Resolving Special Relativity & Quantum Mechanics Pt2

[nstansbury]

This is a simplification of my original ideas presented here: http://www.scienceforums.net/forum/showthread.php?t=33375 with proof from first principles.

 

I suggest the following:

 

1. Gravity, Time & Matter are a result of 3 dimensional superluminal waves of "energy" created by the big bang .
   
2. The missing mass/energy in the Universe is a consequence of the energy in these Space-Time waves.
   
3. These waves can be considered Space-Time itself
   
4. Every particle is a result of a wave partially collapsing, due to a loss of energy.
   
5. Gravity & thus Mass are a function of the remaining un-collapsed wave's amplitude
   
6. Time & thus velocity/distance are a function of the propagation of the wave's frequency/wavelength
   
7. As Time is the propagation of the wave, causality can't be violated.
   

 

• Drop a cork onto a flat pool of water, the waves will propagate radially outwards from it
  
• As the radius (and thus distance) of the wave from the cork increases the amplitude will decrease
  
• Drop a second cork a distance from the first, and as their waves superimpose, constructive interference will increase the combined wave's amplitude, literally pulling the two corks together.
  
• If the corks are analogous to particles, the amplitude of the superimposing waves is analogous to gravity, and the waves amplitude reduces as a function of the distance (radius) from the originating particles.
  

 

 

The particle Energy of a Space-Time wave is given by: [math]E=c2\pi r\lambda f[/math]

 

Where:

[math]c[/math] is its' rate of propagation

[math]r[/math] is the radius of the Space-Time wave front

[math]\lambda f[/math] are it's frequency and wavelength

 

 

I will prove this from first principles using Dirac's constant, and then use this equation to prove both:

 

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

 

[math]F=G\left( \frac{m1*m2}{d^2} \right)[/math]

 

 

 

I define Dirac's constant as: [math]\hbar=c\lambda r[/math]

 

Where:

[math]c[/math] is the wave's rate of propagation

[math]r[/math] is the radius of the Space-Time wave front

[math]\lambda[/math] is its' wavelength

 

 

The Planck constant is given as: [math]h=\hbar 2\pi[/math]

 

Thus: [math]h=c2\pi r\lambda[/math]

 

Given: [math]E=hf[/math] or

 

Then: [math]E=c2\pi r\lambda f[/math]

 

Given that as a radial wave propagates outwards, its' radius and thus its' circumference increases proportionally: [math]c=2\pi r[/math]

 

The amplitude of a propagating radial wave is inversely proportional to the length of its' wavefront. As it's wave front or circumference doubles, the wave's amplitude is halved.

 

Thus: [math]A^2=2\pi r[/math]

 

Where:

[math]r[/math] is the radius of the Space-Time wave front

[math]A^2[/math] is the square of its' amplitude

 

Therefore: [math]h=cA^2\lambda[/math]

 

And: [math]E=cA^2 \lambda f[/math]

 

And the velocity of a wave: [math]v = f \lambda[/math] thus: [math]c = f \lambda[/math]

 

Therefore: [math]E=A^2c^2[/math]

 

If: [math]E=mc^2[/math] then: [math]m=A^2[/math]

 

And: [math]A^2=\frac{E}{c^2}[/math]

 

 

Given: [math]h=cA^2 \lambda[/math]

 

Then: [math]\frac{h}{\lambda} = cA^2[/math]

 

Thus momentum: [math]p = \frac{h}{\lambda}=\frac{E}{c}=cA^2[/math]

 

 

Assuming: [math]E^2=(mc^2)^2 +(pc)^2[/math]

 

Then: [math]m=\sqrt{\frac{E^2-(pc)^2}{c^4}}[/math]

 

 

And: [math]m=\sqrt{\frac{(A^2c^2)^2-(A^2c^2)^2}{c^4}}[/math]

 

 

Then: [math]m \propto \sqrt{\frac{A^2C^2-A^2C^2}{c^2}}[/math]

 

 

Thus: [math]m=0[/math] and [math]A^2=m[/math]

 

 

 

My next goal is to prove Special Relativity & Gravity can be explained as waveforms, and both be affected by the amplitude, frequency & wavelength of these Space-Time waves, using the following equations:

 

• [math]E=mc^{2}[/math]
   
   
  
• [math]F=G\left( \frac{m1*m2}{d^2} \right)[/math]
  

 

 

Mass

I suggest that "mass" is the result of the superimposing of amplitude of multiple Space-Time waves.

