Part 5

Considerations on Gravitation by Lorentz Icon

September 28, 2021

The following are insufficient to account for universal attraction:

  • the states of the aether
  • electromagnetic laws

Yet we can try to establish a theory of gravity similar to electricity with some features in common with it.

I use Mossotti’s idea which has been afterwards accepted by Wilhelm Weber and Zöllner.

They say every particle of ponderable matter consists of two oppositely electrified particles. Between 2 particles of matter, there will be:

  • 4 electric forces
  • 2 attractions between the charges of different
  • 2 repulsions between those of equal signs

Mossotti supposes the attractions to be somewhat greater than the repulsions. Gravity is the difference between the two. Such a difference might exist when an action of a specific electric nature is not exerted.

To harmonize this with electrical science, we must regard the 4 forces of Mossotti as the effect of states in the aether which are called forth by the positive and negative ions.

A positive ion, as well as a negative one, is the centre of a dielectric displacement.

These two displacements have the same nature, so that, if in opposite directions and of equal magnitude, they wholly destroy each other.

If gravitation is to be included in the theory, this view must be modified.

If the actions exerted by positive and negative ions depended on vector-quantities of the same kind, in such a way that all phenomena in the neighbourhood of a pair of ions with opposite charges were determined by the resulting vector, then electric actions could only be absent, if this resulting vector were 0, but, if such were the case, no other actions could exist; a gravitation, i.e. a force in the absence of an electric field, would be impossible.

The two disturbances in the aether, produced by positive and negative ions, are of a somewhat different nature. So that, even if they are represented by equal and opposite vectors, the state of the aether is not the natural one.

This corresponds in a sense to Mossotti’s idea that positive and negative charges differ from each other to a larger extent, than may be expressed by the signs + and .

After having attributed to each of the two states an independent and separate existence, we may assume that, though both able to act on positive and negative ions, the one has more power over the positive particles and the other over the negative ones.

This difference will lead us to the same result that Mossotti attained by means of the supposed inequality of the attractive and the repulsive forces.

§ 6.

Each of the two disturbances of the aether is propagated with the velocity of light. By itself, it obeys the ordinary laws of the electromagnetic field.

These laws are expressed in the simplest form if, besides the dielectric displacement d, we consider the magnetic force H, both together determining, as we shall now say, one state of the aether or one field. In accordance with this, I shall introduce two pairs of vectors, the one d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}}, H {\displaystyle {\mathfrak {H}}} {\displaystyle {\mathfrak {H}}} belonging to the field that is produced by the positive ions, whereas the other pair d ′ {\displaystyle {\mathfrak {d}}’} {\displaystyle {\mathfrak {d}}’}, H ′ {\displaystyle {\mathfrak {H}}’} {\displaystyle {\mathfrak {H}}’} serve to indicate the state of the aether which is called into existence by the negative ions. I shall write down two sets of equations, one for d {\displaystyle {\mathfrak {d}}} {\displaystyle {\mathfrak {d}}}, H {\displaystyle {\mathfrak {H}}} {\displaystyle {\mathfrak {H}}}, the other for d ′ , H ′ {\displaystyle {\mathfrak {d}}’,{\mathfrak {H}}’} {\displaystyle {\mathfrak {d}}’,{\mathfrak {H}}’}, and having the form which I have used in former papers[1] for the equations of the electromagnetic field, and which is founded on the assumption that the ions are perfectly permeable to the aether and that they can be displaced without dragging the aether along with them.

I shall immediately take this general case of moving particles.

Let us further suppose the charges to be distributed with finite volume-density, and let the units in which these are expressed be chosen in such a way that, in a body which exerts no electrical actions, the total amount of the positive charges has the same numerical value as that of the negative charges.

Let ϱ {\displaystyle \varrho } \varrho be the density of the positive, and ϱ ′ {\displaystyle \varrho ‘} {\displaystyle \varrho ‘} that of the negative charges, the first number being positive and the second negative.

Let v {\displaystyle {\mathfrak {v}}} {\displaystyle {\mathfrak {v}}} (or v ′ {\displaystyle {\mathfrak {v}}’} {\displaystyle {\mathfrak {v}}’}) be the velocity of an ion.

