A Charge in a Gravitational Field

[TrappedLight]

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]

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.

I checked to see if anyone had offered this equation in literature - while though the Larmor equation is well-known and is used exactly for this, there has been no attempts to rewrite it for calculations made simultaneously for either red or blue gravitational shift. They are usually calculated separately, but what I have shown is that they are really part of the same equation (or at least, can be). 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.

Edited October 5, 2013 by TrappedLight

[TrappedLight]

Another thing is that both the quantities

 

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

 

To give you a taste of what some of the mathematics of my equation can do, suppose you wanted to calculate the total redshift of a wave: You would actually take an integral on the metric expression

 

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

 

like so

 

[math]\int_{R}^{s} \frac{GM}{R^2c^2} dr[/math]

 

But this would be for the weak scale. It is taken that this approach using the Larmor formula is in fact a universal property for all particles which may experience Synchroton radiation.

Edited October 5, 2013 by TrappedLight

[TrappedLight]

The first expression in post 2, is missing an extra term for the relativistic Larmor equation

 

[math](\frac{dP}{dt})^2[/math]

[TrappedLight]

Found a very relevant paper. http://users.jyu.fi/~ilsamaki/radiation.pdf

 

It also asks where the energy for the radiation comes from. It's conclusion is that it comes from the near fields surrounding the particle. The metric in our case acts as a near field and unifies the quantum with our topology.

[Enthalpy]

How strong could this radiation be? Any hope to measure it?

 

Because, well, last time I hoped to produce extreme UV with an undulator, the figures were so disappointing... And I computed with a short-period magnetic field. Since gravitation fields tend to be vast, I fear the radiated power is below detection.

 

What could be the best case? Electrons orbiting a small black hole near the horizon? Even if the orbital period is 1ms (no idea) radiation will be imperceptible I fear.

 

Could many electrons self-organize in bunches as in an X ray laser to emit more power?

[TrappedLight]

There is an interesting conclusion I reached to.

 

[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]

 

We can also perform an integral on the Larmor formula part of the equation, to find the Abraham-Lorentz force, it becomes the gravitational acceleration performing the jerk. 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.

How strong could this radiation be? Any hope to measure it?

 

Yes, if my equations are correct and using my choice of metric description, for an arbitrary charged mass which is accelerating in some distant gravitational field is

 

The radiation emitted is therefore

[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.

And for how strong the luminosity could be, will depend on more investigation.

The luminosity also depends on the object you are dealing with. This could even be for a black hole.

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. Perhaps we can find some interesting physics when we recognize the metric in the denominator is in fact a product of advanced and retarded waveforms which allows the gravitational red or blue shift. There might be some interesting physics when making time symmetric in this equation.

 

 

 

This is of course a classical field limit. If the classical case does not exist in nature, then the net force will need to be described fully with non-classical quantum mechanics. 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.

 

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}{2 c^5}[/math]

 

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

 

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

 

This would make the equation

 

[math]F = \frac{e^2 \hbar \omega}{2 c^5} \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}}}}[/math]

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

Edited October 7, 2013 by TrappedLight

[imatfaal]

[latex]F = \frac{e^2 \hbar \omega}{2 c^5} \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}}}}
[/latex]

 

There is an interesting conclusion I reached to.

 

[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]

 

We can also perform an integral on the Larmor formula part of the equation, to find the Abraham-Lorentz force, it becomes the gravitational acceleration performing the jerk. 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.

 

Yes, if my equations are correct and using my choice of metric description, for an arbitrary charged mass which is accelerating in some distant gravitational field is

 

The radiation emitted is therefore

[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.

And for how strong the luminosity could be, will depend on more investigation.

The luminosity also depends on the object you are dealing with. This could even be for a black hole.

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. Perhaps we can find some interesting physics when we recognize the metric in the denominator is in fact a product of advanced and retarded waveforms which allows the gravitational red or blue shift. There might be some interesting physics when making time symmetric in this equation.

 

 

 

This is of course a classical field limit. If the classical case does not exist in nature, then the net force will need to be described fully with non-classical quantum mechanics. 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.

 

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}{2 c^5}[/math]

 

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

 

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

 

This would make the equation

 

[math]F = \frac{e^2 \hbar \omega}{2 c^5} \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}}}}[/math]

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

 

 

[latex]F = \frac{e^2 \hbar \omega}{2 c^5} \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}}}}[/latex]

 

I have expanded your 2nd denominator

 

 

[latex]\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}}}[/latex]

 

I cannot see the difference between the top and the bottom - why wouldn't they just cancel? or have you suppressed some indecies/subscripts?

