slides_lecture09_main.tex

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\title[PHYS 201 / Lecture \thislecture]
{
  PHYS 201 / Lecture \thislecture\\
  {\it Electromagnetic Waves}\\
}

\input{slides_author.tex}

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  \titlepage
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\input{slides_lecture08_summary.tex}

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\begin{frame}{Plan for Lecture \thislecture}

\begin{itemize}
   \item Maxwell's equations in vacuum in absence of sources
   \item Reminder on waves
   \item The electric and magnetic fields satisfy the wave equation
   \item Electromagnetic waves and light
   \item Energy carried by electromagnetic waves
   \item The Poynting theorem
\end{itemize}

\end{frame}

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\begin{frame}{Maxwell's equations in vacuum in absence of sources}

\begin{columns}
  \begin{column}{0.40\textwidth}
   {\small
       Let me start from the known Maxwell's equations in vacuum,
       and consider the case where there are:
       \begin{itemize}
       {\small
             \item {\bf no charges} ($\rho = 0$)
             \item and {\bf no currents} ($\vec{j} = \vec{0}$).
       }
       \end{itemize}
  }
  \end{column}
  \begin{column}{0.60\textwidth}
     \begin{equation*}
       \vec{\nabla} \cdot \vec{E} = \cancelto{0}{\frac{\rho}{\epsilon_0}} \Rightarrow
       {\color{magenta} \vec{\nabla} \cdot \vec{E} = 0 }
     \end{equation*}
     \begin{equation*}
        {\color{magenta} \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} }
     \end{equation*}
     \begin{equation*}
        {\color{magenta} \vec{\nabla} \cdot \vec{B} = 0 }
     \end{equation*}
     \begin{equation*}
       \vec{\nabla} \times \vec{B} = \cancelto{0}{\mu_0 j} + \epsilon_0 \mu_0 \frac{\partial \vec{E}}{\partial t} \Rightarrow
        {\color{magenta} \vec{\nabla} \times \vec{B} = \epsilon_0 \mu_0 \frac{\partial \vec{E}}{\partial t} }
     \end{equation*}
  \end{column}
\end{columns}

\vspace{0.5cm}

{\bf A change in one field feeds the other} (even in absence of sources):\\
Faraday's law tells us that a changing magnetic field generates an electric field, and
Maxwell's correction to Ampere's law tells us that a changing electric field generates a magnetic field.\\

\vspace{0.2cm}

{\bf Can this interplay generate waves?}

\end{frame}



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{
\reminderslide

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\begin{frame}{Reminder: Waves}

\begin{center}
 A wave is a {\bf disturbance that travels through space.}
\end{center}

\begin{columns}
  \begin{column}{0.70\textwidth}
    \includegraphics[width=0.98\textwidth]{./images/photos/mech_wave_water_1.jpg}\\
  \end{column}
  \begin{column}{0.30\textwidth}
    \begin{center}
       As this disturbance travels it transfers energy.\\
       \vspace{0.4cm}
       If there is no absorption or dispersion, that disturbance
       propagates with a constant velocity and a fixed `shape'.
    \end{center}
  \end{column}
\end{columns}

\end{frame}

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\begin{frame}{Reminder: Types of waves}

There are {\bf three main types of waves}:

\begin{itemize}

\item {\bf Mechanical waves}\\
    \begin{itemize}
        \item These are the waves you are most familiar with
           \begin{itemize}
                \item For example, sound waves, sea waves, seismic waves.
           \end{itemize}
        \item Mechanical waves {\bf need a medium} to propagate into. They propagate by deforming that medium.
           \begin{itemize}
                 \item When I speak, I disturb the air molecules in front of my mouth.
                 \item They collide with neighbouring molecules and bounce back.
                 \item This disturbance propagates, and it reaches and vibrates your eardrums.
           \end{itemize}
%         \item For mechanical waves, no medium = no propagation.
    \end{itemize}

\item {\bf Electromagnetic waves} {\color{red} $\leftarrow$ The subject of this lecture}\\
    \begin{itemize}
        \item Less familiar, probably, although you interact with them constantly.
        \item They {\bf do not require a medium to propagate.}
    \end{itemize}

\item {\bf Matter waves}  {\color{red} $\leftarrow$ Will study in future modules}\\
    \begin{itemize}
          \item You are probably very unfamiliar with these, but they take centre stage
                    in quantum mechanics ({\em wave-particle duality}).
          \item Waves associated with fundamental particles (such as the electron),
                    composite particles (such as the proton), atoms or even molecules.
    \end{itemize}

\end{itemize}

\end{frame}

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\begin{frame}{Reminder: The wave equation}

A {\bf wave equation} describes how the disturbance propagates in time.\\

\vspace{0.3cm}

Let $\phi(\vec{r}, t)$ be a function that describes that disturbance as a function of position in space and time.
Then it satisfies the following equation:
\begin{equation*}
   \vec{\nabla}^{2} \phi(\vec{r}, t) = \frac{1}{u^2} \frac{\partial^{2} \phi(\vec{r}, t)} {\partial t^{2}}
\end{equation*}
where {\bf u is the wave velocity}.\\

%\vspace{0.3cm}
%This equation can be derived in many different physical situations and I am sure that you have seen it already in other modules.\\
\vspace{0.3cm}

A mathematician would say that the wave equation is a
{\bf 2$^{nd}$ order linear partial differential equation (p.d.e)}.
\begin{itemize}
   \item This particular type of p.d.e. is called {\em hyperbolic}.
\end{itemize}

\end{frame}

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\begin{frame}{Reminder: Waves}

Look carefully at the wave equation:
\begin{equation*}
   \vec{\nabla}^{2} \phi(\vec{r}, t) = \frac{1}{u^2} \frac{\partial^{2} \phi(\vec{r}, t)} {\partial t^{2}}
\end{equation*}

Here, we encounter $\vec{\nabla}^{2}$ again.
This is the so-called {\em Laplace operator} that we have already seen at a previous lecture (Poisson equation).\\

\vspace{0.2cm}

The Laplacian $\vec{\nabla}^{2} \phi$ is defined as
{\bf the divergence of the gradient} of the scalar function $\phi$:
\begin{equation*}
   \vec{\nabla}^{2} \phi(\vec{r}, t) =
   \vec{\nabla} \cdot \Big( \vec{\nabla} \phi(\vec{r}, t) \Big) =
   \frac{\partial^{2} \phi(\vec{r}, t)} {\partial x^{2}} +
   \frac{\partial^{2} \phi(\vec{r}, t)} {\partial y^{2}} +
   \frac{\partial^{2} \phi(\vec{r}, t)} {\partial z^{2}}
\end{equation*}

Recall that it represents a quantity which is important in several physical processes:
The Laplacian $\vec{\nabla}^{2} \phi(\vec{r})$ of a scalar function $\phi$ at a point $\vec{r}$
tells you how much $\phi(\vec{r})$ differs from its average over a small volume around $\vec{r}$.

\end{frame}


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\begin{frame}{Reminder: Solutions of the wave equation}

In one dimension, say x, the wave equation becomes:
\begin{equation*}
    \frac{\partial^{2} \phi(x, t)} {\partial x^{2}} =
    \frac{1}{u^2} \frac{\partial^{2} \phi(x, t)} {\partial t^{2}}
\end{equation*}
where u is the wave velocity.\\

\vspace{0.2cm}

{\bf What are the solutions of the wave equation?}

\vspace{0.2cm}

\underline{\bf Any} (any whatsoever) function $\phi$ of x and t,
{\bf where x and t appear only in the combination of x-ut},
is a solution.
\begin{equation*}
  \phi(x,t) = \psi(z) = \psi(x-ut)
\end{equation*}

\vspace{0.1cm}

For example, the following function is a solution of the wave equation:
\begin{equation*}
  \phi = sin^3\Big( x-ut \Big) - \frac{e^{(x-ut)}}{cos^{\frac{7}{2}}\Big( x-ut\Big)} + \Big( x-ut\Big)^{1.298}
\end{equation*}

\vspace{0.1cm}

We can easily see that using the chain rule.

