Part 1

# Can we ascribe gravity to a certain state of the aether?

by Lorentz

A theory of universal attraction, founded on the state of the aether, would take the simplest form

It is believed that only electrically charged particles or ions, are directly acted on by the aether.

This leads to the idea that every particle of ponderable matter might consist of 2 ions with equal opposite charges. Or at least:

• they might contain 2 such ions
• gravity might be the result of the forces experienced by these ions.

So many phenomena have been explained by a theory of ions. This idea seems to be more admissible than before.

Electromagnetic disturbances in the aether might possibly be the cause of gravitation. This is possible if they could penetrate all bodies without diminishing in intensity.

Electric vibrations of extremely small wavelength possess this property.

But what action there would be between 2 ions, if the aether were traversed in all directions by trains of electric waves of small wavelength?

The above ideas are not new.

Every physicist knows Le Sage’s theory in which innumerable small corpuscula are supposed to move with great velocities. This produces gravitation by their impact against the coarser particles of ordinary ponderable matter.

This theory is not in harmony with modern physical views.

But it was found that a pressure against a body may be produced as well by trains of electric waves and by rays of light (e.g. as by moving projectiles).

When penetrating x-rays were discovered, it was natural to replace Le Sage’s corpuscula by vibratory motions.

There could be radiations, far more penetrating than even the X-rays. This would account for gravity which is independent of all intervening matter.

But this theory of rapid vibrations as a cause of gravitation cannot be accepted.

### Part 2

Let an ion carrying a charge e1 and a mass be at the point P(x,y,z). It may be subject or not to an elastic force, proportional to the displacement and driving it back to P, as soon as it has left this position.

Let the aether be traversed by electromagnetic vibrations, the dielectric displacement being denoted by d , and the magnetic force by H. Then the ion will be acted on by a force.

4 π V 2 e d


whose direction changes continually, and whose components are X = 4 π V 2 e d x , Y = 4 π V 2 e d y , Z = 4 π V 2 e d z (1)

V means the velocity of light.

By the action of the force (1) the ion will be made to vibrate about its original position P, the displacement (x, y,z) being determined by well known differential equations.

We shall confine ourselves to simple harmonic vibrations with frequency n.

All our formulae will then contain the factor cos ⁡ n t or sin ⁡ n t, and the forced vibrations of the ion may be represented by expressions of the form x = a e d x − b e d ˙ x , y = a e d y − b e d ˙ y , z = a e d z − b e d ˙ z (2)

with certain constant coefficients a and b. The terms with d ˙ x , d ˙ y have been introduced in order to indicate that the phase of the forced vibration differs from that of the force (X,Y,Z); this will be the case as soon as there is a resistance, proportional to the velocity, and the coefficient b may then be shown to be positive.

One cause of a resistance lies in the reaction of the aether, called forth by the radiation of which the vibrating ion itself becomes the centre, a reaction which determines at the same time an apparent increase of the mass of the particle. We shall suppose however that we have kept in view this reaction in establishing the equations of motion, and in assigning their values to the coefficients a and b. Then, in what follows, we need only consider the forces due to the state of the aether, in so far as it not directly produced by the ion itself.

Since the formulae (2) contain e as a factor, the coefficients a and b will be independent of the charge; their sign will be the same for a negative ion and for a positive one. Now, as soon as the ion has shifted from its position of equilibrium, new forces come into play. In the first place, the force 4 π V 2 e d {\displaystyle 4\pi V^{2}e{\mathfrak {d}}} {\displaystyle 4\pi V^{2}e{\mathfrak {d}}} will have changed a little, because, for the new position, d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}} will be somewhat different from what it was at the point P. We may express this by saying that, in addition to the force (1), there will be a new one with the components, 4 π V 2 e ( x ∂ d x ∂ x + y ∂ d x ∂ y + z ∂ d x ∂ z ) {\displaystyle 4\pi V^{2}e\left({\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathsf {z}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)} {\displaystyle 4\pi V^{2}e\left({\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathsf {z}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)} etc. (3)

