It is the geodesic because it is merely following the spacetime curvature, while if the metric is constant, this term vanishes and the action reduces to the free particle term for the equation of motion where $\frac{d^{2} x^{\mu}}{ds^{2}} = 0$. (Q.E.D.)

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Last updated: 12/25/2020