A new physics paper

Paper #2 (will have to be cut up again)

Hope you enjoy it, it has taken many months to make this paper.

ABSTRACT

The curved trajectory itself cannot be simply applied to General Relativity geometry, however, this can apply in certain circumstances (as a generalized orbit in near-flat vacuum state neglecting gravitational waves). In this work, we will create a thought-experiment which involves an approximated classical trajectory round a certain center with radius

R

obtained from some calculus. We then apply this to the Larmor equation by noticing some equivalent terms that will be derived. We then later specify we are dealing with a particle of mass

M

and provide the equations which will calculate the charge emitted in a gravitational field and the luminosity value of the system. While the classical (calculus) approach needs a full theory in three-geometry which holds true for relativity to bridge this gap, I do present a method in which we can not only calculate the power radiated by the electron, but we will also find that it's the gravitational field itself (metric) that provides this electromagnetic radiation.

Before we delve into the mathematics of the theory, let's just begin to construct exactly what we are talking about. Anyone who knows Green's Theorem off-by-heart will know of the geometrical consequences of the Larmor equation. A pure mathematician is likely to notice it, but probably had never considered it.

For some particle in an electromagnetic field (thus having a charge) and mass

M

has it's spin

\vec{S}

and its angular momentum coupled as

L\ \cdot S

. Intuitively, from calculus, we know this is an angle between two vectors... the real derivation of this term is slightly more complicated. It involves commutative variables. Let's quickly do it!

J = L + S

is our total angular momentum, it consists of the particles inertial angular momentum to it's intrinsic spin. We then compute a dot product consisting of the sum divided into two paranthesis

J^2 = (L + S)(L + S) = L^2 + S^2 + 2L \cdot S

since

S

and

L

commute, we get

L \cdot S = \frac{1}{2} (J^2 - L^2 - S^2)

However, as I said before, the good mathematical college student should know that the dot product of any two vectors actually yield a geometrical meaning. What we will find out soon, is that a term in the Larmor equation has another hidden geometrical meaning. In fact, the mathematical tools and all of the derivations we use to reforumulate the Larmor equation are the same instruments you use to construct Greene's Theorem.

Remember, Greene's theorem actually describes the perimeter of a curved object, of any shape. It should be no surprise then that Larmor has something of that description hidden in his equation [1]. Let's do the math now.

Deriving The Larmor Equation in a New Form

The Larmor energy is written as a Hamiltonian

\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1}\frac{\partial V(r)}{\partial r}(L \cdot S)

The part we will concentrate on is

\frac{\partial V(r)}{\partial r}

And we will use calculus (that is also found as first principles to Greene's theorem) to derive an equivalence with this expression. We begin with the determinant

\nabla \times F = \begin{vmatrix}\hat{n}_1 & \hat{n}_2 & \hat{n}_3 \\ \partial_x & \partial_y & \partial_z \\F_x & F_y & 0 \end{vmatrix}

You can write this as

\nabla \times F = \frac{\partial F_y}{\partial x}\hat{n}_1 - \frac{\partial F_x}{\partial y}\hat{n}_2 + (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3

The first set of terms cancel out

\nabla \times F = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3

A unit vector squared just comes to unity, so if you multiply a unit vector of both sides we get

\nabla \times F \cdot \hat{n} = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})

Now, a curl of the force equation can be given as

\nabla \times F = \frac{\partial V(r)}{\partial r} \hat{n}

Notice, apart from the unit vector, this is an identical term found in the Larmor energy. Again, if one multiplies the unit vector on both sides we get

\nabla \times F \cdot \hat{n} = \frac{\partial V(r)}{\partial r}

A quick check over the original Larmor energy

\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1}\frac{\partial V(r)}{\partial r}(L \cdot S)

Shows that the Larmor energy can be written as

\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1}\nabla \times F \cdot \hat{n} (L \cdot S)

Here

L \cdot S = |L| |S| \cos \theta

appears like an equation describing an angle between two vector quantities, the momentum and it's spin coupling. But more interesting the force

\vec{F}

will be also in the direction of the spin coupling at a given instant. This not only couples the trajectory and the spin in this instance, but force is also conserved in the equation

\Delta H = \frac{1}{r}\frac{2 \mu}{\hbar Mc^2 e} \nabla \times \vec{F} \cdot \hat{n} (L \cdot S)

There is something fundamental about the relationship, that the force is probably responsible for the motion (resulting in the trajectory round the perimeter of a curved object

\mathcal{C}

) coupled to the spin

S

. Understand, that in this equation, Larmor has defined an inertial energy

Mc^2

so we are talking about particle with mass naturally. Notice though, it does have charge. This means, any classically-charged particle following a curved path will emit radiation. It's uncertain to me whether anyone has made the necessary equations to calculate an electron following a classical path in a gravitational field, but there are ways to do it: Not only ways to do it, but I will present a radically new way to do it!

