Superphysics Superphysics
Part 3

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

by Hendrik Lorentz Icon
11 minutes  • 2284 words

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 (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

(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.

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