## 1 Background

Assume one scattering point and compare it to the line of sight, then one has an additional delay and fringe rate

τ = ½θ2deff/c,

fD = veff·θ/λ

Here,

deff = dpsr(1-β)/β, β=1-ds/dpsr,

veff = viss/β = vs/β - v- vpsr(1-β)/β

Here, viss is the velocity of the scintillation pattern across the line of sight, and all other velocities are relative to the solar system barycentre.

Note that fD is related to the time derivative of τ,

dτ/dt = dθ/dt⋅θ(deff/c) = veff⋅θ/c = λfD/c

Sometimes it is useful to work in terms of the effective proper motion:

μeff = veff/deff = μs - μ - μpsr

with μs = vs/ds, μ = v/deff, and μpsr = vpsr/dpsr.

For a linear structure, where angle α between veff and θ is constant, one has τ=ηfD², with curvature η=deffλ2/2cveff²cos2α. Measuring the curvature thus constrains

λ2/2cη = veff,∥2/deff = deff μeff,∥2.

Inverted arcs are due to interference between scatterers.

Can model whole lot by screen made up of points with phase and velocity offsets given by parabola; Walker+ http://adsabs.harvard.edu/abs/2008MNRAS.388.1214W.

Phase differences between observations at different locations help localize screen; Brisken+ http://adsabs.harvard.edu/abs/2010ApJ...708..232B. Φ(r) = φ + k(x-βr)/2βds, φ is plasma phase delay, x is transverse offset of screen spot, r is transverse offset of obs.

## 2 Information from VLBI of screen and pulsar (proper motion and parallax)

VLBI allows one to measure θ on the sky. Combined with τ, this gives deff, while combined with fD it provides veff,∥, the component of veff parallel to θ.

If we also know the proper motion, μ=vpsr / dpsr, then with the orientation of the screen from VLBI, we know μ = μ⋅θ/|θ|. For this case,

fD = |θ|/λ ( vs,∥/β - v⊕,∥ - μdpsr(1-β)/β ) = |θ|/λ ( vs,∥/β - v⊕,∥ - μdeff )

Since we know deff from τ and can calculate v⊕,∥ (with the known orientation), we thus measure vs,∥/β.

Given also a parallax, i.e., a measure of dpsr, one obtains β and can thus solve for everything.

## 3 Effect of Earth's orbit for a single 1-D screen

If we make multiple measurements of scintillation throughout a year, then, assuming the screen does not change in orientation or velocity, one will see variations in curvature η, or, equivalently, in veff,∥2/deff. Since

veff,∥2 = (veff,∥,SSB + v⊕,∥)2

This will lead to a variation in curvature over the year, the phase of which (combined with the known orientation) allows us to determine the position angle between the screen and Earth's orbit.

Since the curvature scales as v2/deff, the amplitude of the variations allows us infer deff (see also below). Thus, it gives the same information as VLBI, but under the assumption of a one-dimensional screen that does not change orientation.

## 4 Expected orbital velocity changes for a single 1-D screen

### 4.1 Circular orbit

Consider a circular orbit in an coordinate system where X, Y are in the sky, X towards the ascending node, Y perpendicular to it, Z away from the observer. Then as a function of phase φ=2π(t-tasc)/P the positions and velocities are

r = apsr (cos φ, sin φ cos i, sin φ sin i)

v = v0 (-sin φ, cos φ cos i, cos φ sin i)

where apsr is the semi-major axis of the pulsar orbit, and the velocity v0 = 2πapsr/P. The normal observables are the projected semi-major axis x≡apsrsin i and radial-velocity amplitude Kpsr≡v0sin i=2πx/P.

