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#1 Dubbelosix

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Posted 08 January 2019 - 03:39 PM

Let's discuss the physics of a topic I investigated by ''knitting'' together (as I have been accused of before lol) essential parts of questions involving accelerated charges producing Larmor radiation.

 

A while back, I considered the [contribution] of radiation from the accelerated charges inside of the cavity. This was obtained from a Langrangian that I formed from the ordinary Rayleigh-Plesset equation, but motivated in forming the equation to show strong similarities to the expanding-collapsing model known as the Friedmann equation for cosmology:

 

[math]mR \ddot{R} + \frac{3}{2}m\dot{R}^2 + \frac{4 \nu m}{R} \dot{R} + \frac{2S m}{\rho_L R} + \frac{\Delta P(t)m}{\rho} = \mathcal{L}[/math]

 

* Notice, for this to be true, the pressure has units of energy and the denominator is a normal density [math]\frac{\Delta P}{\rho}[/math] with units of velocity squared. We also extract meaning in

 

 

[math]\frac{4 \nu m}{R} \dot{R}[/math]

 

[math]\frac{3}{2}m\dot{R}^2[/math]

 

In the sense that we know from these two expressions that the viscosity divided by the radius is the same as

 

[math]\frac{\nu}{R} = \dot{R}[/math]

 

And notice how similar the form is when dividing through by the radius squared:

 

[math]m \frac{\ddot{R}}{R} + \frac{3}{2}m(\frac{\dot{R}}{R})^2 + \frac{4 \nu m}{V} \dot{R} + \frac{2S m}{\rho V} + \frac{1}{R^2}\frac{\Delta P(t)m}{\rho} = 0[/math]

 

This is of course, no coincidence, but both solutions are very closely linked since both solutions pertain to an expanding or collapsing sphere (though many shapes of the universe have been speculated) in fluid dynamics. Notice we can simplify some terms, mainly the density can be extracted on two of the expressions:

 

[math]m \frac{\ddot{R}}{R} + \frac{3}{2}m(\frac{\dot{R}}{R})^2 + 4 \rho \nu \dot{R} + 2S + \frac{1}{R^2}\frac{\Delta P(t)m}{\rho} = 0[/math]

 

 

 

And so this concludes the first part... even after all this talking, the modified equation still does not describe the radiation given up, the point of this first post was to show the structure of the equation in terms of other things. Next post will describe how charges rotating round a central potential will give up radiation and can be added into the equation as a ''correction term.''


Edited by Dubbelosix, 08 January 2019 - 03:46 PM.


#2 Dubbelosix

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Posted 09 January 2019 - 10:17 AM

Let's take a look at the Friedmann Langrangian, another equation I stumbled upon which has relevance to the understanding of the posts:

 

[math]m\dot{R}^2 - \frac{8 \pi GmR^2}{3}\rho + \frac{\Lambda mc^2}{3}R^2 = \mathcal{L}[/math]

 

The derivatives of the logarithmic spiral giving up radiation is a collection of two terms:

 

[math]m\dot{R}^2 = \frac{e^2}{6 \pi c^3} \dddot{R} + \frac{1}{2}e\mathbf{V}[/math]

 

The last term refers to rotating systems and so will produce Larmor radiation. Since the viscosity actually refers to the motion of the fluid around the bubble then any charged particles compressed to the region [will] exhibit the behaviour of this equation above!

 

The reason why this was stated because it is believed by a number of scientists that Larmor radiation from accelerated charged particles may be a contributor to the phenomenon. Based only on the fact that magnetic fields have been detected around the ‘’star in a jar’’ - I think this is possible since a rotary feature gives rise to a closed current in which charged particles could be bound in a high momentum, giving off Larmor radiation. Certainly the amount of energy from the source, (if it cannot be described alone by nuclear fusion) could have other additional contributors. As stated in previous articles I have written, I explain how a third derivative in time gives rise to a non-conserved form of the Friedmann equation and in a sense, we require the same thing for the Rayleigh-Plesset equation so that we can describe the system under a power equation

 

So let's plug it in, first we rearrange:

 

[math]m \frac{\ddot{R}}{R} + \frac{3}{2}m( \frac{\dot{R}}{R})^2 = - 4 \rho \nu \dot{R} + 2S + \frac{1}{R^2}\frac{\Delta P(t)m}{\rho}[/math]

