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4 and a mechanical force equal to

$c\dfrac{dg}{dt}-b\dfrac{dh}{dt}+g\dfrac{dc}{dt}-h\dfrac{db}{dt},$

if the electrified bodies are at rest.

The first of these corresponds to the well-known expression for the electromotive force on a conductor moving in a magnetic field; the second is the mechanical force on a current in a magnetic field plus the term $$g\dfrac{dc}{dt}-h\dfrac{db}{dt}$$.

We can deduce an important consequence of the assumption, if we consider the case of the æther moving with uniform velocity between two parallel planes charged, the one with positive, the other with negative electricity.

If $$v$$ is the velocity of the æther, $$h$$ the electric displacement at right angles to the planes, the magnetic force between the planes will be parallel to $$x$$, and equal to $$-4\pi vh$$; or if $$\sigma$$ is the surface-density of the electrification on the planes $$-4\pi v\sigma$$, the magnetic force vanishes except between the planes, so that on crossing the positively electrified surface there is an increase in the magnetic force parallel to $$x$$ equal to $$4\pi\sigma v$$. Thus the charged surface acts like a current sheet of intensity $$-\sigma v$$, but $$-v$$ is the velocity of the plane relatively to the æther; so that a charged surface moving with velocity $$v$$ relatively to the æther must act like a current sheet of intensity $$\sigma v$$.

We will now proceed to apply these results to some special cases. Let us suppose that we have a charged sphere moving along the axis of $$z$$ with the velocity $$w_{0}$$, and that it sets the æther around it in motion in the same way as an incompressible fluid is set in motion by a solid sphere of the same radius moving through it with the same velocity. If $$a$$ is the radius of the sphere,

$\begin{array}{l} u=\frac{1}{2}w_{0}a^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r},\\ \\ v=\frac{1}{2}w_{0}a^{3}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r},\\ \\ w=\frac{1}{2}w_{0}a^{3}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}, \end{array}$|undefined

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Rh $f=-\dfrac{e}{4\pi}\dfrac{d}{dx}\frac{1}{r},\quad g=-\dfrac{e}{4\pi}\dfrac{d}{dy}\frac{1}{r},\quad h=-\dfrac{e}{4\pi}\dfrac{d}{dz}\frac{1}{r}:$

hence by equations (1),

$\begin{array}{l} \alpha=ew_{0}\left(1+\frac{1}{2}\dfrac{a^{3}}{r^{3}}\right)\dfrac{y}{r^{3}},\\ \\ \beta=-ew_{0}\left(1+\frac{1}{2}\dfrac{a^{3}}{r^{3}}\right)\dfrac{x}{r^{3}},\\ \\ \gamma=0. \end{array}$|undefined

Thus the lines of magnetic force are circles with their centres along and their planes at right angles to the axis of $$z$$.

At a distance from the centre large compared with the radius of the sphere the magnetic force is the same as that due to a current $$ew_{0}$$, but close to the sphere the relative motion of the sphere and æther causes it to be larger than this, and at the surface of the sphere it is the same as that due to a current $$\tfrac{3}{2}ew_{0}$$.

The energy due to this distribution of currents is $$\tfrac{3}{7}\dfrac{e^{2}w_{0}^{2}}{a}\mu$$.

Another case which can be easily solved is that of a right circular cylinder rotating with an angular velocity $$\omega$$, each unit length of the cylinder being charged with $$\mathrm{E}$$ units of electricity. If $$a$$ is the radius of the cylinder,

$\begin{array}{lll} u=-\omega\dfrac{a^{2}y}{r^{2}}, & & v=\omega\dfrac{a^{2}x}{r^{2}},\\ \\ f=\dfrac{\mathrm{E}}{2\pi}\dfrac{x}{r^{2}}, & & g=\dfrac{\mathrm{E}}{2\pi}\dfrac{y}{r^{2}}; \end{array}$|undefined

and by equations (1).

$\alpha=0,\quad \beta=0,\quad \gamma=-2\mathrm{E}\omega\dfrac{a^{2}}{r^{2}}$|undefined

Thus outside the rotating cylinder there is a magnetic force parallel to the axis of rotation.

If we assume that the æther outside the sphere is at rest, we can find the solution of the case of a charged metal sphere executing harmonic oscillations. Suppose the sphere to be moving parallel to the axis of $$z$$. the velocity at any time $$t$$ being represented by the real part of $$\omega\epsilon^{ipt}$$. Then if we take rectangular axes passing through the centre of the sphere and moving with it, the following equations are true inside the sphere if $$u, v, w$$ are the components of the current, $$a, b, c$$ those of magnetic induction, $$\psi$$ the electrostatic potential,

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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6 the components of the rector potential, and $$\sigma$$ the specific resistance of the metal.