Well... actually, I shouldn't get ahead of myself. I would never have came to this idea if it wasn't for reading several of the late Prof. Lloyd Motz' papers. He was massively underrated yet, never heard a bad word about him really. His papers and books where published over a number of years, involving radically new ways of thinking about particle dynamics, based on Cosmological measurements. He was effectively at this time attempting to bridge a gap between Cosmology and physics, but neither community truly grasped the inspirational idea's that where probably revolutionary for it's time. He in fact, predicted using some fundamental equations that mass can in fact be treated as a charge

Gm^2 = \hbar c

using a gravitational constant coefficient term. In many ways, only 4 years later after his first idea's on the subject that Peter Higgs proposed his... astounding theory of how photons get mass. Interestingly... and it really should be noted, the only particle so far in the Standard Model that have mass and no charge, are presumed Neutrino's. It's believed we know this with some certainty, but the delicacy of measurements and the notion its charge could be infinitesimally-small as it's mass is, should be considered.

(edited 10 years ago)

Reply 1

Graviphoton

OP

It seems in many respects, mass and charge where the same thing. You could say, Motz' approach was in fact a more fundamental prediction to symmetry breaking: Relating the Planck Charge to a Planck Mass related to a gravitational constant, which may even vary upon scale, involving what is known as the strong gravitational constant

G_s

. Then we find all sorts of rather appeasing gravitational relationships to electromagnetic theory such as

e^2 = \sqrt{4 \pi \epsilon Gm^2}

which attempts to measure some fundamental scale to the Heirarchy of particle masses; ie. a particles mass not only depends on the symmetry breaking, but there is an electromagnetic mass present as well

m_{em} = \frac{2}{3} \frac{e^2}{Rc^2}

where

R

is the classical radius

(\frac{e}{Mc^2})

.

As it turns out, this recent ''Higgs Boson'' appears to not quite fit the normal ''standard Higgs Boson.'' It has a greater expectancy to decay into photons... which at the time asked whether the HIggs detected was in fact part of the Standard Model at all.

Now, only a few days ago, it was announced that a four-quark had been detected to some great degree, reported in many articles [2]... which is also not predicted by the Standard Model. It seems like our theories are going ''beyond the standard model'' way.

Leaving alone now, these definitions of charge and of mass, concentrating again on the fact it has an electromagnetic charge

e

must involve giving up radiation. Now, whilst the geometric formulation of the first part cannot always hold true in curved spacetime, it will always hold true in a hypothetical two dimensional situation: in this case, the two dimensional classical motion is defined on a two dimensional plane

R^2

on a closed surface.

To extend properly into a proper three-geometry system involving complete curvature terms

\Gamma_i

will involve a much fuller theory: It is by no regret I do not have that yet, but I do have a full three-geometry theory about how our charged system should behave to some observer watching it traverse a geodesic.

The relativistic Larmour equation that measures an electrons radiation emitted while accelerating is given as

P = \frac{2}{3}\frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3}

Motz takes us to how we present this as a charged particle in a relativistic framework [3]

P = \frac{e^2}{mc^3}(\frac{dP}{dt})^2

He then applies this with a denominator that is the Schwarzschild solution (and is dimensionless)

P = \frac{e^2}{mc^3}\frac{a^{2}_{g}}{1 - \frac{r_s}{r}}

Instead of this approach, I want to make it even more relativistic, instead of measuring a single reference frame from the metric, I would rather wish to formulate a full relativistic situation, which measured the redshift relative

z

to some distant observer. In a pure time-symmetric case, this metric becomes static, in fact, in electromagnetic theory, it is static because of retarded and advanced wave forms travelling through lightcones. If that was the full model presented here today, we'd have a very fundamental picture which unites space, spin and geodesics together: in a sense, we will get this any way later from some classical analysis and calculus.

