Key Outcomes of 5D Relativity

Abstract

Interstellar travel needs enormous amounts of energy to accelerate payload, structure, and propellant to high speeds. For a travel to a distant star all three have to be huge. Chemical propellant needs more than 90% of the whole start weight, even from an orbit around the Moon. Most of the energy is necessary to increase kinetic energy. Because kinetic energy depends on inertial mass, a reduction of effective energy could reduce the amount of energy needed. As inertial mass and gravitational mass are the same following the equivalence principle of General Relativity and gravity depends on the gravitational constant, a local variation of the gravitational constant might result in variation of the effective mass. A varying scaling between mass/energy and spacetime could also cause a new force as known form the Pioneer anomaly. Kaluza-Klein Theory is General Relativity extended to 5 dimensions (5D). The gravitational constant is substituted by a scalar field making it variable. This scalar field is predicted to change under strong dynamic electromagnetic fields. Deriving the equation of motion from this 5D-metric predicts a fifth force.

1. Introduction

The basic idea of classical Kaluza-Theory [1] is briefly presented. Many following introductory and basic passages and also figures are taken from a conference paper published with the ASCEND 2021 conference [2]. The focus here is on the didactic presentation of the results in the form of 10 key statements, which are listed and explained in different sections. This 5D Relativity theory is named “induced matter theory,” because no extra 5D stress-energy-momentum tensor is necessary, but the known 4D stress-energy-momentum tensor is generated or “induced” by curvature of an empty 5D space. The key results are as follows:

1. the 4 off-diagonal elements of the new fifth column $g_{5\alpha }$ represent the 4-vector potential $\mathbf{A}$ of moving charges

2. the 5th diagonal element takes the role of the gravity constant

3. there is an inhomogeneous wave equation for the scalar field with excitation by electromagnetic fields

4. in the vicinity of a charge the gravitational constant becomes larger

5. the electric field lines of a charge are caused by pressure only, because the positive field energy of the electric field and the negative field energy of the scalar field cancel each other out

6. a charged black hole in 5D has only one event horizon in contrast to the Reissner-Nordström-solution, which in 4D has two event horizons and a repulsive gravitational field between them

7. there is a static soliton solution consisting of a cloud of radiation particles of the scalar field (radions, dilatons, scalarons, axions) with kinetic mass around the mass center

8. a dependence of the elements on the 5th coordinate can generate mass, pressure, momentum, and energy

9. there is an inhomogeneous wave equation for the scalar field with excitation by variation of the ten 4D gravitiy-potentials with the 5th dimension

10. there is a new rectilinear component of the Lorentz force that causes acceleration that could look like a reduction in inertia.

2. Classical electrodynamics retrospected

Maxwell’s eqs. (3D plus time) specify sources and curls for electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ by which the whole fields are determined according to Helmholtz’s theorem [3]. As basic mathematical laws they do not reveal which quantity is a cause and which is a consequence. This way the 3rd Maxwell equation

$$\nabla \times \mathbf{E}=-{\mu }_0\frac{\partial B}{\partial t}$$E1

only declares two simultaneously occurring effects always as equal (non-causal). The cause for all electric and magnetic fields are always charges and currents (charges at rest and charges in motion) [4]. The coupling of the fields $\mathbf{E}$ and $\mathbf{B}$ by Maxwell’s equations means that both appear as a dual entity. So both are parts of one electromagnetic field with six components (3 electric and 3 magnetic) and instead of electricity and magnetism we speak about electromagnetism.

2.1 Electric and magnetic field derived from four scalar potentials

From Maxwell’s equations follows that both fields can be derived from potentials. Maxwell’s $2^{\mathrm{nd}}$ equation states that there are no sources and only curls for magnetic fields; thus, the magnetic field B can be written as rotation of another vector field A(r, $t$) which depends on position and time:

$$\nabla \cdot \mathbf{B}=0\iff \mathbf{B}=\nabla \times \mathbf{A}\text{with}\mathbf{A}=\left(A_1,A_2,A_3\right)$$E2

The vector potential A can be used in the $3^{\mathrm{rd}}$ Maxwell equation (Eq. (1)) and we get:

$$0=\nabla \times \mathbf{E}+{\mu }_0\frac{\partial }{\partial t}\nabla \times \mathbf{A}=\nabla \times \left(\mathbf{E}+{\mu }_0\frac{\partial \mathbf{A}}{\partial t}\right)$$E3

Because the term in the brackets has a curl of zero, it can be written as a gradient of a scalar potential $\phi$:

$$\mathbf{E}=-{\mu }_0\frac{\partial \mathbf{A}}{\partial t}+\nabla \phi \left(\mathbf{r},t\right)$$E4

Thus, it is possible to calculate B and E also from A and $\phi$. The electromagnetic potentials introduced in classical theory—formally as helping quantities—appear in other theories as real necessary fields with physical meaning (e.g., Schrödinger-equation, QED, Proca-equation [3]).

2.2 Electromagnetic field tensor: a dual entity derived from 4-dimensional rotation of four vector potential

In special Relativity (4D) the electromagnetic potentials A and $\phi$ are combined to a 4-vector, setting $A_0$ or $A_4$ to $\phi$. The Faraday tensor or field tensor is defined by the 16 components [5]:

$$F_{\mathit{ik}}=\frac{\partial A_k}{\partial x_i}-\frac{\partial A_i}{\partial x_k}\text{with}i,k=\mathrm{1,2,3,4}$$E5

