Article 12:
Deriving Equations for Deflection, Splay and Critical Distances

This is a supplementary article. To go to the main introductory article about Stationary Energy Theory, and its links to other supplementary articles, please click here.

Knowing the amount of deflection of BTTPs at the sun’s photosphere enables us to calculate the deflection caused by any object, at any distance, and this lets us calculate gravitational critical distances.

The deflection of BTTPs at the sun’s photosphere is, as we found in Article 8, α = 0.87555”, or α = (0.87555/(60 x 60))π/180 = 4.24479 x 10-6 radians. Since the deflection is in proportion to the mass of the deflecting object, and, as discussed previously, inversely proportional to the distance from the object, we can derive a general equation for deflection of:

α = Mass in Solar Masses x 4.24479 x 10-6 radians/Distance in solar radii

This equation would be more useful if it were in terms of light years, and since there are 13,598,852.68 solar radii in a light year, we can rewrite the equation as:

α = Mass in Solar Masses x ((4.24479 x 10-6)/(13,598,852.68))/Distance in ly, or:

α = msuns x 3.121432925 x 10-13/dly radians             .            .            .            (22)

As we mentioned in the gravitation section, over considerable distances there is a splay in the direction of BPs moving forward in time due to the curvature of the Universe, and when this splay exceeds the deflection of BTTPs, then the gravitational force becomes repulsive. The angle of deflection, α, when two objects interact, is the deflection on just one side of the interaction, so the corresponding splay that needs to exceed the deflection to turn the gravitational intensity factor, I, negative, is just one half of the total splay angle between two objects. At one radian separation, 46 billion light years, there is one radian (57.29578°) of total splay, so the splay on each side is half this, or half a radian. Based on this we can calculate the splay (σ), as follows, from distance in light years (dly):

σ = 0.5 x dly/4.6000 x 1010 = dly x 1.0870 x 10-11 radians

This splay equation is based on the assumption that the Universe is exactly spherical. If there are areas that are flattened out, and areas that have a greater curvature to make up for the flatter areas, then the splay will change in those areas. To account for these, I will introduce a “flatness factor”, “f”, which equals 1 at the exactly spherical curvature, and increases or decreases with the effective radius of curvature, so that, for instance, if the curvature is flattened out half way to being flat, where its radius of curvature is doubled, then f = 2. The splay will reduce in proportion to how much “f” increases, so the general equation becomes:

σ = dly x 1.0870 x 10-11/f radians      .            .            .            .            .  (23)

What is most useful to know about deflection and splay is the critical distance for any two objects, when gravitational attraction turns to repulsion. This is the distance where the deflection caused by the larger object equals the splay due to the distance — that is when σ = α. In terms of Equations 22 and 23, this is when:

dly x 1.0870 x 10-11/f = msuns x 3.1214 x 10-13/dly

Multiplying both sides by f and dly, and dividing both sides by 1.0870 x 10-11, we get:

dly˛ = (3.1214 x 10-13/1.0870 x 10-11) x msuns x f, or

dly˛ = 2.87157 x 10-2 x msuns x f

Taking the square root of each side we get:

dly = 0.16946 x (msuns x f)˝

Since this is the critical distance in light years, let’s call it “dcly”, so:

dcly = 0.16946 x (msuns x f)˝            .            .            .            .            .          (24)

In standard SI units of meters and kilograms, the equation becomes:

dc = 1.1368 x (mf)˝  .            .             .            .            .            .            .      (25)

Galactic Implications of Critical Distance

Applying Equation 24 to our sun and another object weighing less than or equal to our sun, assuming the average curvature of space where f = 1, we get:

dc = 0.16946 x (1 x 1)˝ = 0.16946 ly

This would mean our sun would only attract objects less than or equal to its own mass if they were less than about one sixth of a light year away! While 1/6 ly is still about 28 times as far as comet Hale-Bopp reaches at aphelion, so such a critical distance might be able account for how the sun attracts everything in the solar system, it would mean the sun, rather than attract nearby stars, would actually repel them! If there were considerable flattening of space in our part of the Universe — our Local Supercluster of galaxies — such that f = 100, the critical distance of our sun would still only be 1.69 ly, and it would still slightly repel nearby stars. Of course, in this situation, our galaxy would still attract the stars in it. Our galaxy and the Andromeda Galaxy each have a mass of about 7.1 x 1011 times the sun, so with f = 100 again, its power of attraction would extend to:

dc = 0.16946 x (7.1 x 1011 x 100)˝ = 0.16946 x (71 x 1012)˝ = 1.4280 x 106 ly

This critical distance for our galaxy of about 1.4 million light years would certainly be sufficient for it to attract all its stars and even its satellite galaxies, such as the Magellanic Clouds, but since the Andromeda Galaxy, M31, is about 2.36 million light years away (and about the same mass as our galaxy), it would mean the Milky Way and the Andromeda Galaxy would be slightly repelling each other. This is rather unlikely (though possible), considering these galaxies are moving toward each other with a speed of 120 km/s. What would be much more likely would be a minimum flattening of about f = 300, which would make the critical distance between the two galaxies 2.47 million light years, so they would be slightly attracting each other at their current distance apart. This would be an absolute minimum, though, and when a variety of factors are considered, especially the interactions of molecules, which we will look at later, it seems likely there would be a flattening of about f = 2,500 within our Local Supercluster.

For f = 2,500, the critical distance for our sun would be close to 8.5 light years. A star one fourth the mass of our sun would then have a critical distance of about 4.25 light years, and one four times the mass of our sun would have a critical distance of about 17 light years. Critical distances of this order would ensure stars in globular clusters, in the central bulges of galaxies, and even in open clusters, would attract each other. Stars in the spiral arms of our galaxy and other spiral galaxies, on the other hand, would be far enough apart that their gravitational attraction would be significantly diminished or they would slightly repel each other, making it less likely they would have the kind of close interactions that are common in clusters, helping to give planetary systems around suitable single stars in the spiral arms the stability needed for life to evolve.

So, it seems like space in the neighborhood of galaxies has a “flat” (Euclidian) geometry to a fairly high degree of flattening, probably about f = 2,500. What, then, could be the cause of this flattening? This will be discussed in the next article in this series: “Structure of the Universe — Local Flattenings.”

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