\end{frame}

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\begin{frame}{Reminder: Solutions of the wave equation}

Any function of x and t, where x and t appear only in the combination of x-ut,
is a solution of the wave equation.\\

\vspace{0.1cm}

Let $\psi(z)$ be a function of x-ut (z = x-ut).
The partial derivative of  $\psi(z)$ with respect to time is:

\begin{equation*}
  \frac{\partial \psi(z)}{\partial t} =
  \frac{\partial z}{\partial t} \cdot \frac{\partial \psi(z)}{\partial z} =
  \bigg\{ \frac{\partial}{\partial t} \Big( x - ut \Big) \bigg\} \cdot \frac{\partial \psi(z)}{\partial z} =
  -u \frac{\partial \psi(z)}{\partial z}
\end{equation*}

Differentiating once again, we have:
\begin{equation*}
  \frac{\partial^{2} \psi(z)}{\partial t^{2}} =
  \frac{\partial}{\partial t} \Big( \frac{\partial \psi(z)}{\partial t} \Big) =
  \frac{\partial}{\partial t} \Big(  -u \frac{\partial \psi(z)}{\partial z} \Big) =
  \frac{\partial z}{\partial t} \frac{\partial}{\partial z} \Big(  -u \frac{\partial \psi(z)}{\partial z} \Big) =
\end{equation*}

\begin{equation*}
 = \bigg\{ \frac{\partial}{\partial t} \Big( x - ut \Big) \bigg\}  \frac{\partial}{\partial z} \Big(  -u \frac{\partial \psi(z)}{\partial z} \Big) =
  -u \frac{\partial}{\partial z} \Big(  -u \frac{\partial \psi(z)}{\partial z} \Big) =
   u^2 \frac{\partial^{2} \psi(z)}{\partial z^{2}} \Rightarrow
\end{equation*}

{\color{magenta}
\begin{equation*}
  \frac{1}{u^2} \frac{\partial^{2} \psi(z)}{\partial t^{2}} = \frac{\partial^{2} \psi(z)}{\partial z^{2}} (*)
\end{equation*}
}

\end{frame}

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\begin{frame}{Reminder: Solutions of the wave equation}

Similarly, the partial derivative of  $\psi(z)$ with respect to x is:

\begin{equation*}
  \frac{\partial \psi(z)}{\partial x} =
  \frac{\partial z}{\partial x} \cdot \frac{\partial \psi(z)}{\partial z} =
  \bigg\{ \frac{\partial}{\partial x} \Big( x - ut \Big) \bigg\} \cdot \frac{\partial \psi(z)}{\partial z} =
  \frac{\partial \psi(z)}{\partial z}
\end{equation*}

Differentiating once again, we have:
\begin{equation*}
  \frac{\partial^{2} \psi(z)}{\partial x^{2}} =
  \frac{\partial}{\partial x} \Big( \frac{\partial \psi(z)}{\partial x} \Big) =
  \frac{\partial}{\partial x} \Big( \frac{\partial \psi(z)}{\partial z} \Big) =
  \frac{\partial z}{\partial x} \frac{\partial}{\partial z} \Big( \frac{\partial \psi(z)}{\partial z} \Big) =
\end{equation*}
\begin{equation*}
  =  \bigg\{ \frac{\partial}{\partial x} \Big( x - ut \Big) \bigg\} \frac{\partial}{\partial z} \Big( \frac{\partial \psi(z)}{\partial z} \Big) =
   \frac{\partial^{2} \psi(z)}{\partial z^{2}} \Rightarrow
   {\color{magenta} \frac{\partial^{2} \psi(z)}{\partial x^{2}} =  \frac{\partial^{2} \psi(z)}{\partial z^{2}} } (**)
\end{equation*}

The right-hand sides of Eqs. (*) and (**) are equal, and so are the left-hand sides:
\begin{equation*}
   {\color{blue} \frac{\partial^{2} \psi(z)}{\partial x^{2}} =   \frac{1}{u^2} \frac{\partial^{2} \psi(z)}{\partial t^{2}} }
\end{equation*}

Therefore,  $\psi(z) =  \psi(x-ut)$ is a solution of the wave equation.

\end{frame}


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\begin{frame}{Reminder: Solutions of the wave equation}

So, any function $\phi$ of x and t,
{\bf where x and t appear only in the combination of x-ut},
is a solution.
\begin{equation*}
  \phi(x,t) = \psi(x-ut)
\end{equation*}

\vspace{0.1cm}

Can you think of {\bf another family of solutions?}\\

\vspace{0.1cm}

Any function $\phi$ of x and t,
{\bf where x and t appear only in the combination of x+ut},
is also a solution.
\begin{equation*}
  \phi(x,t) = \chi(x+ut)
\end{equation*}

\vspace{0.1cm}

The wave equation is {\bf linear}: A superimposition of solutions is a solution.\\

\vspace{0.1cm}

Therefore, most general solution to the wave equation can be written as:\\

\begin{equation*}
  \phi(x,t) = \psi(x-ut) + \chi(x+ut)
\end{equation*}

\end{frame}

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\begin{frame}{Reminder: Direction of wave propagation}

% See the following two snapshots of a wave taken by a small time interval ${\Delta}t$ apart.
% Imagine that at t=0, the disturbance $\phi(x,0)$ is - Draw?
% At a later fixed time t, the disturbance is $\phi(x,t)$
% But the disturbance is still the same, just displaced in x (because the wave has moved)

What does this particular combination of variables (x-ut, x+ut) tell us about
the direction of wave propagation?
\begin{itemize}
       \item $\psi(x-ut)$ describes a wave moving in the positive direction of x.
       \item $\chi(x+ut)$ describes a wave moving in the negative direction of x.
\end{itemize}

\begin{columns}
  \begin{column}{0.60\textwidth}
     \begin{center}
         \includegraphics[width=0.98\textwidth]{./images/schematics/wave_direction.png}\\
     \end{center}
  \end{column}
  \begin{column}{0.40\textwidth}
  {\scriptsize
     Convince yourselves that the guy on the left should ride a
     $\psi(x-ut)$ wave, not a $\chi(x+ut)$ one.
  }
  \end{column}
\end{columns}

\end{frame}

}% ending reminder


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\begin{frame}{The electric field satisfies a wave equation}

Now that we have refreshed our memory on waves, let me return to my
original question. Maxwell's equation in vacuum, in the absence of
charges ($\rho = 0$) and currents ($\vec{j} = \vec{0}$) are shown below:

     \begin{equation*}
       \vec{\nabla} \cdot \vec{E} = 0
     \end{equation*}
     \begin{equation*}
       \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}
     \end{equation*}
     \begin{equation*}
       \vec{\nabla} \cdot \vec{B} = 0
      \end{equation*}
      \begin{equation*}
        \vec{\nabla} \times \vec{B} = \epsilon_0 \mu_0 \frac{\partial \vec{E}}{\partial t}
      \end{equation*}

