[evlatests] D*P contributions to total intensity

[Rick Perley]

    Well, I agree it's complicated, and also that it needs to be written 
down (which I am attempting to do).  But I don't agree that 'I = I'. 

    Starting with the famous general expression by Morris, Rad and 
Seielstad (and reversing to sign of V to align with current 
definitions), one can generate a compact matrix equation relating the 
observed visibility vector (RR, LL, RL, LR) to the 'Stokes' Visibility 
Vector (I+V, I-V, Q+iU, Q-iU).  (I've written these as row vectors, but 
they are actually column vectors).  These are related by the so-called 
Mueller matrix.  (A bad name, in my opinion, but never mind ...).  This 
relation can easily be recovered by use of Jones matrices and the 'outer 
product' rules. 

    The elements of the Mueller matrix are products of the terms 
involving the antenna ellipticities, which I write as Cr, Cl, Sr, and 
Sl.  Formally

    Cr = cos(beta_r)exp(-i phi_r)
    Cl = cos(beta_l)exp(i phi_l)
    Sr = sin(beta_r) exp(i phi_r)
    Sl = sin(beta_l) exp(-i phi_l)

    The 'beta' terms are the deviations of the antenna ellipticities 
from perfect circular, while 'phi' is the orientation of the antenna 
ellipses. 

    In general, the 16 terms in the Mueller matrix are non-zero (but 
some can be made very small by good design), so that the inversion from 
the visibility vector to the Stokes vector involves all observables.  By 
taking shortcuts (like defining I' = RR + LL, and pretending this 
represents true I), we incur errors, whose sizes depend on the size of 
the various products of the Cs and Ss, and on the sizes of the terms in 
the visibility vector. 

    A simple example should help.  Suppose we observe a source with is 
100% circularly polarized.  Then, I = V, and the two parallel hand 
visibilities become very simple:

    RR = Cr1.Cr2* . I
    LL = Sl1.Sl2* . I

    The physical interpretations of the Cs and Ss are simple:  C 
represents the loss of RCP from R_in to R_out of the polarizer, (the 
signal went into the L_out port), while S represents the mixing of LCP 
from L_in to R_out. 

    If we now generate I' by adding RR and LL, we get:

    I' = (Cr1.Cr2* + Sl1.Sl2*).I

    Now -- in general, the term in brackets will not equal 1.  But, 
under closer inspection (I'll leave the details out here), if the two 
antennas involved have their polarization ellipses each equal (beta_r = 
beta_l), and orthogonal (phi_r = phi_l + pi/2) then the magnitude of the 
term in brackets is indeed equal to 1, and we would get the right answer 
for I by taking that 'short-cut'.  Hence, taking the shortcut gives the 
right answer only if each of the antenna feeds is truly orthogonal.  I 
suspect this is true whatever the incoming state of the polarization is, 
but have not attempted to show this.  As Barry noted, this is 
complicated stuff...

    Note in the above that I've assumed the data are properly 
calibrated, in the sense that the amplifier gain terms have been 
properly accounted for without disturbing the polarimetric relations.  
Just how we do this -- since the raw RR and LL outputs are mixtures of 
the incoming polarization states, and our calibrators in general have 
small but non-negligible Q, U, and V contributions, is a bit of mystery 
to me ...

Barry G. Clark wrote:
> All very complicated, and really needs to be written down carefully.
> However, I can perhaps convince you of one simple truth - I is I.
>
> We are used to making the claim that I = RR+LL.  But it is true
> in any orthogonal polarization system.  Suppose we have a system
> P = (a*R + b*L)
> Q = (A*L - b*R)
> where a^2 + b^2 = 1.
> a=1, b=0 is pure circular, and a=b is orthogonal linears, and we are
> somewhere in the middle, around b~0.1 or a bit less.
> If in that system, we define the usual I
> I' = PP + QQ
>    = (a^2) RR + (b^2) LL + ab(RL + LR)
>       + (a^2) LL + (b^2) RR - ab(RL + LR)
>    = I
> Therefore, the 'absolute' D terms, correcting our mean polarization
> to real RR and LL, do not inter into making images of I.  They do
> enter, at the few percent of source polarization level, in imaging
> Q, U, V.  (Watch out for the last of those - it can hurt you.)
>
> What 'D terms' do is so account for antenna polarizations, which differ
> from antenna to antenna and which are not even orthogonal on one
> antenna, correcting the measurements to those that would have been
> made in a system in which all antennas had identical, orthogonalally
> polarized feeds.
>
> I don't see how you can call the fact that q and u can be bigger
> than I remarkable - this sort of thing shows up all the time -
> for instance, you often see stronger fringes in the bottom of
> an absorption line than in the continuum.  If the Q and U maps
> get bigger than the I map, then you can start to worry.
>
>   
>>     Plots of the cross-power visibility spectrum of Cygnus A, in all
>> Stokes parameters have shown the remarkable fact that the Q and U
>> visibilities are often a substantial fraction of -- and can even
>> exceed
>> -- the I visibility.  This situation has long been known for
>> observations of distributed galactic emission.  What I want to
>> emphasize
>> here is that it will be a common situation for observations of highly
>> polarized emission in general.
>>
>>     There's no surprise in this.  But what I want to emphasize here is
>> that this provides another explanation (and a good one!) for our
>> troubles in deriving high-fidelity images of objects like Cygnus A.
>> The
>> reason is the leakage between Q and U into I.  It works like this:
>>
>>     The observed correlation in (say) RR is written (ignoring issues
>> of
>> parallel hand calibration, and assuming that V = 0):
>>
>>        Vrr = (1 + Dr1Dr2*)I + Dr1(Q-iU) + Dr2*(Q+iU).
>>
>>     where I, Q and U are the visibilities for Stokes' I, Q, and U, and
>> Dr1 is (for example) the complex coupling from LCP into RCP for
>> antenna
>> 1.  We normally argue that since the D's are a few percent, and both Q
>> and U are a few percent of I, that the cross products between Ds, and
>> between D and Q (or U) are of order 0.1% or less, and hence
>> negligible.
>>
>>     But for highly polarized extended objects, the argument that the Q
>> or U visibilities are negligible is incorrect -- they are often
>> compariable to, and can on occasion exceed the I visibility.  Take the
>> case where the I visibility hits a null (I = 0), while the Q and U
>> visibilities do not.  (This is a common situation).   The measured
>> Vrr,
>> rather than being zero, becomes a scrambled version of the  polarized
>> flux visibility.   Unless a correction is made, the derived 'I'
>> visibilities will be in error, sometimes by significant amounts.
>> This
>> is a non-self-calibrateable error, which will lead to image
>> degradation
>> in the regime where dynamic ranges of thousands - to - one are
>> desired.
>>
>>     So far as I know, the inversion from the 'RR' and 'LL'
>> visibilities
>> to derive the 'I' visibility takes no account of this leakage.
>> Clearly, for precise imaging of objects like Cygnus A, a fuller
>> inversion will be needed.
>>
>>     It is still unclear to me whether the 'relative' Ds that are
>> determined as a matter of course via standard techniques are
>> sufficient
>> for this application, or whether the true Ds are needed.   I think
>> 'true' Ds are needed, but others are invited to argue otherwise!
>>
>>
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>
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