These are the partial derivatives of 4-dimensional (4D) rotation of $A_i$. They also occur in 3D rotation and in the gradient, that way the Faraday-tensor contains directly electric and magnetic field components. In Cartesian coorinates they represent Eqs. (1) and (2):

$$\begin{array}{l}{B}_x=\frac{\partial A_3}{\partial x_2}-\frac{\partial A_2}{\partial x_3}{E}_x=\mathit{ic}\left(\frac{\partial A_4}{\partial x_1}-\frac{\partial A_1}{\partial x_4}\right)\\ B_y=\frac{\partial A_1}{\partial x_3}-\frac{\partial A_3}{\partial x_1}{E}_y=\mathit{ic}\left(\frac{\partial A_4}{\partial x_2}-\frac{\partial A_2}{\partial x_4}\right)\\ B_z=\frac{\partial A_2}{\partial x_1}-\frac{\partial A_1}{\partial x_2}{E}_z=\mathit{ic}\left(\frac{\partial A_4}{\partial x_3}-\frac{\partial A_3}{\partial x_4}\right)\end{array}$$E6

which were derived from 2nd and 3rd Maxwell’s equation. The tensor $F_{\mathit{ik}}$ is the 4D rotation of four vectors $A_i$. It can be written this way with use of the electromagnetic field quantities:

$$\mathbf{F}=\left(F_{\mathit{ik}}\right)=\left[\begin{array}{cccc}0& B_z& -B_y& -\frac{i}{c}{E}_x\\ -B_z& 0& -B_x& -\frac{i}{c}{E}_y\\ B_y& B_x& 0& -\frac{i}{c}{E}_z\\ -\frac{i}{c}{E}_x& -\frac{i}{c}{E}_y& -\frac{i}{c}{E}_z& 0\end{array}\right]$$E7

This tensor is a physical quantity, which unifies electric and magnetic field components into one dual entity: the electromagnetic field tensor $\mathbf{F}$.

2.3 Stress-energy-momentum tensor

The Maxwell stress tensor $\sigma$ (3D) is used in classical electromagnetism to represent the interaction between electromagnetic forces and mechanical momentum ([6], p. 72, 931). Starting with the Lorentz force law $\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times \mathbf{B}\right)$ the force per unit volume is $\mathbf{f}=\rho \mathbf{E}+\mathbf{j}\times \mathbf{B}$. Replacing sources $\rho$ and curls $\mathbf{j}$ with Maxwell’s equations by functions of the fields $\mathbf{E}$ and $\mathbf{B}$, eliminating the curls by vector calculus identities [7] and writing the result in a more compact way one gets:

$${\sigma }_{\mathit{ij}}={\epsilon }_0{E}_i{E}_j+\frac{1}{{\mu }_0}{B}_i{B}_j-\frac{1}{2}{\delta }_{\mathit{ij}}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right)\text{in}\mathrm{SI}\text{units}$$E8

Which is explicitly in cartesian coordinates

$$\sigma =\left(\begin{array}{lll}{\epsilon }_0{E}_x{E}_x+\frac{1}{{\mu }_0}{B}_x{B}_x-\frac{1}{2}\left(E^2+H^2\right)& {\epsilon }_0{E}_x{E}_y+\frac{1}{{\mu }_0}{B}_x{B}_y& {\epsilon }_0{E}_x{E}_z+\frac{1}{{\mu }_0}{B}_x{B}_z\\ {\epsilon }_0{E}_y{E}_x+\frac{1}{{\mu }_0}{B}_y{B}_x& {\epsilon }_0{E}_y{E}_y+\frac{1}{{\mu }_0}{B}_y{B}_y-\frac{1}{2}\left(E^2+B^2\right)& {\epsilon }_0{E}_y{E}_z+\frac{1}{{\mu }_0}{B}_y{B}_z\\ {\epsilon }_0{E}_z{E}_x+\frac{1}{{\mu }_0}{B}_z{B}_x& {\epsilon }_0{E}_z{E}_y+\frac{1}{{\mu }_0}{B}_z{B}_y& {\epsilon }_0{E}_z{E}_z+\frac{1}{{\mu }_0}{B}_z{B}_z-\frac{1}{2}\left(E^2+B^2\right)\end{array}\right)$$E9

This stress tensor can be used to derive the Lorentz force per unit volume by the divergence of the stress tensor as

$$\mathbf{f}=div\sigma +{\epsilon }_0{\mu }_0\frac{\partial \mathbf{S}}{\partial t}\text{with}\text{Poynting}\text{vector}\mathbf{S}=\frac{1}{{\mu }_0}\mathbf{E}\times \mathbf{B}$$E10

With potentials instead of fields (Eq. (6)) and using the definition of the field tensor $F_{\mathit{\mu \lambda }}$ (Eq. (5)) one gets the relativistic electromagnetic stress–energy-momentum tensor $T_{\mathit{\mu \nu }}$ (4D):

$$T_{\mathit{\mu \nu }}=\frac{1}{{\mu }_0}\left[F_{\mathit{\mu \lambda }}{F}_{\mathit{\lambda \nu }}+\frac{1}{4}{\delta }_{\mathit{\mu \nu }}\left(F_{\mathit{\kappa \lambda }}{F}_{\mathit{\kappa \lambda }}\right)\right]$$E11

which is explicitly in matrix form (in SI units):

$$\mathbf{T}=\left(T_{\mathit{ik}}\right)=\left[\begin{array}{cccc}\frac{1}{2}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right)& S_x/c& S_y/c& S_z/c\\ S_x/c& -{\sigma }_{11}& -{\sigma }_{12}& -{\sigma }_{13}\\ S_y/c& -{\sigma }_{21}& -{\sigma }_{22}& -{\sigma }_{23}\\ S_z/c& -{\sigma }_{31}& -{\sigma }_{32}& -{\sigma }_{33}\end{array}\right]$$E12

where $\mathbf{S}=\frac{1}{{\mu }_0}\mathbf{E}\times \mathbf{B}$ is the Poynting vector, ${\sigma }_{\mathit{ij}}={\epsilon }_0{E}_i{E}_j+\frac{1}{{\mu }_0}{B}_i{B}_j-\frac{1}{2}\left({\epsilon }_0{E}^2+\frac{1}{{\mu }_0}{B}^2\right){\delta }_{\mathit{ij}}$ is the Maxwell stress tensor, and the speed of light is $c=\sqrt{\frac{1}{{\mu }_0{\epsilon }_0}}$. Eq. (10)—the Lorentz force per unit volume - can be rewritten as:

