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# WANT TO TIME TRAVEL...............................

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MASTER THESE EQUATIONS.........

AND USE

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I don't quite get it...

How could it be that F = {[sqrt(1-(v²/c²))]-m}a, if we already know that at low velocities F=ma?

Doesn't seem to make sense.

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that's correct use this data to understand...

f = {[(m/ (1-(v ^2 /c ^2 )) ^(1/2) ]-m}a this equation is based on E=mc^2

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It doesn't matter if you use relativistic mass. My point is that at low speeds you should get F=ma. Because at low speeds, the effects of relativity (as well as any other new theory that may replace it) are negligable.

So what's the relativistic equation for force?

If you plug in the relativstic mass:

M=(m/ (1-(v ^2 /c ^2 ))^(1/2)

in the formula F=Ma,

You'll get:

F=[(m/ (1-(v ^2 /c ^2 )) ^(1/2)]a

Which at low velocities (v<

Sounds right? Well, it isn't. Because F=Ma is valid only when M is constant.

Here is the equation for relativistic force, as given in wikipedia:

The first term is indeed equal to Ma (that funny looking symbol before the "m" is the greek letter gamma, which is shorthand for 1/(1-v^2/c^2)^(1/2)). The second term stems from the fact that M is not constant.

Not surprisingly, this equation also reduced to F=ma at low velocities.

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The relativistic kinetic energy expression can be written as

and the square root expression then expanded by use of the binomial theorem

giving

Substituting gives:

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Indeed. Finally you're using the correct equations. It's about time (pun intended)

Now, remember that force can be defined as dE/ds (the amount of energy given to the moving object per unit distance). Using your equation for E, and realising that v=ds/dt, you can work out the correct equation for the force F which is exactly what I've given above.

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satown,

Isn't this supposed to be a relativistic 4-vector situation where

F = dP/d "tau" where "P" is a 4-vector?

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Good question.

Darby, perhaps you can clear the confusion for me. Should it be F=dP/dTau or F=dp/dt?

If I remember correctly (and I'm not sure about this), F is Loerntz-invariant. Doesn't this mean (assuming it's true) that the two equations in the previous paragraph are equivalent?

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Sigo,

dTau = dt in most, but not all situations in general relativity.

In the case of a rotating surface (like a turntable surface) it does not hold strictly true. But this makes perfect sense.

The idea of proper time ("Tau") was imported directly from special relativity into general relativity. Of course, special relativity deals with uniform motions of translation - unaccelerated and in a stright line with no rotations. General relativity involves curved spacetime and includes rotations. Two observers sitting in two spaceships traveling in parallel and the same unaccelerated velocity will have clocks that pretty well match. If they are not traveling at the same velocity (accelerated frame) the Lorentz transformation can be directly applied.

Two observers sitting on a huge turntable traveling "in parallel", one near the edge of the turntable and the other near the center of the turntable will experience time very differently (assuming that the rotational velocity of the outter rim of the disc approaches light speed).

Here's a good reference:

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here's some Einsteinian work

its like that but not really. imagine the world line being a string and a flow at the same time. now make a loop on the string flow and watch the flow go by with out affecting your time. the beginning is separating time from affected and non affected.

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your current view

the next view

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this is how it works watch........

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Actually, I'm very simple minded. As the song goes: "don't know much about algebra....etc." Anyway, why not time travel through astral travel? Our subconscious minds and superconscious minds know far more than our conscious minds could even hope to. I'm very much a feeling person, but isn't there, within many of you, a gut feeling that somehow combined together we can go back, go forward, etc. and make right the things we did that seem to be so wrong? Now we can't change what other people do, but I feel we can change what WE do.

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that's correct use this data to understand...

f = {[(m/ (1-(v ^2 /c ^2 )) ^(1/2) ]-m}a this equation is based on E=mc^2

satown,Your "equation" states:

Force equals the relativistic mass, less the rest mass, times the acceleration of that mass.

Say what?!!

What does "a" (acceleration) equal? Somewhere I think that I recall that acceleration involes mass times the differential of change in velocity over the differential of change in time ( m dv/dt...something like that) - the same mass that you eliminated in the equation (after stating it in relativistic terms).

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• 2 weeks later...

think of acceleration as an increase of speed or velocity of time. the mechanics definition of acceleeration,"the time rate of change of velocity with respect to magnitude or direction; the derivative of velocity with respect to time," will cause problems in your understanding of time. You can not use the original mechanics onderstanding of time because it is written in a linear state. You will have to convert your current understanding of time into worldlines as a standard.

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• 8 months later...

Liquid Space Theory

F = force

M = mass initial

V = velocity

C = speed of light

A = acceleration

H = Planck's constant

E = energy

F = {[(m/ (1-(v^2/c^2)) ^(1/2)]-m}a

Second law of time

A ={{[(m/ (1-(v^2/c^2)) ^(1/2)]-m}^-1}f

Infinite change of time

M = [f /{[(1/ (1-(v^2/c^2)) ^(1/2)]-1}a

Mass as a vector in a 3- orthogonal space

V = c [- (ma/ma-f )^2 +1]^1/2

Velocity of time

C = [v / [- (ma/ma-f )^2 +1]^1/2

Speed of light as a function of mass

E = {[f /{[(1/ (1-(v^2/c^2)) ^(1/2)]-1}a}{[v / [- (ma/ma-f )^2 +1]^1/2 }^2

Time conservation law

Wave = {{[f /{[(1/ (1-(v^2/c^2)) ^(1/2)]-1}a}{[v / [- (ma/ma-f )^2 +1]^1/2 }^2 } / h

(e / h)

Wave length and energy of the force

Energy of the force = {{[(m/(1-(v^2/c^2)) ^(1/2)]-m}c^2}

Wave = {{[(m/ (1-(v^2/c^2)) ^(1/2)]-m}c^2} / h

Time has an avg. 10^18 â€“ 10^23 Hz

` As you travel the currents of time beware of whirlpools you encounter and the damaging counter current. Your life will be right if you follow God and the never-ending fulfillment of his understanding. `

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Time has an avg. 10^18 â€“ 10^23 Hz

Time is an x-ray - hard gamma ray?
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yes, at the quantum level. time is the relativity of energy.

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did you ever read Plato's allegory of the cave? i think you should read that before you mess with this stuff.

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yes, at the quantum level. time is the relativity of energy.

I'm not quite sure that I follow you here. You've used the equations to define the quanta of time as laying somewhere between 10^18 & 10^23 Hz...which is the frequency of X-ray to hard Gamma Ray radiation.That being said, have you gone back to your equations and plugged in the numbers to verify your results? You've referenced Lorentz transformations, velocity (v), the speed of light © and acceleration (a) - all of which have an implied time component as written above but which also have an explicit time component if you take their derivitives. What happens when you plug and chug the equations by substituting 10^18 or 10^23 Hz in place of (dt) in the equations where they reference time? Do the numbers come out the same and verify the equations?

One problem that you may also have is your theoretical approach. You're relating this issue to QM while using equations from Special Relativity. That's probably not going to help you. SR is still a classical interpretation of physics and QM isn't classical at all. For instance, time is an operator in classical physics but it isn't in QM. There's a Hamiltonian (H) that relates the time and energy of the system under consideration but there's still uncertainty involved. (Even with your approach you've defined the "uncertainty" of the time quanta as being spread over five orders of magnitude.)

You might give a try here:

for some assistance.

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• 2 weeks later...

i understand Mr. Darby, look at it as Einstein's Hidden variable theory and the variable being the void of space. the theory above explains how the void moves. later i will post something you will enjoy that uses the Schrodinger Equation.

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