The geometry of Special Relativity

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I wish to describe how I see the geometry of a photon and an inertial frame to determine frame time. A while ago I tried some simple maths to see how accurate this geometry would be against classic relativity. It works fine.

Let me describe how I see this and perhaps find an SR member who might just see how this works.

First we go to an area of gravity free space and consider the geometry of propagating spheres of light.

From a pinhead event and the release of photons from it, we allow around one meter of propagation and position ourselves centrally and above the propagation for a good view. I see a perfect sphere with each photon having described radians. All photons moved at c on these radians from the event.

Lets consider the frame from which the event was produced. I understand the speed of the frame on which the event was produced as having absolutely no influence on the speed of propagation of the sphere. The photons involved will propagate on their respective sphere radians at c from the event and not the frame they were created on. Lets clarify that slightly and consider a single photon on its radian and thus on a heading. If a photon from another event created anywhere else in the universe, matches this heading and closely follows the single photon, we find the photon from this other event will remain at the same distance and direction to infinity regardless of the difference in speed and direction of the frames they left.

We can say that a photon moves at c from its associated event. The event can be described as the halfway point between two opposite direction photons. This point will probably not be where the point on the frame the event was created on, after some propagation and frame movement.

Now as far as Special Relativity is concerned, frame time does have an important geometrical relationship with associated propagating spheres of light.

If there is any interest perhaps I can clarify these points and then describe further this process of SR time dilation.

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Feel free to correct.

Further to frame time and the propagating sphere. Lets introduce a moving inertial frame to gravity free space. A desk will work fine. The desk will move flat face to the direction of travel and we will move this example at .6c. We create an event at the left hand edge of the desk top and allow about one meter of propagation towards the right hand end. This distance is not important but allow enough propagation so that the sphere does not cross the other end of the desk.

Lets see what we have now. We have a sphere radian which corresponds with the direction the desk is moving and since the radian has increased its length at c from its event, we can say the desk edge has progressed along this radian by .6 of this radian, whatever the radians total length may be.

We will now find another interesting radian. This one is on the edge of the sphere where it intersects the desk top. To identify this one clearly, section the sphere and the desktop to reduce it to two dimensions and we can see clearly a radian which has gone from the event to the desktop, a radian which has also progressed at c.

If we consider the origin of the event and where it is Now!, we can ask ourselves, Where did the photons originate from?. We could say certainly not where the edge of the desk is Now! Propagation is independent of the frame it leaves. The event could be considered as having originated from the intersection of the direction radian and the desktop radian. This is not where the edge of the desk is after any propagation.

Consider the desk top crossing photon on its radian from its event. This photon was originally created at the event and at the left hand edge of the desk. It was present at "The start" If we allow about a centimetre of propagation and movement of the desk at .6 of its directional radian, we find the very same photon is on the surface of the desk. If we allow more propagation and desk movement we will find the very same photon is still on the surface of the desk. This will continue right out to the other end of the desk. It is always the same photon. Where did this photon start from?. Its own event and not the event start point of the desk.

Consider placing a pinhead frame on the event and keeping it concentrically centered on the associated propagating sphere. This pinhead will always remain inertial and the reciprocal of All! photons on the sphere will always head back to this point. Again, where did the photon start from, Not where the edge of the desk is now.

For a moment, lets consider this .6c frame in the context of a 3 4 5 triangle, because it will be the most simple example to understand.

As you will know, a 3 4 5 triangle will always form a right angle triangle with sides of those unit lengths. Lets transpose the figures a little by doubling them to 6 8 and 10. It is still a 3 4 5 triangle. Lets divide 6 8 10 by 10 and make .6 .8 and 1. We still have a 3 4 5 triangle but now with numbers we can use to calculate the time speed of a moving inertial clock.

Our desktop is moving at .6c and the desktop intersecting photon is moving at c from its event. We can represent c as 1.

Ask What is the crossing speed of the photon which touches all parts of the desktop right up to the edge of the propagating sphere. As we can see it is .8c relative to the desktop and c relative to its event. The frames clock will measure the photons speed at c because it is running slow, it must run slow! .8 of time to be precise.

