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the big one
2024-08-27 21:48:20 -05:00
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Translational movement of the light ether
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Author: Wilhelm Vienna
Title: About the questions that concern the translational movement of the light ether
out of: Annals of Physics 301 (supplement), 1898, pp. I-XVIII
Editor: G. and E. Wiedemann
Date of creation: 1898
Publisher: John Ambr. Bart
Place of publication: Leipzig
Source: Gallica , Commons
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Complete! This text has been read twice on the basis of source correction . The notation follows the original text.
Index page
[ i ]
On the questions concerning the translational movement of the light ether; by W. Vienna.
(Representation for the 70th meeting of German natural scientists and doctors in Düsseldorf, 1898; Section Physics.)
The question of whether the light ether takes part in the movements of bodies or not, and whether mobility can be attributed to it at all, has occupied physicists for a long time and there are countless assumptions and conjectures that one has to make about the properties of the carrier of electromagnetic phenomena held. However, there can be no doubt that everything we know about the ether is contained in Maxwell 's theory of electromagnetism and everything else belongs to the realm of pure speculation. Accordingly, I have not set myself the task of providing a literary report on the innumerable theories that have the light ether as their subject, but have endeavored to highlight the questions that we have to answer on the basis of Maxwell 's theory regarding the mobility of the ether have to provide.
If we make the assumption that the ether has mobility, further questions immediately arise, namely whether this movement requires energy expenditure, i.e. whether the ether is to be attributed inert mass, and then whether the ether is also set in motion by the movement of solid bodies . The latter does not appear to be the case according to many experiments, especially after the extensive experiments of Lodge , which were carried out with rapidly rotating metal masses or in the vicinity of high-speed circular saws.
We will first compare the assumptions as to whether mobility can be attributed to the ether or not and then move on to the discussion of the empirical facts.
[ ii ]
The assumption of the mobility of the aether.
The tendency to bring the properties of the ether into agreement with those of ponderable matter has led to the assumption that the ether can carry out movements in the manner of a liquid, although not a single experiment indicates the existence of such movements. But if one ascribes mobility to the ether, then, as Hertz first noted, it follows strictly from Maxwell 's theory that under the influence of the pressure forces generated by a variable electromagnetic system, it must carry out movements that can be calculated, if one makes certain assumptions about the inertia of the ether.
Helmholtz gave the basic principles for the calculation of these flows under the assumption that the inertia and compressibility of the aether is zero. However, he did not give any specific examples that would allow this theory to be tested against experience, and I will therefore give here two examples from which some conclusions can be drawn as to the meaning of these assumptions.
Currents in the ether are only excited by electromagnetic tensions when the field is neither static nor stationary, i.e. when the conditions of time are still changeable.
As a first example I introduce an electrified colon, which carries equal quantities of positive and negative electricity at a very small distance from each other, which increase proportionally with time.
If we designate the coordinates with x, y , z, the time with t , the components of the electrical forces with Maxwell 's differential equations
A d L d t = ∂ Z ∂ y − ∂ Y ∂ e.g A d X d t = ∂ M ∂ e.g − ∂ N ∂ y A d M d t = ∂ X ∂ e.g − ∂ Z ∂ x A d Y d t = ∂ N ∂ x − ∂ L ∂ e.g A d N d t = ∂ Y ∂ x − ∂ X ∂ y A d Z d t = ∂ L ∂ y − ∂ M ∂ x ∂ L ∂ x + ∂ M ∂ y + ∂ N ∂ e.g = 0 ∂ X ∂ x + ∂ Y ∂ y + ∂ Z ∂ e.g = 0. {\displaystyle {\begin{array}{ll}A{\frac {dL}{dt}}={\frac {\partial Z}{\partial y}}-{\frac {\partial Y}{\partial z}}&A{\frac {dX}{dt}}={\frac {\partial M}{\partial z}}-{\frac {\partial N}{\partial y}}\\\\A{\frac {dM}{dt}}={\frac {\partial X}{\partial z}}-{\frac {\partial Z}{\partial x}}&A{\frac {dY}{dt}}={\frac {\partial N}{\partial x}}-{\frac {\partial L}{\partial z}}\\\\A{\frac {dN}{dt}}={\frac {\partial Y}{\partial x}}-{\frac {\partial X}{\partial y}}&A{\frac {dZ}{dt}}={\frac {\partial L}{\partial y}}-{\frac {\partial M}{\partial x}}\\\\{\frac {\partial L}{\partial x}}+{\frac {\partial M}{\partial y}}+{\frac {\partial N}{\partial z}}=0\ &{\frac {\partial X}{\partial x}}+{\frac {\partial Y}{\partial y}}+{\frac {\partial Z}{\partial z}}=0.\end{array}}} {\displaystyle {\begin{array}{ll}A{\frac {dL}{dt}}={\frac {\partial Z}{\partial y}}-{\frac {\partial Y}{\partial z}}&A{\frac {dX}{dt}}={\frac {\partial M}{\partial z}}-{\frac {\partial N}{\partial y}}\\\\A{ \frac {dM}{dt}}={\frac {\partial X}{\partial z}}-{\frac {\partial Z}{\partial x}}&A{\frac {dY}{dt}} ={\frac {\partial N}{\partial x}}-{\frac {\partial L}{\partial z}}\\\\A{\frac {dN}{dt}}={\frac { \partial Y}{\partial x}}-{\frac {\partial X}{\partial y}}&A{\frac {dZ}{dt}}={\frac {\partial L}{\partial y} }-{\frac {\partial M}{\partial x}}\\\\{\frac {\partial L}{\partial x}}+{\frac {\partial M}{\partial y}}+ {\frac {\partial N}{\partial z}}=0\ &{\frac {\partial X}{\partial x}}+{\frac {\partial Y}{\partial y}}+{\ frac {\partial Z}{\partial z}}=0.\end{array}}}
[ iii ] We satisfy these equations using the following expressions:
X = ∂ 2 φ ∂ e.g ∂ x L = − A ∂ 2 φ ∂ y ∂ t Y = ∂ 2 φ ∂ e.g ∂ y M = A ∂ 2 φ ∂ x ∂ t Z = ∂ 2 φ ∂ e.g 2 N = 0. {\displaystyle {\begin{array}{lcl}X={\frac {\partial ^{2}\varphi }{\partial z\ \partial x}}&&L=-A{\frac {\partial ^{2}\varphi }{\partial y\ \partial t}}\\\\Y={\frac {\partial ^{2}\varphi }{\partial z\ \partial y}}&&M=A{\frac {\partial ^{2}\varphi }{\partial x\ \partial t}}\\\\Z={\frac {\partial ^{2}\varphi }{\partial z^{\ 2}}}&&N=0.\end{array}}} {\displaystyle {\begin{array}{lcl}X={\frac {\partial ^{2}\varphi }{\partial z\ \partial x}}&&L=-A{\frac {\partial ^{2 }\varphi }{\partial y\ \partial t}}\\\\Y={\frac {\partial ^{2}\varphi }{\partial z\ \partial y}}&&M=A{\frac { \partial ^{2}\varphi }{\partial x\ \partial t}}\\\\Z={\frac {\partial ^{2}\varphi }{\partial z^{\ 2}}}&&N =0.\end{array}}}
