Question on Special Orthogonal Group SO(3)

[Johnny5]

The point is that' date=' as long as there is rotational invariance, the physics is unchanged after rotating the physical system by an angle A, and that this is no different than rotating the axes by an angle -A.

[/quote']

 

I understand that.

 

The formulas which are being used to express the "laws of physics" are the same. Still thinking about it.

 

I'm not quite sure how I would go about showing "rotation invariance."

 

OK i get everything you said...

 

After you have rotated through some angle, the laws of physics are still the same locally. You aren't saying anything more than that.

 

"the physics is unchanged after being rotated through angle A"

 

yes i get that.

 

I'm not sure it's actually true, but i get it.

 

What I certainly get, is that we want to formulate the laws of physics in such a way, as so that they are the same after the rotation.

 

We don't want rotations to change the laws of physics so to speak.

 

But you know what...

 

A spinning frame is a non-inertial frame.

 

The laws of physics do have a different mathematical form there.

 

What say you to that?

 

PS: Actually, I don't want the thread to digress that much, I would like it to remain basically on track, and it is about SO(3), so I'm going to run through your derivation of v`, using complex numbers.

[Tom Mattson]

But you know what...

 

A spinning frame is a non-inertial frame.

 

We aren't talking about a spinning frame, we're talking about a rotated frame. The angular velocity of the rotated frame is zero.

[Severian]

SO(3) is also isomorphic to SU(2), which is one of the fundamental guage groups of the Standard Model of particle physics.

[Tom Mattson]

SO(3) is also isomorphic to SU(2), which is one of the fundamental guage groups of the Standard Model of particle physics.

 

Are you sure about that? An isomorphism is a homomorphism that is both one-to-one and onto. But there is a two-to-one map of SU(2) onto SO(3). For that reason Sakurai (Modern Quantum Mechanics) says that SU(2) and SO(3) are only locally isomorphic.

[matt grime]

Indeed. SU(2) is the simply connected cover of SO(3). They are not isomorphic as Lie groups (there can be no homeomorphism between them, never mind one that respects the group structure). I doubt they are isomorphic as abstract groups either, though I can think of no compelling reason for that in all honesty. But that possibly reflects the fact I've been to the pub after work.

 

Note it is possible for topological groups to be abstractly isomorphic without being isomorphic as topological groups, eg R with the normal topology and R with the Zariski, discrete or indiscrete topology.

[Johnny5]

I said I was going to run through Tom's argument, and that's what I intend to do here. Let me see if I can reproduce it, without looking.

 

Let Z denote an element of the complex numbers.

 

Hence, for some real numbers x,y we have:

 

Z = x+iy

 

Now, this is the rectangular form of the complex number Z.

 

But, we can also write Z in "polar form."

 

The complex number Z, can be represented as a vector in the Argand plane.

 

Let the magnitude of Z be represented by R. And let the angle from the real axis to the 'vector' be denoted by q.

 

Feel free to correct my terminology.

 

Ok so...

 

For some reason I just remember the answer is:

 

Z = R e^iq

 

Let me see if thats true.

 

e^iq = cosq+i sinq

 

Hence:

 

Re^iq = Rcosq+Ri sinq

 

R cos theta is the component of the vector in the X direction, and Rsin theta is the component of the vector in the Y direction hence:

 

Re^iq = x+iy

 

So it is indeed true.

 

The magnitude of the complex number is ZZ* i think.

 

In other words, it should be the case that ZZ*=R²

 

Z* is the complex conjugate of Z.

 

Check:

 

Let Z = R e^iq

 

hence

 

Z* = R e^-iq

 

The previous statement is true by definition.

 

Hence:

 

ZZ* = (R e^iq )(R e^-iq)

 

Adding exponents, we have:

 

ZZ* = R²e^iq-iq

 

The exponent is zero, hence:

 

ZZ* = R²

 

QED

 

Ok so now we have to figure out the formula for a rotation.