\vspace{0.4cm}

{\bf A change in one field feeds the other} (even in absence of sources).\\

\vspace{0.2cm}

{\bf Can this interplay generate waves?}

\end{frame}

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\begin{frame}{The electric field satisfies a wave equation}

We will start from Faraday's law in vacuum:
\begin{equation*}
  \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}
\end{equation*}

We will take the curl of both sides and use identity shown below.
Then, we will use Gauss's and Ampere's law in vacuum and in absence of sources:
\begin{equation*}
  \vec{\nabla} \cdot \vec{E} =
       \cancelto{0}{\frac{\rho}{\epsilon_0}} = 0
  \;\;\; and \;\;\;
  \vec{\nabla} \times \vec{B} =
        \cancelto{0}{\mu_0 \vec{j}} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} =
        \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
\end{equation*}
to substitute the resulting $\vec{\nabla} \cdot \vec{E}$  and $\vec{\nabla} \times \vec{B}$.

\begin{blockminirem}{\small Mini reminder from calculus}
{\small
For any vector field $\vec{F}$:
$\displaystyle \vec{\nabla} \times \Big( \vec{\nabla} \times \vec{F} \Big) =
  \vec{\nabla} \Big( \vec{\nabla} \cdot \vec{F} \Big) - \vec{\nabla}^{2} \vec{F}$\\
\vspace{0.1cm}
The {\bf curl of the curl} of a vector field is
the {\bf gradient of the divergence} of the field minus
the {\bf divergence of the gradient} (Laplacian) of the field.\\
}
\end{blockminirem}

\end{frame}

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\begin{frame}{The electric field satisfies a wave equation}

\begin{equation*}
  \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \Rightarrow
  {\color{red} \vec{\nabla} \times} \Big( \vec{\nabla} \times \vec{E} \Big) =
  {\color{red} \vec{\nabla} \times} \Big( -\frac{\partial \vec{B}}{\partial t} \Big) \Rightarrow
\end{equation*}

\begin{equation*}
  \vec{\nabla} \times (\vec{\nabla} \times \vec{E}) =
  - \vec{\nabla} \times \frac{\partial \vec{B}}{\partial t}
  \xRightarrow{ \vec{\nabla} \times \Big( \vec{\nabla} \times \vec{F} \Big) =
                \vec{\nabla} \Big( \vec{\nabla} \cdot \vec{F} \Big) - \vec{\nabla}^{2} \vec{F} }
\end{equation*}

\begin{equation*}
  \vec{\nabla} \Big( \vec{\nabla} \cdot \vec{E} \Big) - \vec{\nabla}^{2} \vec{E} =
  - \frac{\partial}{\partial t} (\vec{\nabla} \times \vec{B})
  \xRightarrow{
      \vec{\nabla} \cdot \vec{E} = \cancelto{0}{\frac{\rho}{\epsilon_0}} = 0
       \;\; and \;\;
      \vec{\nabla} \times \vec{B} =
           \cancelto{0}{\mu_0 \vec{j}} +\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} =
           \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
  }
\end{equation*}

\begin{equation*}
  - \vec{\nabla}^{2} \vec{E} =
  - \frac{\partial}{\partial t} \Big( \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} \Big)
  \Rightarrow
  {\color{magenta}
     \vec{\nabla}^{2} \vec{E} =
     \mu_0 \epsilon_0 \frac{\partial^{2} \vec{E}}{\partial t^{2}}
  }
\end{equation*}

\end{frame}


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\begin{frame}{The magnetic field satisfies a wave equation}

The calculation for $\vec{B}$ proceeds along similar lines.\\
\vspace{0.2cm}

We will start from Ampere's law in vacuum and in absence of sources:
\begin{equation*}
  \vec{\nabla} \times \vec{B} =
        \cancelto{0}{\mu_0 \vec{j}} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} =
        \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
\end{equation*}

We will take the curl of both sides and use identity used before.
Then, we will use the fact that there are no magnetic monopoles and Faraday's law:
Gauss's and Ampere's law in vacuum and in absence of sources:
\begin{equation*}
  \vec{\nabla} \cdot \vec{B} = 0
  \;\;\; and \;\;\;
  \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}
\end{equation*}
to substitute the resulting $\vec{\nabla} \cdot \vec{B}$  and $\vec{\nabla} \times \vec{E}$.


\end{frame}


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\begin{frame}{The magnetic field satisfies a wave equation}

\begin{equation*}
  \vec{\nabla} \times \vec{B} =
  \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
  \Rightarrow
  {\color{red} \vec{\nabla} \times} \Big( \vec{\nabla} \times \vec{B} \Big) =
  {\color{red} \vec{\nabla} \times} \Big( \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} \Big)
  \Rightarrow
\end{equation*}

\begin{equation*}
  \vec{\nabla} \times \Big( \vec{\nabla} \times \vec{B} \Big) =
  \mu_0 \epsilon_0  \vec{\nabla} \times \frac{\partial \vec{E}}{\partial t}
  \xRightarrow{ \vec{\nabla} \times \Big( \vec{\nabla} \times \vec{F} \Big) =
                \vec{\nabla} \Big( \vec{\nabla} \cdot \vec{F} \Big) - \vec{\nabla}^{2} \vec{F} }
\end{equation*}

\begin{equation*}
  \vec{\nabla} \Big( \vec{\nabla} \cdot \vec{B} \Big) - \vec{\nabla}^{2} \vec{B} =
  \mu_0 \epsilon_0  \frac{\partial}{\partial t} \Big( \vec{\nabla} \times \vec{E} \Big)
  \xRightarrow{ \vec{\nabla} \cdot \vec{B} = 0 \;\; and \;\;
                \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} }
\end{equation*}

\begin{equation*}
  - \vec{\nabla}^{2} \vec{B} =
  \mu_0 \epsilon_0  \frac{\partial}{\partial t} \Big(  - \frac{\partial \vec{B}}{\partial t} \Big)
  \Rightarrow
  {\color{magenta}
    \vec{\nabla}^{2} \vec{B} = \mu_0 \epsilon_0 \frac{\partial^{2} \vec{B}}{\partial t^{2}}
  }
\end{equation*}

\end{frame}

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\begin{frame}{A most peculiar wave!}

\begin{itemize}

   \item The electric and magnetic fields satisfy a wave equation.

   \vspace{0.2cm}

   \item Electric and magnetic field ``disturbances'' can propagate in space
             in absence of charges or currents.

   \vspace{0.2cm}

   \item That disturbance {\bf can not die off}!
             \begin{itemize}
             {\small
               \item A time-varying electric field generates a time-varying magnetic field.
               \item A time-varying magnetic field generates a time-varying electric  field.
             }
             \end{itemize}
             {\bf The change of one field feeds the other!}

   \vspace{0.2cm}

   \item An electric wave can't exist without a magnetic one (and vice versa)\\
             They are part of the same phenomenon: {\bf electromagnetic waves}.

   \vspace{0.2cm}

   \item Unlike mechanical waves, electromagnetic waves {\bf don't need a medium to propagate into}!

