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bio website motls.blogspot.com
location Czech Republic
age 40
visits member for 3 years, 4 months
seen Aug 4 at 12:45

Hi, I am a string theorist and a publicist.


May
7
awarded  Yearling
Feb
3
awarded  Enlightened
Feb
3
awarded  Nice Answer
Dec
15
answered Fourier transform vs Fourier series
Dec
9
comment Rotation matrix in terms of axis of rotation
It's just a conjugation of the simple matrix for a rotation around the $z$-axis, which is effectively just 2 x 2 matrix, by another rotational matrix that rotates the North pole to the point $V$, which is a product of rotation in the theta-direction and the phi-direction to get where you need to get. Conjugation by $U$ is $URU^{-1}$ where the product is matrix product. It's possible ineffective to write these things without matrices so if you don't know matrices, this is a reason to learn them. At any rate, it's not really physics, it's linear algebra and geometry and a basic one.
Dec
2
comment Kummer's Equation
Dear Alex, the equation for $L_n$ is nothing else than a coefficient in front of $t^n$ in the Taylor expansion of the equation for $g$ with respect to $t$. You just compare the terms term-by-term. Please just expand my equation for $g$ as a Taylor expansion in $t$ and don't ask any more questions.
Dec
2
answered Kummer's Equation
Sep
21
awarded  Custodian
Jul
18
awarded  Good Answer
Jun
22
awarded  Enlightened
Jun
22
awarded  Nice Answer
May
23
comment A general pattern to find the roots of the classical lie algebras
Sorry, your updated question got very confusing. I was answering your original question. The new question talks about "levels" etc. There aren't levels in ordinary Lie algebras. The roots are very simple objects and be sure I could easily enumerate all of them for all 7 classes of the Lie algebras, A,B,C,D,E,F,G. They're just the non-negative integral combinations of the simple roots that have the right length (either the same as simple roots or, in non-simply-laced groups, sqrt2 or sqrt3 times longer). ... If you have an algebra, the roots are just defined as the eigenvalues under the Cartan.
May
23
answered A general pattern to find the roots of the classical lie algebras
May
7
awarded  Yearling
Nov
5
comment How do I determine the measure for a volume integral?
Sorry, this was the integral of $r^2=x^2+y^2+z^2$. If I calculate the correct $r_z^2$, the squared distance from the $z$-axis, it's only $x^2+y^2$ so I only get two terms out of three. Due to the spherical symmetry, it's $2/3$ of the result above, so that the moment of inertia is $(2/5)MR^2$ for a solid ball.
Nov
5
comment How do I determine the measure for a volume integral?
Let me calculate the moment of inertia of a ball of radius $R$ here. I choose spherical coordinatets so $dV=r^2\,dr\,d\Omega$. The integral over $d\Omega$ gives $4\pi$, the integral of $r^4dr$ (there was $r^2$ from the moment) from $0$ to $R$ gives $R^5/5$: I can factorize them here. So the result is $4\pi \rho R^5/5=3/5 MR^2$. Less symmetric shapes are more complicated but it's pure maths.
Nov
5
comment How do I determine the measure for a volume integral?
Also, your question as well as your comments below my answer make it spectacularly self-evident that you don't understand what either of these symbols mean, in a striking contradiction with your assertion.
Nov
5
comment How do I determine the measure for a volume integral?
Dear @jnm2, the first equation in your comment above isn't a "magic transformation": it's a totally trivial identity. $\int$ is the integral sign but it's still a sum of infinitesimal pieces. The infinitesimal pieces are proportional to $dV$ and one may choose cubic shapes of $dV$ to make it clear that it may be written as $dx\,dy\,dz$. I don't understand what's your problem. If you can't understand this simple thing about the volume integral, you should give up studying physics quantitatively. Also, I wrote you how you can calculate the triple integral for any shape you mentioned.
Nov
5
answered How do I determine the measure for a volume integral?
Aug
17
awarded  Nice Answer