{{MathForm2|(2)|$$\left.\begin{alignat}{2} \sigma u & = -b\omega\epsilon^{ipt} && -\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx},\\ \\ \sigma v & = -a\omega\epsilon^{ipt} && -\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy},\\ \\ \sigma w & = && -\dfrac{d\mathrm{H}}{dt}-\dfrac{d\psi}{dz}; \end{alignat}\right\} $$}}

and therefore

$\begin{alignat}{2} -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F} & =-b\omega\epsilon^{ipt} && -\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx},\\ \\ -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{G} & =a\omega\epsilon^{ipt} && -\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy},\\ \\ -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{H} & =-\dfrac{d\mathrm{H}}{dt} && -\dfrac{d\psi}{dz}. \end{alignat}$|undefined

In the dielectric outside the sphere, if $$f, g, h$$ are the electric displacements, $$\mathrm{K}$$ the specific inductive capacity, and if $$\partial/\partial t$$ denote partial differentiation with respect to the time, the equations are

$\begin{array}{l} \dfrac{4\pi}{\mathrm{K}}f=-\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{F}-\dfrac{d\psi}{dx},\\ \\ \dfrac{4\pi}{\mathrm{K}}g=-\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{G}-\dfrac{d\psi}{dy},\\ \\ \dfrac{4\pi}{\mathrm{K}}h=-\dfrac{d\mathrm{H}}{dt}-\dfrac{d\psi}{dz}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{H}-\dfrac{d\psi}{dz}; \end{array}$|undefined

and therefore

with a similar equation for $$\mathrm{G}$$.

From the form of these equations we see that the solution will take the form

$\begin{array}{l} \psi=\psi_{0}+\psi_{1}\epsilon^{ipt}+\psi'_{1}\epsilon^{-ipt}+\psi_{2}\epsilon^{2ipt}+\psi'_{2}\epsilon^{-2ipt}+\dots\\ \\ \mathrm{F}=\mathrm{F}_{1}\epsilon^{ipt}+\mathrm{F}'_{1}\epsilon^{-ipt}+\mathrm{F}_{2}\epsilon^{2ipt}+\mathrm{F}'_{2}\epsilon^{-2ipt}. \end{array}$

If we substitute these values in the above equations, we see that we may put $$\psi'_{1}\ \psi'_{2}\dots\mathrm{F}'_{1},\ \mathrm{F}'_{2}$$ all equal to zero.

If $$e$$ is the quantity of electricity on the sphere,

$\psi_{0}=\dfrac{e}{\mathrm{K}r}.$

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Rh Equating the coefficients of $$\epsilon^{ipt}$$ to zero in equations (2) and (3) we have

$-\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}_{1}=-\dfrac{d\mathrm{F}_{1}}{dt}-\dfrac{d\psi_{1}}{dx}$ inside the sphere, |undefined

with similar equations for $$\mathrm{G}_{1}$$ and $$\mathrm{H}_{1}$$; outside the sphere we have

with similar equations for $$\mathrm{G}_{1}$$ and $$\mathrm{H}_{1}$$.

The form of equation (4) suggests that we should put

$\psi_{1}=B\dfrac{d}{dz}\dfrac{1}{r}$

A particular integral of (4) is then

$\mathrm{F}_{1}=\dfrac{c\mathrm{B}}{p}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}-\dfrac{\omega e}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}.$|undefined

The complementary function is that solution of the differential equation

$\nabla^{2}\mathrm{F}_{1}+\mu\mathrm{K}p^{2}\mathrm{F}_{1}=0$

which, when considered as a function of the angular coordinates of a point, varies as $$r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}$$; this (see Proc. Math. Soc. vol. xv. p. 212) is

$\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}$|undefined

where

$\mathrm{E}_{2}(i\lambda r)=\dfrac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{3}}+\dfrac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{2}}+\dfrac{\epsilon^{-i\lambda r}}{i\lambda r}$|undefined

and

$\lambda^{2}=\mu\mathrm{K}p^{2}$

Thus, outside the sphere,

$\begin{array}{l} \mathrm{F}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}+\dfrac{i\mathrm{B}\epsilon^{pt}}{p}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r},\\ \\ \mathrm{G}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r}+\dfrac{i\mathrm{B}\epsilon^{pt}}{p}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r},\\ \\ \mathrm{H}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}-2\mathrm{C}\mathrm{E}_{2}(i\lambda r)+\dfrac{i\mathrm{B}}{p}\epsilon^{pt}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}, \end{array}$|undefined

where $$\mathrm{E}_{0}(i\lambda r)=\epsilon^{-i\lambda r}/i\lambda r$$, and is introduced into the expression

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8 for $$\mathrm{H}$$ to make