\end{itemize}

\end{frame}


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\begin{frame}{The speed of electromagnetic waves in vacuum}

$\vec{E}$ and $\vec{B}$ satisfy the following wave equations:

\begin{equation*}
  \vec{\nabla}^{2} \vec{E} = \mu_0 \epsilon_0 \frac{\partial^{2} \vec{E}}{\partial t^{2}}
\end{equation*}

\begin{equation*}
  \vec{\nabla}^{2} \vec{B} = \mu_0 \epsilon_0 \frac{\partial^{2} \vec{B}}{\partial t^{2}}
\end{equation*}

Recall that earlier we wrote the wave equation as:
\begin{equation*}
   \vec{\nabla}^{2} \phi(\vec{r}, t) = \frac{1}{u^2} \frac{\partial^{2} \phi(\vec{r}, t)} {\partial t^{2}}
\end{equation*}
where u is the wave velocity.\\

\vspace{0.2cm}

Therefore the electric and magnetic waves propagate in vacuum with the
same speed which, from above, can be identified with:
\begin{equation*}
  u = \frac{1}{\sqrt{\mu_0 \epsilon_0}}
\end{equation*}

\end{frame}

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\begin{frame}{The speed of electromagnetic waves in vacuum}

Can you confirm that $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ has velocity units?\\

\vspace{0.4cm}

In SI:
\begin{itemize}
   \item the permeability $\mu_0$ is given in $\frac{N}{A^{2}}$, while
   \item the permitivity $\epsilon_0$ is given in $\frac{C^{2}}{N \cdot m^{2}}$.
\end{itemize}

\vspace{0.2cm}

The product $\mu_0 \epsilon_0$ has units of:
\begin{equation*}
 \frac{\cancel{N}}{A^{2}} \cdot \frac{C^{2}}{\cancel{N} \cdot m^{2}} =
 \frac{1}{A^{2}} \cdot \frac{C^{2}}{m^{2}} =
 \frac{1}{\Big(\frac{C}{s}\Big)^{2}} \cdot \frac{C^{2}}{m^{2}} =
 \frac{s^{2}}{\cancel{C^{2}}} \cdot \frac{\cancel{C^{2}}}{m^{2}} =
 \frac{s^{2}}{m^{2}}
\end{equation*}

\vspace{0.2cm}

Indeed, $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ has SI units of $\frac{m}{s}$.


\end{frame}

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\begin{frame}{The speed of electromagnetic waves in vacuum}

What is the numerical value of $\displaystyle \frac{1}{\sqrt{\mu_0 \epsilon_0}}$?

\begin{equation*}
    u =
      \frac{1}{\sqrt{\mu_0 \epsilon_0}} =
       \frac{1}{\sqrt{ \Big( 8.854 \cdot 10^{-12} \; \frac{N}{A} \Big) \cdot
                              \Big( 4\pi \cdot 10^{-7} \; \frac{C}{N\cdot m} \Big)
                            }
                   } =
       {\bf 299,792,458 \; m/s}
\end{equation*}

\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.50\textwidth}
    \begin{center}
       \includegraphics[width=0.65\textwidth]{./images/misc/space_sign_speed_limit.png}\\
    \end{center}
  \end{column}
  \begin{column}{0.50\textwidth}
    \begin{center}
       Do you recognize this number?
    \end{center}
  \end{column}
\end{columns}

\end{frame}


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\begin{frame}{Measurements of the speed of light before Maxwell's time}

At the time that Maxwell was developing his theory, the speed of light was already known.
In fact, it was known from the late 17th century.\\

\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.33\textwidth}
    \includegraphics[width=0.90\textwidth]{./images/schematics/illustration_from_1676_article_on_the_speed_of_light.jpg}\\
  \end{column}
  \begin{column}{0.67\textwidth}
      In 1670's, {\bf Olaf Roemer} measured the speed of light by observing the {\bf period of the moons of Jupiter.}\\

      \begin{itemize}
      {\small
        \item The apparent period at Earth is different than true period due to the velocity of the Earth.
        \item The variation of the apparent period was
                  \begin{equation*}
                        {\Delta}T = T \cdot \frac{2u}{c}
                  \end{equation*}
                  where T the true period, u the velocity of the earth and c the speed of light.
        \item This allowed the speed of light to be determined to $\sim$~30\%.\\
      }
      \end{itemize}
  \end{column}
\end{columns}

\end{frame}

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\begin{frame}{Measurements of the speed of light before Maxwell's time}

And in 1848-49, a few years before Maxwell published his famous treatise,
{\bf Hippolyte Fizeau} measured the speed of light to $\sim$5\%.\\

\vspace{0.2cm}

Light passes between two teeth of the rotating gear and gets reflected from a mirror.
     If the reflected light again passes between two teeth of the rotating gear,
     the speed of light can be calculated from the gear-mirror distance and the speed of rotation.\\

\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.40\textwidth}
  {\small
     Fizeau determined the speed of light between an intense light source and a mirror about 8 km distant.
     The light source was interrupted by a rotating cogwheel with 720 notches
     that could be rotated at a variable speed of up to hundreds of times a second.\\
  }
  \end{column}
  \begin{column}{0.60\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/fizeau_experiment_absolute_speed_01.png}\\
   \end{center}
  \end{column}
\end{columns}

\end{frame}

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\begin{frame}{Light is an electromagnetic wave!}

Maxwell (with typical British understatement) said that the result
was {\em ``somewhat of a surprise''} !!\\
\vspace{0.1cm}

\begin{columns}
  \begin{column}{0.40\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/people/maxwell.jpg}\\
     {\small
       James Clerk Maxwell\\
       (1831 - 1879)
     }
   \end{center}
  \end{column}
  \begin{column}{0.60\textwidth}
  {\small
       {\em ``I made out the equations before I had any suspicion of the nearness between
       the two values of the velocity of propagation of magnetic effects and of light, so I think
       I have reasons to believe that the magnetic and luminiferous media are identical''.}\\
       \vspace{0.3cm}
       {\em `We can scarcely avoid the inference that light consists in the transverse undulations
       of the same medium which is the cause of electric and magnetic phenomena.}
       (Maxwell discussed his ideas in terms of a model in which the vacuum was like an elastic solid.)
  }
  \end{column}
\end{columns}

\end{frame}


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\begin{frame}{Light is an electromagnetic wave!}

Let's just think how amazing and unexpected that was!\\

\vspace{0.1cm}

{\bf What was $\epsilon_0$} ? \\
It is the constant of proportionality in Coulomb's law that, connects
charge densities and the resulting electric field $\vec{E}$:
\begin{equation*}
    \vec{E} (\vec{r})= \frac{1}{4 \pi {\bf \epsilon_0}}
       \int d\tau^{\prime} \rho(\pvec{r}') \frac{\vec{r}-\pvec{r}'}{|\vec{r}-\pvec{r}'|^3}
\end{equation*}

\vspace{0.05cm}

{\bf What was $\mu_0$} ? \\
It is the constant of proportionality in the Biot-Savart law that
connects currents and the resulting magnetic field $\vec{B}$:
\begin{equation*}
   \vec{B} (\vec{r}) = \frac{{\bf \mu_0}}{4\pi}
       \int d\tau^{\prime} \vec{j}(\pvec{r}') \times \frac{\vec{r}-\pvec{r}'}{|\vec{r}-\pvec{r}'|^3}
\end{equation*}

The fact that the speed of the speed of light is determined by these two (apparently
unrelated to light) constants must have stunned Maxwell (or, in his
works, it was {\em ``somewhat of a surprise''}).

