Relativistic gravitational force

Armed with a gravitational potential where the derivative has the correct distance dependence, it is time to examine the relativistic force. The electromagnetic Lorentz force density arises from moving charges in an electromagnetic field:

\end{displaymath} (21)

If the sign of the electric charge $q$ where changed, this would change the sign of the 4-force, so there are two distinguishable signs for electric charge. Like electric charges repel because the force has a positive sign.

An analogous gravitational Lorentz force can be created with precisely the same substitutions as were used for the Lagrange density:

\end{displaymath} (22)

If the sign of the mass current were changed, then the sign of the inertial mass density in the force would also need to change, resulting in no change, so there is only one distinguishable sign for mass charge. Like mass charges attract because the force has a negative sign.

Plug the weak-field gravitational approximation (20) into the gravitational Lorentz 4-force equation (22), assuming the change in momentum is the product of mass charge density times velocity:

\end{displaymath} (23)

This is the electrically neutral, weak gravitational field Lorentz 4-force density. These are first-order differential equations. The weak equivalence principle equates inertial and passive gravitational mass densities. This principle will be assumed, but is supported by experiments.[16] So that the equations have the same variable, substitute the interval $c^{2}\tau^{2}$ for the distance $-\sigma^{2}$:

$\displaystyle \frac{dU^{0}}{d\tau}-\frac{k}{c\tau^{2}}U^{0}$ $\textstyle =$ $\displaystyle 0$ (24)
$\displaystyle \frac{d\bi{U}}{d\tau}+\frac{k}{c\tau^{2}}\bi{U}$ $\textstyle =$ $\displaystyle 0.$ (25)

Solve for the velocity, $U^{\mu}$. The solution involves exponentials with velocity constants:

U^{\mu}=(v\: exp(-\frac{k}{\tau}),\bi{V}\: exp(\frac{k}{\tau})).
\end{displaymath} (26)

For flat spacetime, $U^{\mu}=(v,\overrightarrow{V})$. The constraint on relativistic velocities in flat spacetime is:

\end{displaymath} (27)

This will be the case for the velocity solution should the spring constant be zero or the interval infinite. Substitute the solution (26) into the constraint (27) multiplying through by $d\tau^{2}$.

d\tau^{2}=exp(-2\frac{k}{\tau})dt^{2}-\: exp(2\frac{k}{\tau})(dR/c)^{2}.\end{displaymath} (28)

This metric equation represents the weak-field gravitational Lorentz 4-force solution (26) under the constraint that it yields a flat spacetime metric for a zero spring constant $k$ or infinite interval $\tau$. The effect of a weak gravitational force is to bend spacetime.

Assume the spring constant $k$ is due to an active geometric mass, $k=GM/c^{2}$. Assume the field is static, so $\sigma^{2}=R^{2}-(ct)^{2}\cong R'^{2}$. Both sigma and tau must have the same magnitude, $R$. To express the metric in terms of tau real, assume sigma is imaginary, so tau is real:

d\tau^{2}=exp(-\frac{2GM}{c^{2}R})dt^{2}-exp(\frac{2GM}{c^{2}R})(dR/c)^{2}.\end{displaymath} (29)

The exponential metric derived from the gravitational Lorentz force law is the same as the solution to the field equations (14), a test of self-consistency. On aesthetic grounds, it is nice to see that the smallest steps away from flat spacetime uses Nature's favorite function, the exponential.

The exponential metric is not a solution to the Einstein field equations. If one chooses a constant potential, the exponential metric is a singular solution to the GEM field equation that depends only on the Christoffel symbol, 13. If instead one chooses a flat Minkowski metric, the potential solution would be the singular $1/R$ potential. The two types of singular solutions with different gauge choices is a stringent test of the consistency of the model.

doug 2005-11-18