Let a be a constant and r 2 = x 2 + y 2 + e.g 2 {\displaystyle r^{2}=x^{2}+y^{2}+z^{2}} {\displaystyle r^{2}=x^{2}+y^{2}+z^{2}} , ϱ = x 2 + y 2 {\displaystyle \varrho =x^{2}+y^{2}} {\displaystyle \varrho =x^{2}+y^{2}} , φ = a t / r {\displaystyle \varphi =at/r} {\displaystyle \varphi =at/r} . The components of the electrical forces are then the partial derivatives of the function
a t ∂ ∂ e.g ( 1 r ) . {\displaystyle at{\frac {\partial }{\partial z}}\left({\frac {1}{r}}\right).} {\displaystyle at{\frac {\partial }{\partial z}}\left({\frac {1}{r}}\right).}
This is the potential of an electric colon in the point r = 0 {\displaystyle r=0} {\displaystyle r=0} with the positive and negative charge a t / l {\displaystyle at/l} {\displaystyle at/l} . The line connecting both charges l {\displaystyle l} {\displaystyle l} is parallel to the z -axis. The components of Poynting 's energy flow are proportional to the quantities
P = Z M − Y N = A a 2 t x ( 1 r 6 − 3 e.g 2 r 8th ) {\displaystyle {\mathfrak {P}}=ZM-YN=Aa^{2}tx\left({\frac {1}{r^{6}}}-{\frac {3z^{2}}{r^{8}}}\right)} {\displaystyle {\mathfrak {P}}=ZM-YN=Aa^{2}tx\left({\frac {1}{r^{6}}}-{\frac {3z^{2}}{ r^{8}}}\right)}
Q = X N − Z L = A a 2 t y ( 1 r 6 − 3 e.g 2 r 8th ) {\displaystyle {\mathfrak {Q}}=XN-ZL=Aa^{2}ty\left({\frac {1}{r^{6}}}-{\frac {3z^{2}}{r^{8}}}\right)} {\displaystyle {\mathfrak {Q}}=XN-ZL=Aa^{2}ty\left({\frac {1}{r^{6}}}-{\frac {3z^{2}}{ r^{8}}}\right)}
R = Y L − X M = 3 A a 2 t e.g x 2 + y 2 r 8th . {\displaystyle {\mathfrak {R}}=YL-XM=3Aa^{2}tz{\frac {x^{2}+y^{2}}{r^{8}}}.} {\displaystyle {\mathfrak {R}}=YL-XM=3Aa^{2}tz{\frac {x^{2}+y^{2}}{r^{8}}}.}
Now let's set
x = ϱ cos ⁡ ϑ y = ϱ sin ⁡ ϑ d x d t = α = d ϱ d t cos ⁡ ϑ − ϱ sin ⁡ ϑ d ϑ d t d ϑ d t = η d ϱ d t = ζ d y d t = β = d ϱ d t sin ⁡ ϑ + ϱ cos ⁡ ϑ d ϑ d t d e.g d t = γ , {\displaystyle {\begin{array}{lllll}x=\varrho \cos \vartheta &&y=\varrho \sin \vartheta &&{\frac {dx}{dt}}=\alpha ={\frac {d\varrho }{dt}}\cos \vartheta -\varrho \sin \vartheta {\frac {d\vartheta }{dt}}\\\\{\frac {d\vartheta }{dt}}=\eta &&{\frac {d\varrho }{dt}}=\zeta &&{\frac {dy}{dt}}=\beta ={\frac {d\varrho }{dt}}\sin \vartheta +\varrho \cos \vartheta {\frac {d\vartheta }{dt}}\\\\&&&&{\frac {dz}{dt}}=\gamma ,\end{array}}} {\displaystyle {\begin{array}{lllll}x=\varrho \cos \vartheta &&y=\varrho \sin \vartheta &&{\frac {dx}{dt}}=\alpha ={\frac {d\varrho }{dt}}\cos \vartheta -\varrho \sin \vartheta {\frac {d\vartheta }{dt}}\\\\{\frac {d\vartheta }{dt}}=\eta &&{\ frac {d\varrho }{dt}}=\zeta &&{\frac {dy}{dt}}=\beta ={\frac {d\varrho }{dt}}\sin \vartheta +\varrho \cos \ vartheta {\frac {d\vartheta }{dt}}\\\\&&&&{\frac {dz}{dt}}=\gamma ,\end{array}}}
this is what the equation of incompressibility requires
∂ α ∂ x + ∂ β ∂ y + ∂ γ ∂ e.g = 0 , {\displaystyle {\frac {\partial \alpha }{\partial x}}+{\frac {\partial \beta }{\partial y}}+{\frac {\partial \gamma }{\partial z}}=0,} {\displaystyle {\frac {\partial \alpha }{\partial x}}+{\frac {\partial \beta }{\partial y}}+{\frac {\partial \gamma }{\partial z}} =0,}
that
α = ∂ ψ ∂ e.g x ϱ 2 − η y β = ∂ ψ ∂ e.g y ϱ 2 + η x γ = − 1 ϱ ∂ ψ ∂ ϱ {\displaystyle \alpha ={\frac {\partial \psi }{\partial z}}{\frac {x}{\varrho ^{2}}}-\eta y\quad \beta ={\frac {\partial \psi }{\partial z}}{\frac {y}{\varrho ^{2}}}+\eta x\quad \gamma =-{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}} {\displaystyle \alpha ={\frac {\partial \psi }{\partial z}}{\frac {x}{\varrho ^{2}}}-\eta y\quad \beta ={\frac {\ partial \psi }{\partial z}}{\frac {y}{\varrho ^{2}}}+\eta x\quad \gamma =-{\frac {1}{\varrho }}{\frac { \partial \psi }{\partial \varrho }}}
is if we assume that because of the symmetry around the z -axis the sizes η , ζ , γ {\displaystyle \eta ,\zeta ,\gamma } {\displaystyle \eta ,\zeta ,\gamma } independent of ϑ {\displaystyle \vartheta } {\displaystyle \vartheta } are.
The differential equations derived by Helmholtz , which express that the electromagnetic [ iv ] Currents caused by tensions in turn produce electromagnetic forces that balance themselves with those acting from outside
(1) { 0 = ∂ P ∂ x + A [ ∂ P ∂ t + β ( ∂ P ∂ y − ∂ Q ∂ x ) − γ ( ∂ R ∂ x − ∂ P ∂ e.g ) ] 0 = ∂ P ∂ y + A [ ∂ Q ∂ t + γ ( ∂ Q ∂ e.g − ∂ R ∂ y ) − α ( ∂ P ∂ y − ∂ Q ∂ x ) ] 0 = ∂ P ∂ e.g + A [ ∂ R ∂ t + α ( ∂ R ∂ x − ∂ P ∂ e.g ) − β ( ∂ Q ∂ e.g − ∂ R ∂ y ) ] {\displaystyle {\begin{cases}0={\frac {\partial P}{\partial x}}+A\left[{\frac {\partial {\mathfrak {P}}}{\partial t}}+\beta \left({\frac {\partial {\mathfrak {P}}}{\partial y}}-{\frac {\partial {\mathfrak {Q}}}{\partial x}}\right)-\gamma \left({\frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z}}\right)\right]\\0={\frac {\partial P}{\partial y}}+A\left[{\frac {\partial {\mathfrak {Q}}}{\partial t}}+\gamma \left({\frac {\partial {\mathfrak {Q}}}{\partial z}}-{\frac {\partial {\mathfrak {R}}}{\partial y}}\right)-\alpha \left({\frac {\partial {\mathfrak {P}}}{\partial y}}-{\frac {\partial {\mathfrak {Q}}}{\partial x}}\right)\right]\\0={\frac {\partial P}{\partial z}}+A\left[{\frac {\partial {\mathfrak {R}}}{\partial t}}+\alpha \left({\frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z}}\right)-\beta \left({\frac {\partial {\mathfrak {Q}}}{\partial z}}-{\frac {\partial {\mathfrak {R}}}{\partial y}}\right)\right]\end{cases}}} {\displaystyle {\begin{cases}0={\frac {\partial P}{\partial x}}+A\left[{\frac {\partial {\mathfrak {P}}}{\partial t}} +\beta \left({\frac {\partial {\mathfrak {P}}}{\partial y}}-{\frac {\partial {\mathfrak {Q}}}{\partial x}}\right) -\gamma \left({\frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z}}\right) \right]\\0={\frac {\partial P}{\partial y}}+A\left[{\frac {\partial {\mathfrak {Q}}}{\partial t}}+\gamma \ left({\frac {\partial {\mathfrak {Q}}}{\partial z}}-{\frac {\partial {\mathfrak {R}}}{\partial y}}\right)-\alpha \ left({\frac {\partial {\mathfrak {P}}}{\partial y}}-{\frac {\partial {\mathfrak {Q}}}{\partial x}}\right)\right]\ \0={\frac {\partial P}{\partial z}}+A\left[{\frac {\partial {\mathfrak {R}}}{\partial t}}+\alpha \left({\ frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z}}\right)-\beta \left({\ frac {\partial {\mathfrak {Q}}}{\partial z}}-{\frac {\partial {\mathfrak {R}}}{\partial y}}\right)\right]\end{cases} }}
Here P means the hydrostatic pressure.
