[NEW] The Feynman Lectures on Physics Vol. II Ch. 20: Solutions of Maxwell’s Equations in Free Space | maxwell’s equations – Pickpeup

maxwell’s equations: นี่คือโพสต์ที่เกี่ยวข้องกับหัวข้อนี้


Waves in free space; plane waves

In Chapter 18 we had reached the point where we had the
Maxwell equations in complete form. All there is to know about the
classical theory of the electric and magnetic fields can be found in the
four equations:

&\FLPcurl{\FLPE}&&\;=-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&\;\;

&\text{I.}&&\FLPdiv{\FLPE}&\;=&\;\frac{\rho}{\epsO}\\[1ex] &\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex] &\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}

When we put all these equations together, a remarkable new phenomenon
occurs: fields generated by moving charges can leave the sources and
travel alone through space. We considered a special example in which
an infinite current sheet is suddenly turned on. After the current has
been on for the time $t$, there are uniform electric and magnetic
fields extending out the distance $ct$ from the source. Suppose that
the current sheet lies in the $yz$-plane with a surface current
density $J$ going toward positive $y$. The electric field will have
only a $y$-component, and the magnetic field, only a
$z$-component. The field components are given by
E_y=cB_z=-\frac{J}{2\epsO c},

for positive values of $x$ less than $ct$. For larger $x$ the fields
are zero. There are, of course, similar fields extending the same
distance from the current sheet in the negative $x$-direction. In
Fig. 20–1 we show a graph of the magnitude of the
fields as a function of $x$ at the instant $t$. As time goes on, the
“wavefront” at $ct$ moves outward in $x$ at the constant
velocity $c$.

Now consider the following sequence of events. We turn on a current of
unit strength for a while, then suddenly increase the current strength
to three units, and hold it constant at this value. What do the fields
look like then? We can see what the fields will look like in the
following way. First, we imagine a current of unit strength that is
turned on at $t=0$ and left constant forever. The fields for
positive $x$ are then given by the graph in part (a) of Fig. 20–2.
Next, we ask what would happen if we turn on a steady current of two
units at the time $t_1$.

The fields in this case will be twice as high as before, but will
extend out in $x$ only the distance $c(t-t_1)$, as shown in part (b)
of the figure. When we add these two solutions, using the principle of
superposition, we find that the sum of the two sources is a current of
one unit for the time from zero to $t_1$ and a current of three units
for times greater than $t_1$. At the time $t$ the fields will vary
with $x$ as shown in part (c) of Fig. 20–2.

Now let’s take a more complicated problem. Consider a current which is
turned on to one unit for a while, then turned up to three units, and
later turned off to zero. What are the fields for such a current? We
can find the solution in the same way—by adding the solutions of
three separate problems. First, we find the fields for a step current
of unit strength. (We have solved that problem already.) Next, we
find the fields produced by a step current of two units. Finally, we
solve for the fields of a step current of minus three
units. When we add the three solutions, we will have a current which
is one unit strong from $t=0$ to some later time, say $t_1$, then
three units strong until a still later time $t_2$, and then turned
off—that is, to zero. A graph of the current as a function of time
is shown in Fig. 20–3(a). When we add the three
solutions for the electric field, we find that its variation with $x$,
at a given instant $t$, is as shown in Fig. 20–3(b). The
field is an exact representation of the current. The field distribution
in space is a nice graph of the current variation with time—only drawn
backwards. As time goes on the whole picture moves outward at the
speed $c$, so there is a little blob of field, travelling toward positive $x$,
which contains a completely detailed memory of the history of all the
current variations. If we were to stand miles away, we could tell from
the variation of the electric or magnetic field exactly how the current
had varied at the source.

You will also notice that long after all activity at the source has
completely stopped and all charges and currents are zero, the block of
field continues to travel through space. We have a distribution of
electric and magnetic fields that exist independently of any charges
or currents. That is the new effect that comes from the complete set
of Maxwell’s equations. If we want, we can give a complete
mathematical representation of the analysis we have just done by
writing that the electric field at a given place and a given time is
proportional to the current at the source, only not at the same
time, but at the earlier time $t-x/c$. We can write
E_y(t)=-\frac{J(t-x/c)}{2\epsO c}.