 

Given: [math]m = A^{2}[/math]

 

Waves amplitudes are additive

 

So: [math]\frac{m1*m2}{d^2}\Leftrightarrow \frac{A1^2+A2^2}{d^2}[/math]

 

Proof:

 

Considering two identical masses:

 

If: [math]E=mc^{2} [/math]

 

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

 

Therefore: [math]\frac{E}{c^2} * \frac{E}{c^2}

\Leftrightarrow {A^2 + A^2}[/math]

 

Therefore: [math]\frac{E^2}{c^4} \Leftrightarrow 2A^2[/math]

 

Therefore: [math]m \Leftrightarrow \frac{E}{c^2} \Leftrightarrow A^2[/math]

 

Thus: [math]m \propto E \propto A^2[/math]

 

 

Time

I suggest Time can be explained as a consequence of the relative increase and decrease in frequency of these Space-Time waves.

 

Proof:

 

The velocity of a wave is a function of frequency & wavelength: [math]v = f \lambda[/math]

 

Given two waves travelling parallel to each other at velocities [math]c[/math] and [math]\frac{1}{2}c[/math] with an identical wavelength [math]\lambda[/math]

 

If: [math]f=\frac{c}{\lambda}[/math]

 

Then: [math]\frac{\frac{1}{2}c}{\lambda}=2f[/math]

 

Where: [math]f=\frac{1}{T}[/math]

 

Thus: [math]2f=2 \left(\frac{1}{T}\right)[/math]

 

Therefore to an observer travelling at [math]c[/math], time in the other wave appears to run twice as fast, thus relatively time runs slower in the faster wave - as Special Relativity predicts.

 

 

Distance & Velocity

I suggest that the distance & velocity in any frame of reference are a result of the frequency & wavelength of the Space-Time waves.

 

Proof:

 

The velocity of a wave is a function of frequency & wavelength: [math]v = f \lambda[/math]

 

Where: [math]d = \frac{v}{t}[/math]

 

And: [math]f=\frac{1}{t}[/math]

 

Therefore: [math]d = \frac{f \lambda}{\frac{1}{f}}[/math]

 

As the frequency & wavelength of the Space-Time waves change, and that distance is inversely proportional to velocity (frequency/wavelength) as per the Lorenz Contraction.

 

 

Gravity

I suggest that the gravitational force exerted on a body is affected by the frequency/wavelength and most importantly the amplitude of these Space-Time waves.

 

Proof:

 

As a radial wave propagates outwards, its' radius increases, and given that: [math]c=2\pi r[/math]

 

As its' radius doubles, so does it's circumference.

 

The amplitude of a propagating radial wave is inversely proportional to the length of its' wavefront. As it's wave front or circumference doubles, the wave's amplitude is halved.

 

Considering: [math]E \propto A^{2}[/math]

 

Therefore: [math]A \propto r[/math]

 

Where [math]r[/math] is the distance of propagation.

 

The total gravitational potential in a propagating 3 dimensional Space-Time wave would be described as the surface area of an expanding sphere.

 

Given by: [math]a=4\pi r^2[/math]

 

Therefore: [math]A \propto r^2[/math]

 

However, Gravity doesn't act perpendicularly to two masses, and wave Amplitudes only superimpose when in phase, so a 2D "cross-section" of the wave would describe a body's gravitational potential acting in one direction.

 

If: [math]A \propto r[/math]

 

Then: [math]A^2 \propto r^2[/math]

 

Distance [math]r[/math] is a function of a Space-Time wave's frequency [math]f[/math] & wavelength [math]\lambda[/math], to which the Amplitude [math]A[/math] and thus gravitational [math]G[/math] potential of the wave is proportional to.

 

Thus: [math]r^2 = \left(\frac{f \lambda}{\frac{1}{f}}\right)^2[/math]

 

Therefore: [math]F=G\left(\frac{A^2 + A^2}{r^2}\right)[/math]

 

Therefore the gravitational force exerted on a body would be affected by the frequency/wavelength, amplitude and propagation of these Space-Time waves. Relativistic Mass also increases with velocity (frequency/wavelength) as per Relativity.

 

As: [math]E \propto A^2[/math], an increase in Energy would also increase the particle Space-Time wave's Amplitude, thus increasing its' gravitational potential and therefore mass.