$$\mathbf{f}=div\mathbf{T}$$E13

3. Electromagnetism from 5D

Kaluza found that the relativistic electromagnetic stress–energy-momentum tensor T from Eqs. (11) and (12) is mathematically structural equal to a part of a 5D curvature tensor. To see this, the electromagnetic vector potential $\mathbf{A}$, which had initially 3 spatial components $A_x$,$A_y$,$A_z$ and was extended by a fourth (time) component $A_t$, now is extended again by a fifth component $A_l$ to a 5-vector

$$\mathbf{A}=\left(A_t,A_x,A_y,A_z,A_l\right).$$E14

The component along the $\mathrm{\ell }$-axis shall have constant length and may be slanted giving a picture (Figure 1) like the hairs of a horses coat or fur, in a pelt or like the tufts of a carpet:

Figure 1.

4-brane (x,y,z,t) with upright 5th component $\mathrm{\ell }$ (axes of a pyramid).

$\mathrm{\Phi }$ is the component orthogonal to all spacetime components, which are also perpendicular to each other, so the sum

$$A=\sqrt{A_t^2+A_x^2+A_y^2+A_z^2+{\mathrm{\Phi }}^2}$$E15

is the length of A. To show the capabilities of the Ricci curvature tensor in 5D we take the Minkowski metric and add the 5-vector A as fifth row and column.

$$g^{\mathit{AB}}=\left(\begin{array}{ccccc}1& 0& 0& 0& A_t\\ 0& -1& 0& 0& A_x\\ 0& 0& -1& 0& A_y\\ 0& 0& 0& -1& A_z\\ A_t& A_x& A_y& A_z& -{\mathrm{\Phi }}^2\end{array}\right)$$E16

With $A_l=1$ and $A_t=\phi =\frac{q}{r}$ the fifth diagonal element becomes

$${\mathrm{\Phi }}^2=A_l^2-A_t^2=1-{\phi }^2=1-\frac{q^2}{r^2}$$E17

and setting $A_x=A_y=A_z=0$ (no magnetic field) we get with $r=\sqrt{x^2+y^2+z^2}$:

$$g^{\mathit{AB}}=\left(\begin{array}{ccccc}1& 0& 0& 0& \frac{\mathbf{q}}{\mathbf{r}}\\ 0& -1& 0& 0& 0\\ 0& 0& -1& 0& 0\\ 0& 0& 0& -1& 0\\ \frac{\mathbf{q}}{\mathbf{r}}& 0& 0& 0& -\left(1-\frac{{\mathbf{q}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\right)\end{array}\right)$$E18

Applying the 5D Einstein tensor $G^{\mathit{AB}}=R^{\mathit{AB}}-\frac{1}{2}{g}^{\mathit{AB}}R$ multiplied by two we get (Maple-instruction: 2*Einstein[$\sim$]):

$$2\cdot G^{\mathit{AB}}=\left[\begin{array}{ccccc}-\frac{q^2}{r^4}& 0& 0& 0& -\frac{\sqrt{2}{q}^3}{r^5}\\ 0& {\left(\frac{\mathit{qx}}{r^3}\right)}^2-\frac{q^2}{2r^4}& \frac{\mathit{qx}}{r^3}\frac{\mathit{qy}}{r^3}& \frac{\mathit{qx}}{r^3}\frac{\mathit{qz}}{r^3}& 0\\ 0& \frac{\mathit{qy}}{r^3}\frac{\mathit{qx}}{r^3}& {\left(\frac{\mathit{qy}}{r^3}\right)}^2-\frac{q^2}{2r^4}& \frac{\mathit{qy}}{r^3}\frac{\mathit{qz}}{r^3}& 0\\ 0& \frac{\mathit{qz}}{r^3}\frac{\mathit{qx}}{r^3}& \frac{\mathit{qz}}{r^3}\frac{\mathit{qy}}{r^3}& {\left(\frac{\mathit{qz}}{r^3}\right)}^2-\frac{q^2}{2r^4}& 0\\ \frac{\sqrt{2}{q}^3}{r^5}& 0& 0& 0& -\frac{2q^2}{r^3}+1\end{array}\right]$$E19

With: ${\overrightarrow{e}}_r=\frac{x\widehat{i}+y\widehat{j}+z\widehat{k}}{\sqrt{x^2+y^2+z^2}},\overrightarrow{E}=\frac{q}{r}{\overrightarrow{e}}_r=\frac{q\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)}{{\left(x^2+y^2+z^2\right)}^{3/2}}$.

an identification of the electric field components in (19) is possible:

$$E^2=\frac{q^2\left(x^2+y^2+z\right)}{{\left(x^2+y^2+z^2\right)}^3}=\frac{q^2}{r^4}$$

$$E_x=\widehat{i}\cdot \overrightarrow{E}=\frac{\mathit{qx}}{{\left(x^2+y^2+z^2\right)}^{3/2}}=\frac{\mathit{qx}}{r^3}$$

$$E_y=\widehat{j}\cdot \overrightarrow{E}=\frac{\mathit{qy}}{{\left(x^2+y^2+z^2\right)}^{3/2}}=\frac{\mathit{qy}}{r^3}$$E20

$$E_z=\widehat{k}\cdot \overrightarrow{E}=\frac{\mathit{qz}}{{\left(x^2+y^2+z^2\right)}^{3/2}}=\frac{\mathit{qz}}{r^3}$$

$$E_x^2-\frac{1}{2}{E}^2=\frac{q^2{x}^2}{{\left(x^2+y^2+z^2\right)}^3}-1/2\frac{q^2}{{\left(x^2+y^2+z^2\right)}^2}$$

and it becomes obvious, that the first 4 rows and columns represent the electromagnetic energy-stress-momentum tensor $T_{\mathit{ik}}$ of (11) and (12).

