What about f(x,y) = exp(x)exp(i*y)
Then first derivative w.r.t. x is just f(x,y)
First derivative w.r.t. y is i*f(x,y)
Sum of their squares is zero, yet they are not constant.
f is necessarily constant under the constraint that f has first partial derivatives which are functions...
I understand that the magnetic field in a solenoid can be approximated as being constant as the length of the solenoid tends to infinity, but I was wondering if anyone could show me or point me in the direction of a derivation of the precise magnetic field at any point within a single loop of...
So, If you've got two points and a given length of curve to 'hang' between them, what shape is the curve which minimises the area underneath it? For a curve which is almost the same length as the distance between the points, this would be a catenary, I think (a la famous hanging chain problem)...
Naty1 - I'm not sure this accounts for the ring. Look at http://en.wikipedia.org/wiki/Sun_dog , posted by jmatejka. That seems to match up perfectly with what I saw. The similarities of the visual description I gave to the one in the article is actually quite uncanny.
I was in the French Alps the other day, high in the mountains (not sure if this is relevant), and, on the night of a full moon, with mist in the sky, there was a circular ring of light around the moon.
The ring subtended an angle about 5 times that subtended by the moon (that is to say, it...
I'd argue that, classically, neither turbulence nor brownian motion can be considered fundamentally random.
In the classical argument, as long as we know exactly where every particle is at some point, and what it's velocity is at that time, we can predict exactly what will happen, even how...
I find it very interesting trying to solve the 'classic' kind of physics problems. The ones that Euler, the Bernouillis and co. bandied about, I've come across:
The shape of a hanging chain
The shape of a hanging elastic string
The brachistochrone (the shape of a wire such that a bead...
Incidentally, this treatment predicts that the shape of the meniscus is independant of the ambiant pressure, which I like.
It agrees with rater ad hoc experiments I performed with water in a shot glass: covering the glass with my mouth and sucking in to see if the shape of the meniscus...
I think I may have a solution (which might demonstrate to you the problem I had in the first place - I don't think I expressed it particularly well).
Far from the edge of the glass, the surface water pressure is zero.'
'Within' the meniscus, the pressure must be negative (pgh and all...
Because (assuming the glass is hydrophilic), the water / glass interface has a lower surface energy than the glass / air interface, so the water 'creeps up' a little to minimize the energy. It's the same as saying the water's surface tension pulls it up the wall.
That's the thing, though, I think there would have to be a meniscus in a vacuum. The water would still 'creep up' the side of the tube to reduce the surface energy. Wouldn't it?
Studiot, I'm not conviced by your treatment of the meniscus. In a wide beaker, where the surface curvature is zero in most places, some column of water is clearly not held up by the surface interactions, but rather by the normal force of the bottom of the beaker, I would suggest that this is the...
Andy - my problem is in your second sentence. In the water droplet diagram provided by studiot, the curvature clearly counteracts the pressure drop. The water is of a higher pressure than the air around it (or the vacuum in the example given). In the meniscus example, the water surface curves...
If you apply an impulse (magnitude I) to a pencil, the net linear momentum of the pencil must increase, according to Newton's third law, wherever the impulse is applied.
In the example, we push the end of the pencil. The impulse acts along a line perpendicular to the pencil. Taking moments...
It's a question that simply cannot be answered by classical physics.
It was a big problem in the 19th Century that contemporary theory suggested that the most energetically favourable configuration of charged particle would be 'all on top of each other'.
It was only when Heisenberg came...
We are considering a jug of the stuff. Apologies - the droplets were a separate thought illustrating a point - that we could forget about boiling.
Here are the conditions:
We are interested in the profile curve of a liquid at a solid - vacuum boundary (say a water meniscus in a test tube...
I don't think so. Effect of evaporation and boiling, I think, can be ignored. Consider a water droplet in a vacuum - really it would boil away. However, we can still model a non-boiling stable droplet, where the fluid pressure in the droplet equals the net inwards 'curvature force' due to...
Agreed, but it acts on a surface element in the same direction and the net curvature force due to surface tension - away from the water (towards the vacuum) - along the normal of the element. How are these forces balanced?
And less relevantly, but equally interesting:
If you hadn't heard of him, Feynman was one of the great physicists of the 20th century, and won a Nobel Prize for his work on quantum mechanics, so a reliable source!
Consider the free body diagram of a surface element of a water - glass meniscus in a vacuum. Along the line normal to the surface, the water pressure acts towards the vacuum, and the direction of the surface tension 'curvature force' depends on whether the surface curves like a 'u' or like an...
I was struggling with the same problem, and think I have some sort of a remedy. In deriving the E-L equations, by varying a path and evaluating the stationary point of action, it is true that start and end points (define point: time and place) are set. However, it is equally relevant that these...