In the second place, in consequence of the velocity of vibration, there will be an electromagnetic force with the components e ( y ˙ H z − z ˙ H y ) {\displaystyle e\left({\dot {\mathsf {y}}}{\mathfrak {H}}{\mathsf {z}}-{\dot {\mathsf {z}}}{\mathfrak {H}}{\mathsf {y}}\right)} {\displaystyle e\left({\dot {\mathsf {y}}}{\mathfrak {H}}{\mathsf {z}}-{\dot {\mathsf {z}}}{\mathfrak {H}}{\mathsf {y}}\right)}, etc. (4)

If, as we shall suppose, the displacement of the ion be very small, compared with the wave-length, the forces (3) and (4) are much smaller than the force (1); since they are periodic — with the frequency 2n, — they will give rise to new vibrations of the particle. We shall however omit the consideration of these slight vibrations, and examine only the mean values of the forces (3) and (4), calculated for a rather long lapse of time, or, what amounts to the same thing, for a full period 2 π n {\displaystyle {\tfrac {2\pi }{n}}} {\displaystyle {\tfrac {2\pi }{n}}}.

§ 3.

This mean force will be 0 if the ion is alone in a field in which the propagation of waves takes place equally in all directions.

It will be otherwise, as soon as a second ion Q has been placed in the neighbourhood of P.

Then, in consequence of the vibrations emitted by Q after it has been itself put in motion, there may be a force on P, of course in the direction of the line QP.

In computing the value of this force, one finds many terms, which depend in different ways on the distance r.

We shall retain those which are inversely proportional to r or r2.

But we shall neglect all terms varying inversely as the higher powers of r.

The influence of these, compared with that of the first mentioned terms will be of the order λ r {\displaystyle {\tfrac {\lambda }{r}}} {\displaystyle {\tfrac {\lambda }{r}}} , if λ {\displaystyle \lambda } \lambda is the wave-length, and we shall suppose this to be a very small fraction.

We shall also omit all terms containing such factors as cos ⁡ 2 π k r λ {\displaystyle \cos \ 2\pi k{\tfrac {r}{\lambda }}} {\displaystyle \cos \ 2\pi k{\tfrac {r}{\lambda }}} or sin ⁡ 2 π k r λ {\displaystyle \sin \ 2\pi k{\tfrac {r}{\lambda }}} {\displaystyle \sin \ 2\pi k{\tfrac {r}{\lambda }}} (k a moderate number). These reverse their signs by a very small change in r; they will therefore disappear from the resultant force, as soon as, instead of single particles P and Q, we come to consider systems of particles with dimensions many times greater than the wave-length.

In applying the above formulae to the ion P, it is sufficient, to take for d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}} and H {\displaystyle {\mathfrak {H}}} {\displaystyle {\mathfrak {H}}} the vectors that would exist if P were removed from the field. In each of these vectors two parts are to be distinguished. We shall denote by d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}} and H 1 {\displaystyle {\mathfrak {H}}{1}} {\displaystyle {\mathfrak {H}}{1}} the parts existing independently of Q, and by d 2 {\displaystyle {\mathfrak {d}}{2}} {\displaystyle {\mathfrak {d}}{2}} and H 2 {\displaystyle {\mathfrak {H}}{2}} {\displaystyle {\mathfrak {H}}{2}} the parts due to the vibrations of this ion.

Let Q be taken as origin of coordinates, QP as axis of x, and let us begin with the terms in (2) having the coefficient a.

To these corresponds a force on P, whose first component is 4 π V 2 e 2 a ( d x ∂ d x ∂ x + d y ∂ d x ∂ y + d z ∂ d x ∂ z ) + e 2 a ( d ˙ y H z − d ˙ z H y ) {\displaystyle 4\pi V^{2}e^{2}a\left({\mathfrak {d}}{\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathfrak {d}}{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathfrak {d}}{\mathsf {\mathsf {z}}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)+e^{2}a\left({\mathfrak {\dot {d}}}{\mathsf {y}}{\mathfrak {H}}{\mathsf {z}}-{\mathfrak {\dot {d}}}{\mathsf {z}}{\mathfrak {H}}{\mathsf {y}}\right)} {\displaystyle 4\pi V^{2}e^{2}a\left({\mathfrak {d}}{\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathfrak {d}}{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathfrak {d}}{\mathsf {\mathsf {z}}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)+e^{2}a\left({\mathfrak {\dot {d}}}{\mathsf {y}}{\mathfrak {H}}{\mathsf {z}}-{\mathfrak {\dot {d}}}{\mathsf {z}}{\mathfrak {H}}{\mathsf {y}}\right)}. (5)