Key result (i): This was only a demonstration which shows that the 5th row and 5th column are related to the electromagnetic 4-vector. Next has to be taken into account, that energy causes a gravitational field.

4. Field of a charged mass at rest in 5D

In the previous section was shown how the 5D Einstein tensor produces the electromagnetic energy-stress-momentum tensor $T_{\mathit{\mu \nu }}$. Because the 5D Einstein tensor already includes the electromagnetic energy-stress-momentum tensor, there is no need for anything besides the Ricci curvature tensor and solutions shall fulfill (symbols with a tilde mark 5D quantities) the 5D equation ${\tilde{R}}_{\mathit{AB}}=0$ which infers $\tilde{R}=0$ and ${\tilde{G}}_{\mathit{AB}}=0$.

We start with a 5D line element (with capital S as mark for a 5D quantity) in a form which can be found in ([8], p. 128):

$${\mathit{dS}}^2=\left(\frac{B}{E}-{\mathit{EA}}^2\right)d{\left(\mathit{ct}\right)}^2-\frac{1}{B}{\mathit{dr}}^2-r^{2}d{\theta }^2-r^{2}sin{\theta }^2-2\mathit{EA}\mathit{dt}\mathit{dl}-E{\mathit{dl}}^2$$E21

To be a solution of $R_{\mathit{AB}}=0$ the functions $A$, $B$, and $E$ have to be:

$$E\left(r\right)=1+\frac{q^2}{2\mathit{mr}},B\left(r\right)=\mathrm{E}-\frac{2m}{r},A\left(r\right)=-\frac{q}{\mathrm{E}r}$$E22

Inserting the solution into the the line element gives:

$$\begin{array}{l}{\mathit{dS}}^2=\left(\frac{1+\frac{q^2}{2\mathit{Mr}}-\frac{2M}{r}}{1+\frac{q^2}{2\mathit{Mr}}}-\frac{q^2}{1+\frac{q^2}{2\mathit{Mr}}}\right)d{\left(\mathit{ct}\right)}^2-\frac{1}{1+\frac{q^2}{2\mathit{Mr}}-\frac{2M}{r}}{\mathit{dr}}^2-r^{2}d{\theta }^2-r^{2}sin{\theta }^2\\ -2\frac{q}{r}\mathit{dt}\mathit{dl}-\left(1+\frac{q^2}{2\mathit{Mr}}\right){\mathit{dl}}^2\end{array}$$E23

In matrix form this is:

$${\tilde{g}}_{\mathit{AB}}=\left(\begin{array}{ccccc}1-\frac{2M}{r}& 0& 0& 0& \frac{q}{r}\\ 0& \frac{1}{1-\frac{2M}{r}+\frac{q^2}{2\mathit{Mr}}}& 0& 0& 0\\ 0& 0& -r^2& 0& 0\\ 0& 0& 0& -r^2{\left(sin\left(\theta \right)\right)}^2& 0\\ \frac{q}{r}& 0& 0& 0& -1-\frac{q^2}{2\mathit{Mr}}\end{array}\right)$$E24

Calculating the Ricci curvature with computer algebra software one gets:

$${\tilde{R}}_{\mathit{AB}}=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right)$$E25

The 5D-Monopole metric ${\tilde{g}}_{\mathit{AB}}$ of Eq. (24) thus is a solution of equation ${\tilde{R}}_{\mathit{AB}}=0$.

4.1 Projection of 5D solutions onto 4D space

We start with a common fully covariant 5D line element and use a latin letters to mark 5D components ([8], p. 118)

$${\mathit{dS}}^2={\gamma }_{\mathit{AB}}{\mathit{dx}}^A{\mathit{dx}}^B$$E26

To get the 4D line element and 4D metric tensor from 5D, the part along the fifth coordinate is split into an orthogonal part and a tangent part in relation to the 4D hyper-surface. A unit vector along the fifth coordinate is (because ${\gamma }_{55}={\gamma }_{51}^2+{\gamma }_{52}^2+{\gamma }_{53}^2+{\gamma }_{54}^2+{\mathrm{\Phi }}^2$):

$${\mathrm{\Psi }}^A\equiv \frac{{\delta }_5^A}{\sqrt{{\mathit{\epsilon \gamma }}_{55}}}{\gamma }_{\mathit{AB}}=\frac{1}{\sqrt{\mathit{\epsilon \gamma }55}}\left({\gamma }_{51},{\gamma }_{52},{\gamma }_{53},{\gamma }_{54},\mathrm{\Phi }\right)$$E27

This will enable us to split the 5D metric into a part parallel to ${\mathrm{\Psi }}^A$ and a 4D part orthogonal to it. We define a projector:

$$g_{\mathit{AB}}\equiv {\gamma }_{\mathit{AB}}-\epsilon {\mathrm{\Psi }}^A{\mathrm{\Psi }}^B$$E28

where by definition of $\mathrm{\Psi }$ follows

$$g_{\mathit{AB}}={\gamma }_{\mathit{AB}}-\frac{{\gamma }_{5A}{\gamma }_{5B}}{{\gamma }_{55}}$$E29

This has only 4 rows and 4 columns:

$$g_{55}=g_{5A}=0$$E30

so the 5D line element becomes the sum of a 4D line element and an extra part:

$${\mathit{dS}}^2=g_{\mathit{\mu \nu }}{\mathit{dx}}^{\mu }{\mathit{dx}}^{\nu }+{\gamma }_{55}{\left({\mathit{dx}}^5+\frac{{\gamma }_{5\alpha }}{{\gamma }_{55}}{\mathit{dx}}^{\alpha }\right)}^2$$E31

$${\mathit{ds}}^2\equiv g_{\mathit{\mu \nu }}{\mathit{dx}}^{\mu }{\mathit{dx}}^{\nu }$$E32