The orientation of the pulsar orbit on the sky is given by Ωpsr (measured from N through E), and similarly the position angle of the screen is Ωs. The velocity we can measure is parallel to the screen, and hence depends on ΔΩ = Ωspsr, the angle of the screen relative to the ascending node,

v = vx cos ΔΩ + vy sin ΔΩ = v0 (-sin φ cos ΔΩ + cos φ cos i sin ΔΩ)

Defining χ and b by

tan χ = cos i sin ΔΩ / cos ΔΩ = cos i tan ΔΩ

b2 = cos2ΔΩ + cos2i sin2ΔΩ

v = v0 b (- sin φ cos χ + cos φ sin χ) = - Kpsr / sin i * b sin (φ - χ)

So, v depends on two unknowns, i and ΔΩ, through b and χ, which can be solved by measuring its amplitude and phase offset. Unfortunately, we do not measure v directly, but v(1-β)/β, relative to what is caused by linear motions of the pulsar, earth, and screen, i.e., we measure

Δveff = v(1-β)/β = v(deff/dpsr)

Or, with curvature measurements instead of VLBI,

Δveff/deff1/2 = v(1-β)/βdeff1/2 = v(deff1/2/dpsr)

Hence, we end up measuring χ and v0 b deff/dpsr (with VLBI). The measurement of χ on its own is not very useful, as it just gives a constraint on cos i tan ΔΩ.

But given χ, what does the amplitude constrain? With some trigonometry, we can rewrite b2 in terms of χ and the real quantity of interest, sin i:

b2 = (1 - sin2i) / (1 - sin2i cos2χ)

With this our effective velocity amplitude is given by

Δveff = Kpsr / sin i * (1 - sin2i)/(1 - sin2i cos2χ) * deff/dpsr.

Since Kpsr and χ are known, to determine sin i needs a measurement of the pulsar and effective distances. The latter we can in principle get from the Earth's orbit, in which case we still need dpsr (or β). Conversely, if we knew sin i independently and inferred deff from VLBI or the Earth's orbit, in principle we could measure dpsr.

### 4.2 Eccentric orbit

For an eccentric orbit, the motion in the orbital plane in terms of true anomaly θ is,

r = [apsr (1-e2)/(1+e cos θ)] (cos θ, sin θ)

v = [2πapsr / P (1-e2)1/2] (-sin θ, e + cos θ)

Following Rickett et al., we'll call the term in square brackets v0. Then, rotating by the periastron angle ω,

v = v0 (sin θ cos ω - (e + cos θ) sin ω, sin θ sin ω + (e + cos θ) cos ω) = v0 (-sin φ - e sin ω, cos φ + e cos ω)

where we defined φ = ω + θ. Finally, inclining by inclination i, the xyz velocities are,

v = v0 (-sin φ - e sin ω, (cos φ + e cos ω) cos i, (cos φ + e cos ω) sin i)

This is consistent with Eq. 7 of Rickett et al. 2014.

For a screen at an angle ΔΩ from the ascending note, the velocity is

v = vx cos ΔΩ + vy sin ΔΩ = v0 [(-sin φ - e sin ω) cos ΔΩ + ( cos φ + e cos ω) cos i sin ΔΩ] = v0 [(-sin φ cos ΔΩ + cos φ cos i sin ΔΩ) - e (sin ω cos ΔΩ + cos ω cos i sin ΔΩ)]

Defining χ and b as for the circular orbit,

v = v0 b [(- sin φ cos χ + cos φ sin χ) - e (sin ω cos χ + cos ω sin χ)] = - v0 b [sin (φ - χ) - e sin (ω - χ)]

So, the result is the same as for the circular orbit if measured in terms of true anomaly, except for a constant offset. Unfortunately, that carries no new information, as it depends in the same way on v0 b as the variable part.

### 4.3 Earth's orbit revisited

One can write the components of the Earth's position and velocity in the same way as those for the pulsar orbit above, with the z axis towards pulsar, x along the line of nodes of the Earth's orbit, and y perpendicular to x and z, with similar angles φ and i, which are related to ecliptic longitude λ and ecliptic latitude β by i=90ˆpsr and φ = λpsr+90ˆ.

Similarly to the case of the pulsar orbit, one will infer a χ and b, but in this case, since we know the orientation of the Earth's orbit relative to the pulsar, χ gives us the orientation of the screen on the sky, Ωs, and the amplitude will equal vb/deff1/2, i.e., with known v and measured b, we can determine deff.

