Bussard's bremsstrahlung calculation

[icarus]

TRM wrote:

So the only quibble I'd have with your analysis is that, given that the B-field doesn't just end abruptly inside some critical radius, is that the electrons will have a slightly more complex geometry to their bounces--one where angle of reflection is slightly less than than angle of incidence.

I've pondered this a bit. The "reflections" will not be specular, I do not think. Due to the interaction with the strong electric field gradient orthogonal to the magnetic field lines at the wiffle-edge the electrons will return to the interior with greater angles than they arrived on the average, I figure. In other words, they deposit tangential momentum at the interface, normal velocity component is reversed but remains constant magnitude (?), i.e. the kinetic pressure.

[Art Carlson]

TRM wrote:

So the only quibble I'd have with your analysis is that, given that the B-field doesn't just end abruptly inside some critical radius, is that the electrons will have a slightly more complex geometry to their bounces--one where angle of reflection is slightly less than than angle of incidence.

I've pondered this a bit. The "reflections" will not be specular, I do not think. Due to the interaction with the strong electric field gradient orthogonal to the magnetic field lines at the wiffle-edge the electrons will return to the interior with greater angles than they arrived on the average, I figure. In other words, they deposit tangential momentum at the interface, normal velocity component is reversed but remains constant magnitude (?), i.e. the kinetic pressure.

Anything that can be done can be undone. Just reverse the direction of the magnetic field and the reverse trajectory becomes a possibility. If trapping happens, then un-trapping must also happen. If you can deposit tangential momentum sometimes, then at other times you can make a withdrawal. But trapping will not happen at all in a simple geometry. You need something like rapidly varying field strength or direction.

[TheRadicalModerate]

I have never heard R. Nebel chime in with his opinion.

Dr Nebel told me that in the my video Fusion for Dummies, the electron paths of one cusp to the other cusp was wrong. I would then put Dr Nebel in the same cusp group.

Still seems to me that whether they re-enter (fall back) through the same cusp or recirculate around to the the opposite cusp is purely a function of the energy of the electron upon leaving the cusp and the strength of the B-field.

If the escape energy is much less than the injection energy, the electron will fall back, but it probably won't have enough energy to make it back through the cusp. It will therefore oscillate across the magrid until it migrates to the anode.

If the escape energy is approximately the same as the injection energy, the electron may fall back or it may recirculate, if it's got enough energy to bend all the way around. I would expect that these electrons will re-enter the cusp and be hot enough to bounce around inside the sheath for a while.

If the escape energy is much larger than the injection energy, the electron will escape the B-field and hit the wall.

Now, as for the energy distribution of electrons upon escape, I have no clue. Those that aren't trapped in the sheath prior to escape should be pretty close to monoenergetic, but those that migrate through the cusp from the sheath should be pretty well thermalized, I'd think.

[TheRadicalModerate]

I've pondered this a bit. The "reflections" will not be specular, I do not think. Due to the interaction with the strong electric field gradient orthogonal to the magnetic field lines at the wiffle-edge the electrons will return to the interior with greater angles than they arrived on the average, I figure. In other words, they deposit tangential momentum at the interface, normal velocity component is reversed but remains constant magnitude (?), i.e. the kinetic pressure.

Consider a perfectly sharp B-field that is 0 for all r < some critical R, and of uniform strength for all r >= R. If the electron arrives completely normal to the B-field, it will gyrate exactly pi radians and then find itself back in the 0 B-field, with its velocity exactly reversed.

But if B isn't completely discontinuous, then the electron will already have experienced some lorenz force before reaching R. Hence, its total gyration will be > pi radians. So everything will be slightly bent to one side for each face of the wiffle-ball/cube.

[Art Carlson]

I've pondered this a bit. The "reflections" will not be specular, I do not think. Due to the interaction with the strong electric field gradient orthogonal to the magnetic field lines at the wiffle-edge the electrons will return to the interior with greater angles than they arrived on the average, I figure. In other words, they deposit tangential momentum at the interface, normal velocity component is reversed but remains constant magnitude (?), i.e. the kinetic pressure.

Consider a perfectly sharp B-field that is 0 for all r < some critical R, and of uniform strength for all r >= R. If the electron arrives completely normal to the B-field, it will gyrate exactly pi radians and then find itself back in the 0 B-field, with its velocity exactly reversed.