$\dfrac{d\mathrm{F}}{dx}+\dfrac{d\mathrm{G}}{dy}+\dfrac{d\mathrm{H}}{dz}=0.$|undefined

Inside the sphere the differential equations for $$\mathrm{F}_{1}$$ and $$\mathrm{G}_{1}$$ are of the form

$-\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}_{1}=-ip\mathrm{F};$

if $$\lambda_{1}^{2}=-4\pi\mu'ip/\sigma$$, the solution of this equation is

$\mathrm{F}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda_{1}r\right)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r},$|undefined

where

$\mathrm{S}_{2}\left(\lambda_{1}r\right)=\dfrac{3\sin\lambda_{1}r}{\left(\lambda_{1}r\right)^{3}}-\dfrac{3\cos\lambda_{1}r}{\left(\lambda_{1}r\right)^{2}}-\dfrac{\sin\lambda_{1}r}{\lambda_{1}r}.$|undefined

Similarly

$\mathrm{G}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda r\right)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r};$|undefined

the differential equation for $$\mathrm{H}_{1}$$ is

$-\dfrac{\sigma}{4\mu'}\nabla^{2}\mathrm{H}_{1}=-ip\mathrm{H}_{1}+\dfrac{\mathrm{B}}{a^{3}}.$|undefined

So that

$\mathrm{H}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda_{1}r\right)r^{3}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}+2\mathrm{D}\epsilon^{ipt}\mathrm{S}_{0}\left(\lambda_{1}r\right)+\dfrac{\mathrm{B}}{ipa^{3}}\epsilon^{ipt},$|undefined

where

$\mathrm{S}_{0}\left(\lambda_{1}r\right)=\dfrac{\sin\lambda_{1}r}{\lambda_{1}r},$

and is introduced to make

$\dfrac{d\mathrm{F}}{dx}+\dfrac{d\mathrm{G}}{dy}+\dfrac{d\mathrm{H}}{dz}=0.$|undefined

Since $$\mathrm{F}_{1},\mathrm{G}_{1},\mathrm{H}_{1}$$ are continuous when $$r=a$$, if $$a$$ is the radius of the sphere we have

{{MathForm2|(5)|$$\left.\begin{array}{c} \mathrm{C}\mathrm{E}_{2}(i\lambda a)-\mathrm{DS}_{2}\left(\lambda_{1}a\right)+\dfrac{i\mathrm{B}}{pa^{3}}=\dfrac{\omega e}{p^{2}a^{3}k},\\ \\ -2\mathrm{C}\mathrm{E}_{0}(i\lambda a)-2\mathrm{DS}_{2}\left(\lambda_{1}a\right)+\dfrac{i\mathrm{B}}{pa^{3}}=0. \end{array}\right\} $$}}

Since, on the assumption discussed above, the electrification on the surface of the moving sphere is equivalent to a tangential current-sheet whose intensity is $$\omega\epsilon^{ipt}\sin\theta\dfrac{e}{4\pi a^{2}}$$, we have as another surface-condition that the difference between the

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Rh magnetic force outside and inside $$=\omega\epsilon^{ipt}\sin\theta\dfrac{e}{a^{2}}$$; hence

$-3\mathrm{C}\dfrac{d}{da}\mathrm{E}_{0}(i\lambda a)-3\mathrm{D}\dfrac{dS_{0}(\lambda_{1}a)}{da}=\dfrac{\omega e}{a^{2}}.$|undefined

From (5) we have

$\mathrm{C}\big(\mathrm{E}_{2}(i\lambda a)+2\mathrm{E}_{0}(i\lambda a)\big)-\mathrm{D}\big(\mathrm{S}_{2}(\lambda_{1}a)+2\mathrm{S}_{0}(\lambda_{1}a)\big)=\dfrac{\omega e}{p^{2}a^{3}\mathrm{K}}.$|undefined

From these equations we have

where

$\Delta=\Big(\mathrm{E}_{2}(i\lambda a)+2\mathrm{E}_{0}(i\lambda a)\dfrac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)+\dfrac{d}{da}\mathrm{E}_{0}(i\lambda a)\big(\mathrm{S}_{2}(\lambda a)-2\mathrm{S}_{0}(\lambda_{1}a)\big)\Big).$

Let us first consider the case where $$\lambda a$$ and $$\lambda_{1}$$ are both small. In this case $$\mathrm{E}_{2}(i\lambda a)$$ is large compared with $$\mathrm{E}_{0}(i\lambda a)$$ and $$\mathrm{S}_{2}(\lambda_{1}a)$$, very small compared with $$\mathrm{S}_{0}(\lambda_{1}a)$$; hence we see that