\end{frame}


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\begin{frame}{The electromagnetic spectrum}

\begin{center}
\includegraphics[width=0.98\textwidth]{./images/schematics/emspectrum.png}\\
\end{center}

\end{frame}



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% Worked example
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{
\problemslide

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\begin{frame}{Worked example}

\begin{blockexmplque}{Question}
An electric field $\vec{E}$ is given by:
\begin{equation*}
  \vec{E} = E_{0} \cdot sin\Big(k \cdot  (x-ct)\Big) \cdot \hat{y}
\end{equation*}
where k is the propagation number.\\
\vspace{0.3cm}
\begin{enumerate}[(a)]
 \item Show that it fulfils the wave equation.
 \item Find the corresponding magnetic field.
 \item Comment on directions and amplitudes.
\end{enumerate}
\end{blockexmplque}

\end{frame}

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\begin{frame}{Worked example}

First, we will calculate the Laplacian ($\vec{\nabla} \cdot \vec{E}$)
of the given electric field $\vec{E}$:
\begin{equation*}
  \vec{E} = E_{0} sin\Big(k(x-ct)\Big) \hat{y}
\end{equation*}

We have:
\begin{equation*}
  \vec{\nabla}^{2} \vec{E} =
  \frac{\partial^{2} \vec{E}} {\partial x^{2}} +
  \cancelto{0}{\frac{\partial^{2} \vec{E}} {\partial y^{2}}} +
  \cancelto{0}{\frac{\partial^{2} \vec{E}} {\partial z^{2}}} =
  \frac{\partial^{2}} {\partial x^{2}} \bigg\{ E_{0} sin\Big(k(x-ct)\Big) \hat{y} \bigg\} =
\end{equation*}
\begin{equation*}
     = E_{0} \bigg\{ \frac{\partial^{2}} {\partial x^{2}} sin\Big(k(x-ct)\Big) \bigg\} \hat{y} =
        E_{0} \bigg\{ k \frac{\partial} {\partial x} cos\Big(k(x-ct)\Big) \bigg\} \hat{y} =
\end{equation*}
\begin{equation*}
     = E_{0} \bigg\{ - k^2 sin\Big(k(x-ct)\Big) \bigg\} \hat{y} =
      - k^2 \bigg\{ E_{0} sin\Big(k(x-ct)\Big) \hat{y} \bigg\} \Rightarrow
\end{equation*}
\begin{equation*}
  \vec{\nabla}^{2} \vec{E} = - k^2 \vec{E}
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

Then, we will calculate the second-order partial derivative with respect to time
($\partial^2 / \partial t^2$) of the given electric field $\vec{E}$:
\begin{equation*}
  \vec{E} = E_{0} sin\Big(k(x-ct)\Big) \hat{y}
\end{equation*}

\begin{equation*}
  \frac{\partial^{2} \vec{E}} {\partial t^{2}} =
  \frac{\partial^{2}} {\partial t^{2}} \bigg\{ E_{0} sin\Big(k(x-ct)\Big) \hat{y} \bigg\} =
\end{equation*}
\begin{equation*}
  = E_{0} \bigg\{ \frac{\partial^{2}} {\partial t^{2}} sin\Big(k(x-ct)\Big) \bigg\} \hat{y} =
     E_{0} \bigg\{-kc \frac{\partial} {\partial t} cos\Big(k(x-ct)\Big) \bigg\} \hat{y} =
\end{equation*}
\begin{equation*}
   = E_{0} \bigg\{ - k^{2} c^{2} sin\Big(k(x-ct)\Big) \bigg\} \hat{y} =
    - k^{2} c^{2} \bigg\{ E_{0} sin\Big(k(x-ct)\Big) \hat{y} \bigg\} \Rightarrow
\end{equation*}
\begin{equation*}
  \frac{\partial^{2} \vec{E}} {\partial t^{2}} = - k^2 c^2 \vec{E}
\end{equation*}

\end{frame}


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\begin{frame}{Worked example}

So, we found that:
\begin{equation*}
  \vec{\nabla}^{2} \vec{E} = - k^2 \cdot \vec{E}
\end{equation*}

and:
\begin{equation*}
  \frac{\partial^{2} \vec{E}} {\partial t^{2}} = - k^2 \cdot c^2 \cdot \vec{E}
\end{equation*}

Hence:
\begin{equation*}
  \frac{\partial^{2} \vec{E}} {\partial t^{2}} = c^2 \cdot \vec{\nabla}^{2} \vec{E}
\end{equation*}

The electric field $\displaystyle \vec{E} = E_{0} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \cdot \hat{y}$
fullfils a wave equation with velocity c.

\end{frame}


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\begin{frame}{Worked example}

To find the corresponding magnetic field $\vec{B}$, we will use Faraday's law:
\begin{equation*}
  \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}
\end{equation*}

First we will calculate $\vec{\nabla} \times \vec{E}$,
and then integrate the result over time.\\

\vspace{0.2cm}

We have:
\begin{equation*}
  \vec{\nabla} \times \vec{E} =
   \left|
    \begin{array}{ccc}
      \hat{x}                     & \hat{y}                     & \hat{z} \\
      \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
       E_{x}                      &  E_{y}                      &  E_{z} \\
    \end{array}
   \right| =
   \hat{x}
   \left|
    \begin{array}{cc}
      \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
       E_{y}                      &  E_{z} \\
    \end{array}
   \right| -
   \hat{y}
   \left|
    \begin{array}{cc}
      \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\
       E_{x}                      &  E_{z} \\
    \end{array}
   \right| +
   \hat{z}
   \left|
    \begin{array}{cc}
      \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\
       E_{x}                      &  E_{y}                      \\
    \end{array}
   \right| \Rightarrow
\end{equation*}

\begin{equation*}
   \vec{\nabla} \times \vec{E} =
   \hat{x} \cdot \Big( \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} \Big) -
   \hat{y} \cdot \Big( \frac{\partial E_{z}}{\partial x} - \frac{\partial E_{x}}{\partial z} \Big) +
   \hat{z} \cdot \Big( \frac{\partial E_{y}}{\partial x} - \frac{\partial E_{x}}{\partial y} \Big)
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

Recall that $\vec{E}$ has only a y component ($E_{x} = E_{z} = 0$),
which has only x dependence ($E_{y} = E_{y}(x)$ with no y or z dependence).
Therefore:

\begin{equation*}
  \vec{\nabla} \times \vec{E} =
   \hat{x} \cdot \Big( \cancelto{0}{\frac{\partial E_{z}}{\partial y}} - \cancelto{0}{\frac{\partial E_{y}}{\partial z}} \Big) -
   \hat{y} \cdot \Big( \cancelto{0}{\frac{\partial E_{z}}{\partial x}} - \cancelto{0}{\frac{\partial E_{x}}{\partial z}} \Big) +
   \hat{z} \cdot \Big( \frac{\partial E_{y}}{\partial x} - \cancelto{0}{\frac{\partial E_{x}}{\partial y}} \Big) \Rightarrow
\end{equation*}

\begin{equation*}
  \vec{\nabla} \times \vec{E} = \hat{z} \cdot\frac{\partial E_{y}}{\partial x}
\end{equation*}

The first-order partial derivative of $E_{y}(x)$ with respect to x is:
\begin{equation*}
  \frac{\partial E_{y}}{\partial x} =
      \frac{\partial}{\partial x} \bigg\{ E_{0} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \bigg\} =
      E_{0} \cdot k \cdot cos\Big(k \cdot (x-c \cdot t)\Big)
\end{equation*}

Hence:
\begin{equation*}
  \vec{\nabla} \times \vec{E} = \bigg\{  E_{0} \cdot k \cdot cos\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{z}
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

Substituting the previous result in Faraday's law, we have:
\begin{equation*}
  - \frac{\partial \vec{B}}{\partial t} = \bigg\{  E_{0} \cdot k \cdot cos\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{z}
\end{equation*}

