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A point source away from the tracked direction has residual geometric delay $\tau$. Averaging its frequency-dependent phase over a top-hat band gives the displayed sinc attenuation times its central-frequency complex phase. The first bandwidth null is complete cancellation, not a small-loss operating limit. Fine frequency channels preserve the information needed for later frequency-aware imaging.

Electronic compensation of the time-dependent reference geometric delay multiplies each complex visibility by an opposite reference phase. A stationary point source at the phase centre then has constant corrected phase despite Earth rotation. antenna pointing alone does not perform this correction. Off-axis sources retain the relative fringe evolution that encodes their angular offsets.

An incompressible point source of volume flux $Q$ at height $a$ above an impermeable plane has velocity potential $\varphi =-Q(1/r_{+}+1/r_{-})/(4\pi )$, with an equal image source below the plane. On the wall, $u_s=Qs/[2\pi (s^2+a^2{)}^{3/2}]$. The Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{s}^2/[8{\pi }^2(s^2+a^2{)}^3]$. Integrating the pressure deficit over the plane gives the displayed attraction force relative to ambient pressure. If source strength instead denotes the coefficient $Q/(4\pi )$ of radial inverse-square velocity, the force is $\pi \rho m^2/a^2$.

Choose the baseline vector from the first antenna to the second. Apart from a sign convention for receiver delay, the geometric delay is ${\tau }_g=\mathbf{b}\cdot \mathbf{s}/c$, and the phase difference is $\varphi =2\pi \nu {\tau }_g$. For equal received source powers $P_a$, adding the voltages and averaging over many wave periods givesUncorrelated receiver noise adds a mean level but not the source's coherent cross term. For an east-west baseline and one choice of hour-angle sign, ${\tau }_g=(b/c)\cos \delta \sin H(t)$. Earth rotation therefore sweeps the phase; locally the fringe frequency is $(b/\lambda )\cos \delta \cos H\dot{H}$. Tracking the antenna pointing keeps the source in the primary beam. Electronic fringe stopping is a separate operation: it can remove this phase evolution, making a source at the phase centre constant in the corrected complex visibility. The oscillating power above is the geometrical fringe before that correction.

For a small angular displacement $d\theta$ along the projected baseline, $d\varphi =2\pi b_{p}d\theta /\lambda$. One full fringe therefore spans $\Delta \theta \sim \lambda /b_p$. Here $b_p=|\mathbf{b}-(\mathbf{b}\cdot \mathbf{s})\mathbf{s}|$. This is the angular period of the measured Fourier component and the characteristic resolution scale when such long baselines are synthesized; precise beam width and image ambiguity also depend on baseline coverage and weighting. A single pair by itself does not give an unambiguous complete image.

Writing $x=h\nu /(k_{B}T)$, the expansion $e^x-1=x+O(x^2)$ immediately gives the Rayleigh-Jeans law,The factor two counts the two independent modes of polarization of waves. A small brightness-temperature perturbation thus has $\Delta I_{\nu }=2k_B{\nu }^2\Delta T_{\mathrm{RJ}}/c^2$ in this limit.

Hot intracluster electrons inverse-Compton scatter the microwave-background photons. The scattering optical depth is $d\tau =n_e{\sigma }_{T}dl$ and the characteristic thermal fractional energy transfer is of order $k_B{T}_e/(m_e{c}^2)$. The integrated distortion is therefore controlled byScattering conserves photon number while shifting photons toward higher frequencies, giving a low-frequency decrement and a high-frequency increment. Thus the printed estimate correctly gives a negative change of order $y{T}_{\mathrm{CMB}}$. Its numerical coefficient, however, is not the standard Rayleigh-Jeans coefficient. For nonrelativistic electrons and small $y$, the Kompaneets equation gives $\Delta n=y{x}^{-2}{\partial }_x(x^4{\partial }_x{n}_0)$. At small $x$, $n_0\simeq 1/x$, so $\Delta n\simeq -2y/x$ andEquivalently the thermodynamic-temperature spectral factor is $x\coth (x/2)-4\to -2$. The original PDF really prints the estimate with $-y$, not $-2y$; both numerical conventions are retained below. Relativistic corrections at the supplied $T_e$ are small compared with this factor-of-two difference.^[ref]