But if B isn't completely discontinuous, then the electron will already have experienced some lorenz force before reaching R. Hence, its total gyration will be > pi radians. So everything will be slightly bent to one side for each face of the wiffle-ball/cube.

I think you are both wrong. For a B field in the z direction, and an E field in the x direction, with the strength of both B and E arbitrary functions of x alone, the reflections will be "specular", although the position of the "mirror" may depend on the initial velocity of the particle.

The most elegant proof surely uses scalar and vector potentials, but I don't know my stuff well enough to just write it down.

[icarus]

Art said:

The most elegant proof surely uses scalar and vector potentials, but I don't know my stuff well enough to just write it down.

Most elegant I'd imagine. Unfortunately we don't possess such a proof, as yet, but once the Carlson sheath theory is fully developed we'll be in a much better position.

In the meantime, you'd have to agree that the further an electron penetrates the sheath, i.e. in the direction normally outward to the plasma, then the more radial it's velocity would necessarily become due to the purely radial electric field?

In my thinking, the magnetic field would be excluded from the sheath because it is electron-rich and capable of sustaining mag-field repelling loop currents. On this hypothesis, the electric field has a bigger influence on the sheath and thus "reflections than the mag-field .... so I reason non-specular and tangential momentum deposit.

[Art Carlson]

Art said:

The most elegant proof surely uses scalar and vector potentials, but I don't know my stuff well enough to just write it down.

Most elegant I'd imagine. Unfortunately we don't possess such a proof, as yet, but once the Carlson sheath theory is fully developed we'll be in a much better position.

I'm sorry. I thought it was considered good practice here to base arguments on unpublished and possible non-existant proofs and measurements.

In the meantime, you'd have to agree that the further an electron penetrates the sheath, i.e. in the direction normally outward to the plasma, then the more radial it's velocity would necessarily become due to the purely radial electric field?

I agree to no such thing. "In the meantime" consider an electron in crossed E and B fields. The motion is a superposition of the gyromotion and the EXB drift. These are independent, so the tangential velocity at the turning point (v_r=0) can be positive, negative, or zero. Cut off the picture of the cycloid at some radius and you have the orbit for the case that E and B begin abruptly and then remain constant. The velocity entering this "sheath" can take on any value, but when it leaves the sheath again, the velocity in the B direction will be unchanged, the velocity in the E (radial) direction will be reversed, and the velocity in the EXB (tangential) direction will be unchanged. Specular reflection.

In my thinking, the magnetic field would be excluded from the sheath because it is electron-rich and capable of sustaining mag-field repelling loop currents. On this hypothesis, the electric field has a bigger influence on the sheath and thus "reflections than the mag-field .... so I reason non-specular and tangential momentum deposit.

Why should the sheath be electron rich? (And why should it matter?) There is nothing to keep the ions from expanding to the same radius as the electrons (They don't see the B field much, remember?), and their inertia will force them to expand a little bit beyond the electrons. The outermost layers should be ion rich.

[TheRadicalModerate]

Why should the sheath be electron rich? (And why should it matter?) There is nothing to keep the ions from expanding to the same radius as the electrons (They don't see the B field much, remember?), and their inertia will force them to expand a little bit beyond the electrons. The outermost layers should be ion rich.

Shouldn't the E-field generated by the core electrons limit the expansion of the ions?

[Art Carlson]

Why should the sheath be electron rich? (And why should it matter?) There is nothing to keep the ions from expanding to the same radius as the electrons (They don't see the B field much, remember?), and their inertia will force them to expand a little bit beyond the electrons. The outermost layers should be ion rich.

Shouldn't the E-field generated by the core electrons limit the expansion of the ions?

Shouldn't the E-field generated by any core electrons force the expansion of the electrons, eradicating any electron excess in the core?

[Art Carlson]

Art said:

The most elegant proof surely uses scalar and vector potentials, but I don't know my stuff well enough to just write it down.

Most elegant I'd imagine. Unfortunately we don't possess such a proof, as yet, but once the Carlson sheath theory is fully developed we'll be in a much better position.

I'm sorry. I thought it was considered good practice here to base arguments on unpublished and possible non-existant proofs and measurements.

Wait, why don't you let me give it a try.

