## Table of Contents

## 1 Background

Basic idea: summarised in http://adsabs.harvard.edu/abs/2004MNRAS.354...43W and Brisken+ http://adsabs.harvard.edu/abs/2010ApJ...708..232B.

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

τ = ½θ^{2}d_{eff}/c,

f_{D }= v_{eff}·θ/λ

Here,

d_{eff }= d_{psr}(1-β)/β, β=1-d_{s}/d_{psr},

v_{eff }= v_{iss}/β = v_{s}/β - v_{⊕ }- v_{psr}(1-β)/β

Here, v_{iss} 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 f_{D} is related to the time derivative of τ,

dτ/dt = dθ/dt⋅θ(d_{eff}/c) = v_{eff}⋅θ/c = λf_{D}/c

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

μ_{eff} = v_{eff}/d_{eff} = μ_{s }- μ_{⊕} - μ_{psr}

with μ_{s} = v_{s}/d_{s}, μ_{⊕} = v_{⊕}/d_{eff}, and μ_{psr} = v_{psr}/d_{psr}.

For a linear structure, where angle α between v_{eff} and θ is constant,
one has τ=ηf_{D}², with curvature η=d_{eff}λ^{2}/2cv_{eff}²cos^{2}α.
Measuring the curvature thus constrains

λ^{2}/2cη = v_{eff,∥}^{2}/d_{eff} = d_{eff} μ_{eff,∥}^{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)/2βd_{s}, 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
d_{eff}, while combined with f_{D} it provides v_{eff,∥}, the component of
v_{eff} parallel to θ.

If we also know the proper motion, μ=v_{psr} / d_{psr}, then with the
orientation of the screen from VLBI, we know μ_{∥} = μ⋅θ/|θ|. For this
case,

f_{D} = |θ|/λ ( v_{s,∥}/β - v_{⊕,∥} - μ_{∥}d_{psr}(1-β)/β )
= |θ|/λ ( v_{s,∥}/β - v_{⊕,∥} - μ_{∥}d_{eff} )

Since we know d_{eff} from τ and can calculate v_{⊕,∥} (with the known
orientation), we thus measure v_{s,∥}/β.

Given also a parallax, i.e., a measure of d_{psr}, 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
v_{eff,∥}^{2}/d_{eff}. Since

v_{eff,∥}^{2} = (v_{eff,∥,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 v_{⊕}^{2}/d_{eff}, the amplitude of the
variations allows us infer d_{eff} (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-t_{asc})/P the
positions and velocities are

r = a_{psr} (cos φ, sin φ cos i, sin φ sin i)

v = v_{0} (-sin φ, cos φ cos i, cos φ sin i)

where a_{psr} is the semi-major axis of the pulsar orbit, and the velocity
v_{0} = 2πa_{psr}/P. The normal observables are the projected
semi-major axis x≡a_{psr}sin i and radial-velocity amplitude K_{psr}≡v_{0}sin i=2πx/P.

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

v_{∥} = v_{x} cos ΔΩ + v_{y} sin ΔΩ
= v_{0} (-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

b^{2} = cos^{2}ΔΩ + cos^{2}i sin^{2}ΔΩ

v_{∥} = v_{0} b (- sin φ cos χ + cos φ sin χ)
= - K_{psr} / 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

Δv_{eff} = v_{∥}(1-β)/β = v_{∥}(d_{eff}/d_{psr})

Or, with curvature measurements instead of VLBI,

Δv_{eff}/d_{eff}^{1/2} = v_{∥}(1-β)/βd_{eff}^{1/2} = v_{∥}(d_{eff}^{1/2}/d_{psr})

Hence, we end up measuring χ and v_{0} b d_{eff}/d_{psr} (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 b^{2} in terms of χ and the real quantity of
interest, sin i:

b^{2} = (1 - sin^{2}i) / (1 - sin^{2}i cos^{2}χ)

With this our effective velocity amplitude is given by

Δv_{eff} = K_{psr} / sin i * (1 - sin^{2}i)/(1 - sin^{2}i cos^{2}χ) * d_{eff}/d_{psr}.

Since K_{psr} 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 d_{psr} (or β).
Conversely, if we knew sin i independently and inferred d_{eff} from VLBI
or the Earth's orbit, in principle we could measure d_{psr}.