Let us introduce a 4-vector and a scalar field

$$A_{\mu }\equiv \frac{{\gamma }_{5\alpha }}{{\gamma }_{55}}$$E33

$${\mathrm{\Phi }}^2={\mathit{\epsilon \gamma }}_{55}$$E34

With these we get

$${\mathit{dS}}^2={\mathit{ds}}^2+\epsilon {\mathrm{\Phi }}^2{\left({\mathit{dx}}^4+A_{\mu }{\mathit{dx}}^{\mu }\right)}^2$$E35

This line element rewritten in matrix form is ([8], p. 135):

$${\tilde{g}}^{\mathit{AB}}=\left[\begin{array}{cc}{g}^{\mathit{\mu \nu }}& -A^{\mu }\\ -A^{\nu }& -\frac{1}{{\mathrm{\Phi }}^2}+A^{\alpha }{A}_{\alpha }\end{array}\right],{\tilde{g}}_{\mathit{AB}}=\left[\begin{array}{cc}{g}_{\mathit{\mu \nu }}-{\mathrm{\Phi }}^2{A}_{\mu }{A}_{\nu }& -{\mathrm{\Phi }}^2{A}_{\mu }\\ -{\mathrm{\Phi }}^2{A}_{\nu }& -{\mathrm{\Phi }}^2\end{array}\right]$$E36

With the assumption that the ${\tilde{g}}_{\mathit{AB}}$ are functions only of $x^{\mu }$ and do not depend on $x^5$, meaning

$$\frac{\partial }{\partial x^5}{\tilde{g}}_{\mathit{AB}}\left(x^{\mu }\right)=0$$E37

we get the following expression

$${\tilde{G}}^{\mathit{\alpha \beta }}\equiv \left[{\tilde{R}}^{\mathit{\alpha \beta }}-\frac{1}{2}{\tilde{g}}^{\mathit{\alpha \beta }}\tilde{R}\right]$$E38

$$=\left[R^{\mathit{\alpha \beta }}-\frac{1}{2}{g}^{\mathit{\alpha \beta }}R\right]+\frac{1}{2}{\mathrm{\Phi }}^2\left(F_{\lambda }^{\alpha }{F}^{\mathit{\beta \lambda }}-\frac{1}{4}{g}^{\mathit{\alpha \beta }}{F}^{\mathit{\mu \nu }}{F}_{\mathit{\mu \nu }}\right)-\frac{1}{\mathrm{\Phi }}\left({\mathrm{\Phi }}^{\alpha ;\beta }-g^{\mathit{\alpha \beta }}{\mathrm{\Phi }}_{;\mu }^{\mu }\right)$$E39

$$=G^{\mathit{\alpha \beta }}+\frac{1}{2}{\mathrm{\Phi }}^2{T}_{\mathit{em}}^{\mathit{\alpha \beta }}+T_{\mathit{sc}}^{\mathit{\alpha \beta }}$$E40

$$\equiv 0$$E41

Key result (ii): here the scalar field plays the role of the gravity constant via $\frac{1}{2}{\mathrm{\Phi }}^2$.

For the components with $A=5$ or $B=5$ two more equations can be derived: a Maxwell-like equation, where the source depends on the scalar field

$$F_{\alpha ;\lambda }^{\lambda }=-\frac{3}{\mathrm{\Phi }}{\mathrm{\Phi }}^{\lambda }{F}_{\mathit{\lambda \alpha }}$$E42

$${\mathrm{\Phi }}_{;\alpha }^{\alpha }=-\frac{1}{4}{\mathrm{\Phi }}^3{F}^{\mathit{\mu \nu }}{F}_{\mathit{\mu \nu }}$$E43

and a wave-like equation for the scalar field itself—Key result (iii).

5. Projection of a 5D-monopole onto 4D

To get the 4D-metric of the 5D-Monpole solution one has to perform the projection of Eq. (29). Because only the first 4 rows need a projection and components ${\gamma }_{52}$, ${\gamma }_{53}$ and ${\gamma }_{54}$ are zero, the only projection which has to be done is the one with ${\gamma }_{51}$:

$$g_{11}={\gamma }_{11}-\frac{{\gamma }_{51}^2}{{\gamma }_{55}}$$E44

We have ${\gamma }_{11}=\frac{B}{E}-{\mathit{EA}}^2$, ${\gamma }_{51}=-\mathit{EA}$ and ${\gamma }_{55}=-E$, so

$$g_{11}=\frac{B}{E}-{\mathit{EA}}^2-\frac{E^2{A}^2}{-E}=\frac{B}{E}=1-\frac{2m}{\mathit{rE}}=1-\frac{2m}{r\left(1+\frac{q^2}{2\mathit{mr}}\right)}=1-\frac{4m^2}{2\mathit{mr}+q^2}$$E45

The 4D line element is then

$${\mathit{ds}}^2=\frac{B}{E}d{\left(\mathit{ct}\right)}^2-\frac{1}{B}{\mathit{dr}}^2-r^{2}d{\theta }^2-r^{2}sin{\theta }^2$$E46

or as matrix

$$g_{\mathit{\mu \nu }}=\left(\begin{array}{cccc}\frac{-4m^2+2\mathit{mr}+q^2}{2\mathit{mr}+q^2}& 0& 0& 0\\ 0& \frac{2\mathit{mr}}{4m^2-2\mathit{mr}-q^2}& 0& 0\\ 0& 0& -r^2& 0\\ 0& 0& 0& -r^2{\left(sin\left(\theta \right)\right)}^2\end{array}\right)$$E47

Now that the 4D metric for the 5D-Monopole is known, one can calculate the 4D-Einstein tensor:

$$G_{\mathit{\alpha \beta }}=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& \frac{q^2}{r^2\left(2\mathit{mr}+q^2\right)}& 0& 0\\ 0& 0& \frac{q^2\left(20m^{3}r+4m^2{q}^2-4r^2{m}^2-4{\mathit{mq}}^{2}r-q^4\right)}{4\mathit{rm}{\left(2\mathit{mr}+q^2\right)}^2}& 0\\ 0& 0& 0& \frac{q^2{\left(sin\left(\theta \right)\right)}^2\left(20m^{3}r+4m^2{q}^2-4r^2{m}^2-4{\mathit{mq}}^{2}r-q^4\right)}{4\mathit{rm}{\left(2\mathit{mr}+q^2\right)}^2}\end{array}\right)$$E48

which after (38) is equal to $\frac{1}{2}{\mathrm{\Phi }}^2{T}_{\mathit{em}}^{\mathit{\alpha \beta }}+T_{\mathit{sc}}^{\mathit{\alpha \beta }}$ with $\frac{1}{2}{\mathrm{\Phi }}^2=g$.