## 5 Uncertainty in deconvolution

Suppose one measures for any giant pulse

s(ν) = A g(ν) + n(ν)

where s(ν) is the observed fourier component, A the amplitude of the delta function (i.e., proportional to sqrt of the S/N of the giant pulse in power), g(ν) the ism response, which is normalised such that ⟨g2⟩=1, and n(ν) some noise. If we then decorrelate the signal using

d2(ν) = s2(ν) s1*(ν) / |s1(ν)|

we get

d2(ν) = (A2g(ν) + n2(ν)) (A1g(ν) + n1(ν))* / |A1g(ν) + n1(ν)|

Now looking at the various terms, those with n2 will average to zero, so one is left with

⟨d2(ν)⟩ = A2 (A1g² + g n1) / (A12g2 + A1(gn1+g n1) + n12)1/2

Here, the terms with n1 will not average to zero. Writing g n1 = |g||n1|(cosΔφ + i sinΔφ), realising that gn1+g n1 = 2|g||n1|cos(Δφ) and that the integral over sinΔφ will vanish, and dividing by A1, one finds:

⟨d2(ν)⟩ = A2 * (g2 + |g||n1/A1|cosΔφ) / (g2 + 2|g||n1/A1|cosΔφ + (n1/A1)2)1/2

So, as expected, the extent to which one can descatter a pulse depends only on the signal-to-noise ratio of the pulse one uses to descatter. For very high signal-to-noise, ⟨d2⟩≅A2|g|. Given that ⟨g2⟩=1, and a χ2 distribution for |g|, the expectation value ⟨|g|⟩ = (π/4)1/2. Hence, in power, one expects a reduction of π/4=0.785…

For S/N=1 per channel, averaging over frequency should bring the signal further down by ∼1/√2, i.e., one will only be able to decorrelate ∼0.4 of the power.

## 6 Frequency shifts due to viewing angle - application to Crab

General expression for Doppler shift of a moving blob observed under some angle θr towards the receiver (angle in the receiver's frame):

fr / fs = 1 / γ(1-βcosθr)

We have γ=1/√(1-β2) and hence for a blob moving with γ≫1,

β = √(1-1/γ2) ∼ 1 - ½γ-2

Furthermore, for small θr,

cosθr ∼ 1 - ½θr2

Hence,

fr / fs ∼ 1 / γ[1 - (1-½γ-2)(1-½θr2)] ∼ 1 / γ½(γ-2 + θr2) ∼ 2γ / (1 + γ2θr2)

Now consider we observe the blob over a small range of angles from θr to θr+δ. The fractional change in received frequency fr will be,

Δfr / fr = (1+γ2r+δ)2) / (1 + γ2θr2) - 1 = (2γ2θrδ + γ2δ2) / (1 + γ2θr2)

The best possible case might have θr∼1/γ (i.e., observer at the edge of the forward-boosted beam), for which δ≪θr and

Δfr / fr ∼ γδ

For the most pessimistic case, where the blob is initially coming straight at us, i.e., θr=0, the ratio becomes γ2δ2.

To estimate the angle within the Crab nebula: delays are τ ≅ 0.5 ms and scattering in the Crab nebula happens at d ∼ 0.5 pc away from the pulsar. With cτ = d/cosδ-d ∼ ½δ2d, one gets δ2 = 2cτ/d ∼ 2×10-11 and thus δ ∼ 4×10-6 ∼ 1".

For a 5% change in observed frequency, this implies a rather large γ ∼ 0.05/δ ∼ 10000.

### 6.1 Likelihood of observing a given brightness at some angle

The signal will be boosted. Since Sν3 is invariant, for a given spectral index Sν∝ν and energy distribution N(E>E0)∝E0, the rate boost will be

R ∝ D3+α+γ ≡ Dp.

But the chance to see it at a given angle is ∝θr, so the total probability,

P(θr) ∝ θr(2γ/(1+γ2θr2))p

This has a maximum (dP/dθr=0) at

γθr = 1/√(2p-1)

For a typical p=5, one thus has θr,max≅1/3γ.

Created: 2021-08-06 Fri 19:30

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