The Nature Of Gravity And Mass Through Investigation Of Planck-Like Particles

[Aethelwulf]

There may be a negative sign in there

 

[math]-\frac{GM^2}{r_s} = M\phi[/math]

 

depending on what classical textbook you might read. Usually, which I just remembered, there is often a negative sign for

 

[math]-\frac{GM}{R} = \phi[/math]

[Aethelwulf]

Inertia can also be attributed to a rotational system - if we are dealing with a classical radius, then the angular momentum is not an intrinsic property as such when you would find when dealing with point-like systems. I just wanted to note a certain derivation for this post:

 

[math]\frac{2GM^2}{Mc^2} = \frac{E_gr_s}{Mc^2} = \frac{2GM}{c^2} = \frac{E_gr_s}{Mc^2} = \frac{e^2}{Mc^2}[/math]

 

for a classical radius.

 

 

(edited for fixing of a mistake)

Edited August 17, 2012 by Aethelwulf

[Pincho Paxton]

As I said however, you can search for the paper yourself, it is easy to find :)

 

No, I mean that the search goes to a PDF that doesn't work.

[Aethelwulf]

No, I mean that the search goes to a PDF that doesn't work.

 

 

When I search the paper, it works for me... strange. Try some other link --- if not, check your reader.

[Aethelwulf]

It's a good paper, go out your way to find it. Full of good info, especially the part about the self-energy of an electron, it is generally infinite in theory without any renormalization techniques.

[Aethelwulf]

Sorry, latexed a t^2 in there, which shouldn't be there if I am correctly doing these dimensions... I will fix it now.

[Aethelwulf]

nope I have done something wrong somewhere ... sorry about this.

[Aethelwulf]

Oh... I think I might have had it right,

 

My brain is in a muddle right now...

 

I should cancel the mass, if one was to simply multiply it on both sides of

 

[math]\frac{E_gr_s}{Mc^2} = \frac{e^2}{Mc^2}[/math]

 

thus the mass cancels, which leaves

 

[math]\frac{E_gr_s}{c^2} = \frac{e^2}{c^2}[/math]

 

What we really have is

 

[math]\frac{Mc^2r_s}{c^2} \rightarrow Mr = \frac{e^2}{c^2}[/math]

 

multiply the radius on both sides gives a rotational inertia

 

[math]Mr^2 = \frac{e^2r}{c^2}[/math]

 

In previous work, I derived from

 

[math]M = \frac{r^3}{2Gt^2}[/math]

 

one can get

 

[math]2E_gt^2 = Mr^2[/math]

 

This means that

 

[math]2E_gt^2 = \frac{e^2r}{c^2}[/math]

 

Is the gravitational energy rotational inertia in terms of the electric charge and the radius.

Edited August 17, 2012 by Aethelwulf

[Aethelwulf]

Ok, I think that is me. If anyone see's a problem with this, please point it out.

[Aethelwulf]

I had to look for some justification on line there because it didn't look right, but my memory has somewhat refreshed on [math]Et^2[/math] being related to an inertial quantity.

 

According to http://books.google.co.uk/books?id=MTdX4kZVREQC&pg=PA49&lpg=PA49&dq=inertia+-+energy+times+time+squared&source=bl&ots=yjZv6vursA&sig=IpA8tNOEgun1C8akYP6jq8t7XHw&hl=en&sa=X&ei=G5vfT_O-O-LP0QX4j9HLCg&sqi=2&ved=0CFsQ6AEwAQ#v=onepage&q=inertia%20-%20energy%20times%20time%20squared&f=false

 

...energy times [math]t^2[/math] is a thermal inertia parameter. It surely then is no coincidence it can be related directly to a rotational inertial quantity [math]Mr^2[/math], (or moment of inertia), which is usually given as

 

[math]I = \sum_{i=1}^{N} M_ir^{2}_{i}[/math]

[Aethelwulf]

Here is a nice excerpt about the charge quantization method adopted by Motz

 

http://encyclopedia2.thefreedictionary.com/Dirac-Zwanziger-Schwinger+quantization+condition

 

What we have is

 

[math]e\mu = \frac{1}{2}n\hbar c[/math]

 

It seems to say that [math]\mu[/math] plays the role of a magnetic charge - this basically squares the charge on the left handside, by a quick analysis of the dimensions previously analysed.

Edited August 17, 2012 by Aethelwulf

[Aethelwulf]

Here, referenced by Motz, you can see the magnetic charge is given as [math]g[/math], the electric charge of course, still given by [math]e[/math].

 

http://wdxy.hubu.edu.cn/ddlx/UpLoadFiles/File/2011070110535154942.pdf

 

what we have essentially is

 

[math]\frac{e\mu - e\mu}{4\pi} = n\hbar c[/math]

 

where your constants in the paper have been set to natural units. In light of this, one may also see this must be derived from the Heaviside relationship

 

[math]e = \sqrt{4\pi \alpha \hbar c}[/math]

 

I say this, because it picked up a [math]4 \pi[/math] term.