The uniform cube has a constant line-of-sight depth $L=6\times 10^{22}\mathrm{m}$. Using the supplied cross-section rather than replacing it with a more precise value givesAdopt $T_{\mathrm{CMB}}=2.72548\mathrm{K}$, explicitly an external measured constant, not a temperature supplied in the question. The values $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$ set the physical units. Hence the printed relation gives $\Delta T\simeq -22\mu \mathrm{K}$; the standard Rayleigh-Jeans result gives $\Delta T\simeq -44\mu \mathrm{K}$. Keeping the answer as $-8.14\times 10^{-6}{T}_{\mathrm{CMB}}$, or twice that, avoids needing a numerical background temperature.^[ref]

The face is a square, so its solid angle is $\Omega =(6\pi /10800{)}^2=3.046\times 10^{-6}\mathrm{sr}$, not the area of a circular six-arcminute aperture. At $15\mathrm{GHz}$,Thus $\Delta S_{\nu }\simeq -4.67\times 10^{-4}\mathrm{Jy}=-0.47\mathrm{mJy}$ using the printed estimate, or $-0.93\mathrm{mJy}$ using the standard factor two. The sign denotes missing brightness relative to the surrounding microwave background. At this frequency $h\nu /(k_B{T}_{\mathrm{CMB}})\simeq 0.264$; the exact nonrelativistic spectral intensity differs from the strict Rayleigh-Jeans approximation by about one percent. This precision does not affect the estimates.

For the array noise, the system temperature sets the receiver noise scale; let $A_e={\eta }_A\pi d^2/4$ be the effective area of each dish, and let ${\eta }_c$ describe correlation losses. No efficiencies or polarization of waves count are supplied, so first take ${\eta }_A={\eta }_c=1$ and a single independent polarization of waves. An unpolarized point source produces antenna temperature $T_a=S_{\nu }{A}_e/(2k_B)$, and the system-equivalent flux density is $\mathrm{SEFD}=2k_B{T}_{\mathrm{sys}}/A_e$. The interferometric radiometer equation gives one baseline an rms flux error $\mathrm{SEFD}/\sqrt{2\Delta \nu t}$. There are $N(N-1)/2=45$ cross-correlation baselines; averaging their independent thermal noise for a point source givesThe ideal single-dish effective area is $4\pi {\mathrm{m}}^2$, giving $\mathrm{SEFD}=6.59\times 10^3\mathrm{Jy}$ andCombining two independent parallel-hand polarization of waves measurements gives $6.3\mu \mathrm{Jy}$. Nonunit efficiencies increase these estimates by $1/({\eta }_A{\eta }_c)$. Here signal equal to noise means a weak-source signal-to-noise ratio of one after integration; it does not mean equating the source to the entire unaveraged receiver-noise power.^[ref]

At the band centre, $\lambda =0.01999\mathrm{m}$. Taking a representative maximum projected baseline of $20\mathrm{m}$ gives ${\theta }_{\mathrm{res}}\sim 3.4$ arcminutes. A corner-to-corner baseline of $20\sqrt{2}\mathrm{m}$ would give about $2.4$ arcminutes before projection and weighting. The exact antenna locations and source declination are not supplied, so neither a unique synthesized beam shape nor an exact FWHM is determined. The individual-dish primary-beam scale, roughly $\lambda /4\mathrm{m}\sim 17$ arcminutes, is the field of view, not the synthesized resolution.

To extract the cluster signal, retain frequency-resolved complex visibilities and fit an extended SZ template together with contaminating compact sources. For the ideal uniform square centred at the phase centre and with a nearly constant primary beam across it, the Fourier transform isThis is why the integrated decrement must not be treated as a point-source flux on every baseline. The first axial visibility null occurs at $b=\lambda /\Theta \simeq 11.5\mathrm{m}$. Short baselines carry most of the large-scale decrement; the longest baselines are useful for separating unresolved positive radio sources. natural visibility weighting or a suitable taper gives a clearer extended-source image than emphasizing the longest baselines, while a visibility-domain matched filter uses all samples with weights proportional to template amplitude divided by noise variance. Its amplitude estimate is $\hat{a}=\mathrm{Re}(T^{†}{C}^{-1}d)/(T^{†}{C}^{-1}T)$ for a real template amplitude. A joint fit propagates radio-source uncertainty instead of subtracting exact, noise-free source models.

Earth-rotation and multifrequency synthesis improve Fourier coverage, but each channel must use its own baseline in wavelengths, primary beam and SZ spectrum. Use the longer baselines and spectra to identify variable point sources, flag bad data, and test the decrement with time, antenna and frequency splits. Primary-beam and absolute-flux calibration must be included. Large-scale microwave-background fluctuations and residual ground or atmospheric signals can remain even below the thermal-noise estimate. The narrow low-frequency band alone gives limited separation of SZ and primary microwave-background temperature structure. There is no zero-spacing interferometric datum: integrated flux is model dependent unless sufficiently short baselines or additional total-power information are available. Exact cluster detection significance cannot be obtained from the point-source noise alone without the actual projected baseline distribution and weighting.