If you are in the frame and believe you are measuring a photon crossing speed of c, go for your journey and return to the inertial frame which remained inertialy centered on the propagating sphere and lets see how well the clock has been running that you used to measure c. It will have run slower than the sphere centre clock.

Make the desktop move at .8c and we find the desktop photon will cross at .6c and within the desktop frame we will still measure it moving at c using the moving frames clock..

The same photon moves at c from its event.

I have not finished but wish to use a couple of classic relativity examples to show this geometry gives the same answers.

For the moment, consider this classic relativity statement in the above context. "The Aether, must be of the nature of a solid body, because transverse waves are not possible in a fluid, but only in a solid"

I believe as you can see. All! photons are radial from an event and All! transverse waves are made up of photons which move radially from an event. Some events may be continuous but a photons origin and its direction reciprocal will always go back to the event point the photon creation originally took place.

As you may be able to see, the opposite to c is absolute rest. This point holds the fastest clock in the universe.

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Time dilation calculations using the above geometry are simple.

These particular examples are from David Darlings website and are calculated using classic relativity. I hope he is fine with me using his figures. Lets see how my odd geometry stacks up.

The table below gives the travel times as measured by onboard clocks to Arcturus located 36 light years from the sun, for a spacecraft travelling at various speeds (given as fractions of the speed of light)

Speed Travel time in years

.01c 358.2

.5c 62.4

.9c 17.4

.99c 5.13

.999c 1.61

Lets try two examples. .05c

For a start 36 light years divided by speed .5c will tell us his earth trip time which is 72 years.

We need to find the desktop crossing speed of the frame photon to find his time dilation factor.

The spaceship has moved .5 up the directional photon radian.

The desktop photon has moved from the event and is on the surface of the desk and forms the hypotenuse of the geometry and as such is a full radian which is c or 1 for the calculation.

The photon frame crossing speed is now a simple Pythagoras calculation. It is the length of the crossing photon compared to the c hypotenuse.

So . 1 minus .5 squared is 1 minus .25 is .75

The square root of .75 is the distance the photon has moved across the desk compared to c. which is .866025 etc

The trip took 72 years earth time. The speed of the desktop photon was .866025. The frames clock has no option but to run at that speed also so. 72years times .866025 is 62.35years for the onboard clock.

 

Lets try .99c. 36 divided by .99= 36.363636 years earth time

1minus .99 squared is .0199. The square of this is .141067.

This is the tabletop crossing photons speed compared to c.

So 36.363636 times .141067 is 5.12972 years frame travel time. Check the chart.

Not bad. Does a photon cross a moving inertial frame desktop at c-- No.

Draw the geometry accurately and take the dimensions directly off the drawing and you will get the same answers.

Think about a desktop frame where the edge of the desk stays with a desktop crossing photon event. There is no matching directional photon radian. There is no time dilation because the desktop crossing photon crosses at c. The desktop radian matches the desktop, unlike the other examples which in a moving frame become a hypotenuse, from its associated event. This must be absolute rest. The reciprocal of any photon from this event will always return to the edge of the desk. There will not be another clock which will run faster, there will be many running slower on moving frames or at the same speed when at absolute rest.

Does classic relativity have a geometry we can measure directly. I dont think so. The closest may be a space-time diagram. To me, a slow clock measures a slow photon at c. The slow photon is the desktop photon on a moving frame. It cannot cross at c, the geometry will show this. If it did it would exceed c from its event.

Just a couple of words re General Relativity. I believe SR and GR are one and the same. If you can see my intent from the above, consider your desktop and a photon crossing it, Yes you will measure its crossing at c and so will an office above you, at c, but the lower floors of an office frame will have slower clocks and the, crossing speed will be slower at lower altitudes---- in gravity. Does the event stay at the start of the desk, I would like to show that it does not. It will move at c from its event and not cross at c.Perhaps I could expand on this later.

Consider a slow photon example re Dr Shapiro's haystack experiment. Slow relative to its gravity frame, c relative to its event which will not remain at the same altitude of the gravity crossing photon.