Integrating over time:
\begin{equation*}
  \vec{B} = \bigg\{ - E_{0} \cdot k \cdot \int cos\Big(k \cdot (x-c \cdot t)\Big) dt \bigg\} \cdot \hat{z} =
\end{equation*}
\begin{equation*}
   = \bigg\{ - E_{0} \cdot k \cdot \frac{sin\Big(k \cdot (x-c \cdot t)\Big)}{-k \cdot c} \bigg\} \cdot \hat{z} + \vec{C} =
       \bigg\{ \frac{E_{0}}{c} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{z} + \vec{C}
\end{equation*}

In the absence of currents there are no time invariant components of $\vec{B}$ ($\vec{C} = \vec{0}$) hence:
\begin{equation*}
  \vec{B} = \bigg\{ \frac{E_{0}}{c} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{z}
\end{equation*}


\end{frame}

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\begin{frame}{Worked example}

Our electric and magnetic fields are given by:
\begin{equation*}
   \vec{E} = \bigg\{ E_{0} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{y}
    \;\;\;\; and \;\;\;\;
   \vec{B} = \bigg\{ \frac{E_{0}}{c} \cdot sin\Big(k \cdot (x-c \cdot t)\Big) \bigg\} \cdot \hat{z}
\end{equation*}

Notice that the {\bf amplitude of the magnetic field is smaller to that of the electric field
by a factor of c}.\\

\vspace{0.2cm}

Also notice that:
\begin{itemize}
   \item the wave propagates on the x-axis,
   \item the electric field oscillates on the y-axis, and
   \item the magnetic field oscillates on the z-axis
\end{itemize}

Therefore the {\bf electric and magnetic fields are}:
\begin{itemize}
    \item {\bf perpendicular to each other}, and
    \item {\bf perpendicular to the direction of motion}.
\end{itemize}

\end{frame}

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\begin{frame}{Worked example}

\begin{center}
\includegraphics[width=0.80\textwidth]{./images/schematics/emwave_01.jpg}\\
\end{center}

\end{frame}

} % Worked example


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\begin{frame}{Energy carried by electromagnetic waves}

We know that an e/m wave carries energy.\\
\vspace{0.1cm}

How can we express the {\bf energy transported by an e/m wave?}\\
\vspace{0.2cm}

The energy stored in the electric field is:
\begin{equation*}
  U_{e} = \frac{\epsilon_0}{2} \int_{\tau} d\tau |\vec{E}|^2
\end{equation*}
while the energy stored in the magnetic field is:
\begin{equation*}
  U_{m} = \frac{1}{2\mu_0} \int_{\tau} d\tau |\vec{B}|^2
\end{equation*}
Therefore the energy stored in an electromagnetic field is:
\begin{equation*}
  U_{em} = \frac{1}{2} \int_{\tau} d\tau \Big( \epsilon_0 |\vec{E}|^2 + \frac{1}{\mu_0} |\vec{B}|^2 \Big)
\end{equation*}

\end{frame}

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\begin{frame}{Energy carried by electromagnetic waves}

Consider fields $\vec{E}$ and $\vec{B}$ within some volume $\tau$.
A charge q moves within that volume with velocity $\vec{u}$ and it is displaced by a short distance $d\vec{\ell}$.\\
\vspace{0.2cm}
The rate of change of the work W done on the charge q by the
e/m field\\ (see Optional Reading) is given by:
\begin{equation*}
  \frac{dW}{dt} =
     - {\color{magenta} \oint_{S} d\vec{S} \cdot \frac{1}{\mu_0} \Big(\vec{E} \times \vec{B} \Big) }
     - \frac{\partial}{\partial t} \int_{\tau} d\tau  \Big( \frac{1}{2\mu_0} \vec{B}^2 + \frac{\epsilon_0}{2} \vec{E}^2 \Big)
\end{equation*}

As could be anticipated, the {\bf work done on the charge q} is related with the
{\bf decrease of the energy stored in the e/m field} (2nd term on right).\\
\vspace{0.2cm}

The additional {\color{magenta}``surface'' term} above,
can be interpreted as the {\em energy carried away
from the volume $\tau$, through its surface S, by the e/m field}.\\
\vspace{0.2cm}

The above expression is known as the {\bf Poynting theorem}.


\end{frame}


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\begin{frame}{The Poynting vector}

We define the {\bf Poynting vector} $\vec{N}$ as:
\begin{equation*}
  \vec{N} = \frac{1}{\mu_0} \Big( \vec{E} \times \vec{B} \Big)
\end{equation*}

\vspace{0.3cm}

It represents the {\bf energy flux density (rate energy transfer per unit area)} and it has units of $W(att)/m^2$.\\

\vspace{0.3cm}

Recall that the cross product $\vec{A} \times \vec{B}$ of 2 vectors $\vec{A}$ and $\vec{B}$
is perpendicular both to $\vec{A}$ and $\vec{B}$. So, in an electromagnetic wave, the energy
flows in a direction which is perpendicular to both the electric and magnetic field.

\end{frame}

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% Worked example
%

{
\problemslide

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\begin{frame}{Worked example}

An e/m wave is given by the time-dependent fields:
\begin{equation*}
  \vec{E}(\vec{r},t) = E_0 sin\Big( k(z-ct) \Big) \hat{x}
\end{equation*}
\begin{equation*}
  \vec{B}(\vec{r},t) = \frac{E_0}{c} sin\Big( k(z-ct) \Big) \hat{y}
\end{equation*}

This gives the Poynting vector:
\begin{equation*}
  \vec{N} =
    \frac{1}{\mu_0} \Big( \vec{E} \times \vec{B} \Big) =
    \frac{E_0^2}{\mu_0 c} sin^{2}\Big( k(z-ct) \Big) \Big( \hat{x} \times \hat{y} \Big)
\end{equation*}

One can easily see that:
\begin{equation*}
  \hat{x} \times \hat{y} = \hat{z}
\end{equation*}

Therefore, the Poynting vector is:
\begin{equation*}
  \vec{N} =
    \frac{E_0^2}{\mu_0 c} sin^{2}\Big( k(z-ct) \Big) \hat{z}
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

Therefore, the Poynting vector is:
\begin{equation*}
  \vec{N} =
    \frac{E_0^2}{\mu_0 c} sin^{2}\Big( k(z-ct) \Big) \hat{z}
\end{equation*}

The above vector describes the  power transmitted by the e/m wave
per unit area perpendicular to its propagation.

The average power transmitted by the e/m wave, is given by
the {\bf average of $\vec{N}$ over a period T}:
\begin{equation*}
  <\vec{N}> = \frac{1}{T} \int_{0}^{T} \vec{N} dt
\end{equation*}

\begin{equation*}
  <\vec{N}> = \frac{1}{T} \int_{0}^{T} \Big( \frac{E_0^2}{\mu_0 c} sin^{2}\Big( k(z-ct) \Big) \hat{z} \Big) dt
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

\begin{equation*}
  <\vec{N}> = \frac{1}{T} \int_{0}^{T} \Big( \frac{E_0^2}{\mu_0 c} sin^{2}\Big( k(z-ct) \Big) \hat{z} \Big) dt
\end{equation*}

\begin{equation*}
  <\vec{N}> =  \hat{z} \frac{1}{T} \frac{E_0^2}{\mu_0 c} \int_{0}^{T} sin^{2}\Big( k(z-ct) \Big) dt
\end{equation*}

\vspace{0.2cm}

The average of $sin^{2}\theta$ over a period is 1/2, therefore:
\begin{equation*}
  <\vec{N}> =  \hat{z} \frac{E_0^2}{2 \mu_0 c}
\end{equation*}

