The end is the beginning

If you visualize any mass (eg Planet Earth) - and draw a number of concentric circles around it - you will have a sketch showing orbits around the globe. Note that even the surface of the Earth is an orbit with zero altitude. My theory is that all the orbits are the same length. Whilst they may appear to be of different lengths - the density of space lessens as you get further out from the center. This means that whilst a space ship appears to be travelling further on the outer orbits - with the space being thinner - the distance travelled through space is the same along all orbits. In some ways this logic reinforces the Flat Earth argument by throwing in space/time variables to support the claims.


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Jan 27, 2020
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greywolf -

Could you kindly explain what your statement "with the space being thinner" means?

Secondly, using C = 2πR, explain how any set of circumferences with increasing radii, have equal circumferences?

Hartmann352
 
greywolf -

Could you kindly explain what your statement "with the space being thinner" means?

Secondly, using C = 2πR, explain how any set of circumferences with increasing radii, have equal circumferences?

Hartmann352
Hi - "Space being thinner" = If one imagines travelling in an orbit 1000 miles above the Earth and comparing it to swimming in a river and then imagine travelling in an orbit 100,000 miles above the Earth where it would feel like swimming in air. The fabric of space is less dense. So to cover the same distance through that medium you would have to go a lot further in geometrical terms. Note the orbital distances drawn on paper are geometrical distances that assume that the fabric of space is uniformly dense. This should also explain why C = 2πR doesn't work in an environment such as we are discussing. Thanks for your reply Hartman352
 
Mar 4, 2020
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Using a helix for a circumference, allows one to keep an absolute length or amount of travel. By varying the diameter of the helix...OR the number of turns......ANY C/D ratio(pi) can be anything greater than 3.14. A variable (pi) rotation.

Many multiples of diameters can be applied to one circumference. It happens to be the charge particle dynamic.

I don't know what skinny space is.
 
Using a helix for a circumference, allows one to keep an absolute length or amount of travel. By varying the diameter of the helix...OR the number of turns......ANY C/D ratio(pi) can be anything greater than 3.14. A variable (pi) rotation.

Many multiples of diameters can be applied to one circumference. It happens to be the charge particle dynamic.

I don't know what skinny space is.
HI Hayseed - Space that is skinny is where the Helix has a larger diameter than less skinny space. Or moving through dense space is like swimming through crude oil and skinny space is like swimming in air. The constant is how far you have travelled through the fabric of space. Swimming 1 mile through crude oil = the same as swimming 20 miles in air.
 
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If that's true, why don't we see objects shrink and expand with the tides? Do all objects, including me, shrink and expand together, so that I can't notice? Wouldn't we notice that distance between objects get smaller and larger with the tides? As soon as I change size, the distance to every other object, changes. Unless they change opposite at same rate.

Does this space density change the speed c? or only the direction of c? If c is constant, we should be able to measure this respiration. This in and this out.

Let's dig a deep bore hole. 10 miles deep. We insert a 20 mile, precision machined shaft into the bore. 10 deep and 10 high. Rotate shaft one turn.

Did all of that shaft, rotate one turn at the same rate and duration? If you put atomic clocks at the top, middle and bottom...your atomic clocks will tell you different RPMs. The atomic clock makers will tell you that the size and diameter of the shaft is what changed. Their clocks are perfect. And proves that gravity or any acceleration changes not only length but time itself.

What is really changing? Your clock's tic rate is the only thing that is changing. All vibrations and oscillations change with a gravity gradient. Or acceleration.

However, a rotation, incident to a gravity gradient will remain constant. A rotational ticker is needed for a stable clock.

Vibrations and oscillations(usually 3 axis) are the dynamics we use, to measure accelerations, because they are so sensitive to them. And their directions.

Patent a rotational clock and retire. Your great grand kids could retire. To this very day, we still need a stable clock. And until then, you're stuck on space time.

Imagine the accelerations and de-accelerations, the complete stoppage and reversal in directions... necessary for one oscillation. AND the durations of each component.

Now image a dynamic with one or two constant accelerations....with one constant duration.

AN oscillation is a complex ever changing dynamic, a rotation is a simple constant dynamic.

