Supercazzola

Show how to get a conformal map from the region outside a semicircle, C∖SR={(x1,x2):|x1|2+|x2|2=R2,x2≤0} to the region outside a disk D of radius R2–√ centered at the origin, C∖D , ending up with h(u)=iR+u+i⋅R2−u2−−−−−−−√2 , with u=x1+ix2 . I know that the idea is to use a Moebius transform to send the semicircle to two perpendicular lines through the origin (something like az+Rz−R ) keeping track where infinity goes, then compress the three quarters we get the outside sent too to a half plane (again keeping track of infinity), then send that to the unit circle with infinity to 0 and then invert with z→c/z to get the outside circle of right radius and send infinity back to itself, but I didn't manage to get the correct result. Could you help me?

Forgive me, I don't think I understand.

Okay, reviewing the problem I came to these conclusions. Since the surface is perfectly wetting¹, the contact angle in the typical sense does not matter since there is no point at which the portion of the liquid forming the droplet is in contact with your actual surface. The first thing your condensing liquid does is to coat the entire ceiling. Before that has happened, no droplets will form. Afterwards, the layer of liquid coating the surface will just stay as it is, and below droplets will form as goverened by a balance of surface tension and gravity. These droplets will not interact with the actual ceiling but only with the layer of liquid coating it². Here is how I would approach this problem: First fix the volume of your droplet. Consider functions $f: ℝ⁺→ℝ⁺$ that describe half the cross-section of your droplet. These completely determine the droplet’s shape due to rotational symmetry. Use variational calculus or some numerical method to find that function $f$ that minimises the energy of your droplet taking into account the surface tension (as energy per surface of your function) and the gravitational energy of the water descending from the surface. You do not have to consider the adhesion to the ceiling as that concerns the unchanged layer that perpetually wets it. Finally find the volume at which this function becomes singular in a way corresponding to a tearing droplet, and determine the corresponding height. ¹ I here assume that the tiny difference made by gravity doesn’t change that. ² Mind that this is only an approximation based on the assumption that molecules only interact with direct neighbours. If you take into account further ranging molecular interactions, things become a bit more complicated, but then you’ll certainly know these details about your molecular interactions. Could you help me formalize this in terms of physical relations?

I thank you for your response. This problem is taken from Rudolf Ortvay Competition in Physics, 2013. I am not sure what type of solution is best to look for in competitions of this type. I don't think I fully understood. Are the three figures A, B, C three different ways of approaching the problem, or three physical situations consequential to each other? Thank you. Did I not make this assumption in my calculations? Excuse my naivety, I can't understand it. I am not familiar with this theorem. Could you give me some suggestions? Obviously, I don't want the ultimate solution, but I would like to be led to it through some help. It seems to me to be quite a challenging problem. Thank you, but why is my expression of weight force not exact? Is there something hidden in it that I am unaware of? Any clarification is appreciated. How to approach these two methods? Any suggestions? Thanks again.

I was reading a physics problem related to astronomy, and upon re-reading it, I realized that it could be really indicated to extrapolate some really interesting physics-related information. One of the questions I thought of is: How could we measure the ratio of a planet's radius to a star? The only idea I have is to compare them when the planet passes exactly in front of the star (i.e. they are aligned with our view), but this only makes sense if the distance between the two is much smaller than the distance between us and that star system (which I think is true enough for every system except the Solar system) and if it is possible to obtain such high resolutions (and I already had my doubts about the distance between the two, which should be much greater than their radii anyway).

Thanks. Any help is appreciated by me.

Thank you for your response. Could someone give me suggestions on how to calculate this angle? Thanks. However, my procedure that should lead to the water droplet shape for the time being still does not assume that the maximum angle is 90°. So, there must also be something else wrong.... I would be grateful if you could help me out.

I would like to understand whether the physical concepts I have applied and the physical situation I have modeled are reasonable, as I am really afraid that they are not consistent with the result I have obtained, a frankly unsolvable ODE. First consideration) Perfect wetting means the contact angle is zero. The drop won't have constant mean curvature because the pressure inside at the bottom will be more than at the top. Second consideration) Considering the forces acting on the portion of the droplet below some level, we have gravity, surface tension (where the surface of the lower portion meets the surface of the upper portion) and air pressure. So, if we take the portion of the droplet below a cut as the body acted on, there are four forces: pressure from air below pressure from water above weight of portion surface tension, but surface tension will contribute to the pressure in the droplet. Third consideration) The steepest gradient, i.e. the slope of the surface, will be horizontal at the bottom and top (perfect wetting). It reaches a maximum somewhere in between. When the drop becomes unstable, the steepest gradient will be π2 and it cannot accommodate any more weight. We are asked to consider a drip that is as large as it can be without becoming unstable. I reason that instability is when some part of the surface reaches vertical. At that point, the vertical component of the tension force has reached its limit. So we are considering a shape in which the gradient just reaches vertical, then tips out again, forming an S. But we need to find the general equation of shape, so I will consider a horizontal slice at an arbitrary height, so now θ can be any value from 0 to π2 . I would set up and solve the differential force balance on the surface (Young-Laplace equation) in cylindrical coordinates and solve this in conjunction with the hydrostatic equilibrium equation within the drop: I would work in terms of axial contour length along the drop s and the contour angle θ(s) . All derivatives are w.r.t. distance s along the surface from lowest point unless otherwise stated. We consider a horizontal slice through the droplet. θ is the angle of the surface to the horizontal, p is pressure within the droplet, pa is atmospheric pressure. Using the Laplace-Young equation: p=α(1r+θ′)+pa . F is the force exerted by adjacent layers: F=pπr2 T is the vertical component of the force due to tension in the surface T=α2πrsin(θ) The gravitational downward force on an element is πr2sin(θ)ρgΔs The upward vertical force due to air pressure on an element is pa2πrΔr Force balance gives: πr2sin(θ)ρgΔs+ΔT+ΔF=pa2πrΔr πr2sin(θ)ρg+T′+F′=pa2πrr′ r′=cos(θ) T′=2απ(cos(θ)sin(θ)+rcos(θ)θ′) p′=α(r−2cos(θ)+θ′′) F′=p′πr2+p2πrcos(θ)=απ(cos(θ)+r2θ′′)+(pa+α(1/r+θ′))2πrcos(θ) πr2sin(θ)ρg+T′+F′=pa2πrr′ From these equations, I get: r2ρgθ+2α(θ+rθ′)+α(1+r2θ′′)+2r(pa+α(1/r+θ′))=2par. We can also replace r with s , producing an ODE in θ(s) : ρgαs2θ+2θ+4sθ′+s2θ′′+3=0. Ok, this produces an ODE, but there are too many variables. It has both $$r(s)$$ and $$\theta(s)$$. In principle, that can be resolved using $$r'=\cos(\theta)$$, but eliminating $$\theta$$ will result in $$\arccos \theta$$ terms, and to eliminate $$r$$ we would have to differentiate the equation so that it only involves derivatives of $$r$$, not $$r$$ itself. For these reasons, I suspect the problem intends we take the portion below the point where the surface is vertical as being a hemisphere. It won't be far wrong. But even then, it will only let us get a height for that portion. To find the entire height we would still need to solve the shape equation. I would like to understand whether it is my assumptions of the physical situation that lead me to these apparently wrong calculations. I still cannot figure out what my errors are in the physical assumptions and related concepts. Could you help me point them out? The issue regarding the shape of a falling drop seems very difficult to fix. Thanks.