\end{frame}

} % A worked example


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\begin{frame}{Radiation Pressure}

E/M waves {\bf carry momentum}:
Can {\bf exert pressure} on an object when shining on it!\\
\vspace{0.1cm}

Assume that an object
is illuminated by radiation for a time interval ${\Delta}t$ and that
the radiation is entirely {\bf absorbed} by the object.
If its energy gain is ${\Delta}U$, then the magnitude of its momentum
change ${\Delta}p$ is given by:
\begin{equation*}
  {\Delta}p = \frac{{\Delta}U}{c}
  \;\;\; {\color{red} (total \; absorption)}
\end{equation*}
where c is the speed of light.\\
\vspace{0.1cm}

If, instead of being fully absorbed, the radiation is fully reflected back
by the object along its original path, the momentum change of the
object is twice  the one given above:
\begin{equation*}
  {\Delta}p = \frac{2{\Delta}U}{c}
  \;\;\; {\color{red} (total \; reflection \; along \; original \; path)}
\end{equation*}

In realistic situations the momentum change is between $\frac{{\Delta}U}{c}$
and $\frac{2{\Delta}U}{c}$.

\end{frame}

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\begin{frame}{Radiation Pressure}

From Newton's second law, a change in momentum is related to a force
\begin{equation*}
   F = \frac{{\Delta}p}{{\Delta}t}
\end{equation*}

If the radiation has intensity I (power per area, or energy per time
per area), then over a time interval ${\Delta}t$, the energy
intercepted by an area A is:
\begin{equation*}
  {\Delta}U = I A {\Delta}t
\end{equation*}
Therefore, the force exerted by radiation on an object is given by:
\begin{equation*}
  F = \frac{IA}{c}
  \;\;\; {\color{red} (total \; absorption)}
\end{equation*}
\begin{equation*}
  F = \frac{2IA}{c}
  \;\;\; {\color{red} (total \; reflection \; along \; original \; path)}
\end{equation*}

If the radiation is partially absorbed and partially reflected, the
exerted force will be between $\frac{IA}{c}$
and $\frac{2IA}{c}$.

\end{frame}

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\begin{frame}{Radiation Pressure}

The force per unit area on an object due to radiation is the {\bf radiation
pressure} $P_{r}$.\\
\vspace{0.2cm}

From the previous expressions, we obtain:
\begin{equation*}
  P_r = \frac{I}{c}
  \;\;\; {\color{red} (total \; absorption)}
\end{equation*}
\begin{equation*}
  P_r = \frac{2I}{c}
  \;\;\; {\color{red} (total \; reflection \; along \; original \; path)}
\end{equation*}


\end{frame}

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\begin{frame}{Radiation Pressure}

{\bf Solar sail} a form of spacecraft propulsion!
\begin{itemize}
{\small
  \item Using radiation pressure from sunlight (see tomorrow's
    workshop tasks)
  \item See {\color{blue}https://en.wikipedia.org/wiki/Solar\_sail}
}
\end{itemize}

\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.50\textwidth}
   \begin{center}
         \includegraphics[width=0.90\textwidth]{./images/photos/lightsail.jpg}\\
   \end{center}
  \end{column}
  \begin{column}{0.50\textwidth}
       {\bf LighSail 2}
      \begin{itemize}
      {\small
        \item Area: 32 m$^2$
        \item Launch date: March 2017!
        \item See {\color{blue}http://sail.planetary.org}
      }
      \end{itemize}
  \end{column}
\end{columns}

\vspace{0.4cm}

What if we replace sunlight with powerful directed laser beams?\\
See: ``NASA thinks there's a way to get to Mars in three days''\\
{\color{blue}http://phys.org/news/2016-02-nasa-mars-days.html}\\

\end{frame}


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\begin{frame}{Radiation Pressure}

Who cares about Mars!\\
\vspace{0.2cm}

Shoot for {\bf Alpha Centauri} and its recently-discovered planet Proxima b!
\vspace{0.2cm}

See {\bf Breakthrough Starshot}:
{\color{blue}http://breakthroughinitiatives.org/Initiative/3}\\

\end{frame}


%...............................................................................

%
% Worked example
%

{
\problemslide

\begin{frame}{Worked example}

\begin{blockexmplque}{Question}
  A small laser emits light at power $P$ = 5.00 mW and wavelength $\lambda$ = 633 nm.
  The laser beam is focused (narrowed) until its diameter $d$ matches the 1266 nm
  diameter of a sphere placed in its path. The sphere is perfectly absorbing
  and has density $\rho$ = 5.00 $\times$ 10$^3$ kg/m$^3$.
  What are
  \begin{itemize}
  \item the beam intensity $I$ at the sphere's location,
  \item the radiation pressure $P_r$ on the sphere,
  \item the magnitude of the corresponding force $F_r$, and
  \item the magnitude of the acceleration $\alpha$ that force alone would give the sphere?
  \end{itemize}
\end{blockexmplque}

We note that cross-section of the beam is $\pi d^2/4$.
Therefore, the beam intensity is:
\begin{equation*}
   I = \frac{P}{\pi d^2 / 4}
     = \frac{5 \times 10^{-3} \; W}{3.14 (1266 \times 10^{-9} \; m)^2 / 4 }
     = 3.97 \times 10^9 \; W/m^2
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

The rediation pressure is:
\begin{equation*}
   P_r = \frac{I}{c}
       = \frac{3.97 \times 10^9 \; W/m^2}{3.0 \times 10^8 \; m/s}
       = 13.2 \; Pa
\end{equation*}

The corresponding force is:
\begin{equation*}
   F_r = P_r \Big( \pi d^2 / 4 \Big)
       = 13.2 \Big(3.14 (1266 \times 10^{-9} \; m)^2 / 4 \Big)
       = 1.67 \times 10^{-11} \; N
\end{equation*}

We note that volume of the sphere is $4\pi(d/2)^3/3 = \pi d^3/6$ and,
therefore its mass $m$, is given by $m = \rho \pi d^3/6$.
The acceleration due to the force computed previously is
\begin{equation*}
   \alpha = \frac{F_r}{m} = \frac{F_r}{\rho \pi d^3/6} = \frac{6 F_r}{\rho \pi d^3} \Rightarrow
\end{equation*}
\begin{equation*}
   \alpha = \frac{6 (1.67 \times 10^{-11} \; N)}{(5 \times 10^3 \; kg/m^3) 3.14 (1266 \times 10^{-9} \; m)^3}
          = 3.14 \times 10^3 \; m/s^2
\end{equation*}

\end{frame}

} % \problemslide

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{
\problemslide

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\begin{frame}{Worked example}

\begin{blockexmplque}{Question}
  A small space shuttle with a mass of only $1.5 \times 10^3$ kg
  (including crew) is drifting in outer space with negligible
  gravitational forces acting on it. If the astronaut turns
  on a 10 kW laser beam, what speed will the shuttle attain in 1.0 day
  because of the momentum carried away by the beam?
\end{blockexmplque}

Since the total momentum of the spaceship and light is conserved,
this is the magnitude of the momentum acquired by the spaceship
is equal to the magnitude of the momentum carried away by the laser beam.\\
\vspace{0.2cm}
If P is the power of the laser, then the energy U carried away in time t is:
\begin{equation*}
   U = P \cdot t
\end{equation*}
Thus the magntitude of the momentum is given by:
\begin{equation*}
   p = P \cdot t/c
\end{equation*}

\end{frame}

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\begin{frame}{Worked example: Space shuttle propelled by a laser beam /}