 

 

seems that it would just simplify to a zero-point jerk!

[TrappedLight]

[latex]F = \frac{e^2 \hbar \omega}{2 c^5} \frac{\dot{a}_{g}^{2}}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{1 -\sqrt{ 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}}

[/latex]

 

 

 

[latex]F = \frac{e^2 \hbar \omega}{2 c^5} \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}}}}[/latex]

 

I have expanded your 2nd denominator

 

 

[latex]\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}}}[/latex]

 

I cannot see the difference between the top and the bottom - why wouldn't they just cancel? or have you suppressed some indecies/subscripts?

 

 

seems that it would just simplify to a zero-point jerk!

 

 

Dimensionally, yes, they cancel, just like a metric is cancelled on itself by a length divided by a length [math]\frac{r}{r_s}[/math]. As Julian Barbour says, ''the only real physical quantities in physics are those which are dimensionless.''

 

Anyway, what we have is a general case in the metric. What we really have is

 

[math]z = \frac{1}{\sqrt{(1 -\frac{2GM}{rc^2}})} - 1 = \frac{\lambda - \lambda_0}{\lambda}[/math]

 

You simply reconfigure the equation to express the appropriate redshift since [math]\frac{\lambda}{\lambda_0}[/math] is in fact unity when the redshift is zero. You may represent this important ratio as the rewritten metric in terms of the vacuum energy.

Edited October 7, 2013 by TrappedLight

[timo]

I'm pretty sure imatfaal mean cancel in the sense of 2/2=1 or [latex]\frac{5x}{5x} = 1[/latex] or [latex]\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}}}=1[/latex].

 

[TrappedLight]

I'm pretty sure imatfaal mean cancel in the sense of 2/2=1 or [latex]\frac{5x}{5x} = 1[/latex] or [latex]\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}}}=1[/latex].

 

 

Yes, what I said was that it comes to unity only when there is a zero redshift.

[imatfaal]

 

Yes, what I said was that it comes to unity only when there is a zero redshift.

 

And what Timo and I both said is that it ALWAYS cancels out to unity; the top and the bottom are equal and thus the quotient is equal to one. If you believe that there are circumstance in which

 

[latex]\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}}}[/latex]

 

does not cancel (or in some circumstance equal zero) then please post them - ie what possible values provide any quotient of interest?

[TrappedLight]

You're are both mistaken. You should read up on the gravitational redshift for a metric here: http://en.wikipedia.org/wiki/Redshift It is written exactly how I have given it, except I have written it for the charged metric case, but it uses all the same arguments as a Schwarzschild case.

[timo]

You might want to pay more attention to comments like "or have you suppressed some indices/subscripts?".

[TrappedLight]

You might want to pay more attention to comments like "or have you suppressed some indices/subscripts?".

 

missed it. I was in a rush yesterday when I read that comment.

However I thought the difference was obvious since I was dealing with the ratio of wavelengths which were written differently.

That's such an immature thing to do negging a post timo.

 

Get a life.

Edited October 8, 2013 by TrappedLight

[imatfaal]

 

missed it. I was in a rush yesterday when I read that comment.

However I thought the difference was obvious since I was dealing with the ratio of wavelengths which were written differently.

That's such an immature thing to do negging a post timo.

 

Get a life.

 

The neg rep was me! I felt your replies were unhelpful, dismissive, and not very friendly. Your response to the neg rep pretty much confirms my thoughts.

[TrappedLight]

 

The neg rep was me! I felt your replies were unhelpful, dismissive, and not very friendly.

 

 

I am sorry if my posts came across unfriendly, I have no opinion about you either way.

[John Cuthber]

 

And for how strong the luminosity could be, will depend on more investigation.

 

Really?

If we don't do the investigation on which the luminosity depends, does it get brighter or duller?

 

More realistically, perhaps you would like to offer some idea of how much light you actually get from some example system.

[TrappedLight]

Really?

If we don't do the investigation on which the luminosity depends, does it get brighter or duller?

 

More realistically, perhaps you would like to offer some idea of how much light you actually get from some example system.

 

It could calculate the lumonisity of quasars possibly, or a large charge which is accelerating the vicinity of a massively warped space and time. The luminosity will certainly get brighter from the source.

[John Cuthber]

, perhaps you would like to offer some idea of how much light you actually get from some example system.

[TrappedLight]

I'll plug in some hypothetical values later.