Let us put the above values ​​of into these equations P , Q , R , α , β , γ {\displaystyle {\mathfrak {P,\ Q,\ R,}}\alpha ,\beta ,\gamma } {\displaystyle {\mathfrak {P,\Q,\R,}}\alpha ,\beta ,\gamma } one, so we get
0 = ∂ P ∂ ϱ + ϱ A 2 a 2 ( 3 ϱ 2 r 8th − 2 r 6 − 6 e.g t r 8th 1 ϱ ∂ ψ ∂ ϱ ) , {\displaystyle 0={\frac {\partial P}{\partial \varrho }}+\varrho A^{2}a^{2}\left({\frac {3\varrho ^{2}}{r^{8}}}-{\frac {2}{r^{6}}}-{\frac {6zt}{r^{8}}}{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}\right),} {\displaystyle 0={\frac {\partial P}{\partial \varrho }}+\varrho A^{2}a^{2}\left({\frac {3\varrho ^{2}}{r ^{8}}}-{\frac {2}{r^{6}}}-{\frac {6zt}{r^{8}}}{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}\right),}
0 = ∂ P ∂ e.g + e.g A 2 a 2 ( 3 ϱ 2 r 8th − 6 t r 8th − ∂ ψ ∂ e.g ) . {\displaystyle 0={\frac {\partial P}{\partial z}}+zA^{2}a^{2}\left({\frac {3\varrho ^{2}}{r^{8}}}-{\frac {6t}{r^{8}}}-{\frac {\partial \psi }{\partial z}}\right).} {\displaystyle 0={\frac {\partial P}{\partial z}}+zA^{2}a^{2}\left({\frac {3\varrho ^{2}}{r^{8 }}}-{\frac {6t}{r^{8}}}-{\frac {\partial \psi }{\partial z}}\right).}
The angular velocity η {\displaystyle \eta } {\displaystyle \eta } has completely fallen out, so it does not need to have a value other than zero.
If we eliminate P from this , we get
(2) ϱ e.g − t ∂ ψ ∂ ϱ + 8th e.g t r 2 ( e.g ∂ ψ ∂ ϱ − ϱ ∂ ψ ∂ e.g ) = 0. {\displaystyle \varrho z-t{\frac {\partial \psi }{\partial \varrho }}+{\frac {8zt}{r^{2}}}\left(z{\frac {\partial \psi }{\partial \varrho }}-\varrho {\frac {\partial \psi }{\partial z}}\right)=0.} {\displaystyle \varrho zt{\frac {\partial \psi }{\partial \varrho }}+{\frac {8zt}{r^{2}}}\left(z{\frac {\partial \psi } {\partial \varrho }}-\varrho {\frac {\partial \psi }{\partial z}}\right)=0.}
One can see immediately from this equation that ψ {\displaystyle \psi } {\displaystyle \psi } the factor 1 / t {\displaystyle 1/t} {\displaystyle 1/t} must contain. For t = 0 {\displaystyle t=0} {\displaystyle t=0} the charge of the electric colon is zero. So at the moment the charge begins, the currents in the ether would become infinite.
Since Maxwell 's differential equations are completely fulfilled, there is no reason to exclude such a charge that increases from zero in proportion to time.
A solution to the differential equation (2) is
ψ = r 2 e.g 10 t {\displaystyle \psi ={\frac {r^{2}z}{10t}}} {\displaystyle \psi ={\frac {r^{2}z}{10t}}}
It follows
ζ = 1 ϱ ∂ ψ ∂ e.g = ( 2 e.g 2 + r 2 10 t ) 1 ϱ , {\displaystyle \zeta ={\frac {1}{\varrho }}{\frac {\partial \psi }{\partial z}}=\left({\frac {2z^{2}+r^{2}}{10t}}\right){\frac {1}{\varrho }},} {\displaystyle \zeta ={\frac {1}{\varrho }}{\frac {\partial \psi }{\partial z}}=\left({\frac {2z^{2}+r^{2 }}{10t}}\right){\frac {1}{\varrho }},}
− γ = 1 ϱ ∂ ψ ∂ ϱ = 2 e.g 10 t . {\displaystyle -\gamma ={\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}={\frac {2z}{10t}}.} {\displaystyle -\gamma ={\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}={\frac {2z}{10t}}.}
[ v ] So the ether would flow parallel to the streamlines,
in which the planes laid by the z -axis are the surfaces r 2 e.g = c O n s t . {\displaystyle r^{2}z={\rm {const.}}} {\displaystyle r^{2}z={\rm {const.}}} cut. However, such a flow is hydrodynamically impossible because of the speed γ {\displaystyle \gamma } {\displaystyle \gamma } for ϱ = 0 {\displaystyle \varrho =0} {\displaystyle \varrho =0} becomes infinite.
As a second case, we consider an electrified point with the charge e , which moves through space with the constant speed v . This case is completely treated by Heaviside and his solution gives the following values ​​of the electric and magnetic forces, related to a coordinate system fixed in the electrified point in whose x -axis the movement takes place.
X = 1 v ∂ U ∂ x ( 1 − A 2 v 2 ) , Y = 1 v ∂ U ∂ y , Z = 1 v ∂ U ∂ e.g , M = − A ∂ U ∂ e.g , N = A ∂ U ∂ y , L = 0. {\displaystyle {\begin{array}{lclcc}X={\frac {1}{v}}{\frac {\partial U}{\partial x}}\left(1-A^{2}v^{2}\right),&&Y={\frac {1}{v}}{\frac {\partial U}{\partial y}},&&Z={\frac {1}{v}}{\frac {\partial U}{\partial z}},\\\\M=-A{\frac {\partial U}{\partial z}},&&N=A{\frac {\partial U}{\partial y}},&&L=0.\end{array}}} {\displaystyle {\begin{array}{lclcc}X={\frac {1}{v}}{\frac {\partial U}{\partial x}}\left(1-A^{2}v^ {2}\right),&&Y={\frac {1}{v}}{\frac {\partial U}{\partial y}},&&Z={\frac {1}{v}}{\frac { \partial U}{\partial z}},\\\\M=-A{\frac {\partial U}{\partial z}},&&N=A{\frac {\partial U}{\partial y} },&&L=0.\end{array}}}
U = e v r 2 − A 2 v 2 ϱ 2 ϱ 2 = y 2 + e.g 2 {\displaystyle U={\frac {ev}{\sqrt {r^{2}-A^{2}v^{2}\varrho ^{2}}}}\quad \varrho ^{2}=y^{2}+z^{2}} {\displaystyle U={\frac {ev}{\sqrt {r^{2}-A^{2}v^{2}\varrho ^{2}}}}\quad \varrho ^{2}=y ^{2}+z^{2}} .