If m is the mass of the spaceship, its speed u is:
\begin{equation*}
  u = \frac{p}{m} = \frac{P \cdot t}{m \cdot c}
\end{equation*}

There are 86400 seconds in a day, therefore:
\begin{equation*}
  u = \frac{(10 \times 10^3 \; W)
      \cdot (86400 \; s)}{(1.5 \times 10^3 \; kg)
      \cdot (2.998\times 10^8 \; m/s)}
    = 1.9 \times 10^{-3} \; m/s
\end{equation*}

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\begin{blockexmplque}{Question}
  The magnetic field in a traveling EM wave has an rms
  strength of 22.5 nT. How long does it take to deliver 335 J of
  energy to 1 cm$^2$ of a wall that it hits perpendicularly?
\end{blockexmplque}

The energy per unit area per unit time is given by the magnitude of the
Poynting vector $\vec{S}$.
\begin{equation*}
  \vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) \Rightarrow
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

\begin{equation*}
  <S> = \frac{E_0 B_0}{2\mu_0} = \frac{c B_0^2}{2\mu_0} = \frac{c B_{rms}^2}{\mu_0}
\end{equation*}

Let $\Delta U$ represent the energy that crosses
area $A$ in a time $\Delta t$. Then:

\begin{equation*}
  <S> = \frac{\Delta U}{A \Delta t}
\end{equation*}

Therefore:
\begin{equation*}
  \frac{\Delta U}{A \Delta t} = \frac{c B_{rms}^2}{\mu_0} \Rightarrow
  \Delta t = \frac{\Delta U \mu_0}{A c B_{rms}^2} \Rightarrow
\end{equation*}
\begin{equation*}
  \Delta t =
  \frac
  {
    (335 \; \text{J})
    (4\pi \times 10^{-7} \; \text{Tm/A})
  }
  {
    (10^{-4} \; \text{m}^2)
    (3 \times 10^{8} \; \text{m/s})
    (22.5 \times 10^{-9} \; \text{T})^2
  } =
  2.66 \times 10^7 \text{s} ( \approx 321 \; \text{days})
\end{equation*}

\end{frame}

} % \problemslide

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{
\problemslide

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\begin{frame}{Worked example: rms values of $\vec{E}$ in starlight}

\begin{blockexmplque}{Question}
  Estimate the rms electric field in the sunlight that hits Mars,
  knowing that the Earth receives about 1350 W/m$^2$ and that
  Mars is 1.52 times farther from the Sun (on average) than is
  the Earth.
\end{blockexmplque}

The radiation from the Sun is isotropic,
so the rate $P$ at which energy passes through a sphere with radius $R$
centered at the Sun is:
\begin{equation*}
    P = <S> \Big( 4\pi R^2 \Big)
\end{equation*}

This rate is the same at any distance from the Sun,
so the above expression is valid on Earth:
\begin{equation*}
    P = <S_{Earth}> \Big( 4\pi R_{Earth}^2 \Big)
\end{equation*}
and on Mars:
\begin{equation*}
    P = <S_{Mars}> \Big( 4\pi R_{Mars}^2 \Big)
\end{equation*}

\end{frame}

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\begin{frame}{Worked example}

Therefore, from above:
\begin{equation*}
   <S_{Mars}> = <S_{Earth}> \frac{R_{Earth}^2}{R_{Mars}^2}
\end{equation*}

But, we know that:
\begin{equation*}
  <S> = \frac{E_0 B_0}{2\mu_0}
      = \frac{c B_0^2}{2\mu_0} = \frac{c B_{rms}^2}{\mu_0}
      = \frac{E_0^2}{2\mu_0 c} = \frac{E_{rms}^2}{\mu_0 c}
\end{equation*}
Therefore, on Mars:
\begin{equation*}
  <S_{Mars}> = \frac{E_{rms;Mars}^2}{\mu_0 c} \Rightarrow
  E_{rms;Mars} = \sqrt{<S_{Mars}> \mu_0 c}
\end{equation*}

Using the earlier expression we found for $<S_{Mars}>$, we have:
\begin{equation*}
  E_{rms;Mars} = \sqrt{<S_{Earth}> \mu_0 c} \frac{R_{Earth}}{R_{Mars}} \Rightarrow
\end{equation*}

\vspace{-0.2cm}
\begin{equation*}
  E_{rms;Mars} =
   \sqrt{(1350 \; \text{W/m}^2) (4\pi \times 10^{-7} \; \text{Tm/A}) (3\times 10^8 \; \text{m/s})}
   \frac{1}{1.52} = 469 \; V/m
\end{equation*}

\end{frame}

} % \problemslide

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{
\problemslide

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\begin{frame}{Worked example}

  \begin{blockexmplque}{Question}
    When you look at the North Star (Polaris), you intercept light
    from a star at a distance of 431 light years and emitting energy
    at a rate of 2.2 $\times$ 10$^3$ times that of our Sun
    ($P_{sun}$ = 3.9 $\times$ 10$^{26}$ W).\\
    Neglecting any atmospheric absorption, find the rms values of
    the electric and magnetic fields when the starlight reaches you.
  \end{blockexmplque}

  The intensity $I$ of E/M radiation, in terms of the amplitudes $E_0$, $B_0$,
  of the E/M wave is given by:
  \begin{equation*}
    I = \frac{E_{0} B_{0}}{2\mu_0}
  \end{equation*}

  The rms values are related to the amplitudes of the fields, as shown below:
  \begin{equation*}
    E_{rms} = \frac{E_{0}}{\sqrt{2}} \;\; and \;\;
    B_{rms} = \frac{B_{0}}{\sqrt{2}}
  \end{equation*}

\end{frame}

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\begin{frame}{Worked example}

  The amplitude magnetic field is related to amplitude of the
  electric field:
  \begin{equation*}
    B_{0} = \frac{E_{0}}{c}
  \end{equation*}

  Using the above, we can $I$ as a function of $E_{rms}$, as follows:
  \begin{equation*}
    I = \frac{E_{rms}^2}{c\mu_0}
  \end{equation*}

  Since the source is isotropic, we have:
  \begin{equation*}
    I = \frac{P}{4\pi r^2}
  \end{equation*}

  Therefore:
  \begin{equation*}
    \frac{E_{rms}^2}{c\mu_0} = \frac{P}{4\pi r^2} \Rightarrow
     E_{rms} = \sqrt{\frac{P c\mu_0}{4\pi r^2}}
  \end{equation*}

\end{frame}

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\begin{frame}{Worked example}

  Substituting $P = $ (2.2$\times$10$^3$)(3.9$\times$10$^{26}$ W),
  $r = $ 431 ly = 4.08$\times$10$^{18}$ m, and the values of the constants,
  we find:

  \begin{equation*}
     E_{rms} \approx 1.24 \; mV/m
  \end{equation*}

  The rms value, $B_{rms}$, of the magnetic field in light, is related
  to $E_{rms}$ as follows:
  \begin{equation*}
     B_{rms} = \frac{E_{rms}}{c}
  \end{equation*}

  Substituting the values of $E_{rms}$ and $c$, we find:
  \begin{equation*}
     B_{rms} = 4.1 \; pT
  \end{equation*}


\end{frame}

} % Worked example

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\renewcommand{\lecturesummarytitle}{Main points to remember }

\input{slides_lecture09_summary.tex}

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\begin{frame}{At the next lecture (Lecture \nextlecture)}

\begin{itemize}
\item Electromagnetic waves in matter
\end{itemize}

\end{frame}

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\input{slides_lecture09_optional.tex}

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