Mass effect equation, help!?

[arkain101]

d is distance

a is the acceleration vector (as acceleration is a vector, it must be described with both a direction and a magnitude).

v is the velocity function

x is the position function (also known as displacement or change in position)

t is time

 

 

I am trying to elaborate on an equation to describe the effect of mass that occurs between two frames from relativistic effects. It is not neccesaraly measure the value just describe how motion causes force that causes mass. However, I need my math checked.

 

 

The effect known as mass is postulated as a product of force. Force of which is a product of velocity amongst two frames.

 

[math] F = ma[/math]

 

>

 

[math]a = \left( \frac {dv}{dt}\right)[/math]

 

>

 

[math]F = m \left( \frac {dv}{dt}\right) [/math]

 

>

 

[math]m = \frac {F}{a}[/math]

 

>

 

[math]m = \frac{F} {(\frac {dv}{dt})} [/math]

 

 

 

relationship of force with energy and mass.

 

 

[math]E = mc^2[/math]

 

[math]\frac {E}{M} = c^2 = \frac {D^2}{t^2} [/math]

 

[math]M = \frac {F}{a}[/math]

 

[math]a = \left( \frac {Dv}{Dt}\right)[/math]

 

[math]m = \frac{F} {(\frac {Dv}{Dt})}[/math]

 

[math]\frac {E}{M} = \frac {D^2}{t^2}[/math]

 

[math]\frac{E} {\left(\frac{F}{(\frac{Dv}{Dt})}\right)} = \frac {D^2}{t^2} [/math]

 

[math] M = \frac {E}{C^2} [/math]

 

>

 

[math] M = \frac {E}{C^2} = \frac {E}{\left(\frac {D^2}{t^2}\right)}[/math]

 

also >

 

[math]M = \left(\frac{F}{(\frac{Dv}{Dt})}\right)[/math]

 

therefore >

 

[math]M = \left(\frac{F}{(\frac{Dv}{Dt})}\right) = \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

>

 

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

 

[math]M = \gamma m[/math]

 

[math]\gamma = {1 \over {\sqrt{1- \frac{u^2}{c^2}}}[/math]

 

u is the relative velocity between the observer and the object, and

c is the speed of light.

 

[math]M = \gamma \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

>

 

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = \gamma \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

>

 

*[math]F = ma[/math]

 

>

 

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = \gamma \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

*Stopped here for now. First time using latex, it gave me a splitting headache! :doh:

 

 

 

cont'd post #5

[InfiniteNow]

[cough] Latex [/cough] :cup:

[arkain101]

Original post latex 'd

 

[math]\text LaTeX = Hard + to + learn [/math] :lol:

[Qfwfq]

...but worth the effort!

[arkain101]

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = \gamma \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

>

 

[math]\left(\frac{F}{A}\right) = \gamma \frac {E}{\left(\frac {D^2}{t^2}\right)} [/math]

 

>

 

[math]\left(\frac{F}{A}\right) = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)[/math]

 

OR

 

[math]\left(\frac{F}{a}\right) = \gamma \left( \frac {E}{c^2}\right)[/math]

 

>

 

[math]m = \frac {F}{a}[/math]

 

[math]m = \frac {dp}{dt}[/math]

 

[math]p = \frac {mv}{\sqrt{1- \frac{u^2}{c^2}[/math]

 

[math]\frac {\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt} = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)[/math]

 

With this I am trying to show the effect known as mass is shown here to be a reference frame experience force and acceleration that is a product of the lorentz factor [math]\gamma[/math] relativistic effects, multiplied by the energy limit devided by the universal constant.

 

This is, Mass is force, and that force is dependent on accelleration and velocity between a frames velocity in a squared manner... (not quite sure on this)

 

Anyone?

[arkain101]

Is this math correct?

 

Example;

 

a)

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = M[/math]

 

[math]\left(\frac{30}{(\frac{20m/s-10m/s}{5})}\right) = M[/math]

[math]\left(\frac{30}{(\frac{20m/s-10m/s}{5})}\right) = 15kg @ 5s[/math]

 

 

b)

[math]\left(\frac{F}{(\frac{Dv}{Dt})}\right) = M[/math]

 

[math]\left(\frac{30}{(\frac{20m/s-10m/s}{10})}\right) = M[/math]

[math]\left(\frac{30}{(\frac{20m/s-10m/s}{10})}\right) = 30kg @ 10s[/math]

 

 

If time is observed to run slower (take longer) mass increases.

 

So what I am proposing to conclude is that Time is mass, and mass is force and force is accleration which is velocity and velocity is lorentz factor, where lorents factor is energy value and energy is constant and constant is squared.

 

Connecting this to a principle, all events are products of change between two reference frames.

[arkain101]

Questions:

 

What is the purpose?

 

What does this predict?

 

How can it be tested?

 

It suggests that the effect of mass of a body of matter is due to accelerations occuring at the quantum level. The velocity average is close to C. If you accelerate a body of matter, this increases the average accelerations of the quantum (atomic material) the relativistic affects cause its time to slow and mass to increase. The force of the mass comes from velocity relative to the rest frame and vice versa.

 

Prediction:

 

Using this concept, if you reduce the tempeature of matter to absolute zero, a bose-einstein condensate, and control that material with the right magnetic field, you should be able to reduce the internal accelerations so much that the material will become massless. It will defy gravity, and assuming all correctness, act massless. It will become a spacial volume of pure energy per say.

[Qfwfq]

It suggests that the effect of mass of a body of matter is due to accelerations occuring at the quantum level. The velocity average is close to C.