We have, believe it or not, already derived this same equation from
another point of view in Vol. I, when we were dealing with the
theory of the index of refraction. Then, we had to figure out what
fields were produced by a thin layer of oscillating dipoles in a sheet
of dielectric material with the dipoles set in motion by the electric
field of an incoming electromagnetic wave. Our problem was to
calculate the combined fields of the original wave and the waves
radiated by the oscillating dipoles. How could we have calculated the
fields generated by moving charges when we didn’t have Maxwell’s
equations? At that time we took as our starting point (without any
derivation) a formula for the radiation fields produced at large
distances from an accelerating point charge. If you will look in
Chapter 31 of Vol. I, you will see that
Eq. (31.9) there is just the same as the
Eq. (20.3) that we have just written down. Although our
earlier derivation was correct only at large distances from the source,
we see now that the same result continues to be correct even right up to
the source.

We want now to look in a general way at the behavior of electric and
magnetic fields in empty space far away from the sources, i.e., from
the currents and charges. Very near the sources—near enough so that
during the delay in transmission, the source has not had time to
change much—the fields are very much the same as we have found in
what we called the electrostatic or magnetostatic cases. If we go out
to distances large enough so that the delays become important,
however, the nature of the fields can be radically different from the
solutions we have found. In a sense, the fields begin to take on a
character of their own when they have gone a long way from all the
sources. So we can begin by discussing the behavior of the fields in a
region where there are no currents or charges.

Suppose we ask: What kind of fields can there be in regions where
$\rho$ and $\FLPj$ are both zero? In Chapter 18 we saw
that the physics of Maxwell’s equations could also be expressed in
terms of differential equations for the scalar and vector potentials:
\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&=
-\frac{\rho}{\epsO},\\[1ex] \label{Eq:II:20:5}
\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&=
-\frac{\FLPj}{\epsO c^2}.

If $\rho$ and $\FLPj$ are zero, these equations take on the simpler
\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&=0,\\[1ex] \label{Eq:II:20:7}
\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&=\FLPzero.

Thus in free space the scalar potential $\phi$ and each component of
the vector potential $\FLPA$ all satisfy the same mathematical
equation. Suppose we let $\psi$ (psi) stand for any one of the four
quantities $\phi$, $A_x$, $A_y$, $A_z$; then we want to investigate
the general solutions of the following equation:
\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0.

This equation is called the three-dimensional wave
equation—three-dimensional, because the function $\psi$ may depend
in general on $x$, $y$, and $z$, and we need to worry about variations
in all three coordinates. This is made clear if we write out
explicitly the three terms of the Laplacian operator:
\frac{\partial^2\psi}{\partial x^2}+
\frac{\partial^2\psi}{\partial y^2}+
\frac{\partial^2\psi}{\partial z^2}-
\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0.

In free space, the electric fields $\FLPE$ and $\FLPB$ also satisfy
the wave equation. For example, since $\FLPB=\FLPcurl{\FLPA}$, we can
get a differential equation for $\FLPB$ by taking the curl of
Eq. (20.7). Since the Laplacian is a scalar operator, the
order of the Laplacian and curl operations can be interchanged:

Similarly, the order of the operations curl and $\ddpl{}{t}$ can be
\FLPcurl{\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}}=
\frac{1}{c^2}\,\frac{\partial^2}{\partial t^2}(\FLPcurl{\FLPA})=
\frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}.

Using these results, we get the following differential equation
for $\FLPB$:
\frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}=\FLPzero.

So each component of the magnetic field $\FLPB$ satisfies the
three-dimensional wave equation. Similarly, using the fact that
$\FLPE=-\FLPgrad{\phi}-\ddpl{\FLPA}{t}$, it follows that the electric
field $\FLPE$ in free space also satisfies the three-dimensional wave
\frac{1}{c^2}\,\frac{\partial^2\FLPE}{\partial t^2}=\FLPzero.

All of our electromagnetic fields satisfy the same wave equation,
Eq. (20.8). We might well ask: What is the most general
solution to this equation? However, rather than tackling that
difficult question right away, we will look first at what can be said
in general about those solutions in which nothing varies in $y$
and $z$. (Always do an easy case first so that you can see what is going
to happen, and then you can go to the more complicated cases.) Let’s
suppose that the magnitudes of the fields depend only upon $x$—that
there are no variations of the fields with $y$ and $z$. We are,
of course, considering plane waves again. We should expect to get
results something like those in the previous section. In fact, we will
find precisely the same answers. You may ask: “Why do it all over
again?” It is important to do it again, first, because we did not
show that the waves we found were the most general solutions for plane
waves, and second, because we found the fields only from a very
particular kind of current source. We would like to ask now: What is
the most general kind of one-dimensional wave there can be in free
space? We cannot find that by seeing what happens for this or that
particular source, but must work with greater generality. Also we are
going to work this time with differential equations instead of with
integral forms. Although we will get the same results, it is a way of
practicing back and forth to show that it doesn’t make any difference
which way you go. You should know how to do things every which way,
because when you get a hard problem, you will often find that only one
of the various ways is tractable.