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.

Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2{)}^{1/2}$ and $\vartheta =\arg (x+iy)$, superposition of uniform potential flow and a two-dimensional point source giveswhere $(u,v)=\nabla \Phi =({\psi }_y,-{\psi }_x)$. The angular branch cut records the volume flux from the source; the logarithm's arbitrary reference length only changes an additive constant in the velocity potential.

The stagnation point lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid isThe stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing streamlines through it have $\psi =\pm Q/2$ when the angle is taken in $(-\pi ,\pi )$. Far downstream $\vartheta \to 0$, so these streamlines approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the Rankine half-body between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta /\pi )$, $x=y\cot \vartheta$, $0<\vartheta <\pi$.

For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source isThe first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi (x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is thereforeThis is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau -t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote Darcy velocity and bulk-area source flux in a medium of uniform porosity $\varphi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\varphi$. No porosity is specified here, so the displayed time formula fixes the kinematic convention.

The velocity is $\mathbf{u}=\nabla \varphi =q{\mathbf{e}}_r/(2\pi r)$. Its divergence in polar coordinates is zero for $r>0$, while a circle about the origin has fluxBy contour deformation the same holds for any enclosing contour. This is a two-dimensional point source of strength $q$.

For the second potential,which vanishes at $y=0$. In the full plane this is a pair of equal sources of strength $q$ at $(0,\pm a)$. In $y>0$, it is one source at $(0,a)$ beside an impermeable slip wall, with the lower source serving as its image.

Under the Rayleigh criterion, two equally bright, mutually incoherent point sources are just resolved when the central maximum of one diffraction pattern lies at the first minimum of the other. Their intensities, rather than their electric-field amplitudes, add. For an ideal circular aperture of diameter $D$, the first dark ring of the Airy disk givesThe angle is in radians in the small-angle limit. This is a conventional resolution criterion for a specified aperture and wavelength; it is not an absolute prohibition on estimating a source separation by fitting noisy data under additional assumptions.

The region bounded by the dividing streamlines of a two-dimensional point source superposed on uniform flow. Its stagnation point is $-Q/(2\pi U)$ upstream and its far downstream width is $Q/U$. The boundary encloses fluid originating at the source rather than a rigid obstacle; the upstream stagnation point is reached only asymptotically by the upstream source trajectory.

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.

PASCO's documented lamp experiment connects a TD-8553 thermopile sensor to a millivoltmeter and places it level with a lamp filament along a meter stick. Lamp and detector axes are aligned. With the lamp off, readings at several positions establish an average ambient thermal radiation background. With the lamp on at a fixed supply setting, detector readings at different distances are compared after background subtraction. Between readings, a reflective shield prevents the detector warming; its shutter also protects the reference junction when idle. Moving the shutter can move the sensor, so fixed geometry matters. The manual's lamp limits must be followed.^[ref]

This protocol tests whether the background-corrected electric potential difference approximately follows the inverse-square law over a suitable distance range. It is a relative electromagnetic radiation measurement; those readings alone do not calibrate absorbed power in watts. The manual treats the filament as an approximate point source.^[ref]

For an absolute calibration, independently known absorbed power and a monitored reference temperature would be needed. A useful additional control is to repeat a fixed detector position before and after a distance sweep: disagreement reveals possible drift. These additions test the measurement chain rather than assuming that background subtraction fixes every error. A finite source, changing illuminated area, reflections, changing reference temperature, and insufficient settling can alter the measured dependence. In particular, the time constant in thermopile thermal response explains why a quick reading and a settled reading may differ. The detector infers its own absorber temperature difference; inferring the source's temperature additionally requires its emissivity, optical geometry and an appropriate thermal radiation model.

A two-dimensional potential doublet is a localized potential flow singularity with complex potential of a two-dimensional potential flow $W=M/(z-b)$, where $M=m{e}^{i\alpha }$ specifies strength and orientation. It is the limiting field of an equal point source and sink as their separation tends to zero with strength times separation fixed, with the sign absorbed in $M$. Its complex velocity is $-M/(z-b{)}^2$, so the velocity decays as inverse distance squared. The total volumetric flow rate around the singularity is zero. Its two-dimensional decay differs from that of the three-dimensional source dipole.