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## #1 2010-05-22 22:41:02

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

### My own relativity theory.

Imagine the twin paradox, 2 people travel in different speeds, except there is no still reference frame. So naturally no one gets older then the other. In other words, relativity doesn't contract any other coordinate then the path it moves along, time is not along that coordinate.

Imagine now a particle moving in speed of light, a photon:

It moves not lengthwise, because to it, length is contracted, that is callibrateable with pythagoras and the lengthwise, speedwise and the hypothenus is length in the proportion of c*t but not the coordinate time.

Simply, the length it moves in its own reference frame is zero, but the time wise length is not contracted and is c*t, that in its turn is proven with that time = length of the observer divided with lightspeed.

so, the following it the right equation of my relativity theory.

(x^2 + y^2 + z^2)(1-v^2/c^2) + (ct)^2 = moved distance over time and space squared.

So, naturally a particle at near lightspeed cannot decay. Its x, y and z coordinates are zero (since the resultant of the direction has all room coordinates) and hence it cannot fall asunder in any direction.

If a black hole would be created on earth, it would not decay in the same way, since it's affecting resultant is monodirectional and unaffecting the black holes lifelength. It will grow to humongous proportions, and when it has reached a certain size, it will release imaginary matter, which repels eachother among their own kind, and doesn't react to our gravity.

Delighted to hear your responce. Best wishes LQ.

Last edited by LQ (2010-05-23 03:05:09)

I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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## #2 2010-05-23 03:14:02

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

### Re: My own relativity theory.

(x^2 + y^2 + z^2)(1-v^2/c^2) + (ct)^2 = moved distance over time and space squared

per time squared (speed squared)

(x^2 + y^2 + z^2)/t^2(1-v^2/c^2) + c^2 = speed squared = v^2(1-v^2/c^2) + c^2

square root out of both sides:

√(v^2(1 - v^2/c^2) + c^2) = speed = v√(1 - (v/c)^2 + (c/v)^2)

Momentum = mv√(1 - (v/c)^2 + (c/v)^2)

Can someone please simplify that equation, and integrate it?

I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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## #3 2010-05-23 04:58:48

LQ
Real Member
Registered: 2006-12-04
Posts: 1,285

### Re: My own relativity theory.

mv√(1 + (iv/c)^2 + (c/v)^2) = mv√(( iv/c - v/c + c/v + ic/v)^2/√(i)) = mv((iv/c - v/c + c/v + ic/v)/i)^2
= mv((v/c + iv/c - ic/v + c/v))^2

I callibrated that myself.

mv^2/c + imv^2/c - imc + mc = p

E = mv^3/(3c) + imv^3/(3c) - imcv + mcv + constant

the constant must be: imc + mc

So that's the rest energy!

The kinetic energy is: mv(1+i)(v^2/(3c) - c)

Last edited by LQ (2010-05-23 05:22:38)

I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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