:cup:

 

Only the innermost orbitals of heavy atoms need relativistic corrective terms in a non-relativistic treatment. The rest energy of a composite body is not prevalently kinetic, this definitely isn't so for the electrons and I doubt it even for the nucleons.

 

Using this concept, if you reduce the tempeature of matter to absolute zero, a bose-einstein condensate, and control that material with the right magnetic field, you should be able to reduce the internal accelerations so much that the material will become massless.

Classically, 0K means no motion at all so no acceleration to further reduce, but this certainly isn't the correct description. In a proper quantum treatment, 0K means everything in its ground state, I don't see how magnetic fields could further reduce accelerations.

 

It will become a spacial volume of pure energy per say.

If you claim you would have removed all the body's rest energy, which "pure energy" would be left over?

 

I really haven't had the time to follow your algebra exactly but I noticed you seem to be using "E" for energy and also for electric field. I was wondering if you had been careful enough not to confuse these two but I couldn't say. Have you made sure of this?

[arkain101]

Classically, 0K means no motion at all so no acceleration to further reduce, but this certainly isn't the correct description. In a proper quantum treatment, 0K means everything in its ground state, I don't see how magnetic fields could further reduce accelerations.

 

A ground state, it predicts that atoms are no longer as massive as a whole. Up above 0k atomic accelerations occur. The theory predicts these motions of atoms from energy, gives mass or force to the atoms, by accelerating the inner quantum motions. Similar to shaking a bowling ball in your hands, as if the bowling ball was an atom, and the inside 'atoms' were quantum material, as you shake it you increase the velocities that much closer to C, and you are met with the force of mass; momentum and inertia.

 

By putting the atoms to rest. You find a new state of matter. Matter that is less mass.

 

A magnetic field is predicted as relativistic space. By introducing the rest atoms with a magnetic field the quantum material would be placed in a possible expanded space. The atoms would not accelerate, thus the quantum fluctuations are in a enviroment that will force them to act different relative to another frame. This is that they will reduce inner energy, relative to us, by observing space expand in their own frame. Its a prediction of course, and a far stretched one. But logically the frames of the 0k should become potential space or energy once the magnetic field is let off, the quantum fluctuations will observe space contract, which relative to an observer of this 0k state, should cause it to become more energetic.

 

If you claim you would have removed all the body's rest energy, which "pure energy" would be left over?

 

I really haven't had the time to follow your algebra exactly but I noticed you seem to be using "E" for energy and also for electric field. I was wondering if you had been careful enough not to confuse these two but I couldn't say. Have you made sure of this?

 

I may have used E for electric field before but I WAS aware of that. I do not beleive I have used electric field in the recent equations.

 

However I did find an error in the following:

\frac {\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt} = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)

 

[math]\frac {\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt} = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)[/math]

 

it should be

 

[math]\left (\frac {\left ( \frac {\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt}\right )}{a}\right ) = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)[/math]

 

The left side is Total mass, the actions that create mass. The right side is the value of that mass and what forms that value.

 

It comes from this:

 

[math]\left(\frac{F}{a}\right) = \gamma \left( \frac {E}{c^2}\right)[/math]

 

 

It repeditivly expands in this manner.

 

 

[math]\left (\frac {\left ( \frac {\left( \frac {\left(\frac{\left( \frac {\left(\frac{\left( \frac {\left(\frac{\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt}\right )}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt}\right )}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt}\right )}{a}\right )[/math]

 

What the point of this expansion is to show is that the 'a', or the accerlation as a whole of a group of internal velocities(accelerations) affects eachother down the line. So the mass value of an object at rest, is a function of these actions on the left. if you create an accleration in the final 'a' (at the bottom) it affects all the other acelerations and forces down the line, which is met with relavistic changes, further down the line than up the line (down the line being the high energy areas of the atom.

 

Of course this is in respect to this theory, which I neither beleive or disbelieve. I am just expressing a prediction it forms, however I lack the mathatic ability and knowledge to show this in any other form.

[arkain101]

Quote:

Originally Posted by KickAssClown

When you divide a fraction by a fraction you flip and multiply. It is confusing, yes but that is how it goes.

 

Like, nobody here ever knew that!

 

Hey I forgot about that! I told you I am new at math.. which is why I asked for help.

 

I am going to try that on this equation. Self teaching :D

 

[math]m = \frac {E}{c^2}[/math]

 

>

 

[math]\left(\frac{F}{a}\right) = \gamma \left( \frac {E}{c^2}\right)[/math]

 

>

 

[math]\left (\frac {\left ( \frac {\left( \frac {\left(\frac{F}{a}\right)v} {\sqrt{1- \frac{v^2}{c^2}}}\right) }{dt}\right )}{a}\right ) = \left({1 \over {\sqrt{1- \frac{u^2}{c^2}}}} \right) \left( \frac {E}{c^2}\right)[/math]

 

 

>

 

[math]\frac {\sqrt {E^2 - \left (\frac {F}{a}vc \right )^2}}{c^2} = \frac {E}{c^2}[/math]

 

>

 

[math]m = \frac {E}{c^2}[/math]

 

[math] force^2 = \gamma \left (\sqrt {\frac {1-v/c}{1+v/c}}\right ) \left (\frac{\sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}^2 }{\left ( \frac {d}{t}\right )^2\right ) \left ( a \right )[/math]

 

[math]f^2 = \gamma \left (\sqrt {\frac {1-v/c}{1+v/c}}\right ) \frac {e}{c^2} \left ( a \right )[/math]

 

The relationship that developes a force [math] = \gamma (Velocity between two frames)(mass * acceleration)[/math]

 

I'm just posting this to save it but, I need a break.