Since we have only to deal with the mean values for a full period, we may write for the last term

− e 2 a ( d y H ˙ z − d z H ˙ y ) {\displaystyle -e^{2}a\left({\mathfrak {d}}{y}{\mathfrak {\dot {H}}}{z}-{\mathfrak {d}}{}{z}{\mathfrak {\dot {H}}}{y}\right)} {\displaystyle -e^{2}a\left({\mathfrak {d}}{y}{\mathfrak {\dot {H}}}{z}-{\mathfrak {d}}{}{z}{\mathfrak {\dot {H}}}{y}\right)},

and if, in this expression, H ˙ y {\displaystyle {\mathfrak {\dot {H}}}{y}} {\displaystyle {\mathfrak {\dot {H}}}{y}} and H ˙ z {\displaystyle {\mathfrak {\dot {H}}}{z}} {\displaystyle {\mathfrak {\dot {H}}}{z}} be replaced by

4 π V 2 ( ∂ d z ∂ x − ∂ d x ∂ z ) {\displaystyle 4\pi V^{2}\left({\frac {\partial {\mathfrak {d}}{\mathsf {z}}}{\partial x}}-{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)} {\displaystyle 4\pi V^{2}\left({\frac {\partial {\mathfrak {d}}{\mathsf {z}}}{\partial x}}-{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial z}}\right)} and 4 π V 2 ( ∂ d x ∂ y − ∂ d y ∂ x ) {\displaystyle 4\pi V^{2}\left({\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}-{\frac {\partial {\mathfrak {d}}{\mathsf {y}}}{\partial x}}\right)} {\displaystyle 4\pi V^{2}\left({\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}-{\frac {\partial {\mathfrak {d}}{\mathsf {y}}}{\partial x}}\right)},

(5) becomes 2 π V 2 e 2 a ∂ ( b 2 ) ∂ x {\displaystyle 2\pi V^{2}e^{2}a{\frac {\partial \left({\mathfrak {b^{2}}}\right)}{\partial x}}} {\displaystyle 2\pi V^{2}e^{2}a{\frac {\partial \left({\mathfrak {b^{2}}}\right)}{\partial x}}}, (6)

where d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}} is the numerical value of the dielectric displacement.

Now, d 2 {\displaystyle {\mathfrak {d}}^{2}} {\displaystyle {\mathfrak {d}}^{2}} will consist of three parts, the first being d 1 2 {\displaystyle {\mathfrak {d}}{1}^{2}} {\displaystyle {\mathfrak {d}}{1}^{2}}, the second d 2 2 {\displaystyle {\mathfrak {d}}{2}^{2}} {\displaystyle {\mathfrak {d}}{2}^{2}} and the third depending on the combination of d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}} and d 2 {\displaystyle {\mathfrak {d}}{2}} {\displaystyle {\mathfrak {d}}{2}}.

Evidently, the value of (6), corresponding to the first part, will be 0.

As to the second part, it is to be remarked that the dielectric displacement, produced by Q, is a periodic function of the time. At distant points the amplitude takes the form c r {\displaystyle {\tfrac {c}{r}}} {\displaystyle {\tfrac {c}{r}}}, where c is independent of r. The mean value of d 2 {\displaystyle {\mathfrak {d}}^{2}} {\displaystyle {\mathfrak {d}}^{2}} for a full period is 1 2 c 2 r 2 {\displaystyle {\tfrac {1}{2}}{\tfrac {c^{2}}{r^{2}}}} {\displaystyle {\tfrac {1}{2}}{\tfrac {c^{2}}{r^{2}}}} and differentiating this with regard to x or to r, we should get r 3 {\displaystyle r^{3}} {\displaystyle r^{3}} in the denominator.