I'm not done however, I further want this to model an analogous Hawking Radiation of the system; papers later will confirm this approach is dynamically-analogous to intrinsic Hawking radiation, a black-body radiation of the metric itself, which provides the charged system with radiation. I formulated an efficient way to represent this measure of change, by re-writing the metric in terms of it's stationary energy and variate the two in a measure of relativistic redshift.

The key equation which describes all these dynamics is given as:

P = \frac{2}{3} \frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3} \frac{1}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E_{rec}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E_{sou}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}} (\frac{dP}{dt})^2

= \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(\frac{\lambda}{\lambda_0})}

Where the subscripts, {rec} is the reciever and {sou} is the source. We will omit these subscripts for simplicity in the future.

The unique part of this equation is that not only does it describe the energy emitted, but it describes the wavelength emitted

\lambda_0

and an observer at

R

measures the wavelength as

\lambda_0

. If

R < r

the observer will measure it as a blueshift. We have rewritten the metric in terms of the energy, so it does describe the energy received and the energy omitted. Of course, what we have is the shift description

1+z

where

z

is the shift, therefore when there is zero gravitational shift, there remains a unity. These conditions are quite standard and well-known.

Such an equation has not been offered in literature but is completely sound. The radiation emitted and the gravitational shift are usually calculated separately, but what I have shown is that they are really part of the same equation (or at least, can be) since both are related by equivalence. The interesting thing is the energy in the denominator, the metric appears to be providing the energy for the radiation.

http://arxiv.org/pdf/physics/9910019v1.pdf

''Thus we ﬁnd that the work done against the stress force, supplies the energy carried by the radiation.''

''Who is performing this work or, what is the source of the energy of the radiation?''

''It comes out that the energy carried away by the radiation is supplied by the Gravitational Field, that loses this energy.''

As you can see, the paper is saying the gravitational field acts as source for the energy which is carried away by radiation. What we see in my equation, is that the metric describes the gravitational field and now we have rewritten the metric in terms of intrinsic energy, it appears in the equation and from citing the paper, that the radiation is indeed provided from the term

\Delta E

or at least, it may be interpreted that way.

Another thing is that both the quantities

\frac{2}{3} \frac{e^2}{c^3}a_{g}^{2}

and

\frac{1}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}

are natural consequences of the equivalence principle when we consider a mass

M

in a gravitational field described by the metric. The close relationship of gravitational shift and the equivalence principle has been known for a long time now, and the charge accelerating in the gravitational field works from first principles of the same equivalence.

The strength of the gravitational field, is again provided by the metric which we have rewritten in terms of energy. To evaluate how this is featured in the metric, you can show that the radius of curvature is

r_c = \frac{c^2}{g}

Strength of gravitational ﬁeld at the surface of a gravitating object is given as a ratio of the actual radius to the light curve radius.

\frac{R}{r_c} = \frac{GM}{Rc^2}

where we have used

g = \frac{GM}{R^2}

If

\frac{GM}{Rc^2} << 1

then the field is said to be weak. Notice,

\frac{GM}{Rc^2}

is an integral expression of the Schwarzschild metric, therefore it describes the field strength of your metric. From here, you can actually state the ratio of the actual and light curve as a ratio of the gravitational energy and the rest energy by noticing

\frac{R}{r_c} = \frac{GM}{Rc^2} \cdot \frac{m}{m} = \frac{(\frac{GM^2}{R})}{Mc^2} = \frac{E_g}{E_0}

Luminosity Relationship

Based on work by Lloyd Motz for charges in gravitational fields, for an arbitrary charged mass which is accelerating in some distant gravitational field is

P = \frac{2}{3} \frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3} \frac{1}{(1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})} (\frac{dP}{dt})^2

= \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})^2}

The squared metric term in the denominator appears because there is one such factor in

ds^2

. The presence of the square of the metric also ensures that the power emitted by a charge in a gravitational field can increase beyond limit, implying a relationship to the luminosity of such a system. So theoretically we should be able to detect some objects by their luminosity value.