$\begin{array}{l} \mathrm{C}=\dfrac{\omega e}{\mathrm{K}p^{2}a^{3}}\mathrm{E}_{2}(i\lambda a),\\ \\ \mathrm{D}=-\dfrac{-\dfrac{\omega e}{a^{3}}\mathrm{E}_{0}(i\lambda a)}{\mathrm{E}_{2}(i\lambda a)\dfrac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)}. \end{array}$|undefined

Since $$\mathrm{E}_{2}(i\lambda a)=\frac{3\epsilon^{-i\lambda a}}{(i\lambda a)^{3}}$$ approximately, and $$\mathrm{S}_{0}(\lambda_{1}a)=1$$, we have

$\begin{array}{l} \mathrm{C}=-\frac{1}{3}i\omega e\lambda\epsilon^{i\lambda a},\\ \\ \mathrm{D}=\dfrac{\omega e}{a}\dfrac{ip\sigma\mathrm{K}}{4\pi}. \end{array}$|undefined

The magnetic force outside the sphere parallel to the axis of $$x$$ equals

$\dfrac{d\mathrm{H}_{1}}{dy}-\dfrac{d\mathrm{G}_{1}}{dz},$|undefined

or

$\begin{array}{c} -3\mathrm{C}\dfrac{d}{dy}\mathrm{E}_{0}(i\lambda r)\epsilon^{ipt},\\ \\ =i\omega e\lambda\epsilon^{i\lambda a}\dfrac{d}{dy}\dfrac{\epsilon^{-i\lambda a}}{i\lambda r}\cdot\epsilon^{ipt}, \end{array}$|undefined

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10 taking the real part

$=\omega e\dfrac{d}{dy}\dfrac{\cos\big(pt-\lambda(r-a)\big)}{r}.$

Similarly the magnetic force parallel to the axis of $$x$$

$=-\omega e\dfrac{d}{dx}\dfrac{\cos\big(pt-\lambda(r-a)\big)}{r},$

and the magnetic force parallel to $$z$$ vanishes. Thus the magnetic force is the same as that which would be produced by a current-element $$\omega e\cos pt$$ or $$ev$$, $$v$$ being the velocity of the sphere (see Proc. Math. Soc. xv. p. 214).

The magnetic force inside the sphere parallel to $$x$$ equals

$\begin{array}{c} -3\mathrm{D}\epsilon^{ipt}\dfrac{d}{dy}\mathrm{S}_{0}(\lambda_{1}r),\\ \\ =3\mathrm{D}\epsilon^{ipt}\dfrac{d}{dr}\mathrm{S}_{0}(\lambda_{1}r)\dfrac{y}{r}. \end{array}$

Substituting the value for $$\mathrm{D}$$ given by equation (7), this equals

$\dfrac{\omega e}{a}p^{2}\mathrm{K}y\epsilon^{ipt};$

or, taking the real part and writing $$\lambda^{2}$$ for $$\mathrm{K}p^{2}$$,

$\dfrac{\omega e}{a^{3}}\left(\lambda^{2}a^{2}\right)y\cos pt.$|undefined

The component parallel to $$y$$ is

$-\dfrac{\omega e}{a^{3}}\left(\lambda^{2}a^{2}\right)x\cos pt,$|undefined

and the $$z$$-component vanishes. Thus the maximum magnetic force inside the sphere is

$\dfrac{\omega e}{a^{2}}\cos pt(\lambda a)^{2}.$|undefined

If $$\lambda a$$ is very small, this is very small compared with the force outside the sphere. If the velocity is uniform, $$p$$, and therefore $$\lambda=0$$, and the magnetic force inside the sphere vanishes. When there is no magnetic force inside the sphere its energy and the force acting upon it have the values assigned to them by Mr. Heaviside.

Let us next take the case where $$\lambda a$$ is small and $$\lambda_{1}a$$ large: in this case $$\mathrm{C}$$ and $$\mathrm{D}$$ have the same values as before, so that the magnetic force due to the moving sphere is the same.

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Rh We must now consider the case where $$\lambda a$$ and $$\lambda_{1}a$$ are both large; in this case we find from (6) and (7)

$\mathrm{C}=\dfrac{\omega e}{\mathrm{K}p^{2}a^{3}}\dfrac{i\lambda a}{3\epsilon^{-i\lambda a}},$|undefined

if $$\lambda_{1}a/\lambda a$$ and $$\lambda_{1}a/\lambda^{2}a^{2}$$ are both large.