As stated, the scalar field plays the role of the gravity constant $g=\frac{1}{2}\left(1+\frac{q^2}{2\mathit{mr}}\right)$.

Key result (iv): with smaller radius $r$ the gravity constant $g$ increases and the electromagnetic energy generates a stronger gravity field in the vicinity of the charge than from pure 4D theory.

Key result (v): Because $G_{00}$ is the energy component and is zero, this means scalar and electric energy cancel out and there are only pressure terms left to generate force and field lines.

5.1 Comparison with Reissner-Nordström metric

The 4D line element of the Reissner-Nordström metric is known as

$${\mathit{ds}}^2=\left(1-\frac{2m}{r}+\frac{q^2}{r^2}\right)d{\left(\mathit{ct}\right)}^2-\frac{1}{1-\frac{2m}{r}+\frac{q^2}{r^2}}{\mathit{dr}}^2-r^{2}d{\theta }^2-r^{2}sin{\theta }^2$$E49

which is as matrix

$$g_{\mathit{\mu \nu }}=\left(\begin{array}{cccc}1-\frac{2m}{r}+\frac{q^2}{r^2}& 0& 0& 0\\ 0& {\left(1-\frac{2m}{r}+\frac{q^2}{r^2}\right)}^{-1}& 0& 0\\ 0& 0& -r^2& 0\\ 0& 0& 0& -r^2{sin}^2\left(\theta \right)\end{array}\right)$$E50

Both are solution of

$$G_{\mathit{\mu \nu }}=R_{\mathit{\mu \nu }}-\frac{1}{2}R=\frac{q^2}{r^4}{g}_{\mathit{\mu \nu }}\left(=T_{\mathit{\mu \nu }}\right)$$E51

The electromagnetic stress-energy-momentum tensor as matrix:

$$T_{\mathit{\mu \nu }}=\left(\begin{array}{cccc}\frac{-2{\mathit{mq}}^{2}r+q^4+q^2{r}^2}{r^6}& 0& 0& 0\\ 0& \frac{q^2}{2{\mathit{mr}}^3-q^2{r}^2-r^4}& 0& 0\\ 0& 0& -\frac{q^2}{r^2}& 0\\ 0& 0& 0& -\frac{q^2}{r^2}{sin}^2\left(\theta \right)\end{array}\right)$$E52

Key result (vi): The Reissner-Nordström solution has a negative inverse $r$ term $\frac{2m}{r}$ and a positive inverse $r^2$ term $\frac{q^2}{r^2}$ which can result in 2 event horizons. The 4D projection of the 5D Monopole can have only one event horizon.

Both differ only in the near field. The expansion of $g_{00}$ of the 4D projection of the 5D Monopole shows a vanishing difference to $g_{00}$ of the Reissner-Nordström solution with increasing $r$:

$$\mathbf{1}-\frac{\mathbf{2}\mathbf{m}}{\mathbf{r}}+\frac{{\mathbf{q}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}-\frac{1}{2}\frac{q^4}{m{r}^3}+\frac{1}{4}\frac{q^6}{m^2{r}^4}-\frac{1}{8}\frac{q^8}{m^3{r}^5}+O\left(r^{-6}\right)$$E53

6. Neutral soliton solution with a static scalar field

Setting the electromagnetic 4-potential components $g_{5\alpha }=0$ results in an electrically neutral solution. Only $g_{55}$ is nonzero and varies with distance $r$. With a function

$$A\left(r\right)=1-\frac{2m}{r}$$E54

the line-element of the solution is

$${\mathit{ds}}^2=A^a{\mathit{dt}}^2-A^{-\left(a+b\right)}{\mathit{dr}}^2-A^{1-a-b}{r}^2\left(d{\theta }^2+sin{\theta }^2\right)-A^{b}d{\mathrm{\ell }}^2$$E55

which is in matrix form

$$\left(\begin{array}{ccccc}{\left(1-\frac{2m}{r}\right)}^a& 0& 0& 0& 0\\ 0& -{\left(1-\frac{2m}{r}\right)}^{-a-b}& 0& 0& 0\\ 0& 0& -r^2{\left(1-\frac{2m}{r}\right)}^{1-a-b}& 0& 0\\ 0& 0& 0& -r^{2}sin{\left(\theta \right)}^2{\left(1-\frac{2m}{r}\right)}^{1-a-b}& 0\\ 0& 0& 0& 0& {\left(1-\frac{2m}{r}\right)}^b\end{array}\right)$$E56

Calculating the Ricci tensor for this metric results in

$$\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& \frac{2m^2\left(a^2+2\mathit{ab}+b^2-1\right)}{r^2{\left(-r+2m\right)}^2}& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right)$$E57

and requests the condition:

$$a^2+\mathit{ab}+b^2-1=0$$E58

Since $a\ne b$ in general, the metric (52) is bivalent in the sense that it has gravitational and scalar contributions to the energy. Calculating the total energy of a soliton gives (a + b/2) M [7] where M is the mass seen from infinity.

Key result (vii): since there are no electromagnetic sources in the metric (56) and all of its associated matter outside of the central source comes from the last term, the logical consequence is that the cloud of radiation around a soliton is not photons but scalerons, or quanta of the scalar field (radions, dilatons, axions).

