Relativity Geometry Inquiry

[arkain101]

See animation above (post 17). Quite interesting.

 

 

When the Velocity Line nears "C" in length(velocity) it reaches the center of the circle, and the dilation ratio between C and dilation side(:hihi: would hit singularity,infinity.

 

Let radius of circle = hypotenuse of triangle system.

 

However the rate of changes seem relavent to the curvature of dilations calcualted in SR.

[arkain101]

Using this geometry system, I produced some images of various dimension demonstrations. Attatched below, and put an effort into merging it with QM and SR calculations, as one united geometry.

 

Consider it like this.

 

In the animation.

 

Black line = E field = Y axis

 

Red line = B field = X axis

 

Blue Radius line = t = time = Z axis

 

Let us say that each radius is rotates around the center axis (at the same rate)

 

In the animation, we are viewing one axis face on, like a clock, and we will call this t.

 

The black line is rotating vertical on the y axis.

 

The red line is rotating horizontal on the x axis.

 

However, since we view this in 2D we see the black and red lines only change in length while they can be thought to rotate in 3D like the "t" radius..

 

Each time the 't=radius' rotates 180 degrees, or 1/2 a spin it comes in contact with B+(left side) B- (right side) on the x axis (B=radius), and E+(top side) and E-(bottom side) on the y axis (E=radius) when each of the two raidus' (ie; t and B) are at 100% length relative to our perspective of viewing head on.

 

 

illustrated as such:

 

Every 90 degrees the 't=radius' rotates comes meets with a full length of another radius, the moment in our perspective, that particular radius is at full length, perpendicular and horiztonal to our observation, or perpendicular and vertical to our observation.

 

Let these 3 radius' draw a section of area on the sphere in 3d geometric expresion. Where the maximum Surface Area moment is equal to 't radius' @ 45 degrees for each of the 4 the 90 degree sectors of this quarter circle (2D) quater sphere (3D). Which may be equal to (?)

 

[math]E_r[/math] E radius

[math]B_r[/math] B radius

 

[math]Surface Area = A = \frac {\sqrt {(E_r^2+B_r^2)}}{8\pi}[/math]

 

Where E and B are [math]l[/math] length of radius.

Where the minimum radius would be that of [math]l_P[/math] the planck length

 

Example:

At 45* [math]E_r, B_r[/math] should equal 0.707105

 

[math]r_c[/math] = constant radius (frame clock / time radius / t radius)

 

[math] r_c = \sqrt{\left(E_r^2+B_r^2\right)}[/math]

[math] r_c = \sqrt{2\left((0.707105...)^2\right)}[/math]

 

Here is a bit of a visual example.

Where Z radius spins like a clock hand.

 

 

 

Moving into Special Relativity.

 

When we observe one radius as constant in length, which according to this geometry will always be a time radius. The remaining two axis, E and B change length at a specific rate relative to eachother. A Gradient consistent with the gamma factor involved with SR.

 

For example. If you take two rulers of equal length that are connected to a center axis. If ruler (a) is pointing towards you like an arrow, and ruler (b) is pointing perpendicular relative to you.

 

As we begint to set these rulers in equal motion around their axis, the change in length of each ruler relative to this observer will be consistent with the same gradients of SR.

 

Per given change in angle, ruler a's initial position will be the greatest change in length, starting from a 1d infinite point and growing into a 2d perpendicular length.

 

Ruler b, of course will be the equal opposite gradient to ruler a. Ruler b will measured to have an accelerating rate of change, where ruler a will have a deceleration rate of change. Where these gradients are gamma factor consistent.

 

In this right triangle: sin(A) = a/c; cos(A) = b/c; tan(A) = a/b.

 

t = hypotenuse

E_r = side a

B_r = side b

 

[math]t = Tan_t \theta = \frac {E_r}{B_r}[/math]

 

[math][a^2 + b^2 = c^2] = [t^2 = E_r^2 + B_r^2][/math]

 

[math]x[/math] is radius length of E_r or B_r

 

[math]\frac {1}{\gamma} = \frac {t_r}{x_r} = \gamma = \frac {1}{\frac {t_r}{x_r}[/math]

 

[math]\frac {1}{\gamma} = \frac {t_r}{E_r} = \gamma = \frac {1}{\frac {t_r}{E_r}[/math]

 

 

 

 

 

 

 

 

 

 

The prediction of this is such that, space-time (SR) is the system at rest (an observing atom) relative to a system #2, seperated by EMR, where each frame consists of a 1 (face on radius) neutral radius (t) and two active radius (E and B)

 

The geometry of the 3 radi are thought to be consistant with QM.