The terms in (6) which correspond to the part

2 ( d 1 x d 2 x + d 1 y d 2 y + d 1 z d 2 z ) {\displaystyle 2\left({\mathfrak {d}}{1{\mathsf {x}}}{\mathfrak {d}}{2{\mathsf {x}}}+{\mathfrak {d}}{1{\mathsf {y}}}{\mathfrak {d}}{2{\mathsf {y}}}+{\mathfrak {d}}{1{\mathsf {z}}}{\mathfrak {d}}{2{\mathsf {z}}}\right)} {\displaystyle 2\left({\mathfrak {d}}{1{\mathsf {x}}}{\mathfrak {d}}{2{\mathsf {x}}}+{\mathfrak {d}}{1{\mathsf {y}}}{\mathfrak {d}}{2{\mathsf {y}}}+{\mathfrak {d}}{1{\mathsf {z}}}{\mathfrak {d}}{2{\mathsf {z}}}\right)}

in d 2 {\displaystyle {\mathfrak {d}}^{2}} {\displaystyle {\mathfrak {d}}^{2}}, may likewise be neglected. Indeed, if these terms are to contain no factors such as cos ⁡ 2 π k r λ {\displaystyle \cos \ 2\pi k{\tfrac {r}{\lambda }}} {\displaystyle \cos \ 2\pi k{\tfrac {r}{\lambda }}} or sin ⁡ 2 π k r λ {\displaystyle \sin \ 2\pi k{\tfrac {r}{\lambda }}} {\displaystyle \sin \ 2\pi k{\tfrac {r}{\lambda }}}, there must be between d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}} and d 2 {\displaystyle {\mathfrak {d}}{2}} {\displaystyle {\mathfrak {d}}{2}}, either no phase-difference at all, or a difference which is independent of r. This condition can only be fulfilled, if a system of waves, proceeding in the direction of QP, is combined with the vibrations excited by Q, in so far as this ion is put in motion by that system itself. Then, the two vectors d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}} and d 2 {\displaystyle {\mathfrak {d}}{2}} {\displaystyle {\mathfrak {d}}{2}} will have a common direction perpendicular to QP, say that of the axis of y, and they will be of the form d 1 y = q c o s n ( t − x V + ε 1 ) {\displaystyle {\mathfrak {d}}_{1{\mathsf {y}}}=q\ cos\ n\ \left(t-{\frac {x}{V}}+\varepsilon {1}\right)} {\displaystyle {\mathfrak {d}}{1{\mathsf {y}}}=q\ cos\ n\ \left(t-{\frac {x}{V}}+\varepsilon _{1}\right)}

d 2 y = c r c o s n ( t − x V + ε 2 ) {\displaystyle {\mathfrak {d}}_{2{\mathsf {y}}}={\frac {c}{r}}\ cos\ n\ \left(t-{\frac {x}{V}}+\varepsilon {2}\right)} {\displaystyle {\mathfrak {d}}{2{\mathsf {y}}}={\frac {c}{r}}\ cos\ n\ \left(t-{\frac {x}{V}}+\varepsilon _{2}\right)}

The mean value of d 1 y d 2 y {\displaystyle {\mathfrak {d}}{1{\mathsf {y}}}{\mathfrak {d}}{2{\mathsf {y}}}} {\displaystyle {\mathfrak {d}}{1{\mathsf {y}}}{\mathfrak {d}}{2{\mathsf {y}}}} is

1 2 q c r cos ⁡ n ( ε 1 − ε 2 ) {\displaystyle {\frac {1}{2}}{\frac {qc}{r}}\cos n\ \left(\varepsilon _{1}-\varepsilon _{2}\right)} {\displaystyle {\frac {1}{2}}{\frac {qc}{r}}\cos n\ \left(\varepsilon _{1}-\varepsilon _{2}\right)},

and its differential coefficient with regard to x has r 2 {\displaystyle r^{2}} {\displaystyle r^{2}} in the denominator. It ought therefore to be retained, were it not for the extremely small intensity of the systems of waves which give rise to such a result. In fact, by the restriction imposed on them as to their direction, these waves form no more than a very minute part of the whole motion.