Then the equations for the state ( d , H ) {\displaystyle ({\mathfrak {d}},{\mathfrak {H}})} {\displaystyle ({\mathfrak {d}},{\mathfrak {H}})} are[2] D i v d = ϱ D i v H = 0 R o t H = 4 π ϱ v + 4 π d ˙ 4 π V 2 R o t d = − H ˙ ; } {\displaystyle \left.{\begin{array}{c}Div\ {\mathfrak {d}}=\varrho \Div\ {\mathfrak {H}}=0\Rot\ {\mathfrak {H}}=4\pi \varrho {\mathfrak {v}}+4\pi {\mathfrak {\dot {d}}}\4\pi V^{2}Rot\ {\mathfrak {d}}=-{\mathfrak {\dot {H}}};\end{array}}\right}} {\displaystyle \left.{\begin{array}{c}Div\ {\mathfrak {d}}=\varrho \Div\ {\mathfrak {H}}=0\Rot\ {\mathfrak {H}}=4\pi \varrho {\mathfrak {v}}+4\pi {\mathfrak {\dot {d}}}\4\pi V^{2}Rot\ {\mathfrak {d}}=-{\mathfrak {\dot {H}}};\end{array}}\right}} (I) and those for the state ( d ′ , H ′ ) {\displaystyle ({\mathfrak {d}}’,{\mathfrak {H}}’)} {\displaystyle ({\mathfrak {d}}’,{\mathfrak {H}}’)} D i v d ′ = ϱ ′ D i v H ′ = 0 R o t H ′ = 4 π ϱ ′ v ′ + 4 π d ˙ ′ 4 π V 2 R o t d ′ = − H ˙ ′ ; } {\displaystyle \left.{\begin{array}{c}Div\ {\mathfrak {d}}’=\varrho ‘\Div\ {\mathfrak {H}}’=0\Rot\ {\mathfrak {H}}’=4\pi \varrho ‘{\mathfrak {v}}’+4\pi {\mathfrak {\dot {d}}}’\4\pi V^{2}Rot\ {\mathfrak {d}}’=-{\mathfrak {\dot {H}}}’;\end{array}}\right}} {\displaystyle \left.{\begin{array}{c}Div\ {\mathfrak {d}}’=\varrho ‘\Div\ {\mathfrak {H}}’=0\Rot\ {\mathfrak {H}}’=4\pi \varrho ‘{\mathfrak {v}}’+4\pi {\mathfrak {\dot {d}}}’\4\pi V^{2}Rot\ {\mathfrak {d}}’=-{\mathfrak {\dot {H}}}’;\end{array}}\right}}. (II)

In the ordinary theory of electromagnetism, the force acting on a particle, moving with velocity v {\displaystyle {\mathfrak {v}}} {\displaystyle {\mathfrak {v}}}, is

4 π V 2 d + [ v . H ] {\displaystyle 4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}\right]} {\displaystyle 4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}\right]}

per unit charge.[3]

In the modified theory, we shall suppose that a positively electrified particle with charge e experiences a force k 1 = α { 4 π V 2 d + [ v . H ] } e {\displaystyle k_{1}=\alpha \left{4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}\right]\right}e} {\displaystyle k_{1}=\alpha \left{4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}\right]\right}e} (10)

on account of the field ( d , H ) {\displaystyle ({\mathfrak {d}},{\mathfrak {H}})} {\displaystyle ({\mathfrak {d}},{\mathfrak {H}})}, and a force k 2 = β { 4 π V 2 d ′ + [ v . H ′ ] } e {\displaystyle k_{2}=\beta \left{4\pi V^{2}{\mathfrak {d}}’+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}’\right]\right}e} {\displaystyle k_{2}=\beta \left{4\pi V^{2}{\mathfrak {d}}’+\left[{\mathfrak {v\ }}.\ {\mathfrak {H}}’\right]\right}e} (11)

on account of the field ( d ′ , H ′ ) {\displaystyle ({\mathfrak {d}}’,{\mathfrak {H}}’)} {\displaystyle ({\mathfrak {d}}’,{\mathfrak {H}}’)}, the positive coefficients α {\displaystyle \alpha } \alpha and β {\displaystyle \beta } \beta having slightly different values.

For the forces, exerted on a negatively charged particle I shall write k 3 = β { 4 π V 2 d + [ v ′ . H ] } e ′ {\displaystyle k_{3}=\beta \left{4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v}}’\ .\ {\mathfrak {H}}\right]\right}e’} {\displaystyle k_{3}=\beta \left{4\pi V^{2}{\mathfrak {d}}+\left[{\mathfrak {v}}’\ .\ {\mathfrak {H}}\right]\right}e’} (12)

and k 4 = α { 4 π V 2 d ′ + [ v ′ . H ′ ] } e ′ , {\displaystyle k_{4}=\alpha \left{4\pi V^{2}{\mathfrak {d}}’+\left[{\mathfrak {v}}’\ .\ {\mathfrak {H}}’\right]\right}e’,} {\displaystyle k_{4}=\alpha \left{4\pi V^{2}{\mathfrak {d}}’+\left[{\mathfrak {v}}’\ .\ {\mathfrak {H}}’\right]\right}e’,} (13)

expressing by these formulae that e is acted on by ( d , H {\displaystyle {\mathfrak {d}},{\mathfrak {H}}} {\displaystyle {\mathfrak {d}},{\mathfrak {H}}}) in the same way as e’ by ( d ′ , H ′ {\displaystyle {\mathfrak {d}}’,{\mathfrak {H}}’} {\displaystyle {\mathfrak {d}}’,{\mathfrak {H}}’}), and vice versa.

