Time as Related to Moebius Transform E/M

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KiraSjon

Member
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174
Time as Related to Moebius Transform E/M

Calling all physics enthusiasts! I was curious about your opinions on this article.

http://www.rialian.com/rnboyd/time-moebius.htm


Time as Related to Moebius Transform E/M
? Robert Neil Boyd
Minkowski unified space with time in terms of a fourth dimension as related to the speed of light.

But his 4th dimension does not enter into the system in the same manner as the other three dimensions. Minkowski's \"time dimension\" is not written as \"t\", but as \"-ict\", where \"-i\" is the negative square root of minus one, c is light velocity, and t is time. (This in itself holds that time is a complex number system, which implies an orthorotation into a higher physical dimensionality, which seems to be the situation in every case where the square root of -1 comes up in an equation, as in electric engineering.)

When we examine the distance traveled on a journey through this \"space-time\", we can calculate the distance traveled on each axis in terms of a graph which measures the distances traveled along each of the several axes. In normal terms, without time involved, we can measure the distance traveled along the axes in of units along the axes, then calculate the actual distance traveled by use of the Pythagorean theorum. However, when the time axis is included, this calculation based on Pythagoras must be altered to include time. Thus, if we are interested in the space-time distance traveled, to measure along a line AB, relative to two of the axes, say x and y, our equation must look like:

AB^2 = x^2 + (ict)^2

or AB^2 = x^2 - (c^2 t^2) since i squared is -1

According to this, when x = ct the distance along AB = 0 which says that we have not traveled any distance at all! (A conundrum, since we were moving with the velocity of light in the first place.)

This is one of the famous \"null lines\" of relativity.

So when we are moving at speed v, the spacial distance x, traveled in t seconds is simply vt. Substituting vt for x in the previous equation, we have:

SqRt(v^2 t^2 - c^2 t^2) = SqRt(t^2 [v^2 - c^2] ) = distance

This equation indicates that when v = c the space-time length vanishes. By this, no time will have elapsed on our journey between points A and B because there is effectively no distance between A and B. Another conundrum.

Now, on to the actual theme of the title:

When we are looking at the projective non-linear solutions to the Maxwell equations known as the Moebius transforms, we find that the propagation velocity of such discontinuities is allowed to be ANY velocity, from zero velocity to infinite velocity.

Let us examine what happens to the above situation when considering the Moebius transform E/M propagations. In place of our term \"c\" then, let us place a term which can vary from zero to infinity, as do the propagation velocities of the Mobius transform E/M propagations. We shall use the term (0 => v => U), where v is some measureable velocity and U is an unlimited velocity.

Then we have, substituting into the first equation above,

AB^2 = x^2 + ( i [ 0 => v => U ] t )^2

Here the solutions involved with zero velocity and infinite velocity are interesting of themselves, while any numerical velocity between zero and infinity is readily tractible. And also revealing.

What we see from this is that the distance traveled along the space-time from A to B is countable and further, is never zero. That is to say, that for Mobius transform E/M, there are no \"null lines\". Further, we can clearly see that time is always mensurable, as well, with regard to the Mobius transform E/M, except in the case of an infinite velocity, whereupon time ceases to flow

for entity aboard the Mobius transform propagation. At the same time, it is obvious that superluminal velocities are obtained by the Moebius E/M and its passengers, while time flows according to the equation above.

This looks something like

SqRt(t^2 [v^2 - [ 0 => v => U ]^2] ) = distance

What this means is that there are two different kinds of time, depending on whether we are contemplating normal light or the Moebius E/M. In terms of consciousness, an entity traveling along with a Moebius transform E/M propagation will have an awareness of motion, and the passage of time, until the point that an infinite velocity is reached, whereupon time stops. Lorentz contraction does not apply here in the same manner as it does in standard relativity theory. Rather, it proceeds as the function [ 0 => v => U ] , as the propagation velocity approaches infinity.

Pardon my awkward math, but I think these understandings need to be incorporated into relativity theory, as well as QM. There are other implications as well, if you understand what all this means. For example, I think this understanding could be important for interstellar travel.
 