 

The radius' must spin One full revolution to cover a surface area of aprox 1/2 of the sphere, and approx(?) two revolutions to cover an area of the full sphere. (If we consider a triangle forming this "surface area", with its area formed on the surface of the sphere in perportation to the locations of the tip of each radius, drawing a strait line[great circle line] on the surface of the sphere.)

 

 

 

For the images attatched. I took the concept of a sphere with rotating radius, and converted that to a cone geometry, where time can be one radius, and vertical and horiztonal rotations of varying radius lengths can be E and B, and showed many configurations of this geometry. seperate, together, series, etc... Where as one total system we form up a sphere that I assume produces QM measurments. such as electron orbitals. Energy in one packet of EMR equal to C^2, where M, may be equal to radius.

 

 

Animation Concept of geometry.

[arkain101]

Let us imagine this geometry as 3 clocks to form one sphere.

 

 

Let us say C is not a velocity, but Time -the face of one of these 3 clocks.

 

With each one in the same system, One shall represent X, another, Y, and another Z axis.

 

Any of the clocks can be E,B,orT.

 

It is the relative perspective that changes them.

 

However, we will say X axis is horizontal clock, and is B

and Y axis is vertical and is E

and Z axis is Circular(faceing observer) and is t (time)

 

 

If the system is in motion relative to an observer, we can see that Two clocks will have a null effect.

 

For example:

Say the system is a circle 'O'

When it is in motion, O------>

The two clocks that occilate in parrallel <---> have no overall change in their total time for each revolution. Say they rotate at C.

They will rotate slow in the direction of the systems motion, and faster in the direction oppostite of motion.

 

ie:

 

 

However, the clock of which has its face in the direction of travel, must slow down in rotation overall, and let this be the "t" clock. the 't' radius hand.

 

Therefore, the time of the system is slown down depending on its velocity.

[arkain101]

On what grounds? I didn't see anything helpful to clarify this on the page you linked to.

 

How does SR make this claim?

 

Sorry for missing this post.

 

I made the claim that SR claims (based on geometric examples I have studed) a photon Ticks slower in a light clock (moves slower) in the perpendicular direction of travel, relative to an observer. Yet, some examples I have come across (including this animation) show that a pulse of a radial (radiating) wave is quite capable to travel from origin Point (x), perpendicular to motion of the source, and come from that point (x) at rest directly towards observer, and by laws, does so at velocity C.

I am interested, is this claim correct?

 

 

A laser should capable to be used in place of the wave and be expressed as a light clock. If it can, then the same as a photon the laser beam will react the same as the photon. So, if a wave can travel form its origin at C, can a laser do so? Will it contain motion in the direction of the source while also traveling C towards an observer, or does the laser become spread out. That is, a series of pulses seperated by a distance, that forms the visual of an angled shape coming towards the observer as compared to a strait beam (that we would expect from a source at rest).

 

Another animation

(see at bottom of page)

 

Set the speed to "fast". Notice that the path of a photon slows in the direction perpendicular to travel in perpotion to the velocity. This should mean that a photon containing velocity with a source is uncapeable to travel a perpendicular path relative to source direction of motion.

 

This is where I pondered.. If a laser was turned on at the position of perpendiclar to direction observer is facing (north). Relative to the observer would the laser be capable to travel perpendicular (south) at C, when we show that it does not in a light clock geometry.

 

My inquiry is not to find an error in SR. It is to look at the possible interpretations that can be made.

 

There is one perspective I am looking at that would say the light clock is an accurate expression, but it would not express the totality of what is going on, and may have some incompletions.

[arkain101]

Postulate: Turning a moving object 90 degrees requires the same energy as the object that is being turned contains in Ke in motion.

 

Through this I think I may have formulated a process that successfully converts circular to square geometry and applies to the laws of physics.

 

Momentum is a 2d vector. It is never lost. Light or c is momentum, that can not be lost. I am not that familiar with the fine structure constant but, I am presuming it describes the unit of momentum or maybe I mean force?

 

 

 

See attatchment for geometry.