Understanding Cross Products and Angular Momentum in Vector Calculus

Tomsk
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I hope this is the right forum, this is mostly about maths, I'm not looking for a physical interpretation of angular momentum... yet. It also involves *some* calc... anyway...

OK, firstly, I've come to the conclusion I don't get cross products. I understand the properties of them, and can use them OK, there's just something I came across that I don't get. Say you have \vec{a}\times\vec{b}=\vec{c}. Apparently, the magnitude of c is given by the area of the parallelogram formed by a and b. I'm ok with the product axb having units of area, but when you then go and say c has a length that is an area... I get a bit lost. How am I supposed to interpret that?

Actually, scrap the second part, I'm an idiot!

Oh, and my lecturer always seemed to swap between J and L, both apparently for angular momentum. They mean the same thing, right? Or have I completely not understood anything??:cry:

I'll be back. I hate angular momentum.
 
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This kinda looks pointless now, I should always think through a problem thoroughly before looking for help on here! I still don't get the first bit about cross products though.
 
J is the 'total' angular momentum, which is a coupling of the orbital angular momentum and the spin angular momentum.

http://en.wikipedia.org/wiki/Angular_momentum_coupling#LS_coupling

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/vecmod.html

As for the cross product -

http://hyperphysics.phy-astr.gsu.edu/hbase/vvec.html

If vrectors a,b had dimensions of length, then in a x b = c, c would have magnitude of area, and vector would be parallel to the normal of the area.

When we do v x B for the Lorentz force, the resulting vector has units of T-m/s, which have to be equivalent to N/C, since F = q(v x B).

See also - http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning
 
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Think about what quantities angular momentum contains, i.e., what information is the angular momentum vector composed of; the scalar quantity - mass, the vector quantities - velocity and position. Isn't it a intuitive need to know what mass a particle has, where it is located, and what it's velocity is?

Regarding the area/length affare - I wouldn't loose my head thinking about that too much if I were you. The vector c = a x b can have any physical meaning, so it's dimension can be length, velocity, acceleration, force, etc. It's absolute value always equals the area of the a x b paralelogram, but that doesn't mean the dimensions equal, too.
 
Thanks guys. I've not done any QM yet, maybe my lecturer was crossing between J and L subconciously. I think I see the connection, sort of! I can accept the thing about the magnitude of the cross product too, it kinda caught me off guard! Thanks again. :D
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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