The localized Gravitational Jerk for a Distant Charge Radiating

If one took the original equation and performed an integral on the Larmor formula part, we may find the Abraham-Lorentz force, doing so will make it become a gravitational acceleration performing a jerk. We describe this by saying the acceleration is taken with respect to time

F = \frac{e^2}{6 c^3} \frac{\dot{a}_g}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}

The net force consists of separating the force into two parts via a sum: The sum of the radiative force and the external force

F_{net} = F_r + F_e = \frac{e^2}{6 c^3} \frac{\dot{a}_{g}^2}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}+ F_e

The radiative part appears from the Larmor formula part

\frac{2}{3}\frac{e^2}{c^2} \dot{a}_g

and the external field part appears related to the metric (in which the gravitational field supplied the radiation). The two fields are inextricably linked and both are responsible for an overall net force on our system.

A very strange but surprising condition can arise from the new formula

F_{net} = F_r + F_e = \frac{e^2}{6 c^3} \frac{\dot{a}_{g}^{2}}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}+ F_e

and it comes straight out of classical physics. What we really have is

M \dot{v} = F_r + F_e = m t_0 \ddot{v} + F_e

= \frac{e^2}{6 c^3} \frac{\dot{a}_{g}^{2}}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}+F_e

where

t_0 = \frac{e^2}{6mc}

If you integrate the equation once, where an integral extends from

t

to

\infty

we can find the future effecting the past! Signals in an interval of

t_0

into the future affects the acceleration in the present. Whether that has a real world application is another thing, but still interesting to note.

Reply 2

Graviphoton

OP

Zero Point Field Quantum Gravi-Shift

What is interesting is how similar the radiative force

F_r

is to the zero point energy. The recoil then has the appearance of a force exerted on a particle due to zero point fluctuations - in fact, H. E. Puthoff has written a theory in which gravity presents itself as a zero-point force in which the gravitational constant involves proportionally the zero point cut-off

\omega_c

. This all forms from the principle of equivalence again, it predicts the additional zero point contribution to gravitational mass, a necessity of gravitational unification with the ZPF. Basically, spectral lines should be quantized in theory and therefore would be quantities of well-known fundamental constants of nature.

The modelling of the zero point energy would require that we replace the radiative part with the correct zero point expression (again, all done in cgs units)

\frac{e^2 \hbar \omega}{2 c^5}a^{2}_{g}

In a gravitational field, using the equivalence principle, we'd express this as

\frac{e^2 \hbar \omega a_{g}^{2}}{2 c^5} = \frac{e^2 \hbar \omega}{2 c^5} (\frac{GM}{r^2})^2

This would make the equation

P = \frac{e^2 \hbar \omega}{2 c^5} \frac{a^{2}_{g}}{\sqrt{\frac{g_{tt}(r)}{g_{tt}(s)}}}

This describes the theoretical quantum gravitational shift of zero point contribution.

Similarities of Hawking-Unruh Radiation

It might not be enough for me to just say the radiation comes from the metric in the situation of a black hole, a fuller description of the quantum mechanical process is that virtual particles are in fact ''boosted'' by the gravitational field of the system into real particles. We don't see this full description in my equation, but we do see it as an effect provided by the metric. It is interesting though to think of this change in energy in the metric as really an underlying quantum mechanical process of virtual particles in the metric itself, how to implement this isn't entirely clear but will make an interesting investigation.

Someone may even notice how similar this approach is to the temperature equation for a black hole

T = \frac{\hbar}{2\pi ck_B} \frac{GM}{(\frac{2GM}{c^2})^2} = \frac{\hbar c^3}{8 \pi k_B} \frac{1}{GM}

Notice we find the same squared part in the denominator, the length squared (here given as the schwarzschild radius) is the same as finding the extra factor

ds^2

in our equations.

You can actually rewrite the equations I provided to look a bit more like the terms found in this equation above,

P_{per charge} = \frac{2}{3} \frac{e^2}{c^3} \frac{G^2M^2}{R^4 (1 - \frac{2GM}{c^2 R})^2}

.[4]

Final Conclusions

In this work, I've provided a framework, one that needs a lot of work to be fully relativistic, the other is pretty much complete. I've been able to use an idea from Prof. Motz to formulate some equations which are perhaps, more fundamental in terms of measuring a relative shift.

I have been able to show, initially as well, that maybe there is something fundamental to spin, trajectory and even the force directing the system of motion. Classically, the same tools we use to find Greene's Theorem can be used to find the same geometric meaning encased in the Larmor equation of spin-orbit coupling.

