Kinematical Waves in Spacetime

Abstract

We will prove how to create kinematical waves in spacetime. To this end we will combine the newfound technique to change locally the electromagnetic gauge in Minkowsky spacetimes by using ideal solenoids and the Aharonov-Bohm effect. The local kinematical states of spacetime represented by a new kind of local tetrad will be made to oscillate according to preestablished wave equations and we will show how to produce these effects from a mathematical point of view and from a technological point of view. Kinematical waves just to mention one possible application could be used for communication.

Appendices

Appendix I

There is a subtlety related to the choice of gauge vector in the local plane two when the Maxwell equations have a source term. Because in this case there will be no second electromagnetic potential $*A^{\mu }$. This problem arises for instance in the solenoid case where there is a current running through the coil. We recommend reading section Appendix V in order to see how the Maxwell equations or their integrated form take this current into account. We will study in this section how to make a suitable choice for the gauge vector $Y^{\alpha }$ for the Maxwell equations with a source $J^{\mu }$ as it is the case in a solenoid cylindrical geometry. Let us focus for practical purposes in the cylindrical problem as an example that permits a better visualization of this physical situation when the Maxwell equations have sources. The point is that in geometrodynamics, the Maxwell equations,

$$\begin{aligned} f^{\mu \nu }_{\,\,\,\,\,;\nu }= & {} J^{\mu } \end{aligned}$$

(32)

$$\begin{aligned} *f^{\mu \nu }_{\,\,\,\,\,;\nu }= & {} 0 \ , \end{aligned}$$

(33)

tell us about the existence of one potential $X^{\alpha }=A^{\alpha }$. Then the question arises about gauging the vectors in equations (10-11). The tetrad of eigenvectors to the Einstein-Maxwell or Minkowski-Maxwell stress-energy tensor is given by,

$$\begin{aligned} V_{(1)}^{\alpha }= & {} \xi ^{\alpha \lambda }\,\xi _{\rho \lambda }\,X^{\rho } \end{aligned}$$

(34)

$$\begin{aligned} V_{(2)}^{\alpha }= & {} \sqrt{-Q/2} \,\, \xi ^{\alpha \lambda } \, X_{\lambda } \end{aligned}$$

(35)

$$\begin{aligned} V_{(3)}^{\alpha }= & {} \sqrt{-Q/2} \,\, *\xi ^{\alpha \lambda } \, Y_{\lambda } \end{aligned}$$

(36)

$$\begin{aligned} V_{(4)}^{\alpha }= & {} *\xi ^{\alpha \lambda }\, *\xi _{\rho \lambda } \,Y^{\rho }\ . \end{aligned}$$

(37)

The non-zero components of the electromagnetic field are $f_{\theta \,z} = B_{\rho }$ and $f_{\rho \theta } = B_{z}$, with $A_{\theta }$ also given in reference [29]. Then $\xi _{\rho \theta }=f_{\rho \theta }$ and $\xi _{\theta \,z}= f_{\theta \,z}$ in a flat Minkowskian spacetime with signature $(-+++)$ because the complexion is equal to zero. We also know that the metric in cylindrical coordinates $(t, \rho , \theta , z)$ will be diagonal $(-1, 1, \rho ^{2}, 1)$. The metric determinant g will satisfy $\sqrt{-g}=\rho$. For the alternating tensor cylindrical components, section IX. APPENDIX I in reference [1] is useful when considering all these elements. We notice that using the four-dimensional Lorentz flat Minkowski metric tensor in cylindrical coordinates allows the introduction at every point of four orthonormal vectors. Let them be $k^{\mu }_{t}=(1,0,0,0)$, $k^{\mu }_{\rho }=(0,1,0,0)$, $k^{\mu }_{\theta }=(0,0,\frac{1}{\rho },0)$ and $k^{\mu }_{z}=(0,0,0,1)$. Since the vector $k^{\mu }_{t}$ has non-trivial t components, then we can choose the gauge vector $Y^{\alpha }=k^{\alpha }_{t}$. This way, the equation components (38-40) will not be trivial. In the Cylindrical geometry the only non-zero tetrad vector components for the local plane two will be,

$$\begin{aligned} V_{(3)}^{\rho }= & {} \sqrt{Q/2} \,\,*\xi ^{\rho \,t}\,Y_{t}=-\sqrt{Q/2} \,\,*\xi _{t\rho }\, Y^{t} \end{aligned}$$

(38)

$$\begin{aligned} V_{(3)}^{z}= & {} \sqrt{Q/2} \,\,*\xi ^{zt}\,Y_{t}=-\sqrt{Q/2} \,\,*\xi _{tz}\, Y^{t} \end{aligned}$$