 

We can use a trick for the time derivative, with a temperature variation or anistropy:

 

[math]m \frac{\dot{R}}{R} \frac{\ddot{R}}{R} + 3m \frac{\dot{R}}{R}\frac{\ddot{R}}{R} = - d\log_V(4 \rho \nu \dot{R} + 2S + \frac{1}{R^2}\frac{\Delta P(t)m}{\rho} + \frac{e^2}{6 \pi c^3} \dddot{R} + \frac{1}{2}e\mathbf{V})\frac{\dot{T}}{T}[/math]

 

In the past I considered the term [math]\frac{1}{2}eV[/math] to be replaced with the central potential [math]\frac{\partial U}{\partial R}[/math] but it need not be written in such a way, in fact the way it has been presented is arguably the right way. There are other corrections we can make to the equation that we will look at later.


Edited by Dubbelosix, 09 January 2019 - 10:36 AM.


#3 Dubbelosix

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Posted 09 January 2019 - 10:31 AM

Two things to note, how did we get [math]\frac{3}{2}[/math] to become just a factor of [math]3[/math]?

 

This is just a matter of differentiation, for instance, differentiating [math]\dot{R}^2[/math] we bring the power down as a constant and form [math]2 \ddot{R}\dot{R}[/math]. Secondly, the differential logarithm shows up because there was an integration in the original work, this integration can be seen more visibly when written in the following way (ignoring the logarithmic charge and electric potential for now)

 

[math]\frac{\Delta P + \mathbf{P}}{R^2}[/math]

 

[math]= \frac{\ddot{R}}{R}\rho + \frac{3\rho}{2}(\frac{\dot{R}}{R})^2 + \frac{4 m\nu }{V^2}\dot{R} + \frac{2S}{V}  [/math]

 

with

 

[math]\Delta P = P - (P_0 - P(t))[/math]

 

and so integration of the volume element is:

 

[math]\int\ \frac{\Delta P + \mathbf{P}}{R^2}\ dV [/math]

 

[math]= d\log_V(m\frac{\ddot{R}}{R} + \frac{3}{2}m(\frac{\dot{R}}{R})^2 + 4 \rho \nu \dot{R} + 2S)[/math]


Edited by Dubbelosix, 09 January 2019 - 10:41 AM.


#4 Dubbelosix

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Posted 09 January 2019 - 10:47 AM

I really should write the equation in the format given in the last post, [IF] I want the differential logarithm to appear correctly, besides, it's a nicer way to write the equation:

 

 

[math]\int \frac{\Delta P(t)}{R^2}\ \dot{V} = -d\log_V(m \frac{\ddot{R}}{R} + \frac{3}{2}m( \frac{\dot{R}}{R})^2 + 4 \rho \nu \dot{R} + 2S + \frac{e^2}{6 \pi c^3R} \frac{\dddot{R}}{R} + \frac{1}{2R^2}e\mathbf{V})\frac{\dot{T}}{T}[/math]

 

[math]\dot{V} = \frac{dV}{dt}[/math]

 

Which includes the change for the differential volume.

 

(Edit)

 

The voltage is also related to the electric field:

 

[math]+ \frac{1}{2R^2}e\mathbf{V}[/math]

 

[math]\mathbf{V} = \int\ \mathbf{E} \cdot dl[/math]

 

so by simplifying:

 

[math]+ \frac{1}{2R} e\mathbf{E}[/math]


Edited by Dubbelosix, 09 January 2019 - 11:54 AM.


#5 Dubbelosix

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Posted 09 January 2019 - 01:55 PM

Because of the energy weighted down by the inverse length (radius) squared, obtaining the power equation is simple stuff, with dimensions of energy per unit of time:

 

 

 

[math]\int\ \Delta P(t)\ \dot{V} = -d\log_V(m \dot{R}^2 + \frac{3}{2}m R\ddot{R} + 4 m \nu \frac{\dot{R}}{R} + 2SR^2 + \frac{e^2}{6 \pi c^3} \dddot{R} + \frac{1}{2}e\mathbf{V})\frac{\dot{T}}{T}[/math]



#6 Dubbelosix

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Posted 10 January 2019 - 10:37 AM

I have no idea whether these correction terms will work or add any new physics. But this is the ordinary thing to do, to theoretically calibrate equations to intend to correct different circumstances.