Infintesimal transformations and Noether's theorem

[CAF123]

Now, ask yourself this: Didn’t I do all this for you in post #7?

Yes, you did. But what I am doing now is asking a follow up question to try to understand what you wrote. I did not mean to ask you to copy out the derivation again.

The space-time transformation group connects the two observers and their fields at $P$ as follow
$$\bar{ x }^{ \mu } = g^{ \mu }{}_{ \nu } x^{ \nu } \approx x^{ \mu } + \delta x^{ \mu } ,$$
$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } \phi_{ b } ( x ) \approx \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$

I guess my question would be why is it only the sigma matrices which are present and not the whole generator here? For example, on P.5 of this document: http://einrichtungen.ph.tum.de/T30f/lec/QFT/groups.pdf The equation above eqn (4) on page 5 implements the transformation of the coordinates of the field between the two observer frames of reference. The generator present is L and this makes sense (L is orbital, thereby transforming the coordinates). Eqn (5) implements the transformation on both the field and the coordinates by adding a term with generator S that transforms the field spin indices if S ≠0 and I think it resembles your $\Phi'(x') = D(g)\Phi$.

My question is: the only term in your D(g) is the ∑ $\equiv$ S that implements only the change in the field spin indices. So why, conceptually, is there the primed coordinate system on the LHS? I would have expected there to be the orbital and spin part present in D for this to occur.

Thank you.

 

[samalkhaiat]

I guess my question would be why is it only the sigma matrices which are present and not the whole generator here? For example, on P.5 of this document: http://einrichtungen.ph.tum.de/T30f/lec/QFT/groups.pdf The equation above eqn (4) on page 5 implements the transformation of the coordinates of the field between the two observer frames of reference. The generator present is L and this makes sense (L is orbital, thereby transforming the coordinates).

I think, I’ve done better job explaining this.

Eqn (5) implements the transformation on both the field and the coordinates by adding a term with generator S that transforms the field spin indices if S ≠0 and I think it resembles your $\Phi'(x') = D(g)\Phi$.

No, eq(5) is the FINITE version (exponentiation) of the last equation in post #35.

My question is: the only term in your D(g) is the ∑ $\equiv$ S that implements only the change in the field spin indices.So why, conceptually, is there the primed coordinate system on the LHS?

Because the transformation law refers to the SAME geometrical point $P$
$$\bar{ \phi }_{ a } ( P ) = ( e^{ - \omega \ \cdot \ S } )_{ a }{}^{ b } \ \phi_{ b } ( P )$$
As I said in the previous post, the coordinates of point $P$ in the barred system is $\bar{x}$ and $x$ in the unbarred system. So, you can write
$$\bar{ \phi }_{ a } ( \bar{ x } ) = ( e^{ - \omega \ \cdot \ S } )_{ a }{}^{ b } \ \phi_{ b } ( x ) \ \ \ (1)$$
Now, the left hand side can be written as
$$\bar{ \phi }_{ a } ( \bar{ x } ) = ( e^{ \delta x \ \cdot \ \partial } ) \ \bar{ \phi }_{ a } ( x ) \equiv ( e^{ \omega \ \cdot \ L } ) \ \bar{ \phi }_{ a } ( x ) . \ \ \ (2)$$
I explained the meaning of this equation way back in post #5. It simply means this: the barred observer shift the argument of his field to a nearby point $Q$ with coordinates value $\bar{ x }_{ Q }$ equal to the coordinates of $P$ in the unbarred system, which we called $x$.
Now, put (2) back in (1), you find
$$\bar{ \phi }_{ a } ( x ) = ( e^{ - \omega \ \cdot \ L } ) ( e^{ - \omega \ \cdot \ S } )_{ a }{}^{ b } \ \phi_{ b } ( x )$$
This is eq(5).

I would have expected there to be the orbital and spin part present in D for this to occur.

No, the representations of the abstract Lorentz group know absolutely nothing about the spacetime coordinates. It is classified by the spin-matrix only. When we identify the representations with fields on spacetime, then and only then, the orbital generators show up.

Sam

 

[CAF123]

Thanks samalkhaiat! Yes, your derivation is clearer and to see if I understand what you say about the coordinates:

Because the transformation law refers to the SAME geometrical point $P$

In frame $S$, say, we measure $x$ and in frame $\bar S$ we measure $\bar x$ - they are the same point in Minkowski space, it is just their coordinate frame representation of the point is different. For example, relative to frame S, this means the two points will be at different positions, (since the coordinate reps of $x$ in S is not the same as $\bar x$ in S, otherwise the frames would coincide) related by $\bar x = \Lambda x$ if we do a Lorentz transformation (rotation) of the frame S coordinate space. Is that correct? I guess I dragged that on a bit, but I just want to check I understand.

No, the representations of the abstract Lorentz group know absolutely nothing about the spacetime coordinates. It is classified by the spin-matrix only. When we identify the representations with fields on spacetime, then and only then, the orbital generators show up.

Thanks.

 

[samalkhaiat]

In frame $S$, say, we measure $x$ and in frame $\bar S$ we measure $\bar x$ - they are the same point in Minkowski space, it is just their coordinate frame representation of the point is different. For example, relative to frame S, this means the two points will be at different positions, (since the coordinate reps of $x$ in S is not the same as $\bar x$ in S, otherwise the frames would coincide) related by $\bar x = \Lambda x$ if we do a Lorentz transformation (rotation) of the frame S coordinate space. Is that correct? I guess I dragged that on a bit, but I just want to check I understand.


Thanks.

At last. Yes, you always remember this "two pearsons looking at one point is equivalent to one pearson looking at two points"