Radiation Of A Point-Charge-Accelerated Field

[Aethelwulf]

Astract

 

Based on word done by L. Motz, we investigate a possible description of a metric which describes an object accelerating in some distant gravitational field, relative to an observer. We rewrite the metric in terms of energy which further interprets it to mean we have an energy recieved from a source. This energy appears to be related to the gravitational field described by the metric field strength.

 

The Main Equation describing an Accelerated Charge

 

[math]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[/math]

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

 

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

 

The unique part of this equation is that not only does it describe the energy emitted, but it describes the wavelength emitted [math]\lambda_0[/math] and an observer at [math]R[/math] measures the wavelength as [math]\lambda_0[/math]. If [math]R < r[/math] the observer will measure it as a blueshift. We have rewritten the metric in terms of the energy, so it does describe the energy recieved and the energy ommitted.

 

Such an equation has not been offerred 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 [math]\Delta E[/math] or at least, it may be interpreted that way.

 

 

Another thing is that both the quantities

 

[math] \frac{2}{3} \frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3}a_g[/math]

 

and

 

[math]\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}}}}[/math]

 

are natural consequences of the equivalence principle when we consider a mass [math]M[/math] 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

 

[math]r_c = \frac{c^2}{g}[/math]

 

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.

 

[math]\frac{R}{r_c} = \frac{GM}{Rc^2}[/math]

 

where we have used

 

[math]g = \frac{GM}{R^2}[/math]

 

If

 

[math]\frac{GM}{Rc^2} << 1[/math]

 

then the field is said to be weak. Notice,

 

[math]\frac{GM}{Rc^2}[/math]

 

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

 

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

 

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

 

[math]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[/math]

 

[math]= \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}[/math]

 

 

The squared metric term in the denominator appears because there is one such factor in [math]ds^2[/math]. 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

 

[math]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}}}}[/math]

 

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

 

[math]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[/math]

 

The radiative part appears from the Larmor formula part [math] \frac{2}{3}\frac{e^2}{c^2} \dot{a}_g[/math] 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

 

[math]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[/math]

 

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

 

[math]M \dot{v} = F_r + F_e = m t_0 \ddot{v} + F_e[/math]

 

[math]= \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[/math]

 

where

 

[math]t_0 = \frac{e^2}{6mc}[/math]

 

 

If you integrate the equation once, where an integral extends from [math]t[/math] to [math]\infty[/math] we can find the future effecting the past! Signals in an interval of [math]t_0[/math] into the future affects the acceleration in the present. Whether that has a real world application is another thing, but still interesting to note.

 

What is interesting is how similar the radiative force [math]F_r[/math] 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 [math]\omega_c[/math]. 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)

 

[math]\frac{e^2 \hbar \omega}{2m c^5}a_g[/math]

 

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

 

[math]\frac{e^2 \hbar \omega a_{g}^{2}}{2 mc^5} = \frac{e^2 \hbar \omega}{2 mc^5} (\frac{GM}{r^2})^2[/math]

 

This would make the equation

 

[math]P = \frac{e^2 \hbar \omega}{2 mc^5} \frac{a_g}{\sqrt{\frac{g_{tt}®}{g_{tt}(s)}}}[/math]

 

In a sense, the equation would measure a quantum gravi-shift.

Edited October 15, 2013 by Aethelwulf

[Rade]

I formulated an equation which describes the radiation emitted by an accelerated charge in a gravitational field as [math]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} \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}}}} (\frac{dP}{dt})^2[/math][math] = \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(\frac{\lambda}{\lambda_0})} [/math]

 

I have a question. Suppose we have a three mass entity isotope composed of matter [pnp] that is accelerating within a gravitational field...with two protons (p+) and one neutron (n) this would be the stable isotope Helium-3. According to your equation if this [pnp] moved in a gravitational field it would emit radiation given that the two (p+) have charge, plus the two electrons (e-) associated with the protons.

 

So, my question, does your equation apply for this type of more complex situation when the charge associated with an entity accelerating within a gravitational field has both positive (p+) and negative (e-) charge ? If yes, would have the interest and time to solve your equation and determine the amount of radiation predicted to be emitted by a single Helium-3 isotope that accelerates in a gravitational field ? I understand this is a time consuming request, sorry, but no harm to ask.