(39)

$$\begin{aligned} V_{(4)}^{t}= & {} *\xi ^{t\rho }\,*\xi _{t\rho }\,Y^{t} + *\xi ^{tz}\,*\xi _{tz}\,Y^{t} = -(\mid *\xi _{t\rho } \mid ^{2} + \mid *\xi _{tz} \mid ^{2})\,Y^{t} \ . \end{aligned}$$

(40)

where $Q = 2\,f_{\rho \theta }\,f^{\rho \theta }+2\,f_{\theta \,z}\,f^{\theta \,z}={2(B_{z}^{2} + B_{\rho }^{2}) \over \rho ^{2}}$. The difference in sign between $\sqrt{-Q/2}$ in (35-36) and $\sqrt{Q/2}$ comes about because in this geometry $Q>0$. In the Cylindrical geometry the only non-zero tetrad vector components for the local plane one will be,

$$\begin{aligned} V_{(1)}^{\theta }= & {} \xi ^{\theta \rho }\,\xi _{\theta \rho }\,A^{\theta } + \xi ^{\theta \,z}\,\xi _{\theta \,z}\,A^{\theta } \end{aligned}$$

(41)

$$\begin{aligned} V_{(2)}^{\rho }= & {} \sqrt{Q/2} \,\,\xi ^{\rho \theta }\,A_{\theta } \end{aligned}$$

(42)

$$\begin{aligned} V_{(2)}^{z}= & {} \sqrt{Q/2} \,\,\xi ^{z\theta }\,A_{\theta } \ . \end{aligned}$$

(43)

Despite the fact that in this cylindrical geometry we cannot choose the gauge vector $Y^{\alpha }$ to be $A^{\alpha }$ simply because the components of $V_{(3)}^{\alpha }$ and $V_{(4)}^{\alpha }$ will all be zero and nor we can choose $Y^{\alpha }$ to be $*A^{\alpha }$ simply because $*A^{\alpha }$ does not exist in the cylindrical geometry with source, we can choose it to be the vector $Y^{\alpha }=a\,k^{\alpha }_{t}$. It is also possible to choose the following $Y^{\alpha }=a\,k^{\alpha }_{t} + A^{\alpha }$ gauge-vector in plane two. a is an appropriate constant that keeps the units of the first and the second terms equal. We can then always choose another gauge by implementing $Y^{\alpha }=a\,k^{\alpha }_{t} +A^{\alpha } \rightarrow Y^{\alpha }=a\,k^{\alpha }_{t} + A^{\alpha } + \Lambda _{,\beta }\,g^{\alpha \beta }$. It is an electromagnetic potential gauge transformation for a valid choice of gauge-vector and we can study the tetrad eigenvector transformations in the local plane two exactly as in reference [1] and Section 1. There would be no mathematical change in the analysis structure. The whole point of this section is to highlight that when we have in flat-Minkowski spacetime the Maxwell equations with sources, then we have to be careful with our choice for the gauge vector $Y^{\alpha }$. With this observation all the analysis about tetrad vector transformations in the local plane two will follow the same lines as in manuscript [1] and the tetrad study originally made in Section 1 for Einstein-Maxwell curved spacetimes without sources will stand. Let us remember that for Einstein-Maxwell curved spacetimes these steps were not necessary since there was a natural non-trivial choice $Y^{\alpha }=*A^{\alpha }$. In Einstein-Maxwell curved spacetimes with sources we would also have an analogous choice to the one given in this section depending on the case.

Appendix II

1.1 Maxwell Equations in Cylindrical Coordinates

We will study the possible incidence of variable currents through the Aharonov-Bohm solenoid coil I(t) in the calculation of the electromagnetic gauge scalar for a circling electron. We are interested in finding out what possible changes arise when we solve the Maxwell equations in the case where the solenoid coil current is dependent on time. The general expression for the Maxwell equations in the solenoid case is given by,

$$\begin{aligned} -{\partial \overrightarrow{B} \over \partial t}= & {} \nabla \times \overrightarrow{E} \end{aligned}$$

(44)

$$\begin{aligned} \mu _{o}\,\varepsilon _{o}\,{\partial \overrightarrow{E} \over \partial t}= & {} \nabla \times \overrightarrow{B} \end{aligned}$$

(45)

$$\begin{aligned} \nabla \overrightarrow{E}= & {} 0 \end{aligned}$$

(46)

$$\begin{aligned} \nabla \overrightarrow{B}= & {} 0 \ . \end{aligned}$$

(47)

The Maxwell equations in cylindrical coordinates are,

$$\begin{aligned} -{\partial B_{\rho } \over \partial t}\,\widehat{\rho } -{\partial B_{\theta } \over \partial t}\,\widehat{\theta } -{\partial B_{z} \over \partial t}\,\widehat{z}= & {} \nonumber \\ ({1 \over \rho }\,{\partial E_{z} \over \partial \theta }-{\partial E_{\theta } \over \partial z})\,\widehat{\rho } + ({\partial E_{\rho } \over \partial z}-{\partial E_{z} \over \partial \rho })\,\widehat{\theta } + {1 \over \rho }\,({\partial (\rho \,E_{\theta }) \over \partial \rho }-{\partial E_{\rho } \over \partial \theta })\,\widehat{z}{} & {} \end{aligned}$$

