Yeah, I was beginning to realize that was the problem. The relationship between the co-ordinate systems isn't explicitly stated.
What we've got is the usual arbitrary blob in space, which has a dipole moment per unit volume \vec{P} . We want to know what the potential is due to this blob...
Thanks for the help so far.
Christianjb, what you've written is correct, but it proves
\nabla \left( \frac{1}{r} \right) = \frac{- \hat{r}}{r^2}
I'm cool with this, but my problem is that Griffiths is differentiating with respect to a different co-ordinate system (hence the prime on...
Homework Statement
This is from Griffiths' Intro to Electrodynamics. He is discussing the field of a polarized object of dipole moment per unit volume \vec{P} viewed at \vec{r} .
He then states:
\nabla ' \left( \frac{1}{r} \right) = \frac{ \hat{r}}{r^2}
Where \nabla ' denotes...
Thanks for the replies, I was making things way too complicated for myself. I'm not sure if the expression I got was correct but variation's method is far more elegant and uhm... logical. Hope that'll teach me to keep an eye on the bigger picture with simpler problems.
Thanks again.
I've done a countless number of these problems before, so I was quite annoyed when I was unable to do this one. It's from Griffiths' Intro to Elementary Particles.
Homework Statement
A pion traveling at speed v decays into a muon and a neutrino,
\pi^- \rightarrow \mu^- + \bar{\nu}_\mu
If...
When you've got -a|b+c| > -bc, if you want to remove the minus signs you have to flip the inequality sign. So you end up with a|b+c| < bc .
OK, now consider a + b + c > 0 \Rightarrow a > |b + c| .
We had a|b+c| < bc , so now that we know that |b + c| < a , we can replace a with |b + c|...
OK, so I don't know much about rigorous mathematical proofs but I can kinda intuitively prove 6).
abc > 0 says that
-all 3 terms are positive
or
-one is positive and two are negative.
Let's suppose 2 of them are negative.
a + b + c > 0 says that the positive one is greater than...
The reason textbooks seem so dry are because they should really include a lot of information on the topic (e.g. rigorous proofs etc). When professors teach a subject, they choose to emphasize only the things they feel are important (and often omit lengthy proofs etc). Subsequently, when you get...
This is from Kittel's Intro to Solid State Physics:
At temperatures much below both the Debye temperature \theta and the fermi temperature T_F, the heat capacity of metals may be written as the sum of electron and phonon contributions:
C = \gamma T + AT^3
So that explains your gamma...
The first thing you've done is to cube both sides. That's ok but it should give you
( log(x) )^3 = log(x)
Since the whole log(x) is cubed, you can't move the 3 down (that's only if the x was cubed).
But what you can do is take all the terms over to one side and then you just have to solve a...
There's too much geometry in the first one for me at this time of the night.
For the second one, there are 2 variables - the length and the height of the box. The volume has to stay constant and hopefully you can express the volume with your 2 variables. That gives you your first equation...
Homework Statement
\int{ \frac{dx}{Ax^2 + Bx + C}
The Attempt at a Solution
So I can't think of any immediately obvious substitutions. What I've tried is completing the square in the denominator so that the integrand becomes
\frac{1}{(\sqrt{A}x + \frac{B}{2\sqrt{A}})^2 - (\frac{B^2}{4A} -...
You'll probably get a better response if you show more of your work explicitly (i.e. the wavefunctions you've calculated). Anyway, presumably you've matched the two wavefunction amplitudes such that they are continuous across the boundary? Then just find the amplitude ratio so that the new...
Actually, you don't square or add the errors but your answer is still right. You multiply the percentage error by the power (for the one variable case). More generally,
if
A = BC^n
then
(\frac{\Delta A}{A})^2 = (\frac{\Delta B}{B})^2 + n^2 (\frac{\Delta C}{C})^2