 

In a spherical 3D waves propagating in all directions, Amplitude is related: [math]A \propto r^2[/math]

 

Therefore, as the propagating spherical wave increases in area, it's Amplitude decreases proportionally. This explains why the early Universe was so dense and we are exponentially expanding ([math]r^2[/math]) to a Universe with less gravitational potential.

 

Thus: [math]m \propto E \propto A^2 \propto r^2 \propto c^2[/math]

 

Mass [math]m \propto E[/math] because [math]E \propto A^2[/math]

[math]A^2 \propto r^2[/math] because [math]r^2 \propto c^2[/math] which is the speed the 3D wave front is expanding at

[math]c^2 \propto m[/math] because [math]r^2 \propto A^2[/math]

And [math]m \propto A^2[/math] because [math]A^2[/math] determines gravitational potential thus [math]m[/math] etc etc..

 

 

And in the beginning there was Energy...

 

 

 

I will edit this post to fix mistakes.

Edited June 14, 2008 by nstansbury

[insane_alien]

E=mc^{2}

 

please use the full equation as i think it will have a rather large impact on the rest of your maths. the full equation is

 

E^2=(mc^2)^2 +(pc)^2

[nstansbury]

please use the full equation

Happy to be corrected - but Why? I'm working with Relativistic Mass for Gravity, so I can't see how taking momentum into account would change mass as a result of gravitational potential? I'm interested in the total energy of the wave.

[insane_alien]

well, that m you have there is rest mass not relativistic mass. and relativistic mass is generally frowned upon by serious physicists.

 

also your dealing with waves. waves have momentum.

[nstansbury]

Hmmm yes.. very true. Which is interesting because the momentum of a massless object is a wave function with Planck's constant:

 

[math]p = \frac{h}{\lambda}=\frac{E}{c}[/math]

 

And even more interesting as I've already suggested:

 

[math]E \propto A^{2}[/math]

 

and

 

[math]\frac{A^2}{c^2} \propto m[/math]

 

Thus:

 

[math]p = \frac{h}{\lambda}=\frac{E}{c}=\frac{A^2}{c}[/math]

Edited June 10, 2008 by nstansbury
Equation typo

[insane_alien]

as i said, it will have major implications for the maths.

[Klaynos]

Damn I hate relativistic mass... This comes from a lecturer I had who's now a professor of bio physics, but did his PhD in general relativity (nice relationship between the two) and his hatred of the concept of relativistic mass has been passed onto me quite well anyway...

 

http://en.wikipedia.org/wiki/Relativistic_mass#Controversy

 

Enjoy

[nstansbury]

Yes, thanks. So really it's just a synonym for the total energy?

 

So using:

 

[math]E^2=(mc^2)^2 +(pc)^2[/math]

 

Could someone help - have I transformed this correctly:

 

[math]m=\frac{E^2-(pc)^2}{c^2}[/math]

 

 

Having said that, whilst obviously changing the outcome of any values, obtaining momentum from:

 

[math]

p = \frac{h}{\lambda}=\frac{E}{c}=\frac{A^2}{c}

[/math]

 

Would still hold true with my suggestion, that mass (and now momentum which makes sense!) are affected by the amplitude of the wave in question.

Edited June 10, 2008 by nstansbury
Equation typo

[insane_alien]

that rearranging is wrong, it should all be in a square root and that c^2 should be c^4

[swansont]

 

[math]

p = \frac{h}{\lambda}=\frac{E}{c}=\frac{A}{c}

[/math]

 

But E varies as A², not A.

[nstansbury]

Yes apologies - careless typo. Updated above.

 

Umm... would someone be kind enough to check my logic for me please...

 

 

If: [math]E \propto A^{2}[/math]

 

And: [math]p = \frac{h}{\lambda}=\frac{E}{c}[/math]

 

Then: [math]p \propto \frac{A^2}{c}[/math]

 

And: [math]A^2 \propto pc[/math]

 

 

If: [math]E^2=(mc^2)^2 +(pc)^2[/math]

 

 

Then: [math]m=\sqrt{\frac{E^2-(pc)^2}{c^4}}[/math]

 

 

Therefore: [math]m \propto \sqrt{\frac{A^4-A^4}{c^4}}[/math]

 

 

Then: [math]m \propto 0[/math]

 

 

Thus: [math]A^2 \propto m[/math]

Edited June 11, 2008 by nstansbury
Mistakes in equations

[insane_alien]

Therefore: m porportional to sqrt(c^4)

 

thats wrong that 4 should be a negative.