Then the sizes result P , Q , R {\displaystyle {\mathfrak {P,\ Q,\ R}}} {\displaystyle {\mathfrak {P,\Q,\R}}}
P = A ϱ 2 ( r 2 − A 2 v 2 ϱ 2 ) 3 , Q = − A x y ( r 2 − A 2 v 2 ϱ 2 ) 3 , {\displaystyle {\mathfrak {P}}={\frac {{\mathfrak {A}}\varrho ^{2}}{\left(r^{2}-A^{2}v^{2}\varrho ^{2}\right)^{3}}},\quad {\mathfrak {Q}}=-{\frac {{\mathfrak {A}}xy}{\left(r^{2}-A^{2}v^{2}\varrho ^{2}\right)^{3}}},} {\displaystyle {\mathfrak {P}}={\frac {{\mathfrak {A}}\varrho ^{2}}{\left(r^{2}-A^{2}v^{2}\ varrho ^{2}\right)^{3}}},\quad {\mathfrak {Q}}=-{\frac {{\mathfrak {A}}xy}{\left(r^{2}-A ^{2}v^{2}\varrho ^{2}\right)^{3}}},}
R = − A x e.g ( r 2 − A 2 v 2 ϱ 2 ) 3 , {\displaystyle {\mathfrak {R}}=-{\frac {{\mathfrak {A}}xz}{\left(r^{2}-A^{2}v^{2}\varrho ^{2}\right)^{3}}},} {\displaystyle {\mathfrak {R}}=-{\frac {{\mathfrak {A}}xz}{\left(r^{2}-A^{2}v^{2}\varrho ^{2 }\right)^{3}}},}
A = e 2 v A ( 1 − A 2 v 2 ) . {\displaystyle {\mathfrak {A}}=e^{2}vA\left(1-A^{2}v^{2}\right).} {\displaystyle {\mathfrak {A}}=e^{2}vA\left(1-A^{2}v^{2}\right).}
Let's sit again
α = − 1 ϱ ∂ ψ ∂ ϱ , β = ∂ ψ ∂ x y ϱ 2 + η e.g , γ = ∂ ψ ∂ x e.g ϱ 2 − η y , {\displaystyle \alpha =-{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }},\quad \beta ={\frac {\partial \psi }{\partial x}}{\frac {y}{\varrho ^{2}}}+\eta z,\quad \gamma ={\frac {\partial \psi }{\partial x}}{\frac {z}{\varrho ^{2}}}-\eta y,} {\displaystyle \alpha =-{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }},\quad \beta ={\frac {\partial \psi }{ \partial x}}{\frac {y}{\varrho ^{2}}}+\eta z,\quad \gamma ={\frac {\partial \psi }{\partial x}}{\frac {z }{\varrho ^{2}}}-\eta y,} S = A ( r 2 − A 2 v 2 ϱ 2 ) 3 . {\displaystyle {\mathfrak {S}}={\frac {\mathfrak {A}}{\left(r^{2}-A^{2}v^{2}\varrho ^{2}\right)^{3}}}.} {\displaystyle {\mathfrak {S}}={\frac {\mathfrak {A}}{\left(r^{2}-A^{2}v^{2}\varrho ^{2}\right) ^{3}}}.}
so we get from equations (1)
0 = ∂ P ∂ ϱ + A ( − v ∂ S ∂ x x ϱ − ϱ v S + ∂ ψ ∂ ϱ [ 3 S + x ∂ S ∂ x + ϱ ∂ S ∂ ϱ ] ) , {\displaystyle 0={\frac {\partial P}{\partial \varrho }}+A\left(-v{\frac {\partial {\mathfrak {S}}}{\partial x}}x\varrho -\varrho v{\mathfrak {S}}+{\frac {\partial \psi }{\partial \varrho }}\left[3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial x}}+\varrho {\frac {\partial {\mathfrak {S}}}{\partial \varrho }}\right]\right),} {\displaystyle 0={\frac {\partial P}{\partial \varrho }}+A\left(-v{\frac {\partial {\mathfrak {S}}}{\partial x}}x\varrho -\varrho v{\mathfrak {S}}+{\frac {\partial \psi }{\partial \varrho }}\left[3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial x}}+\varrho {\frac {\partial {\mathfrak {S}}}{\partial \varrho }}\right]\right),}
0 = ∂ P ∂ x + A ( − v ∂ S ∂ x ϱ 2 + ∂ ψ ∂ x [ 3 S + x ∂ S ∂ e.g + ϱ ∂ S ∂ ϱ ] ) . {\displaystyle 0={\frac {\partial P}{\partial x}}+A\left(-v{\frac {\partial {\mathfrak {S}}}{\partial x}}\varrho ^{2}+{\frac {\partial \psi }{\partial x}}\left[3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial z}}+\varrho {\frac {\partial {\mathfrak {S}}}{\partial \varrho }}\right]\right).} {\displaystyle 0={\frac {\partial P}{\partial x}}+A\left(-v{\frac {\partial {\mathfrak {S}}}{\partial x}}\varrho ^{ 2}+{\frac {\partial \psi }{\partial x}}\left[3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial z} }+\varrho {\frac {\partial {\mathfrak {S}}}{\partial \varrho }}\right]\right).}
[ vi ] We name U {\displaystyle {\mathfrak {U}}} {\displaystyle {\mathfrak {U}}} the size
3 S + x ∂ S ∂ x + ϱ ∂ S ∂ ϱ , {\displaystyle 3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial x}}+\varrho {\frac {\partial {\mathfrak {S}}}{\partial \varrho }},} {\displaystyle 3{\mathfrak {S}}+x{\frac {\partial {\mathfrak {S}}}{\partial x}}+\varrho {\frac {\partial {\mathfrak {S}}} {\partial \varrho }},}
This results in the elimination of P
(3) 0 = v ϱ ∂ U ∂ x + ∂ ψ ∂ x ∂ U ∂ ϱ − ∂ ψ ∂ ϱ ∂ U ∂ x . {\displaystyle 0=v\varrho {\frac {\partial {\mathfrak {U}}}{\partial x}}+{\frac {\partial \psi }{\partial x}}{\frac {\partial {\mathfrak {U}}}{\partial \varrho }}-{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial {\mathfrak {U}}}{\partial x}}.} {\displaystyle 0=v\varrho {\frac {\partial {\mathfrak {U}}}{\partial x}}+{\frac {\partial \psi }{\partial x}}{\frac {\partial {\mathfrak {U}}}{\partial \varrho }}-{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial {\mathfrak {U}}}{\partial x }}.}
If the speed in the ether is to remain finite everywhere, it must
∂ ψ ∂ x = 0 {\displaystyle {\frac {\partial \psi }{\partial x}}=0} {\displaystyle {\frac {\partial \psi }{\partial x}}=0}
be, then we have
v ϱ = ∂ ψ ∂ ϱ , v = − α . {\displaystyle v\varrho ={\frac {\partial \psi }{\partial \varrho }},\quad v=-\alpha .} {\displaystyle v\varrho ={\frac {\partial \psi }{\partial \varrho }},\quad v=-\alpha .}
So the ether flows with respect to the coordinate system moving with the speed v in the direction x at the same time as the charge with the same speed in the opposite direction, i.e. it rests with respect to a resting coordinate system. This result is remarkable because it shows that the movement of electric quanta is no reason for a movement of the ether, as Helmholtz assumes.
On the other hand, movements can occur if the aether has an inertia other than zero. I give the calculation for this case because it gives an idea of ​​the magnitude of the density that would have to be assigned to the aether in a given case. Then, in addition to the terms of equations (1), there are also the components of the accelerations
s d α d t , s d β d t , s d γ d t {\displaystyle s{\frac {d\alpha }{dt}},\quad s{\frac {d\beta }{dt}},\quad s{\frac {d\gamma }{dt}}} {\displaystyle s{\frac {d\alpha }{dt}},\quad s{\frac {d\beta }{dt}},\quad s{\frac {d\gamma }{dt}}}
to add, where s denotes the density of the ether and