 

Next, suppose the entity was the anti-matter of Helium-3. Would your equation then predict the same numerical emitted radiation value for P only with opposite sign ?

[Aethelwulf]

Though it can describe a single particle, it might have better applications to celestial bodies like a star which is accelerating round a gravitationally-warped spacetime. You can plug in the values yourself and find out the luminosity of a system! And yes, it doesn't matter whether it is a negative or positive charge.

[Aethelwulf]

A good thing is that the equation is fully relativistic.

[Aethelwulf]

I had a few messages from someone on another site about the presentation. There was some confusion about the metric ratio, so I have rewritten the thing in the OP... just added a few extra details thus not to cause future confusion.

[Aethelwulf]

Ok, so why is the equation important?

 

1. As I stated before, this equation is fully relativistic, even more relativistic in structure than the Larmor equation because it takes into account the redshift (a consequence itself of first principles involving equivalence)

 

 

2. It is a good equation because it unifies in some sense, the quantum behaviour of accelerated charges with the metric. The metric turns out to provide the quantum mechanical effects of this charge accelerating. It unifies then, the gravitational physics with standard quantum mechanical behaviour!

 

3. The third good thing, is that it may have diverse applications, or perhaps diverse connections between classical and quantum mechanics. It can even be applied to cosmological bodies accelerating in gravitational fields, where you would treat such a body with a shell in which the radiation per charge is emitted.

[Aethelwulf]

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

 

[math]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}[/math]

 

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 [math]ds^2[/math] in our equations.

 

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

 

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

 

(where I have omitted the charged metric for simplicity for now). In fact, if my equations did describe say, a charged black hole accelerating in a strong gravitational field, the radiation emitted by the metric is pretty much the same arguments in the Unruh-Hawking radiation.

Edited October 12, 2013 by Aethelwulf

[Aethelwulf]

In this paper, it argues that the equivalence principle are in fact precursors to Unruh-Hawking radiation

 

http://www.hep.princeton.edu/~mcdonald/accel/unruhrad.pdf

 

In my equations, the metric is the source of the radiation, in which the contribution is a ratio of

 

[math](\frac{\Delta E_{rec}}{\Delta E_{sou}})[/math]

Edited October 12, 2013 by Aethelwulf

[Aethelwulf]

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.

 

The charged part of this metric appeared because I was considering charged bodies like black holes. For a quasar, the charged part of the metric would vanish to a Schwarzschild metric.

Edited October 12, 2013 by Aethelwulf

[Aethelwulf]

Because today, most black holes are expected to be very cool, the detection of a black hole seems uncertain. Only a relatively small black hole would have a reasonably large contribution from [math]\Delta E[/math] and so might be detectable. In the case of quasars, we would not have the metric contribute the energy. Instead, we would resolve the equation to the normal Schwarzschild arguments, the accretion disk around the centre black hole, where a quasar is acting as a shell of mass of many charged ion particles would give off the synchroton radiation as it accelerated towards the gravitational centre. We would find an extremely high redshift also in this case, as is all cases of quasars in our universe.

Edited October 12, 2013 by Aethelwulf

[Aethelwulf]

Even though we know charges do in fact radiate proportionally to the square of the acceleration there does exist some problems, problems involving conservation.

 

This article writes a little about them http://physics.fullerton.edu/~jimw/general/radreact/

 

Here is a good reading source http://prd.aps.org/abstract/PRD/v27/i8/p1715_1

[Aethelwulf]

My realizing of the quantum zero point fluctuation appears to be written to satisfy the quantum equivalence principle and can be detected in spectral lines in quantities of the constants that composes the energy radiative term. The energy term which describes the ZPF is all about the vacuum fluctuations. For an excited atom, this may have the appearance as fluctuation of the electromagnetic field rather than provided as a radiation runaway of the gravitational field.

 

https://digitallibrary.sissa.it/handle/1963/188

 

I'd advise people to take a copy of this article most sites want you to pay for it!

Edited October 14, 2013 by Aethelwulf

[Aethelwulf]

Trying to find new ways to write my metric to describe the functions properly.

