Bussard's bremsstrahlung calculation

[Art Carlson]

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?

yes

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?

I don't know where you got that. I made no assumptions about how big the region is. The argument applies whether it is infinitesimal or infinite or anything in between. I also made no assumption about which parts may have a B field, an E field, both, or neither.

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?

I made no assumptions about how the fields are produced.

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).

That doesn't change anything. I just showed that an electron leaving the sheath will have the same angle as it had entering the sheath. If it goes in at a steep angle, then it bounces back at a steep angle.

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?

Correct, but not relevant to the point I was making.

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

I assumed that the only variation in any quantity was in the x-direction. In that case dA_x/dy vanishes, and B and A are orthogonal.

[Art Carlson]

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.

A potential well requires an excess of electrons, but that excess can be practically anywhere. My picture is a squarish well formed by a shell of excess electrons near the boundary, with not even 1ppm excess of electrons over most of the volume.

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.

It is correct that a distribution can be isotropic without being Maxwellian. But I am referring to isotropy in velocity space. That has nothing to do with spatial asymmetries in the electron density.

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

Dan Tibbets

Yes, at least roughly.

[icarus]

Art:

I just showed that an electron leaving the sheath will have the same angle as it had entering the sheath. If it goes in at a steep angle, then it bounces back at a steep angle.

Okay, we clearly have a misunderstanding here by one or both of us. If you are saying that the "electron enters the sheath" you must be referring to the x=a postion in my above argument. I think we need to define exactly what is the interface region we are discussing to avoid circumlocution.

As I see it, the sheath encompasses a region where the isotropic electron velocities and quasi-neutral plasma assumptions break down locally.
We have assumed it to be locally planar (dependent on x-only), time independent. Also the E-field and B-field are everywhere orthogonal; the B-field being tangential to the plane and the E-field normal to the plane.

Somwehere inside this region, at x=a, is the beta=1 tangent surface ( I erroneously called this the B=0 plane but it is actually where B=sqrt(2 mu P), P being plasma kinetic pressure.

(Now this raises interesting side question; if the electron velocities are having a larger normal component (x-direction) at this surface due to interaction with the E-field immediately interior to the beta=1 surface, an enhanced kinetic pressure is experienced because of the additional normal component of the molecules reflecting off this surface.)

Back to the main thread though, I'm assuming a region interior to the beta=1 surface, i.e., x<a, exists where there is no B-field but where the E-field is influencing electrons away from isotropic and quasi-neutral conditions in the far-field from the interface. Am I way off target with this assumption? My x=0 point would be where the isotropic and quasi-neutral conditions are recovered locally.

[Art Carlson]

Art:

I just showed that an electron leaving the sheath will have the same angle as it had entering the sheath. If it goes in at a steep angle, then it bounces back at a steep angle.

Okay, we clearly have a misunderstanding here by one or both of us. If you are saying that the "electron enters the sheath" you must be referring to the x=a postion in my above argument. I think we need to define exactly what is the interface region we are discussing to avoid circumlocution.

As I see it, the sheath encompasses a region where the isotropic electron velocities and quasi-neutral plasma assumptions break down locally.
We have assumed it to be locally planar (dependent on x-only), time independent. Also the E-field and B-field are everywhere orthogonal; the B-field being tangential to the plane and the E-field normal to the plane.

Somwehere inside this region, at x=a, is the beta=1 tangent surface ( I erroneously called this the B=0 plane but it is actually where B=sqrt(2 mu P), P being plasma kinetic pressure.

(Now this raises interesting side question; if the electron velocities are having a larger normal component (x-direction) at this surface due to interaction with the E-field immediately interior to the beta=1 surface, an enhanced kinetic pressure is experienced because of the additional normal component of the molecules reflecting off this surface.)

Back to the main thread though, I'm assuming a region interior to the beta=1 surface, i.e., x<a, exists where there is no B-field but where the E-field is influencing electrons away from isotropic and quasi-neutral conditions in the far-field from the interface. Am I way off target with this assumption? My x=0 point would be where the isotropic and quasi-neutral conditions are recovered locally.

