Infintesimal transformations and Noether's theorem

[samalkhaiat]

...Anyway, do you mean to say what I wrote is incorrect?

No, but your remarks were utterly trivial. Of course if you do rotation in the xy-plane, the z-component of a vector will not change. There is no mystery about this.

Before you go, one last question
Let $V^{\mu}$be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu}$. We can write this like $V'^{\mu} = (e^{i\omega})^{\mu}_{\,\,\,\nu}V^{\nu}$ I think where $\omega$denotes a rotation in some plane spanned by indices $\left\{\rho \sigma\right\}$, say. In 2D Euclidean space time, we can write the matrix representation of $\Lambda$ as $$\begin{pmatrix} \cos \omega & \sin \omega\\-\sin \omega&\cos \omega\end{pmatrix}$$ and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields $$\begin{pmatrix}1&\omega\\-\omega&1\end{pmatrix} = \text{Id} + \begin{pmatrix} 0&\omega\\-\omega&0\end{pmatrix} = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0\end{pmatrix}$$

The more general treatment gave $S$ to be $\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu}$. I am wondering how this agrees with the matrix I obtained above multiplying $\omega$. The matrix above is a rep of the generator of the SO(2) rotation group when acting on 2D vectors.

Many thanks.

Again, you are asking me something which I have already answered. This is exactly what I have done in post #25, but instead of one parameter and one 2x2 generating matrix, we had 6 parameters and 6 4x4 generating matrices.
( I + \omega )^{ \mu }{}_{ \nu } = \delta^{ \mu }_{ \nu } + \omega^{ \mu }{ }_{ \nu } = \delta^{ \mu }_{ \nu } + \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma } )^{ \mu }{}_{ \nu }
( \mu , \nu ) give you the different matrix elements of \Sigma
So, for boosts, we have the 3 matrices ( \Sigma_{ 0 1 } )^{ \mu }{}_{ \nu }, ( \Sigma_{ 0 2 } )^{ \mu }{}_{ \nu } and ( \Sigma_{ 0 3 } )^{ \mu }{}_{ \nu }
And for 3 rotations, we have the generating matrices ( \Sigma_{ 1 2 } )^{ \mu }{}_{ \nu }, ( \Sigma_{ 2 3 } )^{ \mu }{}_{ \nu } and ( \Sigma_{ 3 1 } )^{ \mu }{}_{ \nu }

 

[CAF123]

Hi Sam,

The finite components field, \phi_{ a }( x ), transforms by finite-dimensional (matrix) representation, D( g ), of the group in question:

\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } (g) \phi_{ b } ( x )

Infinitesimally, we write this (to first order in \omega)

<br /> \bar{ \phi }_{ a } ( \bar{ x } ) = \left( \delta^{ b }_{ a } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \right) \phi_{ b } ( x ) \ \ (1)<br />

I understand the derivation you did here, however, conceptually $\Sigma$ is the spin matrix for the field $\phi$, which does not transform the coordinates so why is the LHS of the equation in the barred system?

Also here:

<br /> \delta \phi_{ a } ( x ) = \bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = ( \omega \cdot M )_{ a }{}^{ b } \ \phi_{ b } ( x )<br />

I realize M here is the full symmetry transformation, so it is composed of a piece transforming the coordinates and the field. But the LHS of the equation does not have the coordinate x shifted, we remain at the same coordinate x throughout (i.e we do not have any instances of the barred system in this equation).

Could you provide some thoughts here?
Thanks.

 

[samalkhaiat]

Hi Sam,

I realize M here is the full symmetry transformation, so it is composed of a piece transforming the coordinates and the field. But the LHS of the equation does not have the coordinate x shifted, we remain at the same coordinate x throughout (i.e we do not have any instances of the barred system in this equation).

Could you provide some thoughts here?
Thanks.

Spacetime transformations transform the argument ( x ) of the field and mix its components. I explained all this to you in post#7.

 

[CAF123]

Spacetime transformations transform the argument ( x ) of the field and mix its components. I explained all this to you in post#7.

I understood the mathematical derivation, it is just conceptually I don't see why the equations are the way they are. E.g to be more clear, eqn (1) in post #7. The field is acted on upon by $\Sigma$ which transforms the spin indices of the field. It does nothing to the coordinates as far as I understand. But the LHS of that equation has the field and the coordinates in the barred system, where $\bar x = x + \delta x$. So it seems to be the case that the coordinates have changed. If the $\Sigma$ generators only transform the fields shouldn't we only transform to the barred system of the field?
Thanks in advance for clearing my confusion.

 

[samalkhaiat]

I understood the mathematical derivation, it is just conceptually I don't see why the equations are the way they are. E.g to be more clear, eqn (1) in post #7. The field is acted on upon by $\Sigma$ which transforms the spin indices of the field. It does nothing to the coordinates as far as I understand. But the LHS of that equation has the field and the coordinates in the barred system, where $\bar x = x + \delta x$. So it seems to be the case that the coordinates have changed. If the $\Sigma$ generators only transform the fields shouldn't we only transform to the barred system of the field?
Thanks in advance for clearing my confusion.

Observer O uses system x and calls his fields \phi_{ a } ( x ). Observer \bar{ O } uses \bar{ x } and \bar{ \phi }_{ a } ( \bar{ x } ). The very same point P in space has two set of coordinates values x and \bar{x}.
Is this new to you? If yes, then you need to learn about the use of transformations in physics. If no, then the rest is explained in post#7.

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 ) .
Substitute the first equation in the second, you get
\bar{ \phi }_{ a } ( x + \delta x ) = \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .
Now, expand the left-hand-side to first order in \delta x and use the fact that \delta x \bar{ \phi } \approx \delta x \phi:
\bar{ \phi }_{ a } ( x ) + \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) = \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .
Arrange this to
\bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = - \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .
Call the left-hand-side \delta \phi_{ a } ( x ) and rewrite the above as
\delta \phi_{ a } ( x ) = - \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .
Now, ask yourself this: Didn’t I do all this for you in post #7? When I said “I have done every thing you need in this thread” I knew what I was talking about.

Sam

 

[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"