Thus the magnetic force parallel to $$x$$ outside

$=\dfrac{1}{3}\dfrac{\omega e}{\lambda^{2}a^{2}}\dfrac{d}{dy}\dfrac{\cos\left\{ pt-\lambda(r-a)\right\} }{r},$|undefined

and that parallel to $$y$$

$=-\dfrac{1}{3}\dfrac{\omega e}{\lambda^{2}a^{2}}\dfrac{d}{dx}\dfrac{\cos\left\{ pt-\lambda(r-a)\right\} }{r}.$|undefined

Thus, since $$\lambda^{2}a^{2}$$ is large, the magnetic force, though in the same direction as that due to a current $$\omega v$$, is very much smaller in magnitude, and fades away to zero as $$\lambda a$$ increases without limit.

$\mathrm{D}=-\dfrac{\omega e}{3a^{2}\dfrac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)},$

The maximum magnetic force inside the sphere

$3\mathrm{D}\dfrac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)\epsilon^{ipt}=-\dfrac{\omega e}{a^{2}}\cos pt.$|undefined

Thus in this case the magnetic force just inside the sphere is equal to $$-ve$$, while that outside the sphere is very much smaller. This is a striking contrast to the previous cases, where the magnetic force inside the sphere is very small compared with that outside. Thus, in this case, when the time of the oscillation is small compared with that of the electrical oscillations the distribution of magnetic force is turned inside out. The magnetic force diminishes very rapidly as we recede from the surface of the sphere. In this case the total current parallel to the axis of $$z$$ inside the sphere is finite, for this by equation (2) equals

$\begin{array}{l} -\dfrac{ip}{\sigma}2\mathrm{D}\displaystyle\int_0^a\mathrm{S}_{0}(\lambda_{1}r)4\pi r^{2}dr,\\ \\ =\dfrac{ip}{\sigma}\dfrac{2\mathrm{D}4\pi}{\lambda_{1}^{2}}\epsilon^{ipt}a^{2}\dfrac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)\\ \\ =-\tfrac{2}{3}e\omega\cos pt. \end{array}$|undefined

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 So that if the sphere is placed in a magnetic field the force acting upon it is the same as that on a current $$\tfrac{1}{3}e\omega\cos pt$$.

When the sphere is moving with a uniform velocity $$\omega$$, equations (3) become

$-\frac{1}{\mu\mathrm{K}}\nabla^{2}\mathrm{F}=-\omega^{2}\frac{d^{2}\mathrm{F}}{dz^{2}}+\omega\frac{d^{2}\psi}{dx\ dz};$|undefined

whence

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz^{2}}\left(1-\frac{\omega^{2}}{v^{2}}\right)=0,$|undefined

where $$v$$ is the velocity of propagation of electrodynamic action through the dielectric. If we put

$z=\left\{ 1-\frac{\omega^{2}}{v^{2}}\right\} ^{\frac{1}{2}}z',$|undefined

this equation becomes

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz_{1}^{2}}=0.$|undefined

With similar equations for $$b$$ and $$c$$, we see that a solution of these equations is

$\begin{array}{l} a=k\dfrac{d}{dy}\dfrac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ b=-k\dfrac{d}{dx}\dfrac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ c=0; \end{array}$|undefined

where $$k$$ is a constant. Since, if $$\mu=1$$,

$4\pi\frac{df}{dt}=\frac{dc}{dy}-\frac{db}{dz},$

and

$\frac{df}{dt}=-\omega\frac{df}{dz},$

we have

$-4\pi\omega f=-b=k\frac{d}{dx}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$|undefined

Similarly

$-4\pi\omega g=+a=k\frac{d}{dy}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}},$|undefined

and

$-4\pi\omega\frac{dh}{dz}=\frac{db}{dx}-\frac{da}{dy}=k\frac{d}{dz_{1}^{2}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};$|undefined

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x


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Rh so that

$-4\pi wh=k\sqrt{1-\frac{\omega^{2}}{v^{2}}}\frac{d}{dz_{1}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$|undefined

The displacement across any spherical surface must $$=e$$, so that

$\int(xf+yg+zh)d\mathrm{S}=ae;$

and therefore, if $$u<v$$,

$\frac{k}{2\omega}\int_{0}^{\pi}\frac{\sin\theta\ d\theta}{\left(1-\dfrac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}} ;$|undefined

$k=\omega e\left(1-\frac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}.$|undefined

Thus the lines of magnetic forces are circles round the axis of $$z$$ and the magnitude of the force equals

$\frac{\omega e\sin\theta\left(1-\dfrac{\omega^{2}}{v^{2}}\right)}{r^{2}\left\{ 1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\theta\right\} ^{\frac{3}{2}}}$|undefined

which is Mr. Heaviside's result. If $$\omega>v$$, the integral becomes infinite, the displacement will be within a cone of semi-vertical angle $$\sin^{-1}\dfrac{v}{\omega}=\beta$$; we must therefore only integrate within this cone, and the equation to determine $$k$$ is,