We could consider directly the solution of the wave equation for some
electromagnetic quantity. Instead, we want to start right from the
beginning with Maxwell’s equations in free space so that you can see
their close relationship to the electromagnetic waves. So we start
with the equations in (20.1), setting the charges and
currents equal to zero. They become
&\text{I.}&&\FLPdiv{\FLPE}&\;=&\;0\\[1ex] &\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex] &\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\ddp{\FLPE}{t}

We write the first equation out in components:

We are assuming that there are no variations with $y$ and $z$, so the
last two terms are zero. This equation then tells us that

Its solution is that $E_x$, the component of the electric field in the
$x$-direction, is a constant in space. If you look at IV
in (20.12), supposing no $\FLPB$-variation in $y$ and $z$
either, you can see that $E_x$ is also constant in time. Such a field
could be the steady dc field from some charged condenser plates
a long distance away. We are not interested now in such an uninteresting
static field; we are at the moment interested only in dynamically
varying fields. For dynamic fields, $E_x=0$.

We have then the important result that for the propagation of plane
waves in any direction, the electric field must be at right
angles to the direction of propagation. It can, of course, still vary
in a complicated way with the coordinate $x$.

The transverse $\FLPE$-field can always be resolved into two
components, say the $y$-component and the $z$-component. So let’s
first work out a case in which the electric field has only one
transverse component. We’ll take first an electric field that is
always in the $y$-direction, with zero $z$-component. Evidently, if we
solve this problem we can also solve for the case where the electric
field is always in the $z$-direction. The general solution can always
be expressed as the superposition of two such fields.

How easy our equations now get. The only component of the electric
field that is not zero is $E_y$, and all derivatives—except those
with respect to $x$—are zero. The rest of Maxwell’s equations then
become quite simple.

Let’s look next at the second of Maxwell’s equations [II of
Eq. (20.12)]. Writing out the components of the curl
$\FLPE$, we have
&(\FLPcurl{\FLPE})_x&&=\ddp{E_z}{y}&&-\ddp{E_y}{z}&&=0,\\[1.5ex] &(\FLPcurl{\FLPE})_y&&=\ddp{E_x}{z}&&-\ddp{E_z}{x}&&=0,\\[1.5ex] &(\FLPcurl{\FLPE})_z&&=\ddp{E_y}{x}&&-\ddp{E_x}{y}&&=\ddp{E_y}{x}.

The $x$-component of $\FLPcurl{\FLPE}$ is zero because the derivatives
with respect to $y$ and $z$ are zero. The $y$-component is also zero;
the first term is zero because the derivative with respect to $z$ is
zero, and the second term is zero because $E_z$ is zero. The only
components of the curl of $\FLPE$ that is not zero is the
$z$-component, which is equal to $\ddpl{E_y}{x}$. Setting the three
components of $\FLPcurl{\FLPE}$ equal to the corresponding components
of $-\ddpl{\FLPB}{t}$, we can conclude the following:
&\ddp{B_x}{t}=0,\quad\ddp{B_y}{t}=0.\\[1ex] \label{Eq:II:20:16}

Since the $x$-component of the magnetic field and the $y$-component of
the magnetic field both have zero time derivatives, these two
components are just constant fields and correspond to the
magnetostatic solutions we found earlier. Somebody may have left some
permanent magnets near where the waves are propagating. We will ignore
these constant fields and set $B_x$ and $B_y$ equal to zero.

Incidentally, we would already have concluded that the $x$-component
of $\FLPB$ should be zero for a different reason. Since the divergence
of $\FLPB$ is zero (from the third Maxwell equation), applying the
same arguments we used above for the electric field, we would conclude
that the longitudinal component of the magnetic field can have no
variation with $x$. Since we are ignoring such uniform fields in our
wave solutions, we would have set $B_x$ equal to zero. In plane
electromagnetic waves the $\FLPB$-field, as well as the $\FLPE$-field,
must be directed at right angles to the direction of propagation.