(48)

$$\begin{aligned} -\mu _{o}\,\varepsilon _{o}\,{\partial E_{\rho } \over \partial t}\,\widehat{\rho } -\mu _{o}\,\varepsilon _{o}\,{\partial E_{\theta } \over \partial t}\,\widehat{\theta } -\mu _{o}\,\varepsilon _{o}\,{\partial E_{z} \over \partial t}\,\widehat{z}= & {} \nonumber \\ ({1 \over \rho }\,{\partial B_{z} \over \partial \theta }-{\partial B_{\theta } \over \partial z})\,\widehat{\rho } + ({\partial B_{\rho } \over \partial z}-{\partial B_{z} \over \partial \rho })\,\widehat{\theta } + {1 \over \rho }\,({\partial (\rho \,B_{\theta }) \over \partial \rho }-{\partial B_{\rho } \over \partial \theta })\,\widehat{z}{} & {} \end{aligned}$$

(49)

$$\begin{aligned} {1 \over \rho }\,{\partial (\rho \,E_{\rho }) \over \partial \rho } + {1 \over \rho }\,{\partial E_{\theta } \over \partial \theta } + {\partial E_{z} \over \partial z}= & {} 0 \end{aligned}$$

(50)

$$\begin{aligned} {1 \over \rho }\,{\partial (\rho \,B_{\rho }) \over \partial \rho } + {1 \over \rho }\,{\partial B_{\theta } \over \partial \theta } + {\partial B_{z} \over \partial z}= & {} 0 \ . \end{aligned}$$

(51)

Let us see now what would be the (48-51) first adapted to the solenoid with azimuthal symmetry in the finite case and then with symmetry in the z direction along the axis for an infinite ideal solenoid.

1.2 Infinite Ideal Solenoid

If the solenoid is infinite ideal we can make several assumptions. Inside and outside the solenoid the only component of the electric field will be $E_{\theta }$ as a function only of the coordinate $\rho$. The only component of the magnetic field inside the solenoid will be $B_{z}=\mu _{o}\,n\,I(t)$. n is the number of coil turns per unit length. I(t) is the current through the coil.

Then the equations (44), (48) will become when integrated into the Faraday’s law,

$$\begin{aligned} -{d \over dt}{\int \int }_{S}\,\overrightarrow{B}\,\widehat{z}\,dS= & {} \oint _{C}\,\overrightarrow{E}\, . \,d\overrightarrow{l} \ . \end{aligned}$$

(52)

The closed circle C is centered at the solenoid axis and on a plane perpendicular to this axis. The radius $\rho$ is inside the solenoid radius. The surface S has this circle as a border. The result of integrating (52) provides the $E_{\theta }$ tangent to this circle as,

$$\begin{aligned} E_{\theta }= & {} -\mu _{o}\,n\,\dot{I}(t)\,{\rho \over 2} \ . \end{aligned}$$

(53)

Outside the ideal solenoid we will assume that the magnetic field components are zero and $\dot{I}(t)={dI(t) \over dt}$. Once more the only non-zero component will be $E_{\theta }$ and using again Faraday’s law we can find,

$$\begin{aligned} E_{\theta }= & {} -\mu _{o}\,n\,\dot{I}(t)\,{R^{2}_{sol}\over 2\rho } \ . \end{aligned}$$

(54)

Therefore, the Aharonov-Bohm effect in reference [12] will only include the variable currents in the solenoid coil when we calculate the magnetic field flux through a surface that has as a border the circular trajectory of an electron circling around the solenoid at a radius $R_{e}>R_{sol}$. The magnetic field non-trivial component $B_{z}=\mu _{o}\,n\,I(t)$ will be parallel to the solenoid axis inside the solenoid. Let us remember that we also included in our experiments in references [11, 12] an independent constant magnetic field external to the solenoid and parallel to its axis in order to make the electron circle around this axis of symmetry. It is also clear that for the electron to move in circles $\dot{\rho }(t)=0$ with $\rho (t)=R_{e}=constant$ where the radial and tangential accelerations would be given by $a_{\rho }=-R_{e}\,\dot{\theta }^{2}$ and $a_{\theta }=R_{e}\,\ddot{\theta }$ and we also assume $R_{e} \gg R_{sol}$. Therefore, the electromagnetic gauge transformation scalar might include this magnetic field flux through a surface that has as a border the circular trajectory of an electron circling around the solenoid. Therefore the result of the calculation of the electromagnetic gauge scalar will be similar as for the case where the current does not vary with time.