Equation of Interest, Ratio of Total Energy to Rest Mass

[IDMclean]

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

 

[math]E_n = \sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}^2[/math]

 

[math]\frac{E_n}{m} = c^2 =\frac{\epsilon\mu}{\epsilon_0\mu_0}[/math]

 

the symbols are:

[math]E_n[/math], Energy

[math]m[/math], rest mass

[math]c[/math], speed of light

[math]\epsilon[/math], Permittivity of the medium

[math]\mu[/math], Permeability of the medium

[math]\epsilon_0[/math], Permittivity of freespace

[math]\mu_0[/math], Permeability of freespace

[math]\epsilon_r[/math], Relative Permittivity

[math]\mu_r[/math], Relative Permeability

 

I was hoping I might get some help with the mathematics of this. It is simple Algebraic manipulations.

 

The supposition that might take research is the energy is equivalent to the product of relative permittivity and relative permeability. That one I am not sure how to Justify. Not to say It can't be justified, just that I am not sure how I arrived at that conclusion.

 

A note. I use the symbol I do for energy because latter I plan on introducing Parts of Maxwell's Equations.

[IDMclean]

Supposing that the above is correct, the family of equations that fall out of this are thus:

 

[math]\frac{\epsilon\mu}{\epsilon_0\mu_0} = \epsilon_r\mu_r = \frac{E_n}{m} = c^2 =(\frac{E}{B})^2[/math]

 

We can then suppose, assuming the conjecture of [math]m = \sqrt{\epsilon\mu}[/math], this is what I think of as the index of mass as it is directly related to the index of refraction which is [math]n = \sqrt{\epsilon_r\mu_r}[/math]:

 

[math]K = mc^2 - \gamma mc^2[/math]

[math]K = mc^2 - \frac{m}{\sqrt{\frac{1}{c^4} - \frac{\nu^2}{c^{6}}}}[/math]

 

[math]K = \frac {\epsilon\mu} {\epsilon_0\mu_0} - \frac {\epsilon\mu} { \sqrt {(\epsilon_0 \mu_0)^2 - \frac {(\epsilon_0\mu_0)^3} {\epsilon\mu}}}[/math]

 

Of the Kinetic Energy to mass ratio, or the relative permittivity and permeability.

 

[math]\frac{K}{m} = c^2 - \gamma c^2 = \frac{1}{\epsilon_0\mu_0} - \frac{1}{\sqrt{(\epsilon_0\mu_0)^2 - \frac{(\epsilon_0\mu_0)^3}{\epsilon\mu}}} [/math]

 

Comments?

[Erasmus00]

The supposition that might take research is the energy is equivalent to the product of relative permittivity and relative permeability. That one I am not sure how to Justify. Not to say It can't be justified, just that I am not sure how I arrived at that conclusion.

 

The supposition is incorrect. [math] \frac{1}{\epsilon\mu} = c^2 [/math] where c is the speed of light in your material. This does not have the right units to be energy. Its velocity squared.

 

I've suggested this before to no avail, but you should try using a more natural set of units for Maxwells equations

 

[math] \nabla \cdot E = 4\pi\rho[/math]

[math]\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}[/math]

[math]\nabla \cdot B = 0[/math]

[math]\nabla \times B = \frac{4\pi}{c}(J+\frac{1}{4\pi}\frac{\partial E}{\partial t} )[/math]

 

In this system, we don't have the constants [math] \epsilon_0[/math] and [math]\mu_0[/math], just the physically meaningful speed of light in vaccuum.

 

Edit: latex glitch

-Will

[arkain101]

Kick,

 

Here's a relationship that may help you along. You may already know it, hope it helps

 

 

 

 

 

 

Where:

 

E is the total energy of the system

p is the total momentum of the system

c is the speed of light

[IDMclean]

There is no supposition in this of [math]c = \epsilon\mu[/math], this would be because c is known to be [math]c = \frac{1}{\sqrt{\epsilon_0\mu_0}}[/math].

 

There is however the supposition that [math]\nu = \frac{1}{\sqrt{\epsilon\mu}}[/math].

 

the relationship then is:

[math]\frac{\nu^2}{1} \cdot \frac{1}{c^2} = \frac{1}{\epsilon\mu} \cdot \frac {\epsilon_0\mu_0} {1}[/math]

 

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

 

Also, I can't meaningfully read the maxwell equations at this time. I've tried, it doesn't work to well and I can't imagine how they work. So eventually I will worry about them when I am capable of comprehending the repercussions of the set of equations. Until then I am content to work with that which I understand.

[Qfwfq]

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! :evil: I betcha Will is gonna be so confused by it, he'll be really stumped, he'll just never get it, someone fetch the smelling salts......;)

 

You may find the way he wrote the Maxwell equations confusing, yes but that is how it goes. The permittivity of a material is due to it's polarizability, essentially it means the charges it's made up of will shift a little when there's a fiald on the large scale, the overall effect is to change the ratio of field and charge. Relative permittivity is a handy macroscopic representation for most practical purposes but it can be broken down to that, with the equations of the vacuum applying. The dielectric certainly comes to have an energy density greater than the vacuum would, with the same field intensity, but that's just the potential energy of the charges being shifted from equilibrium. What do you find confusing?

[IDMclean]

The calculus involved. Also, don't get me wrong, I don't mean to patronize. The whole reciprocal thing confuses the hell out of me. We'll not even go near the higher maths. On some level I understand some of it but I still can't read it. If I can't read or write it (which requires reading) then I do not consider myself "understanding" it.

 

I don't know what I mean when I read or write things like:

[math] \bigtriangledown \cdot E [/math] what the hell does that mean in plain English? If were going to explain that to a non-technical person what would the basic sentence start with?

 

Also the note on confusion was directed at the apparent confusion of Will in regards to c's relation in the set of questions. No venom intended, just trying to cover the basis and be clear as possible.

[Qfwfq]

OK, but it should have been clear that you were referring to yourself. :)

 

[math]\norm\nabla\cdot V[/math] is called the divergence of the vector field V. The two equations using it are differential versions of the rules about flux going in and out of enclosing surfaces.

 

[math]\norm\nabla\times V[/math] is called the rotor of the vector field V. The two equations using it are differential versions of the circuit, or closed loop, rules for current and changing flux.