Assume B is in the z-direction, E is in the x-direction, and the magnitude of B and E are functions of x only (no dependence on y, z, or t). Then B = curl A implies A is in the y-direction and B = (dA/dx).

the Lorentz force law is (m/q)*(dv/dt) = E + vXB. Writing E as -(dPhi/dx), B as (dA/dx), and (d/dt) as v_x*(d/dx), the three components of are:
x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx) + v_y*(dA/dx)
y: (m/q)*v_x*(dv_y/dx) = -v_x*(dA/dx)
z: (m/q)*v_x*(dv_z/dx) = 0

It follows immediately from z that v_z is constant.
It follows nearly as quickly from y that v_y = -q*A/m is constant. (Think "canonical momentum", m*v+q*A, at this point.). Therefore, whenever the particle is at some given position x, it always has the same value of v_y.
Solving the last equation is left as an exercise for the reader. Those who like to read the last page of mystery novels first can just apply conservation of energy. Since the total energy is constant, and the potential energy is a known function of x, and v_z is constant, and v_y is a known function of x, there is nothing left for v_x to do but be a known function of x as well. The only twist is that v_x enters into equation x quadratically, so sometimes the particle will be traveling in the x-direction and sometimes in the minus x-direction.

QED

[D Tibbets]

Why should the sheath be electron rich? (And why should it matter?) There is nothing to keep the ions from expanding to the same radius as the electrons (They don't see the B field much, remember?), and their inertia will force them to expand a little bit beyond the electrons. The outermost layers should be ion rich.

Shouldn't the E-field generated by the core electrons limit the expansion of the ions?

Shouldn't the E-field generated by any core electrons force the expansion of the electrons, eradicating any electron excess in the core?

Excess electrons in the core is one of the primary requirements for the machine to work. My layman's impression is that the inertia of the electrons provided by the drive potential pushes the electrons to near the center before mutual repulsion slows them to a stop,and then they accelerate to the B field border. So long as the electrons motion remains dominatly radial (enter the arguments about thermalization times here) they will be slow in the central region and fast near the B- field (Wiffleball) border. Any calculation of electron density in any region of the machine has to include the residence time in the various regions/layers . Then there is the ~1 ppm excess in electrons, that apparently help to maintain(?)/enhance the potential well.

It would be nice if there was aviable expermental data concerning densities, and potential well measurements in Elmore Tuck Watson (ETW) Fusors, like there is for Hirsch Farnsworth Meeks type Fusors (sparse as it is). I did a Google search on the ETW and found discussion about them, but I did not find a single expermental paper.

Dan Tibbets

[Art Carlson]

Shouldn't the E-field generated by the core electrons limit the expansion of the ions?

Shouldn't the E-field generated by any core electrons force the expansion of the electrons, eradicating any electron excess in the core?

Excess electrons in the core is one of the primary requirements for the machine to work.

Why do you say that? You need a potential difference to confine the ions, but that will do the job whether it is near the edge, near the middle, or gradual over the whole radius.

So long as the electrons motion remains dominatly radial (enter the arguments about thermalization times here) they will be slow in the central region and fast near the B- field (Wiffleball) border.

Which it won't be. Rick has agreed that the electrons will be isotropic. We have some differences over the question of whether the electrons will be nearly monoenergetic, and whether the ions velocity will be predominantly radial, but not over the isotropic electron distribution.

If the electrons are thermalized, they will have the same energy distribution everywhere. If they are not, then at the center they will have a lower energy than at the edge, but this says nothing about the radius at which this transition occurs.

[D Tibbets]

Shouldn't the E-field generated by any core electrons force the expansion of the electrons, eradicating any electron excess in the core?

Excess electrons in the core is one of the primary requirements for the machine to work.

Why do you say that? You need a potential difference to confine the ions, but that will do the job whether it is near the edge, near the middle, or gradual over the whole radius.

So long as the electrons motion remains dominatly radial (enter the arguments about thermalization times here) they will be slow in the central region and fast near the B- field (Wiffleball) border.

Which it won't be. Rick has agreed that the electrons will be isotropic. We have some differences over the question of whether the electrons will be nearly monoenergetic, and whether the ions velocity will be predominantly radial, but not over the isotropic electron distribution.

If the electrons are thermalized, they will have the same energy distribution everywhere. If they are not, then at the center they will have a lower energy than at the edge, but this says nothing about the radius at which this transition occurs.

