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EliminatingInertia

Stephen Crowley edited this page Jul 21, 2023 · 17 revisions

Following Sciama (4) and Jefimenko (5,6), the vector potential $A(\mathbf{r}, t) \forall r \in \mathbb{R}^3,t \in \mathbb{R}$ is given by

$$A(\mathbf{r}, t) = \frac{G}{c^2}\int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{\rho(\mathbf{q}, s) \mathbf{v}(\mathbf{q}, s)}{|\mathbf{r}-\mathbf{q}|} dV_q ds$$

where $\rho(\mathbf{q}, s)$ is the mass density and $\mathbf{v}(\mathbf{q}, s)=-ν$ is the (constant) velocity (function) at a position $\mathbf{q} \in \mathbb{R}^3$ in the universe and at a specific time $s \in \mathbb{R}$.

In the integral, $dV_q$ denotes a small volume element located at position $\mathbf{q} \in \mathbb{R}^3$ in the universe. It is used when integrating a function over a region of space, which in this case is the entire universe $\mathbb{R}^3$. The small volume element $dV_q$ represents an infinitesimally small part of this space. By integrating over all such volume elements, we're effectively summing the contribution from every point in the universe.

The vector potential $A(\mathbf{r}, t)$ created by the remote stars postulated by Mach’s principle is obtained by integrating over the mass distribution of the entire universe, assuming the universe to be receding with velocity $-\nu$. The velocity being constant comes out of the integral

$$A(\mathbf{r}, t) = \frac{(-ν)G}{c^2} \left(\int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{\rho(\mathbf{q}, s)}{|\mathbf{r}-\mathbf{q}|} dV_q ds\right)$$

and with

$$\Phi(\mathbf{r}, t) = G\left(\int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{\rho(\mathbf{q}, s)}{|\mathbf{r}-\mathbf{q}|} dV_q ds\right)$$

we are left with

$$A(\mathbf{r}, t) = \frac{-\nu\Phi(\mathbf{r}, t)}{c^2}$$

If we look at the definition of $A(\mathbf{r}, t)$, we see that there is a possible multiplier, the velocity $-\nu$.

$$A(\mathbf{r}, t) = \int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{\rho(\mathbf{q}, s) \mathbf{v}(\mathbf{q}, s)}{|\mathbf{r}-\mathbf{q}|} dV_q ds$$

Therefore, these equations indicate that if a negative ring is spun a magnetic dipole field $A$ is created. If we spin the ring faster and faster, over a million revolutions per second, we can get a finite field. Therefore we create a bubble in space of negative vector potential. If an object were to be within this bubble, it would be shielded from the positive vector potential arising out of the universe's background scalar potential. In other words, the object would be shielded from the fields that create inertia; it would not be subject to inertia.

Constant Velocity Assumption

There is an assumption that the universe is receding with a constant velocity, which we can denote as $-\nu$ where $\nu$ is a positive constant. This assumption simplifies the expression for the vector potential $A(\mathbf{r}, t)$, allowing $\nu$ to be factored out of the integral:

$$A(\mathbf{r}, t) = \frac{(-\nu) G}{c^2} \left(\int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{\rho(\mathbf{q}, s)}{|\mathbf{r}-\mathbf{q}|} dV_q ds\right)$$

This model posits a universe in a state of uniform recession, consistent with some simplified cosmological models. In more comprehensive models, the velocity could potentially vary with both time and position, and the integral expression for $A(\mathbf{r}, t)$ would need to account for these variations.

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