Equation (20.16) gives us the additional proposition that
if the electric field has only a $y$-component, the magnetic field
will have only a $z$-component. So $\FLPE$ and $\FLPB$ are at
right angles to each other. This is exactly what happened in the
special wave we have already considered.

We are now ready to use the last of Maxwell’s equations for free space
[IV of Eq. (20.12)]. Writing out the components, we have
c^2\,\ddp{B_z}{y}&&-c^2\,\ddp{B_y}{z}&&=\ddp{E_x}{t},\\[1ex] &c^2(\FLPcurl{\FLPB})_y&&=
c^2\,\ddp{B_x}{z}&&-c^2\,\ddp{B_z}{x}&&=\ddp{E_y}{t},\\[1ex] &c^2(\FLPcurl{\FLPB})_z&&=

Of the six derivatives of the components of $\FLPB$, only the
term $\ddpl{B_z}{x}$ is not equal to zero. So the three equations give us

The result of all our work is that only one component each of the electric and
magnetic fields is not zero, and that these components must satisfy Eqs.
(20.16) and (20.18). The two equations can be combined
into one if we differentiate the first with respect to $x$ and the second with
respect to $t$; the left-hand sides of the two equations will then be the same
(except for the factor $c^2$). So we find that $E_y$ satisfies the equation
x^2}-\frac{1}{c^2}\,\frac{\partial^2E_y}{\partial t^2}=0.

We have seen the same differential equation before, when we studied
the propagation of sound. It is the wave equation for one-dimensional

You should note that in the process of our derivation we have found
something more than is contained in Eq. (20.11).
Maxwell’s equations have given us the further information that
electromagnetic waves have field components only at right angles to the
direction of the wave propagation.

Let’s review what we know about the solutions of the one-dimensional
wave equation. If any quantity $\psi$ satisfies the one-dimensional
wave equation
x^2}-\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0,

then one possible solution is a function $\psi(x,t)$ of the form

that is, some function of the single variable $(x-ct)$. The
function $f(x-ct)$ represents a “rigid” pattern in $x$ which travels
toward positive $x$ at the speed $c$ (see Fig. 20–4). For
example, if the function $f$ has a maximum when its argument is zero,
then for $t=0$ the maximum of $\psi$, will occur at $x=0$. At some later
time, say $t=10$, $\psi$ will have its maximum at $x=10c$. As time goes
on, the maximum moves toward positive $x$ at the speed $c$.

Sometimes it is more convenient to say that a solution of the
one-dimensional wave equation is a function of $(t-x/c)$. However,
this is saying the same thing, because any function of $(t-x/c)$ is
also a function of $(x-ct)$:

Let’s show that $f(x-ct)$ is indeed a solution of the wave
equation. Since it is a function of only one variable—the
variable $(x-ct)$—we will let $f’$ represent the derivative of $f$ with
respect to its variable and $f”$ represent the second derivative
of $f$. Differentiating Eq. (20.21) with respect to $x$, we

since the derivative of $(x-ct)$ with respect to $x$ is $1$. The
second derivative of $\psi$, with respect to $x$ is clearly
\frac{\partial^2\psi}{\partial x^2}=f”(x-ct).

Taking derivatives of $\psi$ with respect to $t$, we find
&\ddp{\psi}{t}=f'(x-ct)(-c),\notag\\[1.5ex] \label{Eq:II:20:23}
&\frac{\partial^2\psi}{\partial t^2}=+c^2f”(x-ct).

We see that $\psi$ does indeed satisfy the one-dimensional wave

You may be wondering: “If I have the wave equation, how do I know
that I should take $f(x-ct)$ as a solution? I don’t like this backward
method. Isn’t there some forward way to find the solution?”
Well, one good forward way is to know the solution. It is possible to
“cook up” an apparently forward mathematical argument, especially
because we know what the solution is supposed to be, but with an
equation as simple as this we don’t have to play games. Soon you will
get so that when you see Eq. (20.20), you nearly
simultaneously see $\psi=f(x-ct)$ as a solution. (Just as now when you
see the integral of $x^2\,dx$, you know right away that the answer
is $x^3/3$.)

Actually you should also see a little more. Not only is any function
of $(x-ct)$ a solution, but any function of $(x+ct)$ is also a
solution. Since the wave equation contains only $c^2$, changing the
sign of $c$ makes no difference. In fact, the most general
solution of the one-dimensional wave equation is the sum of two
arbitrary functions, one of $(x-ct)$ and the other of $(x+ct)$:

The first term represents a wave travelling toward positive $x$, and
the second term an arbitrary wave travelling toward negative $x$. The
general solution is the superposition of two such waves both existing
at the same time.