Perhaps I will learn something new again. By increased electrons in the core, I mean on average more electrons nearer the center ( not nessisarily at the center) than the average ion distance at the top of the potential well, otherwise there would not be any potential well and the machine would just be a bag of thermalized plasma - game over.
In my simple view there may be a small increased percentage of electrons nearer the center compared to the ions, and the excess injected electrons may only boost the effect, or for all I know the effective potential well may be entirely due to that 1 ppm excess of electrons over ions.

I'm going to claim that isotropic does not equate to thermalized. It just means that the electrons are not bunching on one side of the machine or forming beams that shoot out only through a portion of the spherical surface (obvously assumming the cusp losses are occuring well outside the fusion/ potential well dominate area that I guess Nebel was referring to.). A thermalized plasma would do this, but so would electrons in purely or mostly radial transits or anywhere in between so long as this relaxation (?) was ocurring symetrically throughout the machine, even with different densities or gradiants along those radial lines.

When you say near the edge (not at the edge) are you thinking of square, as opposed to eliptical potential wells?


Dan Tibbets

[TallDave]

Still seems to me that whether they re-enter (fall back) through the same cusp or recirculate around to the the opposite cusp is purely a function of the energy of the electron upon leaving the cusp and the strength of the B-field.

The field lines on the outside of the Magrid aren't compressed the same way they are on the inside, so they go out to the wall. An electron would have slice across them to get back in.

If the electrons are thermalized, they will have the same energy distribution everywhere. If they are not, then at the center they will have a lower energy than at the edge, but this says nothing about the radius at which this transition occurs.

I wonder if they'll have measurements on the well geometry for WB-8? I suppose we can guess a lot from the neutron rate when they run D-D. A laser fluouroscopy picture, like in the Japanese study, would be nice but I won't hold my breath for that.

[icarus]

Art, you said:

Assume B is in the z-direction, E is in the x-direction, and the magnitude of B and E are functions of x only (no dependence on y, z, or t). Then B = curl A implies A is in the y-direction and B = (dA/dx).

Just so that we're on the same page, we're talking about the interface region between plasma and B-field here, correct?

If so, your statement above seems to imply that you think the B-field will penetrate quite some way into the plasma, i.e. to the bottom of this interface region, such that at x=0, B is non-zero?

How would you justify that? I think that in this interfacial region, B will be function of electron density (and therefore also a function of x), so that as you move in towards the plasma there will be some threshold value of electron density that is sufficient to sustain mag-field-repelling loop currents. My contention is that the steep electric field gradient associated with the interface penetrates further than the x position where the B-field disappears. The bunching of electrons at the interface (sheath) ensures it, no?

In this picture, the electrons will have been aligned (by the electric field) to already have largely only radial velocity components before encountering the B-field (on an outwardly directed path).

So we add to your equations above the following;

for x<a, a - being the radius/altitude at which the B-field goes to zero,

x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx)
y: (m/q)*v_x*(dv_y/dx) = 0
z: (m/q)*v_x*(dv_z/dx) = 0

and for x>a, the above become (your original);

x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx) + v_y*(dA/dx)
y: (m/q)*v_x*(dv_y/dx) = -v_x*(dA/dx)
z: (m/q)*v_x*(dv_z/dx) = 0

Solving for x<a;
v_y and v_z are now both constant, and v_x = sqrt(2*q*phi/m) (again quadratic but this time only phi dependent). So as phi increases with increasing x up to a, v_x increases but v_y and v_z remain constant. The angle that the electrons enter the B-field region of the interface, at x=a, given by the ratio of tangential to normal velocity components

alpha =atan(sqrt(v_y^2 + v_z^2)/v_x)

and is equivalent to a loss cone angle. We see then that it is solely dependent upon the value of the voltage in the region just beneath the point where the B-field begins. QED?

Aside: I don't think that you can assume that the curl of the vector potential is orthogonal to the vector potential generally, as you have done. The result that I'm after doesn't change just that your equations should have begun;
x: (m/q)*v_x*(dv_x/dx) = -(dPhi/dx) + v_y*(dA_x/dy- dA_y/dx)
y: (m/q)*v_x*(dv_y/dx) = v_x*(dA_y/dx - dA_x/dy)
z: (m/q)*v_x*(dv_z/dx) = 0