Study of Pseudo-magnetic fields in NSR Exotic 5 th force experiment. Axial-axial potential 1. gives rise to pseudo-magnetic field via

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1 Study of Pseudo-magnetic fields in NSR Exotic 5 th force experiment Axial-axial potential V AA ppppp r = g A 2 ħc 2 16 π mc 2 σ v c r r 1 λ c + 1 r e r/λ c r gives rise to pseudo-magnetic field via V = 1 2 γ nħσ B

2 Solving for pseudo-magnetic field by equating above equations we get a differential field due to a bulk material db AA = g A 2 ħc N γ n 8 π mc 2 v y z z s v z (y y s ) v z x x s v x (z z s ) v x y y s v y (x x s ) 1 λ c + 1 r z e r/λ r 2 dddddd y x Matlab code sums over slab source points [x s y s z s ] to calculate pseudo magnetic field due to finite slab. Note: For semi-infinite slab with beam parallel to surface, let v = v z. This eliminates B z and symmetry of semi-infinite slab eliminates B y, leaving only a horizontal pseudo-magnetic field. Using cylindrical coordinates to integrate over the semi-infinite slab, where r 0,, θ 0,2π, y, 0, yields the equation from Piegsa and Pignol. [JoP: Conf. Series 340 (2012) , eq. 7], where Δy is the vertical distance from the slab surface to the field point. B AA = 1 γ n g A 2 4 N ħc mc 2 λ c(v e y ) e Δy/λ c

3 Semi-infinite slab approximation with N Cu =1.9*10 29 /m 3, λ c =1mm at y=0.7mm, v=660m/s gives B=1.8pT which agrees with a 10-mm thick slab away from edges. 1-mm thick slab gives slightly smaller field. Note that B z = 0 when v=v z. Code focuses on corner region of slab, x~- 25mm, z~0mm.

4 λ c =1mm Cu (N=1.9E29/m 3 ) 1 slab versus 4 glass (N=0.75E29/m 3 )

5 λ c =1mm Cu (N=1.9E29/m 3 ) 1 slab

6 λ c =1mm 4 slabs glass (N=0.75E29/m 3 ) +1.1pT -1.8pT

7 λ c =3mm Cu (N=1.9E29/m 3 ) 1 slab

8 λ c =3mm 4 slabs glass (N=0.75E29/m 3 ) +1.2pT -1.8pT

9 λ c =3.0mm 8 slabs +1.43pT -2.25pT

10 λ c =0.3mm 4 slabs glass (N=0.75E29/m 3 ) +0.27pT -0.51pT

11 B x versus x, y, z Eight 1-mm thick slabs with 2-mm vacuum gaps v=[0,0,660m/s], λ c =1.0mm Note: Because of slabs above and below the gap, the y-dependence is almost linear rather than exponential.

12 B x versus x, y, z Eight 1-mm thick slabs with 2-mm vacuum gaps v=[0,0,660m/s], λ c =3.0mm

13 B x versus x, y, z Eight 1-mm thick slabs with 2-mm vacuum gaps v=[0,0,660m/s], λ c =0.3 mm Note: Since λ c is 1/3 the thickness of the slab, the slab approximates the semi-infinite slab case in the bulk region away from the edges.

14 Averaging over volumes in vacuum gap. These micro-radian rotations, φ, are done separately (φ=-γ n BL/v) for x- and y- components. φ x means rotation about B x g A =1E-8 v=v z =660m/s λ c =0.3 mm λ c =1.0 mm λ c =3.0 mm Avg B x (pt) corner* Avg B x (pt) bulk Avg B x (pt) Semi-Inf Avg B y (pt) corner Avg φ x (µrad) bulk Avg φ x (µrad) full Avg φ x (µrad) Semi-Inf Avg φ y (µrad) full *corner width = 5λ c

15 v x =7.3m/s, v z =660m/s (m=0.65 glass), λ c =1.0 mm Eight 1-mm thick slabs with 2-mm vacuum gaps v=[2.6,0,660]m/s, λ c =1.0 mm Note: B x and B y are essentially unchanged. B z is quite small.

16 v x =7.3m/s, v z =660m/s (m=0.65 glass), λ c =1.0 mm

17 No-bounce acceptance is 2mm/500mm=4mrad, so expect lots of bounces. Angular acceptance from multiple bounces is limited by glass m ggggg = 0.65, θ c ggggg 10A = 11mrad v x =7.3m/s, v z =660m/s g A =1E-8 v x =7.3m/s v z =660m/s λ c =1.0 mm λ c =3.0 mm Avg B x (pt) corner* Avg B x (pt) bulk Avg B x (pt) Semi-Inf Avg B y (pt) corner Avg B z (pt) bulk Avg φ x (µrad) bulk Avg φ x (µrad) full Avg φ x (µrad) Semi-Inf Avg φ y (µrad) full Avg φ z (µrad) full So with maximum divergence, z-rotation component is still quite small. Y-rotation component is sizable, but beam polarized along y.

18 Since all rotations are small and if field ~constant along trajectory, treat as one classical rotation of the polarization vector (for µrad, order doesn t matter): p = R φ y R φ z R φ x = cos φ y sin φ z cos φ x + sin φ y sin φ x cos φ z cos φ x sin φ y sin φ z cos φ x + cos φ y sin φ x sin φ z 1 sin φ x p φ z 1 φ x ; where φ z is at most 1% of φ x. The z-rotation is very small for even most divergent neutrons and since we start polarized in y-direction and y-rotation is also small, the net rotation is essentially only dependent on rotation from B x. So, the simulation code can focus only on size of B x (x, y, z) in vacuum gap region.

19 Field is not constant. How to deal with in simulation (nsimdark) Fringing on front and back (z): small ~4mm/500mm<1% effect Ignore Fringing on sides (x): not too small, 1 10% depending on λ c scale field size by an average of polynomial fit of B x (x) within x-range of trajectory if trajectory is mostly in side region. Moving through different y-values, field changes significantly. Use analytic integration of polynomial fit of B x (y) along trajectory to determine φ. For example, for λ c = 1.0mm B x y = 0.28y 3 y y 2.4 (pt) for y[0,2], where y = y o + y f y i z f z i (z z i ) So z f z i φ = γ n B v x y z dd= n

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