Of course, only in trivial 2-dimensional cases does this fully satisfy, whilst more complicated geometric structures will involve a more precise relativistic reformulation of the reformulated Larmor Hamiltonian to relativistic curved complete three-geometries.

The addition of the speculations of a zero-point energy redshift was just as analogous to Unruh-Hawking radiation and so I felt it should be mentioned as a side-note of theoretical possibilities.... So what are the consequences of everything then? So what if we can describe a trajectory classically that fits into the Larmor spin-coupling equation, what does it all prove?

In it's fullest form, it doesn't ''prove'' a thing, it presents a theoretical model which makes predictions. One of these predictions is that for electrons bound in a very tight geodesics round black holes should, according to the equation, emit radiation beyond limit, which will over time accumulate to form a ''glow'' round black holes. This should be a prediction using the laws of quantum mechanics to be able to detect them in space from a reasonable distance. Classically, any system of electrons that are bound to a center and following a curved perimeter, will give up radiation an collapse to the center.

To someone outside the black hole, this takes almost forever to happen, while intrinsically, the event does indeed happen. But this phenomenon outside the black holes frame of reference should be accounted for and the equations given should predict a way to measure that luminosity produced.

Ref.

[1]. A vectorial representation of Greens Theorem is now given to show you how it is related. I can't actually write up graphs here, but it's all about a curve in the x-y plane, in which there can be a force associate to it.

The tangential force equations is written as

\int F_T ds = \int F_T d\ell \int F_T dR

which is another way to describe the line integral.

dR

is the differential displacement.

Working in a two dimensional place, we have out force vector

F

\bar{F} = P_{\hat{i}} + Q_{\hat{j}}

(P,Q)

can be functions of

x

and

y

. The position vector is simply

\bar{R} = x_{\hat{i}} + y_{\hat{j}}

(dropping the hat notation on my components for ease, anyway...

dR = dx_i + dy_j

The line integral can be written as a dot product

\int \bar{F} \cdot \bar{R} = \int Pdx + Qdy

such that

\oint Pdx + Q dy = \int \int [\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}] dx dy

the right handside describes the interior of the curve, the left the perimeter of the curve

Say we want to express this in vector notation, we need to write it as a curl of vector

F

which allows us to write it in a determinent form

\bar{\nabla} \times \bar{F} = \begin{vmatrix} i & j & k \\ \partial_x & \partial_y & \partial_z \\P_{(x,y)} & Q_{(x,y)} & 0 \end{vmatrix}

(that is zero for the z-component - use something to cover up the first row and first column, what we have is

- \frac{\partial}{\partial z} Q_{(x,y)}i

(partial derivetive to z is zero)

(0 - \frac{\partial P_{(x,y)}}{\partial z}) j

the second term is obtain by covering the second column and first row - notice you cannot take the partial derivative with z because it is zero so some terms cancel out

So the curl has no

i

or

j

component

cover up the last column and the first row, we will have the partial of

Q

with respect to

x

minus the partial

P

with respect to

y

so we have a non-zero expression

+ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}k = \bar{\nabla} \times \bar{F}

what happens if we take the dot product with unit vector

k

on both sides? we have

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}k \cdot k = k \cdot(\bar{\nabla} \times \bar{F})

a unit vector squared is just unity, so we have

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = k \cdot(\bar{\nabla} \times \bar{F})

This is the same as

Unparseable latex formula:

\oint Pdx + Q dy = \int \int [\frac{\partial Q}{\patial x} - \frac{\partial P}{\partial y}] dx dy

you can rewrite this in an equivalent form

\oint \bar{F} \cdot d\bar{R} = \int \int (\bar{\nabla} \times \bar{F}) \cdot k dx dy

This is a general equation for Greens theorem in vector form.

dx dy = d\bar{A}

\oint \bar{F} \cdot d\bar{R} = \int \int (\bar{\nabla} \times \bar{F}) \cdot k d\bar{A}

This is more often the form seen in vector notation.