§ 4. So, it is only the terms in (2), with the coefficient b, with which we are concerned. The corresponding forces are − 4 π V 2 e 2 b ( d ˙ x ∂ d x ∂ x + d ˙ y ∂ d x ∂ y + d ˙ z ∂ d x ∂ z ) {\displaystyle -4\pi V^{2}e^{2}b\left({\mathfrak {\dot {d}}}{\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathfrak {\dot {d}}}{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathfrak {\dot {d}}}{\mathsf {z}}{\frac {\partial {\mathfrak {d}}{x}}{\partial z}}\right)} {\displaystyle -4\pi V^{2}e^{2}b\left({\mathfrak {\dot {d}}}{\mathsf {x}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial x}}+{\mathfrak {\dot {d}}}{\mathsf {y}}{\frac {\partial {\mathfrak {d}}{\mathsf {x}}}{\partial y}}+{\mathfrak {\dot {d}}}{\mathsf {z}}{\frac {\partial {\mathfrak {d}}{x}}{\partial z}}\right)} (7)

and − e 2 b ( d y ¨ H z − d z ¨ H y ) {\displaystyle -e^{2}b\left({\ddot {{\mathfrak {d}}{\mathsf {y}}}}\ {\mathfrak {H}}{\mathsf {z}}-{\ddot {{\mathfrak {d}}{\mathsf {z}}}}\ {\mathfrak {H}}{\mathsf {y}}\right)} {\displaystyle -e^{2}b\left({\ddot {{\mathfrak {d}}{\mathsf {y}}}}\ {\mathfrak {H}}{\mathsf {z}}-{\ddot {{\mathfrak {d}}{\mathsf {z}}}}\ {\mathfrak {H}}{\mathsf {y}}\right)}. (8) If Q were removed, these forces together would be 0, as has already been remarked. On the other hand, the force (8) taken by itself, would then likewise be 0. Indeed, its value is n 2 e 2 b ( d y H z − d z H y ) {\displaystyle n^{2}e^{2}b\left({\mathfrak {d}}{\mathsf {y}}\ {\mathfrak {H}}{\mathsf {z}}-{\mathfrak {d}}{\mathsf {z}}\ {\mathfrak {H}}{\mathsf {y}}\right)} {\displaystyle n^{2}e^{2}b\left({\mathfrak {d}}{\mathsf {y}}\ {\mathfrak {H}}{\mathsf {z}}-{\mathfrak {d}}{\mathsf {z}}\ {\mathfrak {H}}{\mathsf {y}}\right)}. (9)

or, by Poynting’s theorem n 2 e 2 b V 2 S x {\displaystyle {\tfrac {n^{2}e^{2}b}{V^{2}}}S_{\mathsf {x}}} {\displaystyle {\tfrac {n^{2}e^{2}b}{V^{2}}}S_{\mathsf {x}}}, if S x {\displaystyle S_{\mathsf {x}}} {\displaystyle S_{\mathsf {x}}} be the flow of energy in a direction parallel to the axis of x. Now, it is clear that, in the absence of Q, any plane must be traversed in the two directions by equal amounts of energy.

We conclude that the force (7), in so far as it depends on the part, ( d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}}), is 0, and from this it follows that the total value of (7) will vanish, because the part arising from the combination of ( d 1 {\displaystyle {\mathfrak {d}}{1}} {\displaystyle {\mathfrak {d}}{1}}) and ( d 2 {\displaystyle {\mathfrak {d}}{2}} {\displaystyle {\mathfrak {d}}{2}}), as well as that which is solely due to tho vibrations of Q, are 0. As to the first part, this may be shown by a reasoning similar to that used at the end of, the preceding §. For the second part, the proof is as follows.

