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I am the first in the world

Time travel

With my equation I am certain that time travel to the future is possible.
I can fly to planet Jupiter in less than 1 hour:
My equation:

m = m0( 1 / ( 1 – v^2 / c^2 )^1/2 – (v / c) / ( 1 – v^2 / c^2 )^1/2)

This means after two hours flight to Jupiter you will find your children older than you!

An abstract of my theory on my blog: http://​adrianferent.blogspot.com/
 
Time travel

With my equation I am certain that time travel to the future is possible.
I can fly to planet Jupiter in less than 1 hour:
My equation:

m = m0( 1 / ( 1 – v^2 / c^2 )^1/2 – (v / c) / ( 1 – v^2 / c^2 )^1/2)

This means after two hours flight to Jupiter you will find your children older than you!

An abstract of my theory on my blog: http://​adrianferent.blogspot.com/

Please clarify something for me, when you say that the trip to Jupiter will take one hour ...

Is that one real hour?
Or one imaginary hour?
Or some combination of the the two?

;)
 
Time travel

With my equation I am certain that time travel to the future is possible.
I can fly to planet Jupiter in less than 1 hour:
My equation:

m = m0( 1 / ( 1 – v^2 / c^2 )^1/2 – (v / c) / ( 1 – v^2 / c^2 )^1/2)

This means after two hours flight to Jupiter you will find your children older than you!

An abstract of my theory on my blog: http://​adrianferent.blogspot.com/

Wow! I just thought of something else.

Does this mean that I will have to get my watch and calendar recalibrated to complex time?

And this spring, will have to adjust to Complex Daylight Saving Time?

And what about my cats? They expect fresh food every morning, so does that mean that will have to adjust to a complex morning time feeding cycle?

Gee whiz! It sure it going to be complicated getting in tune with this complex time thingy.

;)
 
Time travel

With my equation I am certain that time travel to the future is possible.
I can fly to planet Jupiter in less than 1 hour:
My equation:

m = m0( 1 / ( 1 – v^2 / c^2 )^1/2 – (v / c) / ( 1 – v^2 / c^2 )^1/2)

[...]
[latex]$m=m_0 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - \frac{(v/c)}{\sqrt{1-\frac{v^2}{c^2}}}\right) = m_0 \frac{1-\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} = m_0 \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}} = m_0 \sqrt{\frac{c-v}{c+v}} \leq m_0}$[/latex]

Please correct me if I am wrong, but if my math is correct than what you are saying is that a moving object has less mass than the same stationary object... :confused:
 
[latex]$m=m_0 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - \frac{(v/c)}{\sqrt{1-\frac{v^2}{c^2}}}\right) = m_0 \frac{1-\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} = m_0 \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}} = m_0 \sqrt{\frac{c-v}{c+v}} \leq m_0}$[/latex]

Please correct me if I am wrong, but if my math is correct than what you are saying is that a moving object has less mass than the same stationary object... :confused:

Thanks much!

I just checked your math, and it is correct. So I guess that means that I will not have to reclabirate my various time pieces and such for complex number standard time after all.
 
That is my point you can go to Jupiter in less than 1 hour!

It is good because at least are some people who understand!
 
[latex]\dots = m_0 \sqrt{\frac{c-v}{c+v}} \leq m_0}$[/latex]

That relationship is only true for positive velocities. Negative velocities will have an increase in mass. So photons impinging on one side of a surface will have zero momentum, whereas those impinging on the other side will have an infinite momentum. That probably explains how Radiometers work.
 
That is my point you can go to Jupiter in less than 1 hour!

It is good because at least are some people who understand!

Quite true! There are some people here (quite a few people in fact) who understand quite well that your ideas about time are very wrong.
 
That relationship is only true for positive velocities. Negative velocities will have an increase in mass. So photons impinging on one side of a surface will have zero momentum, whereas those impinging on the other side will have an infinite momentum. That probably explains how Radiometers work.

Personally, I'm not all that certain that alfa1's equation is based on a rational derivation. I can see a semblance of special relativity there, but that is all. Regardless of whether you have negative velocity in this equation, positive velocity is saying that your mass is lowered, which is exactly counter to observations. Making any references from this was not the intention: just that it leads to a contradiction.

BTW: photons cannot have infinite momentum. Radiation pressure is how radiometers work.
 
You are right: mass is lowered, which is exactly counter to observations.

If your conclusions contradict common sense, then so much for common sense; if they conflict with received philosophical opinion, then too bad for received opinion; but if they deny the very facts of our experience, then you must consign your conclusions to the flames.

Richard Feynman paraphrased by Baggini & Fosl in "The Philosophers Toolkit".
 

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