Recent content by Jack Davies

Exactly the same. Apologies.

I had a play with the expression. It comes down to writing that expression f(\theta_i , \phi_i ) in the form: 1-f^2 = g^2 \iff f^2=1-g^2 For some 'nice' function g(\theta_i , \phi_i ). I don't think it's possible here either, and see no reason why it 'should' be.

There are infinitely many planes orthogonal to a given vector, so you would also need to specify a point on the plane to calculate its equation. You can write a plane with normal \bf{n} in vector notation as the set of all \bf{x} such that \bf{x \: . n} = d Where d is a scalar determined by...

The potential of a point charge q at a point \bf{x}_0 is given by V(\bf{x})=\frac{q}{4 \pi \epsilon_0 r} where r=|\bf{x}-\bf{x}_0| (assuming the standard case where we take potential vanishing at infinity). Keep in mind that for two positive charges, this is always positive (since r is always...

I think that it's just saying that, with that special condition on a_i(x), the differential equation becomes: a_0 y'' -a_0 '' y+a_1 y' + a_1 ' y =0 Which we can write as: \frac{d}{dx}(a_0 y' -a_0' y) + \frac{d}{dx} (a_1 y) = 0 So with that very useful condition we can write it as a total...

This is true. It is also interesting to consider the dimension of the antisymmetric matrices, A^T=-A. In general for the space of n \times n matrices, you can write A=\frac{1}{2} (A+A^T)+\frac{1}{2}(A-A^T) for any matrix A (i.e 'decompose' into symmetric and antisymmetric parts). Furthermore...

Since each term is squared, it is positive definite. So for the entire sum to be equal to 0, each individual term has to be 0. Does that help?

Hi everyone, I was recently talking to someone with a non-maths background about rotational stability, in particular how rotation is stable around the largest and smallest principal moments but not the intermediate one. He asked me if there was any 'obvious' reason for this, but one didn't...

For a given position x and time t, you evaluate the function f(x,t) at those values to determine the height of the wave. This is different from v, the speed of propagation of the wave. Consider yourself moving to the right with displacement defined by x(t)=v' t +x_0. Then, plugging this into...