Ok, I don't feel like playing around, I just want to run through the argument, so here it is:

 

Neither classical trigonometry nor quaternions is the way to go. The quickest' date=' most direct route to the rotation matrices is to use complex exponentials. This is the case for 2 reasons:

 

1. Complex numbers in C1 can be represented as vectors that obey the same algebra as real vectors in R2.

2. Exponentials are easier to work with than trigonometric functions by several orders of magnitude.

 

That said let's develop our formula.

 

Let z=x+iy be a complex number with the usual vector representation. Its polar form is z=r exp(iB), where B is the angle the vector makes with the polar axis. Now rotate the vector clockwise (remember this is the same as rotating the axes counterclockwise) by an angle A. The new complex number is z'=r exp[i(B-A)']=r exp(iB)exp(-iA)=x'+iy'.

 

Now the question is: What are x' and y'? Noting that r exp(B)=x+iy, we have:

 

x'+iy'=(x+iy)[cos(A)-i sin(A)]

x'+iy'=[x cos(A)+y sin(A)]+i[-x sin(A)+y cos(A)]

 

or...

 

x'=x cos(A)+y sin(A)

y'=-x sin(A)+y cos(A)

 

Tom, you said that B is the angle that the vector makes with the polar axis. Is that what the real axis is called?

 

You say rotate the vector clockwise... by angle A.

 

So we have a new complex number Z`, and the angle is A+B.

 

Wait i did that wrong, you said to rotate the vector clockwise.

 

Ok, so then the new angle is B-A, like you have.

 

Therefore:

 

Z`=R e^{i (B-A)}

 

Which we can write as:

 

Z`=R e^{i B} e^{i (-A)}

 

Let the rectangular form of Z` be x`+iy`

 

Therefore:

 

Z`= R e<sup>i B</sup> e<sup>i (-A)</sup>= x`+iy`

 

Now, we just want to write x`,y` in terms of x,y, and we are done.

 

Recall that:

 

Z= R e^{i B}

 

Thus, we have:

 

Z`= Z e<sup>i (-A)</sup>= x`+iy`

 

Thus, we have:

 

(x+iy) e^{i (-A)}= x`+iy`

 

Thus, we have:

 

(x+iy) ( cos(-A) + isinA) = x`+iy`

Hence we have:

 

(x+iy) ( cos A + isinA) = x`+iy`

 

From which it follows that:

 

xcosA+xisinA+iycosA+iyisinA=x`+iy`

 

Hence;

 

xcosA+xisinA+iycosA-ysinA=x`+iy`

 

Hence:

 

(xcosA-ysinA) + i(xsinA+ycosA) = x`+iy`

 

Hence it follows that:

 

xcosA - ysinA = x`

 

and that

 

xsinA+ycosA = y`

 

QED

 

Now, here is what caused this slight digression:

 

Start by looking at a vector v=<x' date='y> in the xy-plane. Then rotate the axes by an angle A and see what the new components are. You should get:

 

[b']v[/b]'=<xcos(A)+ysin(A),-xsin(A)+ycos(A)>

 

You should be able to write the RHS of the above equation as a matrix times the original vector v. That matrix is:

 

[cos(A) sin(A)]

[-sin(A) cos(A)]

 

So now for the matrix...

 

We started off with

 

V=<x,y> in the complex plane.

 

Then were told to rotate that vector clockwise to obtain

 

V` = <x`,y`>

 

and discovered that:

 

V` = (xcosA - ysinA, xsinA+ycosA)

 

Tom there's a disagreement in our minus sign, how come?

[Tom Mattson]

Tom there's a disagreement in our minus sign' date=' how come?[/quote']

 

Right here:

 

Thus, we have:

 

(x+iy) ( cos(-A) + isinA) = x`+iy`

 

You forgot to put -A into the sine function.

 

Hence we have:

 

(x+iy) ( cos A + isinA) = x`+iy`

 

So the second factor on the RHS should be:

 

( cos A - isinA)

[Johnny5]

Right here:

 

 

 

You forgot to put -A into the sine function.

 

 

 

So the second factor on the RHS should be:

 

( cos A - isinA)

 

Ok thanks, i'll go back and fix it, also I'd like you to have a look at something, I'm going to put in the general mathematics category. It's about vector division... again.