6.1 Neutral soliton solution with a dynamic scalar field

There is also a dynamic (isotropic) solution of a soliton, where the scalar field quanta vanish to infinity whith an analog of a Hubble-constant leading to a Black Hole without a scalar field:

$$\begin{array}{l}{\mathit{ds}}^2={\left(\frac{\mathit{ar}-1}{\mathit{ar}+1}\right)}^{\frac{4}{\sqrt{3}}}{\mathit{dt}}^2-\left(\frac{a^2{r}^2-1}{a^2{r}^2}\right){\left(\frac{\mathit{ar}-1}{\mathit{ar}+1}\right)}^{\frac{2}{\sqrt{3}}}\left(1+\mathit{Ht}\right)d{\sigma }^2\\ -\left(\frac{a^2{r}^2-1}{a^2{r}^2}\right){\left(\frac{\mathit{ar}-1}{\mathit{ar}+1}\right)}^{\frac{2}{\sqrt{3}}}{\left(1+\mathit{Ht}\right)}^{-1}d{\mathrm{\ell }}^2\end{array}$$E59

7. Induced matter

In the preceding section, there was no explicit dependence on $\mathrm{\ell }$. Now, we allow all potentials $g_{\mathit{AB}}$ to depend on $\mathrm{\ell }$ and again work with electric neutral matter (see matrix form in Eq. (36) with condition Eq. (37)):

$$g_{\mathit{\alpha \beta }}=g_{\mathit{\alpha \beta }}\left(x^A\right)g_{5\alpha }=0g_{55}=\mathrm{\Phi }{\left(x^A\right)}^2$$E60

$${\mathit{dS}}^2=g_{\mathit{AB}}{\mathit{dx}}^A{\mathit{dx}}^B$$E61

Next, we can split the 5D Ricci tensor in its 4D analog and terms belonging to the 5th dimension, which can be taken as terms of a stress-energy-momentum tensor. Derivations of the 5th dimension with respect to $\mathrm{\ell }$ are denoted with a star:

$$\begin{array}{c}{\tilde{R}}_{\mathit{\alpha \beta }}=R_{\mathit{\alpha \beta }}+{\left({\mathrm{\Gamma }}_{\mathit{\alpha \beta }}^4\right)}_{,4}-{\left({\mathrm{\Gamma }}_{\alpha 4}^4\right)}_{,\beta }+{\mathrm{\Gamma }}_{\mathit{\alpha \beta }}^{\lambda }{\mathrm{\Gamma }}_{\lambda 4}^4+{\mathrm{\Gamma }}_{\mathit{\alpha \beta }}^4{\mathrm{\Gamma }}_{4D}^D-{\mathrm{\Gamma }}_{\mathit{\alpha \lambda }}^4{\mathrm{\Gamma }}_{\beta 4}^{\lambda }-{\mathrm{\Gamma }}_{\alpha 4}^D{\mathrm{\Gamma }}_{\mathit{\beta D}}^4\\ =R_{\mathit{\alpha \beta }}-\frac{{\mathrm{\Phi }}_{\alpha ;\beta }}{\mathrm{\Phi }}+\frac{1}{2{\mathrm{\Phi }}^2}\left(\frac{\stackrel{\ast }{\mathrm{\Phi }}\stackrel{\ast }{g}{}_{\mathit{\ast \alpha \beta }}}{\mathrm{\Phi }}-{\stackrel{\ast \ast }{g}}_{\mathit{\alpha \beta }}+g^{\mathit{\mu \nu }}{\stackrel{\ast }{g}}_{\mathit{\alpha \lambda }}{\stackrel{\ast }{g}}_{\mathit{\beta \mu }}-\frac{g^{\mathit{\mu \nu }}{\stackrel{\ast }{g}}_{\mathit{\mu \nu }}{\stackrel{\ast }{g}}_{\mathit{\alpha \beta }}}{2}\right)\\ =R_{\mathit{\alpha \beta }}-T_{\mathit{\alpha \beta }}+\frac{1}{2}{g}_{\mathit{\alpha \beta }}T\equiv 0\end{array}$$E62

Key result (viii): the dependence of the elements on the coordinate can generate mass, pressure, momentum and energy (tensor ${\mathbf{T}}_{\mathit{\alpha \beta }}$).

The fifth diagonal element of the 5D-Ricci tensor is also required to be zero:

$${\tilde{R}}_{55}=\mathrm{\Phi }\text{□}\mathrm{\Phi }-\frac{{\stackrel{\ast }{g}}^{\stackrel{\ast }{\lambda }\beta }{\stackrel{\ast }{g}}_{\mathit{\lambda \beta }}}{2}-\frac{g^{\stackrel{\ast \ast }{\mathit{\lambda \beta }}}{\stackrel{\ast }{g}}_{\mathit{\lambda \beta }}}{2}-\frac{\stackrel{\ast }{\mathrm{\Phi }}{g}^{\mathit{\lambda \beta }}{\stackrel{\ast }{g}}_{\mathit{\lambda \beta }}}{2\mathrm{\Phi }}-\frac{g^{\mathit{\mu \beta }}{g}^{\mathit{\lambda \sigma }}{\stackrel{\ast }{g}}_{\mathit{\lambda \beta }}{\stackrel{\ast }{g}}_{\mathit{\mu \sigma }}}{4}\equiv 0$$E63

Key result (ix): and yields a wave equation for the gravitational constant.

