Recent content by opsb

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 did, but you get this really tricky integral. I was wondering if there might be a smarter way to do it than brute force.

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.

Thanks, that's pretty interesting.

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...

Incidentally, this treatment predicts that the shape of the meniscus is independent 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 into 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...

We are, and I've neglected pressure effects in my discussion above - the argument remains the same in a vacuum (I think).

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?

In a vacuum, though, the pressure in the water cannot be lower than the pressure outside it. that was what puzzled me.

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...