d α d t = ∂ α ∂ t + α d α d x + β ∂ α ∂ y + γ ∂ α ∂ e.g , d β d t = ∂ β ∂ t + α d β d x + β ∂ β ∂ y + γ ∂ β ∂ e.g , d γ d t = ∂ γ ∂ t + α d γ d x + β ∂ γ ∂ y + γ ∂ γ ∂ e.g {\displaystyle {\begin{array}{l}{\frac {d\alpha }{dt}}={\frac {\partial \alpha }{\partial t}}+\alpha {\frac {d\alpha }{dx}}+\beta {\frac {\partial \alpha }{\partial y}}+\gamma {\frac {\partial \alpha }{\partial z}},\\\\{\frac {d\beta }{dt}}={\frac {\partial \beta }{\partial t}}+\alpha {\frac {d\beta }{dx}}+\beta {\frac {\partial \beta }{\partial y}}+\gamma {\frac {\partial \beta }{\partial z}},\\\\{\frac {d\gamma }{dt}}={\frac {\partial \gamma }{\partial t}}+\alpha {\frac {d\gamma }{dx}}+\beta {\frac {\partial \gamma }{\partial y}}+\gamma {\frac {\partial \gamma }{\partial z}}\end{array}}} {\displaystyle {\begin{array}{l}{\frac {d\alpha }{dt}}={\frac {\partial \alpha }{\partial t}}+\alpha {\frac {d\alpha }{dx}}+\beta {\frac {\partial \alpha }{\partial y}}+\gamma {\frac {\partial \alpha }{\partial z}},\\\\{\frac { d\beta }{dt}}={\frac {\partial \beta }{\partial t}}+\alpha {\frac {d\beta }{dx}}+\beta {\frac {\partial \beta }{\partial y}}+\gamma {\frac {\partial \beta }{\partial z}},\\\\{\frac {d\gamma }{dt}}={\frac {\partial \ gamma }{\partial t}}+\alpha {\frac {d\gamma }{dx}}+\beta {\frac {\partial \gamma }{\partial y}}+\gamma {\frac {\partial \gamma }{\partial z}}\end{array}}}
are. [ vii ] In the case just considered, the system is stationary with respect to the moving coordinate system. So it is
∂ α ∂ t = ∂ β ∂ t = ∂ γ ∂ t = 0. {\displaystyle {\frac {\partial \alpha }{\partial t}}={\frac {\partial \beta }{\partial t}}={\frac {\partial \gamma }{\partial t}}=0.} {\displaystyle {\frac {\partial \alpha }{\partial t}}={\frac {\partial \beta }{\partial t}}={\frac {\partial \gamma }{\partial t}} =0.}
Let's set the values ​​of α , β , γ {\displaystyle \alpha ,\beta ,\gamma } {\displaystyle \alpha ,\beta ,\gamma } one, the elimination of P
1 A ∂ ψ ∂ x ∂ ∂ ϱ [ 1 ϱ 2 ( ∂ 2 ψ ∂ ϱ 2 − 1 ϱ ∂ ψ ∂ ϱ + ∂ 2 ψ ∂ x 2 ) ] {\displaystyle {\frac {1}{A}}{\frac {\partial \psi }{\partial x}}{\frac {\partial }{\partial \varrho }}\left[{\frac {1}{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi }{\partial \varrho ^{2}}}-{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}+{\frac {\partial ^{2}\psi }{\partial x^{2}}}\right)\right]} {\displaystyle {\frac {1}{A}}{\frac {\partial \psi }{\partial x}}{\frac {\partial }{\partial \varrho }}\left[{\frac {1 }{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi }{\partial \varrho ^{2}}}-{\frac {1}{\varrho }}{ \frac {\partial \psi }{\partial \varrho }}+{\frac {\partial ^{2}\psi }{\partial x^{2}}}\right)\right]}
− 1 A ∂ ψ ∂ ϱ ∂ ∂ x [ 1 ϱ 2 ( ∂ 2 ψ ∂ ϱ 2 − 1 ϱ ∂ ψ ∂ ϱ + ∂ 2 ψ ∂ x 2 ) ] {\displaystyle -{\frac {1}{A}}{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial }{\partial x}}\left[{\frac {1}{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi }{\partial \varrho ^{2}}}-{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}+{\frac {\partial ^{2}\psi }{\partial x^{2}}}\right)\right]} {\displaystyle -{\frac {1}{A}}{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial }{\partial x}}\left[{\frac { 1}{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi }{\partial \varrho ^{2}}}-{\frac {1}{\varrho }} {\frac {\partial \psi }{\partial \varrho }}+{\frac {\partial ^{2}\psi }{\partial x^{2}}}\right)\right]}
+ v ϱ s ∂ U ∂ x + ( ∂ ψ ∂ x ∂ U ∂ ϱ − ∂ ψ ∂ ϱ ∂ U ∂ x ) 1 s = 0. {\displaystyle +{\frac {v\varrho }{s}}{\frac {\partial {\mathfrak {U}}}{\partial x}}+\left({\frac {\partial \psi }{\partial x}}{\frac {\partial {\mathfrak {U}}}{\partial \varrho }}-{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial {\mathfrak {U}}}{\partial x}}\right){\frac {1}{s}}=0.} {\displaystyle +{\frac {v\varrho }{s}}{\frac {\partial {\mathfrak {U}}}{\partial x}}+\left({\frac {\partial \psi }{ \partial x}}{\frac {\partial {\mathfrak {U}}}{\partial \varrho }}-{\frac {\partial \psi }{\partial \varrho }}{\frac {\partial { \mathfrak {U}}}{\partial x}}\right){\frac {1}{s}}=0.}
This equation is satisfied if
1 ϱ ∂ ψ ∂ ϱ = v + 1 ϱ ∂ ψ 1 ∂ ϱ , 1 ϱ ∂ ψ ∂ x = 1 ϱ ∂ ψ 1 ∂ x , {\displaystyle {\begin{array}{lr}{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}=&v+{\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial \varrho }},\\\\{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial x}}=&{\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial x}},\end{array}}} {\displaystyle {\begin{array}{lr}{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial \varrho }}=&v+{\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial \varrho }},\\\\{\frac {1}{\varrho }}{\frac {\partial \psi }{\partial x}}=&{\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial x}},\end{array}}}
1 ϱ 2 ( ∂ 2 ψ 1 ∂ ϱ 2 − 1 ϱ ∂ ψ 1 ∂ ϱ + ∂ 2 ψ 1 ∂ x 2 ) = − U A s {\displaystyle {\frac {1}{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi _{1}}{\partial \varrho ^{2}}}-{\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial \varrho }}+{\frac {\partial ^{2}\psi _{1}}{\partial x^{2}}}\right)=-{\frac {{\mathfrak {U}}A}{s}}} {\displaystyle {\frac {1}{\varrho ^{2}}}\left({\frac {\partial ^{2}\psi _{1}}{\partial \varrho ^{2}}}- {\frac {1}{\varrho }}{\frac {\partial \psi _{1}}{\partial \varrho }}+{\frac {\partial ^{2}\psi _{1}}{ \partial x^{2}}}\right)=-{\frac {{\mathfrak {U}}A}{s}}}
is. It is
U = − 3 A ( x 2 + ϱ 2 ( 1 − A 2 v 2 ) ) 3 . {\displaystyle {\mathfrak {U}}=-{\frac {3A}{\left(x^{2}+\varrho ^{2}\left(1-A^{2}v^{2}\right)\right)^{3}}}.} {\displaystyle {\mathfrak {U}}=-{\frac {3A}{\left(x^{2}+\varrho ^{2}\left(1-A^{2}v^{2}\ right)\right)^{3}}}.}
To integrate the differential equation, we set
ψ 1 = ϱ φ . {\displaystyle \psi _{1}=\varrho \varphi .} {\displaystyle \psi _{1}=\varrho \varphi .}
Then it will be
∂ 2 φ ∂ ϱ 2 + 1 ϱ ∂ φ ∂ ϱ − 1 ϱ 2 φ + ∂ 2 φ ∂ x 2 = − ϱ U A s {\displaystyle {\frac {\partial ^{2}\varphi }{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial \varphi }{\partial \varrho }}-{\frac {1}{\varrho ^{2}}}\varphi +{\frac {\partial ^{2}\varphi }{\partial x^{2}}}=-{\frac {\varrho {\mathfrak {U}}A}{s}}} {\displaystyle {\frac {\partial ^{2}\varphi }{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial \varphi }{\ partial \varrho }}-{\frac {1}{\varrho ^{2}}}\varphi +{\frac {\partial ^{2}\varphi }{\partial x^{2}}}=-{ \frac {\varrho {\mathfrak {U}}A}{s}}}