$\frac{k}{2\omega}\int_{\pi}^{\sin^{-1}\frac{\omega}{v}}\frac{\sin\theta\ d\theta}{\left(1-\dfrac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}};$|undefined

or

$k=\frac{\omega e\left(1-\dfrac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}}{\cos\beta}\left(1-\frac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}.$|undefined

Thus the magnetic force

$=\frac{\omega e\left(1-\dfrac{u^{2}}{v^{2}}\right)^{\frac{1}{2}}\left(1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}}{\cos\beta\ r^{2}\left(1-\dfrac{\omega^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}$.|undefined

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 Since $$\sin \beta=v/\omega$$, this expression vanishes unless $$\theta=\beta$$, when it becomes infinite, so that the magnetic force and the electric displacement seem confined to the surface of a cone of semi-vertical angle $$\beta$$, the vertex pointing in the direction of motion.

Trinity College, Cambridge,
 * April 24, 1889.

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text: On the Magnetic Effects produced by Motion in the Electric Field


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The stories in this collection were taken down from the lips of over sixty negro story-tellers in the remote country districts of Jamaica during two visits to the island, one of six weeks in the summer of 1919, the other of five weeks in the winter of 1921. The music was all recorded during the second visit by Miss Helen Roberts, either directly from the story-teller or from a phonographic record which I had made. In this way the original style of the story-telling, which in some instances mingles story, song and dance, is as nearly as possible preserved, although much is necessarily lost in the slow process of dictation. The lively and dramatic action, the change in voice, even the rapid and elliptical vernacular, can not appear on the printed page. But the stories are set down without polish or adornment, as nearly as possible as they were told to me, and hence represent, so far as they go, a true folk art.

Although some story-tellers claimed to know "more than a hundred" stories, no one narrator gave me more than thirty, and usually not more than four or five at one interview.

To all such story-telling, as to riddling and song, the name of "Anansi story" is applied,—an appellation at least as old as 1816, when Monk Lewis in his journal describes the classes of "Nancy stories" popular in his day among the negroes as the tragical witch story and the farcical "neger-trick." The "neger-trick" harks back to slave times and is rarely heard to-day; tales of sorcery, too, are heard best from the lips of older narrators. Modern European fairy tales and animal stories (evidently unknown to Lewis) have taken their place. Two influences have dominated story-telling in Jamaica, the first an absorbing interest in the magical effect of song which, at least in the old witch tales, far surpasses that in the action of the story; the second, the conception of the spider Anansi as the trickster hero among a group of animal figures. Anansi is the culture hero of the Gold Coast,—a kind of god—, just as Turtle is of the Slave coast and Hare (our own Brer Rabbit) of the Bantu people "Anansi stories" regularly form the entertainment during wake-nights, and it is difficult not to believe that the


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xii vividness with which these animal actors take part in the story springs from the idea that they really represent the dead in the underworld whose spirits have the power, according to the native belief, of taking animal form. The head-man on a Westmoreland cattle-pen even assured me that Anansi, once a man, was now leader of the dead in this land of shades. However this may be, the development of Jamaican obeah or witchcraft has been along the same two lines of interest. Magic songs are used in communicating with the dead, and the obeah-man who sets a ghost upon an enemy often sends it in the form of some animal; hence there are animals which must be carefully handled lest they be something other than they appear.

Riddling is a favorite pastime of the Jamaica negro. Much is preserved from old African originals in the personification of common objects of yard and road-side, much is borrowed also from old English folk riddling. That this spread has been along the line of a common language is proved by the fact that only a dozen parallels occur in Mason's Spanish collection from Porto Rico, at least ten of which are quoted by Espinosa from New Mexico, while of collections from English-speaking neighbors, fourteen out of fifty-five riddles collected in South Carolina and nine out of twenty-one from Andros Island are found also in Jamaica. Particular patterns are set for Jamaica riddling into which the phrasing falls with a rhythmical swing careless of rhyme,—"My father has in his yard" and "Going up to town." The giving of a riddle is regularly preceded by a formula drawn from old English sources—

generally abbreviated into

The art is practised as a social amusement, groups forming in which each person in the circle must propound riddles until his supply is exhausted or his riddle unguessed.