The vibrations excited by Q in any point A of the surrounding aether are represented by expressions of the form

1 r ϑ cos ⁡ n ( t − r V + ε ) {\displaystyle {\frac {1}{r}}\vartheta \cos n\left(t-{\frac {r}{V}}+\varepsilon \right)} {\displaystyle {\frac {1}{r}}\vartheta \cos n\left(t-{\frac {r}{V}}+\varepsilon \right)},

where ϑ {\displaystyle \vartheta } {\displaystyle \vartheta } depends on the direction of the line QA, and r denotes the length of this line. If, in differentiating such expressions, we wish to avoid in the denominator powers of r, higher than the first — and this is necessary, in order that (7) may remain free from powers higher than the second — 1 r {\displaystyle {\tfrac {1}{r}}} {\displaystyle {\tfrac {1}{r}}} and ϑ {\displaystyle \vartheta } {\displaystyle \vartheta } have to be treated as constants. Moreover, the factors ϑ {\displaystyle \vartheta } {\displaystyle \vartheta } are such, that the vibrations are perpendicular to the line QA. If, now, A coincides with P, and QA with the axis of x, in the expression for bx we shall have ϑ = 0 {\displaystyle \vartheta =0} {\displaystyle \vartheta =0}, and since this factor is not to be differentiated, all terms in (7) will vanish.

Thus, the question reduces itself to (8) or (9).

If, in this last expression, we take for d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}} and H {\displaystyle {\mathfrak {H}}} {\displaystyle {\mathfrak {H}}} their real values, modified as they are by the motion of Q, we may again write for the force

n 2 e 2 b V 2 S x {\displaystyle {\frac {n^{2}e^{2}b}{V^{2}}}S_{\mathsf {x}}} {\displaystyle {\frac {n^{2}e^{2}b}{V^{2}}}S_{\mathsf {x}}};

this time, however, we have to understand by S x {\displaystyle S_{\mathsf {x}}} {\displaystyle S_{\mathsf {x}}} the flow of energy as it is in the actual case.

Now, it is clear that, by our assumptions, the flow of energy must be symmetrical all around Q; hence, if an amount E of energy traverses, in the outward direction, a spherical surface described around Q as centre with radius r, we shall have

S x = E 4 π r 2 {\displaystyle S_{\mathsf {x}}={\frac {E}{4\pi r^{2}}}} {\displaystyle S_{\mathsf {x}}={\frac {E}{4\pi r^{2}}}},

and the force on P will be

K = n 2 e 2 b E 4 π V 2 r 2 {\displaystyle K={\frac {n^{2}e^{2}b\ E}{4\pi V^{2}r^{2}}}} {\displaystyle K={\frac {n^{2}e^{2}b\ E}{4\pi V^{2}r^{2}}}}.

It will have the direction of QP prolonged.

In the space surrounding Q the state of the aether will be stationary; hence, two spherical surfaces enclosing this particle must be traversed by equal quantities of energy. The quantity E will be independent of r, and the force K inversely proportional to the square of the distance.

If the vibrations of Q were opposed by no other resistance but that which results from radiation, the total amount of electro-magnetic energy enclosed by a surface surrounding Q would remain constant; E and K would then both be 0.

If, on the contrary, in addition to the just mentioned resistance, there were a resistance of a different kind, the vibrations of Q would be accompanied by a continual loss of electro-magnetic energy; less energy would leave the space within one of the spherical surfaces than would enter that space. E would be negative, and, since b is positive, there would be attraction.

It would be independent of the signs of the charges of P and Q.

The circumstance however, that this attraction could only exist, if in some way or other electromagnetic energy were continually disappearing, is so serious a difficulty, that what has been said cannot be considered as furnishing an explanation of gravitation.

Nor is this the only objection that can be raised. If the mechanism of gravitation consisted in vibrations which cross the aether with the velocity of light, the attraction ought to be modified by the motion of the celestial bodies to a much larger extent than astronomical observations make it possible to admit.

#### Latest Articles

Supersociology
##### The Age of the Universe
Material Superphysics
Supereconomics
##### The Elastic Theory of Gravity
Material Superphysics