An atomic oscillation is even worse, it has many more components than our single frequency oscillators. It's a spectrum of vibrations. Complex with many interference paths.

These atomic vibrations are now being used for "quantum detectors" to sense extremely small accelerations and fields. And measure gravity fields.

So, what do you think about it? Do you believe in local time? Mass "time zones". If I installed a shaft to the moon, would it take time for the rotation to get there?
 
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Would it be possible to instantly communicate with the moon thru a shaft rotation? Information transfer, faster than light? Can an instant transfer be made? Does one end of a long shaft lag the other end? Can one end, start to turn, before the other?

Aren't both ends turned at the same time?

So, with a shaft, I can put my time, anywhere. But they say with light, the receiver must change time zones, in order to catch c's velocity.

Ah, the mysteries of space time. Is a rotating shaft immune to space time?

Instead of rotation, let's use push/pull. Would that be instant? Could I actually push a rod, faster than light?
 
Jan 27, 2020
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Hi - "Space being thinner" = If one imagines travelling in an orbit 1000 miles above the Earth and comparing it to swimming in a river and then imagine travelling in an orbit 100,000 miles above the Earth where it would feel like swimming in air. The fabric of space is less dense. So to cover the same distance through that medium you would have to go a lot further in geometrical terms. Note the orbital distances drawn on paper are geometrical distances that assume that the fabric of space is uniformly dense. This should also explain why C = 2πR doesn't work in an environment such as we are discussing. Thanks for your reply Hartman352
Greywolf -

I understand the idea of varying densities and that atmospheric density is inversely proportional to altitude.

However, one can't simply say "space" alone, because it is called "space-time". When Einstein concocted his general theory of relativity, one of the great advances was to recognize that space and time were combined into a single entity: spacetime. Another was that the presence of matter and energy curved the very fabric of this spacetime, and that curved spacetime, in turn, dictated how matter moved.

If you’ve ever seen a picture of a bent, two-dimensional grid with masses sitting on it representing space, you’ll know this type of illustration is extremely common. It appears to depict the "fabric of space" as being curved by the presence of mass, and therefore, any other particle traveling along this fabric will have its path bent towards this gravitational source. The larger the mass and the closer you get to it, the larger the curvature, and therefore, the larger the observable bending.

It appears as though a mass somehow gets pulled “down” onto the fabric, and then the other particles traveling through that space are pulled “down” by some unseen, mysterious force as well. Additionally, the grid lines curve away from, rather than towards, the mass, which also can’t be right, especially if gravity is attractive.

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The spacetime curvature around any massive object is determined by the combination of mass and distance from the center-of-mass. However, this two-dimensional grid-like depiction of spacetime isn’t necessarily the most accurate way to perceive it. (T. PYLE/CALTECH/MIT/LIGO LAB)

Gravity simply exists, while the equations that describe General Relativity are geometric in nature. The idea that mass-and-energy curves space is correct, even though this naive visualization is wrong when applied to real world space-time.

Greywolf says, "Swimming 1 mile through crude oil = the same as swimming 20 miles in air."

I think you have confused matter density and energy usage as opposed to matter density and distance. While the energy expended by swimming a mile through oil will be greater than swimming through distilled water or spring water, you should use apples to apples, etc., by using the same distance in each case when discussing the energy expended through actions carried out in different densities.

When speaking of distance in space-time, I don't see where simple geometry would not apply. But regardless of what fluid density one swims in or through, the set distances covered will, in each case, be the same.

Concentric orbital rings around a black hole of a million Solar masses, outside of the Schwarzchild radius, will still equate to C = 2πR, regardless of the amount of infalling matter which must be plowed through while navigating a particular circumference.

The effect that such matter density encountered will have is to increase the time the distance is travelled, thus determining speed, not distance. Speed = Distance/Time, and it's iterations (Speed is directly Proportional to Distance and Inversely proportional to Time. Distance = Speed X Time, and
Time = Distance / Speed) which tell us how slow or fast an object, and we, move. (See: https://testbook.com/learn/maths-speed-time-and-distance/)

Thus, the distance travelled divided by the time taken to cover the distance equals the speed of travel. Matter density within that latter distance will affect the time needed to travel a particular distance, not increase the absolute distance.
Hartmann352
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