My own work as a collector in this engrossing field of Jamaican folk-lore owes much to those collectors who have preceded me and who have enjoyed a longer and more intimate acquaintance than has been possible for me with the people and their idiom;—to Monk Lewis, a true folk-lorist, whose "Journal" of 1816 is of the greatest interest to-day, to Mr. Walter Jekyll and


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Rh his excellent volume of songs and stories in the Folk-lore Publications of 1907, and to the writers of nursery tales, Mrs. Milne-Home, Pamela Smith, and Mrs. W. E. Wilson (Wona). I take this opportunity also to acknowledge most gratefully the many courtesies for which I am indebted during my visits to the island. I particularly wish to thank Professor Frank Cundall for his advice and cooperation, and for the use of the invaluable West India library connected with the Jamaica Institute in Kingston where I was able to consult books not easily to be found in library collections. To the Hon. and Mrs. Coke-Kerr, to Mrs. Harry Farquharson and to the Rev. and Mrs. Ashton I am gratefully indebted for many courtesies in the task of finding reliable native informants. To these informants themselves,—to Simeon Falconer, William Forbes, George Parkes, and a score of others I owe thanks for their ready response to my interest. In America also I wish to thank Mrs. Elsie Clews Parsons for suggestions as to method and for the use of her valuable bibliography and Mrs. Louise Dennis Hand for help with Spanish collections, and to express my grateful obligations to Professor Franz Boas for his patient editing and valuable bibliographical suggestions.


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Rh CHAPTER VIII.

. Dodge was about forty years of age, although he looked ten years older; he stood six feet in his mocassins; as for boots or shoes, his feet had long been deficient of those luxuries. He was spare and wiry, his features sharp, decided, and angular; his nose aquiline and slightly drooping at the point; his deep-set eyes were grey and piercing. His invariable dress was a blue flannel shirt, a pair of loose canvass trousers, which had once been white, but were something of the color of mahogany, with a fine polish about the knees, they were studded with stains of blood and a few burnt holes. Round his waist was a broad belt fastened with a massive silver buckle: suspended from the belt was a knife, a pouch made from the skin of a platybus, containing a pipe, tobacco, and a tinder-box. His head was covered by a broad-brimmed Manilla hat, such as a quaker might have envied. His beard was long and straggling, but the moustache which mingled with it was unexceptionable. He was altogether a strange mortal even for the bush, but he was a universal favorite, and a welcome guest wherever and whenever he appeared.

Dodge's father was an English gentleman, and he himself might have been one, had he not, imbued with the spirit of adventure, left his home too early in life to render permanent and decided the advantages derivable from his position. Through the rust of bush habits and feelings, which had grown about him during many years spent in the woods, there were still to be discovered traits which clearly distinguished him from ordinary bushmen. The misfortunes which beset him he accounted for,

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Rh ''dat. of the rel. particle þe, “because of which,” antecedent'' ǣg-wearde).

þes, þēos, þis, demon, adj.,, 411, 484, etc.; ''inst. neut. þȳs, 1395; acc. sg. m., þisne, 75, þysne, 1771; gen. sg. m. and neut. þisses, 1216, þysses, 790, 806; dat. sg. neut.'' þissum, 1169,

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 но небо коптить, — пора приниматься за дело! Let's stop loafing and get busy!

нёбо palate.

небольшой rather small. У меня небольшая комната и малюсенькая кухня. I have a rather small room and a tiny kitchen. • not great. Я небольшой охотник до классической музыки. I'm not a great lover of classical music. • short. Нам оставалось проехать очень небольшое расстояние. There was only a short distance left for us to travel. • little. Теперь у нас остановка за небольшим. There's little to stop us now. □ Все, за небольшим исключением, участвовали в сборе. With few exceptions everyone contributed to the collection. • Ему лет пятьдесят с небольшим. He's just over fifty.

небось probably. В ваших краях, небось, сейчас жарко. It's probably very hot now where you come from. • surely. Он, небось, знает, что мы его ждём. He surely must know we're waiting for him. • I suppose. Небось, устали с дороги? I suppose you're tired after the trip.

небрежность (F) negligence. Это просто непростительная небрежность. That's unforgivable negligence. • carelessness. Все её ошибки — от небрежности. All her mistakes are due to carelessness.

небрежный careless. У него небрежный стиль. He has a careless style. • casually. Она мне это сказала самым небрежным тоном. She told it to me very casually. □ небрежно carelessly. Это сделано очень небрежно. This work was done very carelessly. — Она одевается очень небрежно. She dresses very carelessly.

неважный (sh -жна) poor. Апетит у меня неважный. I have a poor appetite. • indifferent. Ученик он неважный. He's an indifferent student. □ неважно not well. Он неважно себя чувствует. He doesn't feel very well. • not good. Дела идут неважно. Things are not so good. • unimportant. Не беспокойтесь: это неважно. Never mind; it's unimportant. • it doesn't matter. Неважно, что он плохо говорит по-русски, если он хороший инженер. It doesn't matter that he doesn't speak Russian well as long as he's a good engineer. очень неважно bad. Кормят тут очень неважно. The food is bad here.

невежа (M, F) crude person. Господи, что за невежа! God, what a crude person he is!