We will leave the following amusing question for you to think
about. Take a function $\psi$ of the following form:
\psi=\cos kx\cos kct.

This equation isn’t in the form of a function of $(x-ct)$ or
of $(x+ct)$. Yet you can easily show that this function is a solution of
the wave equation by direct substitution into Eq. (20.20).
How can we then say that the general solution is of the form of
Eq. (20.24)?

Applying our conclusions about the solution of the wave equation to
the $y$-component of the electric field, $E_y$, we conclude that $E_y$
can vary with $x$ in any arbitrary fashion. However, the fields which
do exist can always be considered as the sum of two patterns. One wave
is sailing through space in one direction with speed $c$, with an
associated magnetic field perpendicular to the electric field; another
wave is travelling in the opposite direction with the same speed. Such
waves correspond to the electromagnetic waves that we know
about—light, radiowaves, infrared radiation, ultraviolet radiation,
x-rays, and so on. We have already discussed the radiation of light in
great detail in Vol. I. Since everything we learned there applies to
any electromagnetic wave, we don’t need to consider in great detail
here the behavior of these waves.

We should perhaps make a few further remarks on the question of the
polarization of the electromagnetic waves. In our solution we chose to
consider the special case in which the electric field has only a
$y$-component. There is clearly another solution for waves travelling
in the plus or minus $x$-direction, with an electric field which has
only a $z$-component. Since Maxwell’s equations are linear, the
general solution for one-dimensional waves propagating in the
$x$-direction is the sum of waves of $E_y$ and waves of $E_z$. This
general solution is summarized in the following equations:
\FLPE&=(0,E_y,E_z)\\[.5ex] E_y&=f(x-ct)+g(x+ct)\\[.5ex] E_z&=F(x-ct)+G(x+ct)\\[1ex] \FLPB&=(0,B_y,B_z)\\[.5ex] cB_z&=f(x-ct)-g(x+ct)\\[.5ex] cB_y&=-F(x-ct)+G(x+ct).

Such electromagnetic waves have an $\FLPE$-vector whose direction is
not constant but which gyrates around in some arbitrary way in the
$yz$-plane. At every point the magnetic field is always perpendicular
to the electric field and to the direction of propagation.

If there are only waves travelling in one direction, say the positive
$x$-direction, there is a simple rule which tells the relative
orientation of the electric and magnetic fields. The rule is that the
cross product $\FLPE\times\FLPB$—which is, of course, a vector at
right angles to both $\FLPE$ and $\FLPB$—points in the direction in
which the wave is travelling. If $\FLPE$ is rotated into $\FLPB$ by a
right-hand screw, the screw points in the direction of the wave
velocity. (We shall see later that the vector $\FLPE\times\FLPB$ has a
special physical significance: it is a vector which describes the flow
of energy in an electromagnetic field.)

Physics – E\u0026M: Maxwell’s Equations (1 of 30) What are the Maxwell equations? Introduction

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In this video I will introduction to Maxwell’s equations.

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Physics - E\u0026M: Maxwell's Equations (1 of 30) What are the Maxwell equations?  Introduction

Maxwell’s Equations: Crash Course Physics #37

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In the early 1800s, Michael Faraday showed us how a changing magnetic field induces an electromotive force, or emf, resulting in an electric current. He also found that electric fields sometimes act like magnetic fields, and developed equations to calculate the forces exerted by both. In the mid1800s, Scottish physicist James Maxwell thought something interesting was going on there, too. So he decided to assemble a set of equations that held true for all electromagnetic interactions. In this episode of Crash Course Physics, Shini talks to us about Maxwell’s Equations and how important they are to our understanding of Physics.

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Maxwell's Equations: Crash Course Physics #37

What’s a Tensor?

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What's a Tensor?

Divergence and Curl

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Thực Tế 1 Tấm Pin Năng Lượng Mặt Trời 100w Làm Được Những Gì

นอกจากการดูบทความนี้แล้ว คุณยังสามารถดูข้อมูลที่เป็นประโยชน์อื่นๆ อีกมากมายที่เราให้ไว้ที่นี่: ดูบทความเพิ่มเติมในหมวดหมู่Music of Turkey

ขอบคุณที่รับชมกระทู้ครับ maxwell’s equations

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