[ydoaPs]

Therefore: [math]m \propto \sqrt{\frac{A^4-A^4}{c^4}}[/math]

 

 

Therefore: [math]m \propto \sqrt{{c^4}}[/math]

 

NO!!!!!!! A⁴-A⁴=0 => [math]m\propto{0}[/math]

[Klaynos]

Am I the only one who knows that if

 

[math] a\propto b[/math]

[math] c\propto d[/math]

 

[math] e = a + c[/math]

is NOT the same as

[math] e = b + d[/math]

 

Although in this case of the sqrt you can set it proportial too as you can split the squarroot correctly, ok I'll be quiet now...

 

Or maybe not... one sec...

 

 

[math]

m=\sqrt{\frac{E^2-(pc)^2}{c^4}}

[/math]

 

[math]

E = CA^{2}

[/math]

 

[math]

KA^2 = pc

[/math]

 

All from above where C and K are some value of proportionality...

 

[math]

m=\sqrt{\frac{(CA^{2})^2-(KA^2)^2}{c^4}}

[/math]

 

So...

 

 

[math]

m=\sqrt{ \frac{ A^{4} (©^2-(K)^2) }{c^4}}

[/math]

 

[math]

m=\sqrt{ \frac{ A^{4} }{c^4}} \sqrt{(©^2-(K)^2)}

[/math]

 

[math]

m \propto \sqrt{ \frac{ A^{4} }{c^4}}

[/math]

Edited June 10, 2008 by Klaynos

[insane_alien]

NO!!!!!!! A⁴-A⁴=0 => [math]m\propto{0}[/math]

never even noticed that one. lol.

[nstansbury]

no!!!!!!! [math']a4-a4=0 => M\propto{0}[/math]

 

Yes!!!!!!

 

what if the speed of light is just the fastest energy state a particle can exist at?

 

If gravity is the result of wave interference patterns, gravity would exist without matter because the Space-Time waves are still superimposed

 

The effect of gravity is instant because the space-time wave is travelling faster than the speed of light.

 

I think it's interesting...

 

(I did warn you my maths sucked! Embarrassing logic updated above)

 

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 

 

I actually don't get this:

 

If: [math]p = \frac{E}{c}[/math]

 

Then: [math]E = pc[/math]

 

And if: [math]m=\sqrt{\frac{E^2-(pc)^2}{c^4}}[/math]

 

Surely: [math]m=\sqrt{\frac{E^2-E^2}{c^4}}[/math] ....

 

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 

 

Actually... yes I do because p is the momentum of a massless object so [math]m\propto{0}[/math] must always hold true for any massless wave, so that equation was already proved by: [math]A^2 \propto E \propto m[/math]

 

All I've managed to do here is prove that for a given electromagnetic wave propagating at c, the square of its' amplitude is proportional to its' energy whose mass must be 0.

 

I always suggested these Space-Time waves would have no mass, so I am absolutely no further forward! I need to some how relate http://en.wikipedia.org/wiki/Gravitational_energy to the amplitude and interference of a wave.

 

 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 

 

 

Ok, this is incomplete, but this is my first stab at it:

 

According to my idea, there should be an association with the distance a Space-Time wave propagates from its' source, its' amplitude and the force exerted by gravity.

 

I'm hoping to be able to relate it to:

 

[math]F=G\frac{m1*m2}{d^2}[/math]

 

 

So... as a radial wave propagates outwards, its' radius increases, and given that: [math]c=2\pi r[/math]

 

As its' radius doubles, so does it's circumference.

 

The amplitude of a propagating radial wave is inversely proportional to the length of its' wavefront. As it's wave front or circumference doubles, the wave's amplitude is halved.

 

And, given that: [math]E \propto A^{2}[/math] - so does it's energy.

 

Therefore: [math]A \propto r[/math]

 

 

 

Now, more specifically these Space-Time waves are 3D, which means I need to consider any wavefront as a spherical surface area.

 

The surface of a sphere is given by: [math]a=4\pi r^2[/math]

 

The amplitude of a 3D radial wave is again inversely proportional to its wavefront, and as it's wave front or spherical area doubles the wave's amplitude is reduced by a quarter.