 

In order to understand how the metric contributes energy is to revisit the terms which describe it, namely this term

 

[math]-(m\cdot\frac{GM}{r}) = m\phi[/math]

 

This part describes the potential energy in our gravitational field. What we find are two ratio's

 

[math]\sqrt{1-2\frac{m\phi}{\Delta E_{r}}}[/math]

 

[math]\sqrt{1-2\frac{m\phi}{\Delta E_{s}}}[/math]

 

The ratio of the potential to the energy recieved and the ratio of the potential energy to the source of the energy. Therefore, a change in the energy [math]dE[/math] would imply a direct change with the potential energy, which depends on the radius of the system,

 

[math]\sqrt{1 - 2 \frac{dm \phi®}{dE_{r}}}[/math]

 

[math]\sqrt{1 - 2 \frac{dm \phi®}{dE_{s}}}[/math]

 

If this energy is given up in quanta, then energy and momentum is carried off in the form of radiation

 

[math]P^{\mu} = (dp, dE/ c)[/math]

 

To write this in the metric, we state

 

[math]\gamma(u) \sqrt{1-2\frac{d m\phi®}{d E_{r} + u dp}} = \sqrt{1-2\frac{d m\phi®}{\gamma dE_r}}[/math]

 

[math]\gamma(u)'\sqrt{1-2\frac{d m\phi®}{d E_{s} + u dp}} = \sqrt{1-2\frac{d m\phi®}{\gamma dE_s}}[/math]

 

where one momenta (the source) happens at [math]t=0[/math] and the momentum recieved is made at a later time [math]t = dt[/math] respectively.

 

Because our example is about virtual particles boosted in the frame of the black hole into real particles, the usual energy-momentum formula will not obeyed precisely until the boost is performed, some velocities can be negative for the virtual particles [math]\pm v[/math].

Edited October 18, 2013 by Aethelwulf

[Aethelwulf]

Let's take a look at the new equation

 

[math]P = \frac{2}{3} \frac{e^2}{m^2c^3} \frac{1}{\frac{\sqrt{1 - 2\frac{d m\phi®}{d(\gamma E_r)} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{d m\phi®}{d (\gamma E_s)} + \frac{GQ^2}{c^4 R^2}}}}(\frac{d}{dt}\gamma M u)^2 = \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(\frac{\lambda}{\lambda_0})}[/math]

 

I can't think of anything else I need to add to it. The potential supports the physical nature of the energy from the source to the receiver, defining a change in the shift [math]z[/math] if there is indeed one. The energy of course, appears from our metric.

Edited October 18, 2013 by Aethelwulf

[Aethelwulf]

Calculating the Bolometric Flux Density

 

Consider the case for luminosity perhaps for a quasar now, where the squared metric in the denominator implied that the charge could emit a power beyond limit. We may also find a similar representation for the bolometric flux density, by noticing a few relationships.

 

The bolometric flux density is

 

[math]f = \frac{L}{4\pi R^2 (1+z)^2}[/math] 1.

 

Notice we have the term [math]1+z[/math] squared, this is the same thing as what was in our original denominator

 

[math](1+z) = (\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}}})[/math] 2.

 

We should also recognize that the luminosity is in fact

 

[math]L = 4 \pi R^2 \sigma T^4[/math] 3.

 

Plugging this into equation 1 we have

 

[math]f = \frac{4 \pi R^2 \sigma T^4}{4\pi R^2 (1+z)^2}[/math]

 

Cancelling out like terms, we find our relationship for the flux density

 

[math]f = \frac{\sigma 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}[/math]

Edited October 19, 2013 by Aethelwulf

[Aethelwulf]

A more modern representation of the previous equation is

 

[math]j^{*} = \frac{\epsilon(\lambda) \sigma 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}[/math]

 

[math]\epsilon[/math] is the emissivity.

 

We can therefore, using the last equation obtain a second representation of the power equation

 

[math]P = Aj^{*} = \frac{A \epsilon(\lambda) \sigma 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}[/math]

 

with [math]A[/math] being of course our surface area.

Edited October 19, 2013 by Aethelwulf

[Aethelwulf]

The simplicity of the derivation is rather beautiful. It retains all of the important features of the gravi-shift as it did in our original equation for power. The first method can describe luminosity per charge, applications such as virtual particles being boosted in the gravitational field of a black hole. The second power equation looks like it has best applications to large celestial bodies, since we are involving concepts of surface area.