8. Geodesic motion in 5D and extension of Lorentz force

In the preceding section, a new energy-momentum-tensor $T_{\mathit{sc}}^{\mathit{\alpha \beta }}$ occurred. Its divergence should yield in force density (force per unit inertial mass). For that reason, the equation of motion is interesting. Using the Lagrangian approach minimizing the distance between two points in 5D, the equation of geodesic motion becomes ([7], p. 154):

$$\frac{{\mathit{du}}^{\mu }}{\mathit{ds}}+{\mathrm{\Gamma }}_{\mathit{\beta \gamma }}^{\mu }{u}^{\beta }{u}^{\gamma }+\left(g^{\mathit{\mu \alpha }}-\frac{1}{2}\frac{{\mathit{dx}}^{\mu }}{\mathit{ds}}\frac{{\mathit{dx}}^{\alpha }}{\mathit{ds}}\right)\frac{d\mathrm{\ell }}{\mathit{ds}}\frac{{\mathit{dx}}^{\beta }}{\mathit{ds}}\frac{\partial g_{\mathit{\alpha \beta }}}{\mathrm{\partial \ell }}=0$$E64

The first two terms describe geodesic motion in 4D, so we split the equation and define a force density:

$$\frac{{\mathit{du}}^{\mu }}{\mathit{ds}}+{\mathrm{\Gamma }}_{\mathit{\beta \gamma }}^{\mu }{u}^{\beta }{u}^{\gamma }=f^{\mu }$$E65

$$f^{\mu }\equiv \left(-g^{\mathit{\mu \alpha }}+\frac{1}{2}\frac{{\mathit{dx}}^{\mu }}{\mathit{ds}}\frac{{\mathit{dx}}^{\alpha }}{\mathit{ds}}\right)\frac{d\mathrm{\ell }}{\mathit{ds}}\frac{{\mathit{dx}}^{\beta }}{\mathit{ds}}\frac{\partial g_{\mathit{\alpha \beta }}}{\mathrm{\partial \ell }}=0$$E66

The new force density is nonzero if the 4D metric depends on the fifth coordinate $\left(\frac{\partial g_{\mathit{\alpha \beta }}}{\mathrm{\partial \ell }}\right)$ and there is motion in the fifth dimension $\left(\frac{d\mathrm{\ell }}{\mathit{ds}}\right)$. A force $F^{\mu }$ can be split into a normal component $N^{\mu }$ in relation to the 4D velocity and a tangent or parallel component $P^{\mu }$ in relation to the 4D velocity, thus $F^{\mu }=N^{\mu }+P^{\mu }$ giving force densities:

$$n^{\mu }=\left(-g^{\mathit{\mu \alpha }}+u^{\mu }{u}^{\alpha }\right)u^{\beta }\frac{\partial g_{\mathit{\alpha \beta }}}{\mathrm{\partial \ell }}$$E67

$$p^{\mu }=-\frac{1}{2}{u}^{\mu }\left(u^{\alpha }{u}^{\beta }\frac{\partial g_{\mathit{\alpha \beta }}}{\mathrm{\partial \ell }}\right)\frac{d\mathrm{\ell }}{\mathit{ds}}$$E68

Key result (x): the force $\mathbf{P}$ parallel to 4D velocity $\mathbf{u}$ is new and occurs only if there is fifth dimension dependency ([7], p. 157). In the preceding section, we had condition (37) wherefore there a fifth force does not occur.

9. Conclusions

From five dimensional metrics with unrestricted dependence from all five coordinates resolving the equation $R_{\mathit{AB}}=0$ many inductively gained physical laws can be deduced: Newton’s gravity law, Coulomb law, Faraday induction law, Ampère’s circuital law, Maxwell Equations, Einstein’s Gravity law, Einstein-Maxwell equation. Because of further degrees of freedom a scalar wave equation for the gravity constant is gained, yielding in an additional energy and an additional force. For 4D assemblies a reduction of inertia is predicted. The gravity constant corresponds to a scalar field which is part of the 5D metric as the 15th potential.

This 15th component of the metric varies in the vincinity of a charge and seems to behave like the potentials of the electromagnetic 4-vector-potential which is able detach from the sources, giving reason to the assumption that changes of the gravity constant could emerge into the surrounding spacetime.

Aiming at the reduction of inertia and an accelerating fifth force, for interstellar travel or asteroid deflection missions a change from analytical to numerical calculation methods will be necessary.

Recent experiments investigate the possible effects of a fifth force with neutron beam scattering on silicon crystals [9]. Neutrons have internally two negative charges and a double positive charge in the smallest space and in the near field the scalar potential causes the greatest deviations.

Acknowledgments

The author would like to thank the American Institute of Aeronautics and Astronautics (AIAA) for kindly allowing parts of this chapter to be adapted from a conference paper published in the ASCEND 2021 conference. Since most of the content presented is introductory and basic, many passages and figures are taken from that paper. The focus here is on the didactic presentation of the results in the form of 10 key statements.

Nomenclature

a	acceleration
A	electromagnetic vector potential
Aμ	component of electromagnetic 4-vector
B	magnetic field
β	rapidity vc
c	velocity of light, displacement of a field source
ds	line element, 4D
dS	line element, 5D
dt	time step multiplied with c (velocity of light)
E	electric field
ε	sign + or – (factor + 1 or − 1)
ε0	dielectric constant
f	force density (force per unit inertial mass)
F	force, electromagnetic Field tensor, Farady tensor
Fμν	component of the Farady tensor
φ	electric potential
Φ	component along the 5th coordinate
g	metric tensor, 4D
gμν	component of 4D metric tensor
gAB	component of 5D metric tensor
γ	gravity constant, index for 4D, Lorentz factor 1/1−β2, 5D metric tensor
Γkli	Christoffel symbol of the second kind
î,ĵ,k̂	Cartesian unit basis vector (Maple notation)
j	current density
ℓ	fifth coordinate (besides x, y, z, ct)
m	mass
μ0	magnetic permeability
q	charge
r	radial distance
R	curvature scalar
Rμν	Ricci curvature tensor, 4D
RAB	Ricci curvature tensor, 5D
Rμνκσ	Riemann curvature tensor
ρ	charge density
S	Poyting vector
σ	Maxwell stress tensor
Tμν	electromagnetic stress energy tensor
v	velocity
x,y,z	Cartesian component indexes (of a force)
xi	covariant coordinates
xi	contravariant coordinates