§ 7.

Let us next consider the actions exerted by a pair of oppositely charged ions, placed close to each other, and remaining so during their motion. For convenience of mathematical treatment, we may even reason as if the two charges penetrated each other, so that, if they are equal, ϱ ′ = − ϱ {\displaystyle \varrho ‘=-\varrho } {\displaystyle \varrho ‘=-\varrho }.

On the other hand v ′ = v {\displaystyle {\mathfrak {v}}’={\mathfrak {v}}} {\displaystyle {\mathfrak {v}}’={\mathfrak {v}}} ; hence, by (I) and (II),

d ′ = − d {\displaystyle {\mathfrak {d}}’=-{\mathfrak {d}}} {\displaystyle {\mathfrak {d}}’=-{\mathfrak {d}}} and H ′ = − H {\displaystyle {\mathfrak {H}}’=-{\mathfrak {H}}} {\displaystyle {\mathfrak {H}}’=-{\mathfrak {H}}}.

Now let us put in the field, produced by the pair of ious, a similar pair with charges e and e ′ = − e {\displaystyle e’=-e} {\displaystyle e’=-e}, and moving with the common velocity v {\displaystyle {\mathfrak {v}}} {\displaystyle {\mathfrak {v}}}. Then, by (10) — (13),

k 2 = − β α k 1 , k 3 = − β α k 1 , k 4 = k 1 . {\displaystyle k_{2}=-{\frac {\beta }{\alpha }}k_{1},\ k_{3}=-{\frac {\beta }{\alpha }}k_{1},\ k_{4}=k_{1}.} {\displaystyle k_{2}=-{\frac {\beta }{\alpha }}k_{1},\ k_{3}=-{\frac {\beta }{\alpha }}k_{1},\ k_{4}=k_{1}.}

The total force on the positive particle will be

k 1 + k 2 = k 1 ( 1 − β α ) {\displaystyle k_{1}+k_{2}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle k_{1}+k_{2}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}

and that on the negative ion

k 3 + k 4 = k 1 ( 1 − β α ) {\displaystyle k_{3}+k_{4}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle k_{3}+k_{4}=k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}.

These forces being equal and having the same direction, there is no force tending to separate the two ions, as would be the case in an electric field. Nevertheless, the pair is acted on by a resultant force

2 k 1 ( 1 − β α ) {\displaystyle 2k_{1}\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle 2k_{1}\left(1-{\frac {\beta }{\alpha }}\right)}.

If now β be somewhat larger than α {\displaystyle \alpha } \alpha, the factor 2 ( 1 − β α ) {\displaystyle 2\left(1-{\frac {\beta }{\alpha }}\right)} {\displaystyle 2\left(1-{\frac {\beta }{\alpha }}\right)} have a certain negative value –ε, and our result may be expressed as follows:

If we wish to determine the action between two ponderable bodies, we may first consider the forces existing between the positive ions in the one and the positive ions in the other. We then have to reverse the direction of these forces, and to multiply them by the factor ε. Of course, we are led in this way to Newton’s law of gravitation.

The assumption that all ponderable matter is composed of positive and negative ions is no essential part of the above theory. We might have confined ourselves to the supposition that the state of the aether which is the cause of gravitation is propagated in a similar way as that which exists in the electromagnetic field.

Instead of introducing two pairs of vectors ( d , H {\displaystyle {\mathfrak {d,\ H}}} {\displaystyle {\mathfrak {d,\ H}}}) and ( d ′ , H ′ {\displaystyle {\mathfrak {d’,H’}}} {\displaystyle {\mathfrak {d’,H’}}}), both of which come into play in the electromagnetic actions, as well as in the phenomenon of gravitation, we might have assumed one pair for the electromagnetic field and one for universal attraction.

For these latter vectors, say d , H {\displaystyle {\mathfrak {d,\ H}}} {\displaystyle {\mathfrak {d,\ H}}}, we should then have established the equations (I), ϱ {\displaystyle \varrho } \varrho being the density of ponderable matter, and for the force acting on unit mass, we should have put

− η { 4 π V 2 d + [ v . H ] } {\displaystyle -\eta \left{4\pi V^{2}{\mathfrak {d}}+{\mathfrak {\left[v.\ H\right]}}\right}} {\displaystyle -\eta \left{4\pi V^{2}{\mathfrak {d}}+{\mathfrak {\left[v.\ H\right]}}\right}},

where η {\displaystyle \eta } \eta is a certain positive coefficient.

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