### 4.2 Eccentric orbit

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

r = [a_{psr} (1-e^{2})/(1+e cos θ)] (cos θ, sin θ)

v = [2πa_{psr} / P (1-e^{2})^{1/2}] (-sin θ, e + cos θ)

Following Rickett et al., we'll call the term in square brackets v_{0}.
Then, rotating by the periastron angle ω,

v = v_{0} (sin θ cos ω - (e + cos θ) sin ω,
sin θ sin ω + (e + cos θ) cos ω)
= v_{0} (-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 = v_{0} (-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_{∥} = v_{x} cos ΔΩ + v_{y} sin ΔΩ
= v_{0} [(-sin φ - e sin ω) cos ΔΩ + ( cos φ + e cos ω) cos i sin ΔΩ]
= v_{0} [(-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_{∥} = v_{0} b [(- sin φ cos χ + cos φ sin χ) - e (sin ω cos χ + cos ω sin χ)]
= - v_{0} 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
v_{0} 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=90^{ˆ}+β_{psr} and φ = λ_{⊕}-λ_{psr}+90^{ˆ}.

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, Ω_{s}, and the amplitude will equal
v_{⊕}b_{⊕}/d_{eff}^{1/2}, i.e., with known v_{⊕} and measured b_{⊕},
we can determine d_{eff}.

## 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
⟨g^{2}⟩=1, and n(ν) some noise. If we then decorrelate the signal
using

d_{2}(ν) = s_{2}(ν) s_{1}^{*}(ν) / |s_{1}(ν)|

we get

d_{2}(ν) = (A_{2}g(ν) + n_{2}(ν)) (A_{1}g(ν) + n_{1}(ν))^{*} / |A_{1}g(ν) + n_{1}(ν)|

Now looking at the various terms, those with n_{2} will average to zero,
so one is left with

⟨d_{2}(ν)⟩ = A_{2} (A_{1}g² + g n_{1}^{∗}) / (A_{1}^{2}g^{2} + A_{1}(g^{∗}n_{1}+g n_{1}^{∗}) + n_{1}^{2})^{1/2}

Here, the terms with n_{1} will not average to zero. Writing
g n_{1}^{∗} = |g||n_{1}|(cosΔφ + i sinΔφ), realising that
g^{∗}n_{1}+g n_{1}^{∗} = 2|g||n_{1}|cos(Δφ) and that the integral over
sinΔφ will vanish, and dividing by A_{1}, one finds:

⟨d_{2}(ν)⟩ = A_{2} * (g^{2} + |g||n_{1}/A_{1}|cosΔφ) /
(g^{2} + 2|g||n_{1}/A_{1}|cosΔφ + (n_{1}/A_{1})^{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, ⟨d_{2}⟩≅A_{2}|g|. Given that
⟨g^{2}⟩=1, and a χ_{2} 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 θ_{r} towards the receiver (angle in the receiver's frame):

f_{r} / f_{s} = 1 / γ(1-βcosθ_{r})

We have γ=1/√(1-β^{2}) and hence for a blob moving with γ≫1,

β = √(1-1/γ^{2}) ∼ 1 - ½γ^{-2}

Furthermore, for small θ_{r},

cosθ_{r} ∼ 1 - ½θ_{r}^{2}

Hence,

f_{r} / f_{s} ∼ 1 / γ[1 - (1-½γ^{-2})(1-½θ_{r}^{2})]
∼ 1 / γ½(γ^{-2} + θ_{r}^{2})
∼ 2γ / (1 + γ^{2}θ_{r}^{2})

Now consider we observe the blob over a small range of angles from θ_{r}
to θ_{r}+δ. The fractional change in received frequency f_{r} will be,

Δf_{r} / f_{r} = (1+γ^{2}(θ_{r}+δ)^{2}) / (1 + γ^{2}θ_{r}^{2}) - 1
= (2γ^{2}θ_{r}δ + γ^{2}δ^{2}) / (1 + γ^{2}θ_{r}^{2})

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

Δf_{r} / f_{r} ∼ γδ

For the most pessimistic case, where the blob is initially
coming straight at us, i.e., θ_{r}=0, the ratio becomes γ^{2}δ^{2}.

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 ∼ ½δ^{2}d, one gets
δ^{2} = 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 S_{ν}/ν^{3} is invariant, for a given
spectral index S_{ν}∝ν^{-α} and energy distribution
N(E>E_{0})∝E_{0}^{-γ}, the rate boost will be

R ∝ D^{3+α+γ }≡ D^{p}.

But the chance to see it at a given angle is ∝θ_{r}, so the total
probability,

P(θ_{r}) ∝ θ_{r}(2γ/(1+γ^{2}θ_{r}^{2}))^{p}

This has a maximum (dP/dθ_{r}=0) at

γθ_{r} = 1/√(2p-1)

For a typical p=5, one thus has θ_{r,max}≅1/3γ.