We first consider the differential equation
∂ 2 φ 1 ∂ ϱ 2 + 1 ϱ ∂ φ 1 ∂ ϱ + 1 ϱ 2 ∂ 2 φ 1 ∂ ϑ 2 + ∂ 2 φ 1 ∂ x 2 = − ϱ U A s sin ⁡ ϑ . {\displaystyle {\frac {\partial ^{2}\varphi _{1}}{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial \varphi _{1}}{\partial \varrho }}+{\frac {1}{\varrho ^{2}}}{\frac {\partial ^{2}\varphi _{1}}{\partial \vartheta ^{2}}}+{\frac {\partial ^{2}\varphi _{1}}{\partial x^{2}}}=-{\frac {\varrho {\mathfrak {U}}A}{s}}\sin \vartheta .} {\displaystyle {\frac {\partial ^{2}\varphi _{1}}{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial \ varphi _{1}}{\partial \varrho }}+{\frac {1}{\varrho ^{2}}}{\frac {\partial ^{2}\varphi _{1}}{\partial \ vartheta ^{2}}}+{\frac {\partial ^{2}\varphi _{1}}{\partial x^{2}}}=-{\frac {\varrho {\mathfrak {U}} A}{s}}\sin \vartheta .}
whose integral is
φ 1 = A 4 π s ∫ ∫ ∫ d ϱ ′ d ϑ ′ d x ′ ϱ ′ 2 U ′ sin ⁡ ϑ ′ ( x − x ′ ) 2 + ϱ 2 + ϱ ′ 2 − 2 ϱ ϱ ′ cos ⁡ ( ϑ − ϑ ′ ) , {\displaystyle \varphi _{1}={\frac {A}{4\pi s}}\int \int \int {\frac {d\varrho '\ d\vartheta '\ dx'\ \varrho '^{2}\ {\mathfrak {U}}'\sin \vartheta '}{\sqrt {\left(x-x'\right)^{2}+\varrho ^{2}+\varrho '^{2}-2\varrho \varrho '\cos(\vartheta -\vartheta ')}}},} {\displaystyle \varphi _{1}={\frac {A}{4\pi s}}\int \int \int {\frac {d\varrho '\ d\vartheta '\ dx'\ \varrho '^ {2}\ {\mathfrak {U}}'\sin \vartheta '}{\sqrt {\left(xx'\right)^{2}+\varrho ^{2}+\varrho '^{2}- 2\varrho \varrho '\cos(\vartheta -\vartheta ')}}},}
= S sin ⁡ ϑ {\displaystyle =S\sin \vartheta } {\displaystyle =S\sin \vartheta } ,
S = A 4 π s ∫ ∫ ϱ ′ 2 d ϱ ′ d x ′ U ′ R , {\displaystyle S={\frac {A}{4\pi s}}\int \int \varrho '^{2}d\varrho 'dx'{\mathfrak {U}}'R,} {\displaystyle S={\frac {A}{4\pi s}}\int \int \varrho '^{2}d\varrho 'dx'{\mathfrak {U}}'R,}
R = 2 ϱ ′ ϱ ( ( 2 x − ϰ ) K − 2 ϰ E ) , {\displaystyle R={\frac {2}{\sqrt {\varrho '\varrho }}}\left(\left({\frac {2}{x}}-\varkappa \right)K-{\frac {2}{\varkappa }}E\right),} {\displaystyle R={\frac {2}{\sqrt {\varrho '\varrho }}}\left(\left({\frac {2}{x}}-\varkappa \right)K-{\frac {2}{\varkappa }}E\right),}
[ viii ]
ϰ 2 = 4 ϱ ′ ϱ ( e.g ′ − e.g ) 2 + ( ϱ + ϱ ′ ) 2 , K = ∫ 0 π 2 d φ 1 − ϰ 2 sin 2 ⁡ φ , E = ∫ 0 π 2 d φ 1 − ϰ 2 sin 3 ⁡ φ . {\displaystyle \varkappa ^{2}={\frac {4\varrho '\varrho }{(z'-z)^{2}+(\varrho +\varrho ')^{2}}},\quad K=\int \limits _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {1-\varkappa ^{2}\sin ^{2}\varphi }}},\quad E=\int \limits _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {1-\varkappa ^{2}\sin ^{3}\varphi }}}.} {\displaystyle \varkappa ^{2}={\frac {4\varrho '\varrho }{(z'-z)^{2}+(\varrho +\varrho ')^{2}}},\quad K=\int \limits _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {1-\varkappa ^{2}\sin ^{2}\varphi }}},\quad E=\int \limits _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {1-\varkappa ^{2}\sin ^{3}\varphi }}}.}
Then S satisfies the differential equation
∂ 2 S ∂ ϱ 2 + 1 ϱ ∂ S ∂ ϱ − 1 ϱ 2 S + ∂ 2 S ∂ x 2 = − ϱ U A s {\displaystyle {\frac {\partial ^{2}S}{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial S}{\partial \varrho }}-{\frac {1}{\varrho ^{2}}}S+{\frac {\partial ^{2}S}{\partial x^{2}}}=-{\frac {\varrho {\mathfrak {U}}A}{s}}} {\displaystyle {\frac {\partial ^{2}S}{\partial \varrho ^{2}}}+{\frac {1}{\varrho }}{\frac {\partial S}{\partial \ varrho }}-{\frac {1}{\varrho ^{2}}}S+{\frac {\partial ^{2}S}{\partial x^{2}}}=-{\frac {\varrho {\mathfrak {U}}A}{s}}}
and so it is φ = S {\displaystyle \varphi =S} {\displaystyle \varphi =S} .
These are the same expressions that give the velocities of the circular vortex rings in a liquid, where the x -axis is the axis of the vortex rings when the rotation speed of the liquid particles is about the circular rotation axis
3 A ϱ A 2 s [ x 2 + ϱ 2 ( 1 − A 2 v 2 ) ] 3 {\displaystyle {\frac {3{\mathfrak {A}}\varrho A}{2s\left[x^{2}+\varrho ^{2}\left(1-A^{2}v^{2}\right)\right]^{3}}}} {\displaystyle {\frac {3{\mathfrak {A}}\varrho A}{2s\left[x^{2}+\varrho ^{2}\left(1-A^{2}v^{2 }\right)\right]^{3}}}} is.
The magnitude of the movement that occurs depends primarily on the size
3 v e 2 A 2 ( 1 − A 2 v 2 ) ϱ 2 s [ x 2 + ϱ 2 ( 1 − A 2 v 2 ) ] 3 {\displaystyle {\frac {3ve^{2}A^{2}(1-A^{2}v^{2})\varrho }{2s\left[x^{2}+\varrho ^{2}\left(1-A^{2}v^{2}\right)\right]^{3}}}} {\displaystyle {\frac {3ve^{2}A^{2}(1-A^{2}v^{2})\varrho }{2s\left[x^{2}+\varrho ^{2 }\left(1-A^{2}v^{2}\right)\right]^{3}}}}
away. With constant e and s it has a maximum for
v = 1 3 A {\displaystyle v={\frac {1}{{\sqrt {3}}A}}} {\displaystyle v={\frac {1}{{\sqrt {3}}A}}}
and is the same
e 2 A 3 s . {\displaystyle {\frac {e^{2}A}{{\sqrt {3}}s}}.} {\displaystyle {\frac {e^{2}A}{{\sqrt {3}}s}}.}
In cathode rays we have electrical charges that fly through space at almost as high a speed.
Let's assume there would be 6 ⋅ 10 4 {\displaystyle 6\cdot 10^{4}} {\displaystyle 6\cdot 10^{4}} electrostatic units are transported per second and if we take the speed to be a third of the speed of light, then a tube 50 cm long would constantly have a charge of 3 ⋅ 10 − 4 {\displaystyle 3\cdot 10^{-4}} {\displaystyle 3\cdot 10^{-4}} moving. The size of the rotation speed would then be for x = 0 {\displaystyle x=0} {\displaystyle x=0} and ϱ = 1 m m {\displaystyle \varrho =1{\rm {mm}}} {\displaystyle \varrho =1{\rm {mm}}} nearly
1 2 10 − 13 1 s . {\displaystyle {\frac {1}{2}}10^{-13}{\frac {1}{s}}.} {\displaystyle {\frac {1}{2}}10^{-13}{\frac {1}{s}}.}
Outside the tube, noticeable movements would only occur if the ether density was extremely low. [ ix ] Nothing definite can be said about what happened in the immediate vicinity of the cargo. [1]
Reflection on moving transparent media.