невежда (M) ignoramus. Он — круглый невежда. He's a complete ignoramus.

невежественный ignorant. Ну, добро бы это говорил невежественный человек, а ведь он университет кончил. It would be bad enough for an ignorant person to have said it, but he's a college graduate.

невежливый impolite.

невероятный incredible. Они мчались с невероятной быстротой. They were moving at incredible speed. □ невероятно unbelievably. Ему невероятно везёт. He's unbelievably lucky.

невеста bride. Невеста была в белом платье. The bride wore a white gown. • fiancee. Где сейчас ваша невеста? Where is your fiancee now? □ Как ваша сестрёнка выросла, совсем невеста! Well, your little sister is quite a young lady now!

невестка sister-in-law (brother's wife), daughter-in-law.

невиданный (/as if ppp of видать/) not seen. Это было совершенно невиданное зрелище. You've never seen anything like it.

невинный innocent. Он оказался тут невинной жертвой. He turned out to be an innocent victim. • harmless. Это была невинная ложь. It was a harmless he. — Это совсем невинный напиток! That's a perfectly harmless drink.

невозможный impossible. Вы меня поставили в невозможное положение. You put me in an impossible situation. — У него совершенно невозможный характер. He's absolutely impossible. • the impossible. Не требуйте от него невозможного. Don't ask him to do the impossible. □ У нас тут стоит невозможная жара. We're having a terrible heat wave.

невообразимый inconceivable. У них дома невообразимый беспорядок. The disorder at their place is inconceivable.

невпопад. □ Он на всё отвечал невпопад. His answers were way off the point.

невредимый safe. Это просто чудо, что он остался цел и невредим. It's nothing short of a miracle that he came out of it, safe and sound.

невыносимый unbearable. От невыносимой жары у него разболелась голова. The unbearable heat gave him a headache.

негде (§2S) no room. У нас негде поставить столько чемоданов. We have no room for so many valises. • no place. Тут негде переночевать. There's no place to spend the night here. • not … anywhere, nowhere. Здесь негде достать машину, придётся пойти пешком. You can't get a car anywhere around here, so we'll have to walk. □ Ну и комната — повернуться негде. What a room! You can't even turn around in it.

него See он.

негодный worthless. Материал-то, оказывается, совершенно негодный. It turns out that the material is altogether worthless. • unfit. Эта вода негодна для питья. This water is unfit for drinking. • naughty. Ах ты, негодная девчонка, опять все конфеты съела. You naughty girl, you ate up all the candy again. □ Он совершенно негодный работник. As a worker he's a total loss.

негодяй rascal. Он оказался отъявленным негодяем. He turned out to be an out-and-out rascal.

неграмотность (F) illiteracy.

неграмотный illiterate. Моя мать была неграмотная. My mother was illiterate. □ Он медицински неграмотный человек. He doesn't know anything about medicine.

недавний recent. Недавние события показали всю серьёзность положения. Recent events have shown the seriousness of the situation. — Я расскажи вам про один недавний сличай. I'll tell you about something that happened recently. □ недавно recently. Мы познакомились с ним совсем недавно. We only met him recently. — Он недавно приехал в Москву. He came to Moscow recently. • a while back. Ещё недавно тут никто не жил. Only a while back there was no one living here. совсем недавно the other day. Ещё совсем недавно он говорил иначе. Only the other day he spoke differently.

недалеко not far. Я живу недалеко от завода. I don't live


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nytte tealde.Þær genehost brægd eorl Bēowulfesealde lāfe, wolde frēa-drihtnesfeorh ealgian, mǣres þēodnes,ðǣr hīe meahton swā. Hīe þæt ne wiston,þā hīe gewin drugon, heard-hicgendehilde-mecgas, ond on healfa gehwonehēawan þōhton, sāwle sēcan:þone syn-scaðan ǣnig ofer eorþanīrenna cyst, gūð-billa nān,grētan nolde; ac hē sige-wǣpnum*forsworen hæfde, ecga gehwylcre.Scolde his aldor-gedāl on ðǣm dægeþysses līfes earmlīc wurðan,ond se ellor-gāst on fēonda gewealdfeor sīðian. Ðā þæt onfunde,sē þe fela ǣror mōdes myrðemanna cynne fyrene gefremede,hē fāg wið God, þæt him se līc-homalǣstan nolde, ac hine se mōdegamǣg Hygelāces hæfde be honda;wæs gehwæþer ōðrum lifigende lāð.Līc-sār gebād atol ǣglǣca;him on eaxle wearð syn-dolh sweotol;seonowe onsprungon, burston bān-locan.Bēowulfe wearð gūð-hrēð gyfeþe;scolde Grendel þonan feorh-sēoc flēonunder fen-hleoðu, }}

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