Recent content by maverick6664

Thank you! I've got the idea. -Tetsuji

Thank you for your reply. r is a primitive root for m. Sorry I wrote "primitive factor" because I didn't know the math nomenclature (I'm Japanese in Japan). This is a Japanese book for preparation for IMO. I'm too old for IMO, but just interested in it and reading. My 1st daughter was...

I have a question; Let m have a prime factor p \equiv 1 (mod 4). Then Euler function \varphi(m) is divisible by 4. Let x = r^{\varphi(m)}, then m|(x^4-1) and x^4-1=(x^2-1)(x^2+1). As gcd(x^2-1,x^2+1)|2, either x^2-1 or x^2+1 is divisible by m. My book says here because of the nature of a...

I've seen the Fourier representation of the Coulomb potential is -\frac {Ze} {|\mathbf{x}|} = -Ze 4\pi \int \frac {d^3q} {(2\pi)^3} \frac {1} { |\mathbf{q}|^2} e^{i\mathbf{q}\cdot\mathbf{x}} Will anyone show me how to prove it? (yes, it's the Coulomb potential around an atomic nucleus.)...

I want vector calculus formulae tables, such as \mathbi{a}\times(\mathbi{b}\times\mathbi{c}) = \mathbi{b}(\mathbi{a} \cdot \mathbi{c} ) - \mathbi{c} (\mathbi{a}\cdot \mathbi{b}) and \nabla \cdot (\mathbi{a} \times \mathbi{b}) = \mathbi{b} \cdot (\nabla \times \mathbi{a}) - \mathbi{a} \cdot...

Thank you for the reply. I don't understand exactly what you mean right now (why ispin doublet for the antiquarks is (\overline u,-\overline d). In my understanding, it's (\overline u, \overline d) and it must make difference), but I'll proceed keeping it in mind, because I don't have a book...

In my book (Greiner's Quantum Mechanics, vol2, symmetries) says after calculation with Clebsch-Gordan coefficients, |\pi^0\rangle = \frac 1 2 (u\uparrow \overline{u} \downarrow + d\uparrow \overline{d} \downarrow - u\downarrow \overline{u} \uparrow - d\downarrow \overline{d} \uparrow ), And I...

oh..sorry for that!

Question c and d are quite separate ones, and you can answer quickly 0, because in question a you (are supposed to) have got the right radius with which the car won't slip outside or inside regardless of friction. So questions c and d are given to tell you when friction comes in or does not...I...

Easier way is (as others say above) put S(x) = \int e^x \cos x dx = e^x \sin x - \int e^x \sin x dx Now \int e^x \sin x dx = -e^x \cos x +\int e^x \cos x dx Inserting this into the first formula, S(x) = e^x \sin x + e^x \cos x - S(x) and we have S(x)=\frac {e^x} 2 (\sin x + \cos x)

It's just \int_0^1 (x+1-e^{-x}) dx = (\frac 1 2 x^2 + x + e^{-x})\vert_0^1 = 1/2+e^{-1}

I use scilab and octave on Linux. They are not only graphics plotting software, but calculation software similar to MATLAB. And they can be compiled with Intel compilers for faster speed.

My calculation for potential is (as the question requests :)) d \phi = \frac {2\pi a^2 \sigma \sin \theta} x d \theta so it follows that: \phi = \int_0^\pi \frac {2 \pi a^2 \sigma \sin \theta} x d \theta = \int_0^\pi 2\pi a^2 \sigma (r^2+a^2 -2ar \cos \theta)^{-1/2} \sin \theta d \theta =...

Yeah! If you calculate potential, you don't have to consider \cos \angle QPO, so it's much shorter and easier! This cosine part makes the calculation complicated, because of this part, partial integral is necessary.

oh yeah, it says calculate potential first! lol.