Penning Traps Contents Introduction Clasical picture Radiation Damping Number density B and E fields used to increase time that an electron remains within a discharge: Penning, 936. Can now trap a particle 'indefinitely in a combined homogeneous B and electrostatic quadrupole (now known as a Penning trap). A small cloud of such trapped particles is like a many electron atom, with the nucleus replaced by a trapping field. Electrons introduced by applying HV to the field emmission point. The beam of energetic e's collide with sparse neutral, to produce slow electrons, which are captured in the trap. Electrodes are hyperboloids of revolution which produce a quadrupole field. Superimpose a uniform B. Resultant motion is fast circular cyclotron motion with a small radius carried along by a slow circular magnetron drift motion in a large orbit. i.e. an epicyclic motion in the x-y plane. Radius of cyclotron motion shrinks as synchrotron radiation is emitted. The axial oscillation is coupled to an external detector at low temperature. The large magnetron motion is a circle about an effective potential hill. p 7.
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Classical picture charge e, mass m. z axis parallel to B, positive so that cyclotron motion is right handed rule. i.e. z axis is indirection -eb. Cyclotron frequency is ω c = eb mc ˆ z = ω cˆ z p 7.3
e.g. 6T ω c = 64 GHz, wavelength is mm. Proton would oscillate at 89 MHz (radio). Charged particle is bound radially, but not axially. In an ideal Penning trap, superimpose a restoring force to any small perturbation by a quadruple field Write potential as V = V 0 z ρ / d p 7.4
This function satisfies Laplace. Can be produced by placing electrodes along equipotential contours. Three are required. Two endcaps and a ring. p 7.5
Endcaps: z = z 0 + ρ / ring electrode: z = ρ ( ρ 0 ) Constants z 0 and r 0 are the minimum axial and radial distances to the electrodes. Choose characteristic trap dimension d as d = ( z 0 + ρ 0 / ) Then V 0 is the potential difference between endcap and electrodes Axial motion is de coupled from B field, and is a simple harmonic motion. z + ω z z = 0; ω z = ev 0 md Usual to have ω z << ω c. Typically might have V 0 = 0 V, and d = 0.3 cm, so that ν z = 6 MHz. But can easily get 0 kev, so ν z = 6 GHz. Radial motion: m ρ = e Ε + ρ c B E = V 0 d ρ p 7.6
In terms of axial and cyclotron frequency ρ ω c ρ ω zρ = 0 See repulsive last term from electrostatic potential. Find two consequences. First cyclotron frequency reduced, as repulsive radial potential reduces centrifugal force. Second, fast cyclotron orbit is superimposed upon a slower circular magnetron orbit, angular frequency ω m. Resultant is as shown in figure b below, but now add axial harmonic motion. p 7.7
To understand magnetron motion, note that in perpendicular E and B fields, a charged particle with a drift velocity u = cε B/ B will move unimpeded, because in the basic radial motion equation u/cxb will cancel E (a velocity filter). Strictly true only for constant fields, i.e. a constant drift velocity u. But approximately true anyway. The filter velocity does not depend on m or e. Subst for E p 7.8
E = V 0 ρ/ d ( ) to get ω m = u / ρ = E / B / ρ = V 0 /( Bd ) = ω z /ω c shows that the drift or magnetron motion is a circular motion with the same sense of rotation as the cyclotron orbit, independent of e and m, but ω m << ω z << ω c. Note: cyclotron motion is almost exclusively kinetic. Axial motion alternates between kinetic and potential. Reducing energy in either reduces amplitude: they are stable. magnetron motion is almost exclusively potential. i.e. it is an orbit about the top of a potential hill. exciting the magnetron motion makes the particle roll down the hill.. The motion is unbounded: any dissipative process which removes energy from the magnetron motion increases the magnetron radius until the particle hits the ring and is lost But typical damping time is of order years! Radiation damping Accelerated charge radiates em waves. Therefore motion is damped. Transition probability for such an electric dipole is proportional to high power of transition frequency; appreciable radiative decay occurs only for high transition frequencies/ For protons or heavier the frequencies of motion are in radio frequency range, and radiative decay is ignorable. Also true for axial an magnetron motion of electrons, but not cyclotron motion. Energy in a cyclotron orbit is decreased by power radiated, according to d dt E = e 3c 3 r = w c r E = m r r de dt = γ c E γ c = 4e ω c 3mc 3 ( ) = E 0 e γ c t E t Convenient to write in terms of classical radius r 0 of charged particle: r 0 = e mc γ c = 4rω 0 c 3c ω c p 7.9
for electron r 0 =.8x0-3 cm. So if ω c /(π) = 60 GHz, γ c = 80 ms. Note /m 3 dependence. Number density Canonical Angular momentum. Look at old paper by Brillouin (Phys Rev 945) Define momentum p by a standard Lagrangian. For a charge e, e/m vector potential A, electrostatic scalar potential Φ, with independent variable x, v and t: p k = mx k + ea k = L L x x x 3 x x x 3 t x k ( ) = m v eφ+ e( v A) v A = x A + x A + x 3 A 3 Note L(v,x,t). Note A, A, A 3 are components in a rectangular coord system x x x 3. The E and B components are given by E k = Φ k t k =,,3 B = A B 3 = x x Then Lagrange's eqn of motion is dp dt = p = L x dp k dt = p k = L = e Φ + e x x + x k x k + x 3 3 Remember that So that d dt = t + x + x x + x x 3 x 3 m x k + e k t + e x k + x k x x + x k 3 = e Φ + e x x 3 x + x k x k + x 3 3 p 7.0
Take k =, note first terms in each brackets cancels, and we get m x = e e Φ + ex t x which can be arranged as m x k = ee k + e( v B) i.e. the equation of motion. x x + e x 3 3 x x 3 Now the Lagrangian appears in the principle of least action. We can now build up the Hamiltonian ( ) = p k H p p p 3 x x x 3 t H = mv + eφ = m and the eqns of motion are x k = H p k p k = H k = ( m p ea k k ) k x k L ( ) p k ea k + eφ For a single electron, in a z field, conservation of canonical momentum leads to p θ = rmv θ + rea θ = rmv θ + re B r r i.e. p θ = m( rv θ r ω c / ) where ω c = eb / m, v θ is azimuthal velocity, consisting of ExB and thermal (cyclotron) motions. If there are no external torques, and the electrons are introduced without any canonical angular momentum, we have θ =ω c / Note: For rω c >> v θ, then p 7.
p θ = ( mω c / )r g with r g the radius of the guiding center. Note that in absence of external torques P = p θ = ( mω c / ) r g = const i.e. if some electrons increase their radius, other must decrease theirs. Since initial average radius is less than the wall radius,, only small losses can occur. Now radial component of force balance in equilibrium is ee r + ebr θ + mr θ = 0 (E+vxB balances centrifugal). Substitute θ =ω c / = ω H and use conservation of energy (actually H) and let initial velocity be very small (0). Direction 3 is along B field then U = e m Φ = e m m e v = r + r θ + v 3 E r = Φ r = m U e r = m e rω v H + v 3 3 r but need v 3 r = 0, so Poisson's equation becomes r ( r ee r ) = ρ = m ε 0 e ω H Now note that ρ = ne, and get n = ε 0 B m = B µ 0 mc p 7.