[2] http://physics.aps.org/articles/v6/69

[3] http://link.springer.com/article/10.1007%2FBF02735508#page-1

[4] (A paper which directly confirms my prediction, that a fully relativistic equation that must involve a true ratio of the metric terms which measures the gravitational redshift

z

.

http://arxiv.org/pdf/physics/9910019v1.pdf

[5] The equations I derived which calculated the luminosity is given as

L = 4 \pi (R (1 + z))^2 f

As I said before, the objects being spoken about would have what I called a ''luminosity value.'' The value is obtained by noticing
the squared redshift in the paranthesis

(1 + z)^2

. If

R

is the radius of the object, the luminosity value of that
same object is a redshift decided by the ratio of the reciever and the source, in the schwarzchild geometry becomes the redshift ratio
reached by our original equation

L = 4 \pi R( \frac{\sqrt{1 - 2\frac{Gm}{\Delta E_{rec}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E_{sou}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}})^2 f

For the black body limit, this can be set to

\sigma R T^4 = (\frac{\sqrt{1 - 2\frac{Gm}{\Delta E_{rec}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E_{sou}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}})^2 f

where

\sigma

is the Bolztmann constant and

f

is the bolometric flux, where

T

is simply the temperature of our system.

Solving for the temperature, we find the same equation we arrived at before from our power equation

T = \sqrt[4]{\frac{f(\frac{\sqrt{1 - 2\frac{Gm}{\Delta E_{rec}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E_{sou}} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}})^2}{\epsilon(\lambda) \sigma}}

where we have taken into account our dimensionless emissivity which depends on the wavelengths, given as a metric ratio.

extra's

http://www.researchgate.net/publication/234503063_On_the_Radius_of_Gyration_of_Stars
http://www.worldscientific.com/doi/abs/10.1142/S0217732398000887?journalCode=mpla

*

\Delta H = \frac{1}{r}\frac{2 \mu}{\hbar Mc^2 e} \vec{\nabla} \times \vec{F} \cdot \hat{k} (L \cdot S)

Keep a note of this term

\vec{\nabla} \times \vec{F} \cdot \hat{k} = \frac{\partial V(r)}{\partial r}

We might even simplify this in as simply an energy

E

and write it with partial derivative like so

E = \frac{1}{r}\frac{2 \mu}{\hbar Mc^2 e} (L \cdot S) \int \int [\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}] \bar{A}

This leads to new ways to write the magnetic field component as a dot product

\mathbf{B} \cdot \bar{A} = \frac{1}{Mc^2 e} \frac{1}{r} \mathbf{L} \oint \bar{F} \cdot d\bar{R}

= \frac{1}{Mc^2 e} \frac{1}{r} \mathbf{L}(\vec{\nabla} \times \vec{F})\hat{k}d\bar{A}

because

k(\vec{\nabla} \times \vec{F})d\bar{A} = \oint \bar{F} \cdot d\bar{R}

We are only able to derive this from knowing the magnetic field equation

\mathbf{B} = \frac{1}{Mc^2 e} \frac{1}{r} \frac{\partial V(r)}{\partial r}\mathbf{L}

It's maybe illuminating to understand what

\mathbf{B} \cdot \bar{A} = \frac{1}{Mc^2 e} \frac{1}{r} \mathbf{L} \oint_{\mathcal{S}} \bar{F} \cdot d\bar{R}

...physically means. A magnetic field operating on an area like this gives us the magnetic flux over a surface/area. We can thus replace this term with the definition of the flux

\oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \frac{1}{Mc^2 e} \frac{1}{r} \mathbf{L} \oint_{\mathcal{S}} \bar{F} \cdot d\bar{R}

where

\Phi_{\mathbf{B}} = \mathbf{B} \cdot \bar{A} = \oint_{\partial \mathcal{S}} \mathbf{A}\ dr = \mathbf{B}\bar{A} \cos \theta

* Written in a more modern form, we reach the full relativistic Dirac equation

F_{rad} = \frac{\mu_0 e^2}{6\pi Mc}\frac{[\frac{d^2p_{\mu}}{d \tau^2} - \frac{p^{\mu}}{m^2c^2} (\frac{dp_{\nu}}{d \tau} \frac{dp^{\nu}}{d \tau})]}{(1 - \frac{g_{tt}(r)}{g_{tt}(s)})}

I argue it is more relativistic than Diracs original form based on arguments of the presentation of using the redshift, which itself a consequence of the equivalence principle. Motz used it to calculate theoretical luminosities for quasars.

Just to leave you with a ''little'' mathematical gem, we can simplify all the complicated terms and approximate the gravitational shift accordingly so one may make an accurate measurement as possible.

Neglecting all the fancy complicated Einstein notation in Dirac's form, we resume back to our original equations form and rewrite the denominator as