 

Therefore: [math]A \propto r^2[/math]

 

So far so good.

 

However, as Gravity doesn't act perpendicularly to two masses, I think the spherical area would describe the total energy in the Space-Time wave propagating in all directions, and a 2D wave would describe its' gravitational potential acting in one direction... I think!

 

Anyway, for the moment, assuming that to be true, using 2D propagation, and that: [math]A \propto r[/math]

 

 

If: [math]E=mc^{2}[/math] (intentionally no momentum)

 

And: [math]E \propto c^2[/math]

 

 

Then: [math]m \propto E \propto A^2 \propto r^2[/math]

 

 

Which is already interesting because of the Gravity equation above.

 

 

Anyway... thoughts welcome, but TBC

 

 

BTW - How can I switch off this "multiple post merged" thing - I'm joining up disparate ideas here?

Edited June 12, 2008 by nstansbury
multiple post merged

[iNow]

BTW - How can I switch off this "multiple post merged" thing - I'm joining up disparate ideas here?

 

You can't. It's a hard programmed feature of the site itself, nothing specific to your account.

 

What I usually do is go back and edit in a few spaces or a break of some sort (like a line of ~~~~~~~~~~~~) after the posts get merged.

[insane_alien]

if you're posting properly the posts merged shouldn't affect you except in unordinary circumstances like when the forums mess up a bit and you have a double post.

[nstansbury]

How does one "post properly"? I just click reply

 

It seems you cant follow a post by yourself with another.

 

 

 

Guys, could you do a sanity check on the equations & assertions I've changed in my first post above please. I've been over them a dozen times now so I hope I haven't made any embarrassing blunders!

 

 

Interesting what [math]\pi[/math] has to do with it all....

 

The difference between the the Dirac and Plank constants appears to be related to the amplitude defined by [math]2 \pi[/math] which is very close to [math]G=(2\pi )10^{-11}[/math]

 

So it appears [math]G=2 \pi \propto f[/math]

 

What I find really intriguing and elegant about this is, it suggests the Universe is effectively a circular feed-back loop. This would very neatly explain why the Big Bang occurred initially and why it has propagated the way it has.

Edited June 14, 2008 by nstansbury
multiple post merged

[Klaynos]

I define Dirac's constant as: [math]\hbar=c\lambda r[/math]

 

Where:

[math]c[/math] is the wave's rate of propagation

[math]r[/math] is the radius of the Space-Time wave front

[math]\lambda[/math] is its' wavelength

 

 

I've only just noticed this, when scanning, but do you have a reason for this? Or can you cite a reference, I've had a quick look around I can't find any mention of "radius of space-time wavefront" or similar.

 

Nor what that would really mean...

[nstansbury]

I've only just noticed this, when scanning, but do you have a reason for this? Or can you cite a reference, I've had a quick look around I can't find any mention of "radius of space-time wavefront" or similar.

 

Yes, by defining [math]\hbar[/math] as [math]c \lambda r[/math] in terms of a 3D Space-Time wave I can prove [math]E=mc^2[/math] in terms of a Planck quanta.

 

3D Space-Time waves are my proposal of how [math]m \propto E \propto A^2 \propto r^2 \propto c^2 \propto m[/math] in both the Quantum and Relativity worlds.

 

This effectively allows you to describe 4 dimensional Space-Time as a wave

Edited June 14, 2008 by nstansbury

[Klaynos]

But if I define it as a turnip then I can show that you're wrong....

 

You need physical reasons for defining it as such.

[nstansbury]

But if I define it as a turnip then I can show that you're wrong

 

...but if you define it as a turnip can you prove a wave can have zero mass itself and yet explain gravity and mass in Relativity, because to my knowledge no-one else can.

 

All I'm doing is using 3 properties of a wave (speed, distance & wavelength) to describe mass as well.

[Klaynos]

Maybe... but you still need some physical reason for doing it...

 

Also here I recreate you substitution using proportionality constants http://www.scienceforums.net/forum/showpost.php?p=414723&postcount=14 and get a different answer, so I still don't believe your substitution.

[nstansbury]

Also here I recreate you substitution using proportionality constants http://www.scienceforums.net/forum/s...3&postcount=14[/url'] and get a different answer, so I still don't believe your substitution.

 

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

 

But you didn't prove that