An example where the tensions in the aether would cause movement is the reflection of electromagnetic plane waves at the boundary of moving insulators. Let us denote the angle of incidence by φ {\displaystyle \varphi } {\displaystyle \varphi } , with the index e the incident components, with r the reflected components, is according to the known laws
Y e = sin ⁡ ( x sin ⁡ φ + e.g 1 cos ⁡ φ λ − t T ) 2 π , {\displaystyle Y_{e}=\sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t}{T}}\right)2\pi ,} {\displaystyle Y_{e}=\sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t}{T}}\ right)2\pi ,}
L e = cos ⁡ φ sin ⁡ ( x sin ⁡ φ + e.g 1 cos ⁡ φ λ − t T ) 2 π , {\displaystyle L_{e}=\cos \varphi \sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t}{T}}\right)2\pi ,} {\displaystyle L_{e}=\cos \varphi \sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t}{ T}}\right)2\pi ,}
N e = − sin ⁡ φ sin ⁡ ( x sin ⁡ φ + e.g 1 cos ⁡ φ λ − t T ) 2 π . {\displaystyle N_{e}=-\sin \varphi \sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t}{T}}\right)2\pi .} {\displaystyle N_{e}=-\sin \varphi \sin \left({\frac {x\ \sin \varphi +z_{1}\cos \varphi }{\lambda }}-{\frac {t} {T}}\right)2\pi .}
If we move the plate with the speed v in the direction z , we have to use Lorentz for the reflected waves
Y r = R sin ⁡ ( x sin ⁡ φ − e.g 1 cos ⁡ φ λ + A 2 v e.g 1 T − t T ) 2 π , {\displaystyle Y_{r}=R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2}vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi ,} {\displaystyle Y_{r}=R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2} vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi ,}
L r = − R sin ⁡ ( x sin ⁡ φ − e.g 1 cos ⁡ φ λ + A 2 v e.g 1 T − t T ) 2 π , {\displaystyle L_{r}=-R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2}vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi ,} {\displaystyle L_{r}=-R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2 }vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi ,}
N r = − R sin ⁡ ( x sin ⁡ φ − e.g 1 cos ⁡ φ λ + A 2 v e.g 1 T − t T ) 2 π {\displaystyle N_{r}=-R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2}vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi } {\displaystyle N_{r}=-R\ \sin \left({\frac {x\ \sin \varphi -z_{1}\cos \varphi }{\lambda }}+{\frac {A^{2 }vz_{1}}{T}}-{\frac {t}{T}}\right)2\pi }
based on a coordinate system that moves with the plate. If we relate everything to a fixed coordinate system, we have e.g 1 = e.g − v t {\displaystyle z_{1}=z-vt} {\displaystyle z_{1}=z-vt} to set. The factor R is not exactly the same as when the system is at rest. The electromagnetic pressure causes work to be done or consumed and this reduces or increases the energy of the radiation. However, the boundary conditions can only be fulfilled if one assumes that this change in energy is distributed between reflected and refracted rays as if the incoming wave already carried with it the energy increased or decreased in proportion to this work. [ x ] From the given values ​​it follows if we neglect quantities of the order Av
T = ∂ R ∂ x − ∂ P ∂ e.g = 2 sin ⁡ 2 φ λ R { 1 λ sin ⁡ ( 2 e.g cos ⁡ φ λ − 2 v t cos ⁡ φ λ ) 2 π } , {\displaystyle {\mathfrak {T}}={\frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z}}={\frac {2\sin 2\varphi }{\lambda }}R\left\{{\frac {1}{\lambda }}\sin \left({\frac {2z\ \cos \varphi }{\lambda }}-{\frac {2vt\ \cos \varphi }{\lambda }}\right)2\pi \right\},} {\displaystyle {\mathfrak {T}}={\frac {\partial {\mathfrak {R}}}{\partial x}}-{\frac {\partial {\mathfrak {P}}}{\partial z }}={\frac {2\sin 2\varphi }{\lambda }}R\left\{{\frac {1}{\lambda }}\sin \left({\frac {2z\ \cos \varphi }{\lambda }}-{\frac {2vt\ \cos \varphi }{\lambda }}\right)2\pi \right\},}
Let's sit
0 = ∂ α ∂ x + ∂ γ ∂ e.g , {\displaystyle 0={\frac {\partial \alpha }{\partial x}}+{\frac {\partial \gamma }{\partial z}},} {\displaystyle 0={\frac {\partial \alpha }{\partial x}}+{\frac {\partial \gamma }{\partial z}},}
so
α = ∂ ψ ∂ e.g γ = − ∂ ψ ∂ x , {\displaystyle \alpha ={\frac {\partial \psi }{\partial z}}\quad \gamma =-{\frac {\partial \psi }{\partial x}},} {\displaystyle \alpha ={\frac {\partial \psi }{\partial z}}\quad \gamma =-{\frac {\partial \psi }{\partial x}},}
this is how the equations result
0 = ∂ T ∂ t + ∂ ψ ∂ e.g ∂ T ∂ x − ∂ ψ ∂ x ∂ T ∂ e.g , {\displaystyle 0={\frac {\partial {\mathfrak {T}}}{\partial t}}+{\frac {\partial \psi }{\partial z}}{\frac {\partial {\mathfrak {T}}}{\partial x}}-{\frac {\partial \psi }{\partial x}}{\frac {\partial {\mathfrak {T}}}{\partial z,}}} {\displaystyle 0={\frac {\partial {\mathfrak {T}}}{\partial t}}+{\frac {\partial \psi }{\partial z}}{\frac {\partial {\mathfrak {T}}}{\partial x}}-{\frac {\partial \psi }{\partial x}}{\frac {\partial {\mathfrak {T}}}{\partial z,}}}
so
∂ ψ ∂ x = − v γ = v . {\displaystyle {\frac {\partial \psi }{\partial x}}=-v\quad \gamma =v.} {\displaystyle {\frac {\partial \psi }{\partial x}}=-v\quad \gamma =v.}
The tension in the aether would only stop when it moves at the same speed as the moving plate. But this only applies to low speeds. For larger ones, rather complicated values ​​would arise, depending on the period of oscillation.
The fact that the co-movement of the ether only eliminates the tensions in the ether to a first approximation is due to the fact that the movement causes aberration of the ray, which, as is well known, cannot be easily explained by the assumption of moving ether.
It does not seem entirely hopeless to carry out experiments to see whether the aether is carried along in the direction of movement when reflected from rapidly moving plates.
The assumption of dormant aether.
From the foregoing, we cannot entirely deny the possibility that the aether is moving. But the difficulties of carrying out such an assumption should already be sufficiently apparent in the examples outlined. As soon as it is possible to do justice to all the facts observed so far, if one considers the ether to be at rest, this path will initially be recommended for its simplicity. However, we then violate a very general mechanical principle from the outset, [ xi ] that the equality of effect and counter-effect, if we do not want to assume that the electromagnetic tensions that want to set the ether in motion are canceled out by a certain rigid structure. And in general, if we deny the ether mobility, it becomes a substrate with highly indeterminate properties that we actually only use to make the finite value of the speed of light more understandable.
But this path will be particularly recommended to anyone who is initially only interested in the most general presentation of the facts.
The assumption of a resting aether was actually that advocated by Fresnel , although there is still talk of a partial continuation of the aether. However, this continuation only takes place inside the weighable bodies as soon as they themselves are moved and can be completely replaced by the view that what is continued is not the ether itself, but the part of the electromagnetic energy that is present in ponderable bodies liable. This emerges very clearly in the calculation by Reiff [2] , which shows that the Fresnel coefficient of continuation for a light ray in a moving medium results when the ether itself is at rest, the electromagnetic energy partly in the ether, partly in of the ponderable substance is present.
A precise implementation of the theory based on the assumption of resting ether and unchangeably charged ions as well as a complete discussion of all essential observation results is contained in the work of HA Lorentz [3] . E. Wiechert [4] starts from very similar points of view .
From his assumption, Lorentz directly obtains the Fresnel coefficient of the propagation of light through moving media, the aberration and the Doppler principle. All three are directly related and result from a general theorem according to which all are for bodies at rest [ xii ] valid equations of small oscillations can be transferred to moving ones if instead of the time t the variable t − t 1 v A {\displaystyle t-t_{1}vA} {\displaystyle t-t_{1}vA} introduces where t 1 {\displaystyle t_{1}} {\displaystyle t_{1}} means the time that light takes to travel in the free ether from a fixed point to an arbitrarily viewed point, and above all is the ratio of the speed of the body to the speed of light.
There is a further correction element for the continuation coefficient, which is caused by the fact that the movement also causes a change in the oscillation period according to the Doppler principle. It also immediately follows that the influence of the earth's motion is only shown in the aberration and that the prismatic deflection and the observation of the wavelength are not influenced by gratings. It also follows that a stationary current does not have an inductive effect on another wire due to the movement of the earth, because the movement creates an electrostatic charge which compensates for the effect.
With the induction effect, the influence of the earth's movement only occurs in proportion to the size v 2 A 2 {\displaystyle v^{2}A^{2}} {\displaystyle v^{2}A^{2}} so there is no prospect of experimental confirmation.
After Lorentz 's theory, which has been worked out in detail, proves that the assumption of immobile ether is completely sufficient to interpret a number of the diverse and hitherto little-explained phenomena of the influence of movement on electromagnetic processes, we must now point out a difficulty of a fundamental nature with consistent implementation of this theory.