That is (nearly) 100% correct and clearly stated. beta is sometimes defined in different ways, so we might want to watch out for future misunderstandings. An electric field can exist in a quasi-neutral plasma (that's the quasi- part), but it can't be too strong or too extensive, and at some point you have to explain why a current doesn't flow to eliminate the E field. This is why I expect most of the action to be near the edge, but I am willing to entertain the notion of a significant E field even within the radius where B goes to zero.

[D Tibbets]

Art:

I just showed that an electron leaving the sheath will have the same angle as it had entering the sheath. If it goes in at a steep angle, then it bounces back at a steep angle.

Okay, we clearly have a misunderstanding here by one or both of us. If you are saying that the "electron enters the sheath" you must be referring to the x=a postion in my above argument. I think we need to define exactly what is the interface region we are discussing to avoid circumlocution.

As I see it, the sheath encompasses a region where the isotropic electron velocities and quasi-neutral plasma assumptions break down locally.
We have assumed it to be locally planar (dependent on x-only), time independent. Also the E-field and B-field are everywhere orthogonal; the B-field being tangential to the plane and the E-field normal to the plane.

Somwehere inside this region, at x=a, is the beta=1 tangent surface ( I erroneously called this the B=0 plane but it is actually where B=sqrt(2 mu P), P being plasma kinetic pressure.

(Now this raises interesting side question; if the electron velocities are having a larger normal component (x-direction) at this surface due to interaction with the E-field immediately interior to the beta=1 surface, an enhanced kinetic pressure is experienced because of the additional normal component of the molecules reflecting off this surface.)

Back to the main thread though, I'm assuming a region interior to the beta=1 surface, i.e., x<a, exists where there is no B-field but where the E-field is influencing electrons away from isotropic and quasi-neutral conditions in the far-field from the interface. Am I way off target with this assumption? My x=0 point would be where the isotropic and quasi-neutral conditions are recovered locally.

That is (nearly) 100% correct and clearly stated. beta is sometimes defined in different ways, so we might want to watch out for future misunderstandings. An electric field can exist in a quasi-neutral plasma (that's the quasi- part), but it can't be too strong or too extensive, and at some point you have to explain why a current doesn't flow to eliminate the E field. This is why I expect most of the action to be near the edge, but I am willing to entertain the notion of a significant E field even within the radius where B goes to zero.

Perhaps I understand your reasoning better. A couple of questions. Is your assertion that the electrons remain dominatly in a sheath near the B- field border due to the need to avoid the 'current flow' and/or because the noncaptured electrons are bouncing off the B-field at angles concistent with the relevent incomeing angle (like in billiards) - but the surface of each lobe is convexly curved to the center so that subsequent bounces would increase nonaxial directions, till all the electrons are traveling at nearly lateral directions? But, with the multiple lobes of the Wiffleball- an electron bouncing off of a lobe so that it has a more lateral orentation will bounce off a neighboring lobe (as opposed to a lobe on the oppposite side (or reflecting from the center in a similar maner) of the machine) and be deflected back to a more radial direction towards the center.
Perhaps ther
e is current flow, either smooth or bunched (waves or shock fronts) which would essentially neutralize the plasma locally and globally if there was an equal number of oppositly charged particles. In that case wouldn't the excess numbers of electrons still maintain a working potential despite this current flow. Just like in a normal electrical circuit, there can be current flow through a wire so long as you provide an external power input to maintain the potential. If this current flow was to a ground the energy losses would presumably be bad, but this current flow would be in a closed system, essentially back and forth like the analagy I've made before - a capacitive/resistant resonate circuit.

[EDIT] My conceptual model above is based on a single electron without collisions except with B- field borders. Does it breakdown (other effects become dominate) when more electrons and finally ions are introduced? I don't think the global repulsion of the electrons approaching each other near the center would change things (forces would be radial), but sideways scattering from local electron collisions may be significant.

Dan Tibbets