StarLord

Senior Member
Messages
3,187
Time as Related to Moebius Transform E/M

Imho, from what little I can understand of the concept, that makes alot more sense and would substantiate or validate the possibility for visitation of neighbors Lightyears from our planet. Now, if they can link this theory to manipulating gravity with magnetics, we get free energy and FTL capabilities.
 

iooqxpooi

Member
Messages
173
Time as Related to Moebius Transform E/M

OK, first off, there is no measurable distance because we are using less than light mathematics. At c, there would be different dimensions or whatever, changing the mathematics. This is why the relativity equations would be correct. By the way, it is Moebius Transformations...not transforms. Unless he doesn't speak English very well, I don't understand why he said Moebius transforms. E\m=c^2. Thus there is NO need to show E/m. He also could have said v, and at the end of the eqauation have said [v={0,1,2...}]. That implies v goes from 0 to infinity. And that's the end of my critique.
 

StarLord

Senior Member
Messages
3,187
Time as Related to Moebius Transform E/M

Iggy, *that* would only work for you. The rest of us need all that other stuff. :)
 

Harte

Senior Member
Messages
4,507
Re: Time as Related to Moebius Transform E/M

Most of you had to know that I would not let this one pass.



Thus, if we are interested in the space-time distance traveled, to measure along a line AB, relative to two of the axes, say x and y, our equation must look like:

AB^2 = x^2 + (ict)^2

or AB^2 = x^2 - (c^2 t^2) since i squared is -1

According to this, when x = ct the distance along AB = 0 which says that we have not traveled any distance at all! (A conundrum, since we were moving with the velocity of light in the first place.)

If we are traveling a distance relative to two axes, where is the "Y" component here?

There is nothing in this equation or the statements preceding it about the speed of travel. This equation represents only a distance measurement.


So when we are moving at speed v, the spacial distance x, traveled in t seconds is simply vt. Substituting vt for x in the previous equation, we have:

SqRt(v^2 t^2 - c^2 t^2) = SqRt(t^2 [v^2 - c^2] ) = distance

This equation indicates that when v = c the space-time length vanishes. By this, no time will have elapsed on our journey between points A and B because there is effectively no distance between A and B. Another conundrum.


No conundrum here at all. At C, time stops according to special relativity. When no time flows, it is invalid to calculate distance based on vt.

Additionally, let v=0, t=1 in the above equation. (Say you are sitting in your chair for 1 second). Then :

distance = sqrt(-c^2), a large imaginary number. But the first equation already took care of the imaginary element in the time dimension "ict", see?:

AB^2 = x^2 + (ict)^2

or AB^2 = x^2 - (c^2 t^2) since i squared is -1



This would indicate that space is also a "complex number system." Somehow, I doubt that.

?
We shall use the term (0 => v => U), where v is some measureable velocity and U is an unlimited velocity.

Then we have, substituting into the first equation above,

AB^2 = x^2 + ( i [ 0 => v => U ] t )^2


What we see from this is that the distance traveled along the space-time from A to B is countable and further, is never zero. That is to say, that for Mobius transform E/M, there are no \"null lines\". Further, we can clearly see that time is always mensurable, as well, with regard to the Mobius transform E/M, except in the case of an infinite velocity, whereupon time ceases to flow


Can't believe Iggy missed this one.
Far from being "never zero, the distance from A to B is zero when the value of v is equal to x/t.



for entity aboard the Mobius transform propagation. At the same time, it is obvious that superluminal velocities are obtained by the Moebius E/M and its passengers, while time flows according to the equation above.

This looks something like

SqRt(t^2 [v^2 - [ 0 => v => U ]^2] ) = distance

What this means is that there are two different kinds of time, depending on whether we are contemplating normal light or the Moebius E/M. In terms of consciousness, an entity traveling along with a Moebius transform E/M propagation will have an awareness of motion, and the passage of time, until the point that an infinite velocity is reached, whereupon time stops.


This equation above is a combination of the two previous ones with the quantity [0=>v=>u] substituted for the constant C. How do you get away with substituting a variable for a constant? The equation implies that the speed of light varies from 0 to U. Invalid all around.


I think these understandings need to be incorporated into relativity theory, as well as QM.

Maybe I'm wrong here, but I think I can see why they haven't been.:)

Harte
 

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