This difficulty is closely related to the fact that changing electromagnetic states give rise to forces that would set the aether in motion if it were mobile. Let us imagine a body in the free ether in the form of a thin plate, which has different radiating capacities for heat rays on both sides. Since, according to Maxwell 's theory, the emitted rays exert a pressure on the surface, this pressure would predominate on the side of the greater radiation and set the body in motion. [ xiii ] So here we have the case that a body sets its center of gravity in motion through its own internal energy. So if we assume the ether to be immobile, then there would be a violation of the general principle of the center of gravity. On the other hand, the assumption of mobile ether, which has inertia, would avoid this objection.
However, the principle of the center of gravity may possibly be of a special nature and be limited to certain groups of effects in which no moving forces occur in the ether, as is actually the case with the usually observed ponderomotive effects.
Under all circumstances, this point should be kept in mind for further theoretical training.
The test results.
After we have discussed the two theoretical setups that are to be separated from each other, let us take a look at the experiments that have been carried out so far.
The main experiments that relate to our question are as follows:
A. Experiments with positive results.
1. The aberration of the light of the fixed stars. As is well known, the aberration found a simple explanation through the emission hypothesis of light. The difficulties in the undulation theory have only recently been eliminated by HA Lorentz by assuming a ether at rest .
2. The Doppler principle is of general kinematic importance in its nature, but must still be taken into account when considering the question of moving or resting aether.
3. Fizeau 's experiment and its repetition by Michelson and Morley . A ray of light passing through flowing water in the direction of movement experiences an acceleration of the passage in proportion 1 + v ( 1 − ( 1 / n 2 ) ) {\displaystyle 1+v(1-(1/n^{2}))} {\displaystyle 1+v(1-(1/n^{2}))} , where v denotes the speed, n denotes the refractive index of the water. This result finds its complete explanation in the assumption of resting aether.
[ xiv ]
B. Experiments with negative results.
1. Arago's experiment as to whether the movement of the earth influences the refraction of the light coming from the fixed stars.
2. Ketteler ’s interference experiment. The two beams of an interferential refractor are sent through two tubes filled with water and inclined towards each other in such a way that one beam hits one tube after the first reflection (on one glass plate), the other beam hits the second tube after the second reflection (on the other glass plate), i.e. runs in the opposite direction. Although both tubes are carried along by the earth's movement, there is no change in the interference fringes, although one beam is accelerated and the other is delayed.
Both results follow directly from the assumption of resting aether.
3. Klinkerfues ' experiment to determine whether the absorption line of sodium vapor was influenced by the movement of the earth.
Klinkerfues ' positive result would be incompatible with the theory of resting aether. However, the shift found is so small that observation errors cannot be ruled out.
4. Des Coudres 's experiment as to whether the induction effect of two coils of wire on a third is influenced by the fact that the direction of the induction of each coil falls once in the direction of the earth's movement, then in the direction perpendicular to it.
HA Lorentz has proven that when the aether is at rest, this influence only depends on the square of the ratio of the speed of the earth to the speed of light, and is therefore not observable because the movement of the earth creates an electrostatic charge on the current conductors, which cancels out the first-order effect.
5. Lodge 's experiments to investigate to what extent the surrounding aether is carried along by the movement of heavy or magnetizable masses.
6. Zehnder 's experiments as to whether the ether through the [ xv ] Movement of a piston is moved in an air-diluted space.
The experiments of both observers were carried out with sensitive interference methods and gave negative results, so they are in perfect agreement with the assumption of a resting aether.
7. Mascart 's experiments on the rotation of the plane of polarization in quartz. There was no change in rotation when the light rays were once in the direction of the earth's movement and then in the opposite direction.
HA Lorentz gave the theory of this phenomenon and found that, assuming the aether is at rest, the earth's movement changes the existing rotation and independently adds a second one.
The negative result of Mascart 's observations would show that in quartz these two rotations caused by the influence of the earth's movement just cancel each other out.
8. Roentgen 's experiment to determine whether magnetic forces are generated by the movement of the earth from a charged capacitor.
The negative result of this experiment is not compatible with the assumption of a resting aether.
Electric charges and magnets would also have to produce magnetic or electrical forces through the movement of the earth. The absence of these forces would also be incompatible with the requirement of a resting aether.
9. Fizeau 's experiment on the influence of the earth's movement on the rotation of the plane of polarization through glass columns. The positive result of this experiment has recently been questioned. It would not be compatible with the assumption of resting aether according to HA Lorentz 's investigations .
10. The Michelson and Morley experiment . If the aether is at rest, the time it takes for a ray of light to travel back and forth between two plates of glass must change as the plates move. The change depends on the size v 2 A 2 {\displaystyle v^{2}A^{2}} {\displaystyle v^{2}A^{2}} but should be observable when interference is used.
[ xvi ] The negative result is incompatible with the assumption of resting aether. This assumption can only be maintained by the hypothesis that the length dimensions of solid bodies are changed in the same proportion by the movement through the resting ether in order to compensate for the lengthening of the path of the light ray.
The assumption of moving aether would give rise to the possibility that the aether is carried along by the movement of the earth and rests relative to it. This would explain all negative test results. But then the explanation of the aberration would remain.
Gravity and inertia.
The fact that gravity occupies an exceptional position and has no noticeable relationship to the other natural phenomena has often been emphasized. Its attribution to pressure forces is made more difficult by the fact that the energy reserve of a gravitating system has its greatest value when the individual parts of the mass are at an infinite distance. However, it is not always emphasized clearly enough that the acceleration of heavy masses is most likely related to gravity, because two independent definitions of mass are obtained through acceleration and gravity, which, as far as the very precise observations here go, are perfect to match. If one demands a further explanation of gravity, it would also have to give an account of why work is required to accelerate heavy masses. The fact that the two definitions of mass agree would then have to be a consequence of this explanation. It cannot be said with certainty whether such a theory can also be based on the ether, but it is probable.
However, it must also be emphasized here that it is by no means certain whether all effects can be traced back to tensions in the ether, just as it remains doubtful whether the processes in the ether can be completely satisfactorily represented by the laws of mechanics.
[ xvii ] If we now summarize the results, the impression is that there are still a number of questions to be answered before we can decide on the path that science should take.
As we have seen, the assumption of moving ether without inertia leads to improbable consequences.
As an experiment that would be important for this assumption, we recommend trying to see whether the ether is set in motion by the movement of reflecting transparent media.
But since the ether is not set in motion by the movement of solid bodies, as far as is known so far, a negative result is likely.
The following difficulties stand in the way of the assumption that the ether is completely at rest:
1. Violation of the principle of the center of gravity (regarding the equality of effect and counteraction).
2. The negative results of the experiments of Michelson and Morley , that of Roentgen and possibly the experiments of Mascart and Fizeau .
It would therefore be urgently desirable to repeat the following experiments or to carry out new ones.
1. Does the earth's movement affect the rotation of the plane of polarization
a) naturally rotating substances,
b) through glass columns.
2. Does the movement of the earth cause the magnetic forces required by the theory through the movement of electrical charges and the corresponding electrical forces through the movement of magnets?
When the results of these experiments are completely clear, it will become clear whether the otherwise simple theory of resting aether should be retained or abandoned. Should it have to be abandoned, it seems to me that only the way out indicated by Des Coudres would remain; namely influence of gravity on the light ether. This assumption seems to me to be equivalent to the assumption of a low inertial mass of the light ether.
[ xviii ] It would then be explained that the earth pulls the ether with it due to its significant gravity, while the movement of small solid bodies on the earth has no influence. The negative result of the experiments mentioned would be easily explained.
But then the difficulties in explaining the aberration to which HA Lorentz drew attention would essentially remain . However, whether these cannot be overcome if the co-movement of the ether under the influence of gravity is taken into account requires a special investigation. For this purpose, the hydrodynamic problem would have to be solved: determining the movements of a liquid through which a point moves at a constant speed, which attracts the individual parts of the liquid according to Newton 's law.
The Maxwell 's tensions that would set the ether in motion are always so small, because they appear multiplied by the reciprocal speed of light, that the movements generally become imperceptible even with a very small inertial mass.
The task of the theory would then be to look for examples where the movement of the ether could actually be observed.
↑ I used Jamin 's interferential refractor to test whether a light beam passing through a vacuum tube is accelerated by the cathode rays; but the result was definitely negative.
↑ Reiff , Wied. Ann. 50. p. 367. 1893
↑ Lorentz , attempt at a theory of electrical and optical phenomena in moving bodies . Leyden 1895 [WS:Template:1893].
↑ Wiechert , Theory of Electrodynamics. Konigsberg 1896.
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