Master Index Notation: How to Ensure Accuracy in Calculations

[Petraa]

I'm not sure if this step on my calculation is correct or not.

$$A=i\left(M_{\alpha}^{\nu}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}\delta_{\alpha}^{\beta}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}M_{\alpha\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=0$$

I just change the label in the first generator with the delta and then contract the delta with the other generator. Is it 100% rigorous ?

Thank you!

 

[Orodruin]

I'm not sure if this step on my calculation is correct or not.

$$A=i\left(M_{\alpha}^{\nu}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}\delta_{\alpha}^{\beta}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}M_{\alpha\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=0$$

I just change the label in the first generator with the delta and then contract the delta with the other generator. Is it 100% rigorous ?

Thank you!

You will have to modify this, you are not allowed to call a summation index the same thing as a free index and then change which is the free index. Generally, if you have more than two of the samme index, you have done something naughty.

 

[Petraa]

Thank you for your fast response. The problem is that this thing should be zero ... if it is not zero then there must be an error in the previous calculations. Is A=0 ? And why?

 

[Orodruin]

What is M? Does it have some additional properties such as symmetry (as suggested by not caring too much about index ordering with the raised index)? If it is, what stops you from simply raising the nu on one M and lowering it on the other?

 

[Petraa]

M is total antisymetric. M are the generators of the lorentz group. The index order does matter but I have written it too fast in tex.

 

[Orodruin]

I still suspect it is a matter of simply changing which summation index is up and which is down ...

 

[Petraa]

The expression is $$A=i\left(M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right);\thinspace\thinspace\thinspace\thinspace\thinspace\left(M_{\alpha\beta}\right)_{\thinspace\thinspace\nu}^{\mu}=\delta_{\alpha}^{\mu}g_{\beta\nu}-\delta_{\beta}^{\mu}g_{\alpha\nu}$$

I don't see how this two terms equal to zero

 

[Orodruin]

Are you familiar with how to raise and lower indices?

 

[Petraa]

Yes with the metric. I've done it to get to this last result.
Do you mean things like ...

$$A^{i}=g^{ij}A_{j};A_{i}=g_{ij}A^{j};g^{ij}g_{ib}=g_{\thinspace\thinspace b}^{j}=\delta_{b}^{j}$$

and this sort of stuff ?

 

[Orodruin]

Exactly, so how does $M^\nu_{\phantom\nu \alpha}$ relate to $M_{\nu\alpha}$?

 

[Petraa]

I think its

$$M_{\nu\alpha}=g_{\nu\gamma}M_{\thinspace\thinspace\alpha}^{\gamma};M_{\thinspace\thinspace\alpha}^{\nu}=g^{\nu\gamma}M_{\gamma\alpha}$$

 

[Orodruin]

So what do you get if you apply the latter of these relations to your expression? (And switch the names of the summation indices in one of the terms...)

 

[Petraa]

I know what are trying to suggest me but I've tried that and I don't see why it should cancel. I feel so stupid right now :P. For example

$$A=i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)$$

Pffff Sorry for wasting ur time

 

[Orodruin]

Do it for both terms!

Edit: Or not even that. Rename the summation index in the first term from $\xi$ to $\nu$ and use the antisymmetry of M.

 

[Petraa]

Yes, this is exactly the part that I do not understand. I thought that this procedure was incorrect...

M is antisymmetric so
$$M_{ij}=-M_{ji}$$
I guess that
$$M_{\thinspace\thinspace j}^{i}=-M_{\thinspace\thinspace i}^{j}$$
$$A=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\nu\alpha}M_{\beta}^{\thinspace\thinspace\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(-M_{\alpha\nu}M_{\beta}^{\thinspace\thinspace\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right)$$

And this is the farthest I could arrive before posting this thread and I don't get why this is zero. M generators does not even commute so ...

One thing that I'm not 100% sure arei the indexs on the generators. They are not matrix indexs so it does not matter the order, right ?

$$M_{\thinspace\thinspace j}^{i}=M_{ j}^{\thinspace\thinspace i}$$

 

[samalkhaiat]

The expression is $$A=i\left(M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right);\thinspace\thinspace\thinspace\thinspace\thinspace\left(M_{\alpha\beta}\right)_{\thinspace\thinspace\nu}^{\mu}=\delta_{\alpha}^{\mu}g_{\beta\nu}-\delta_{\beta}^{\mu}g_{\alpha\nu}$$

I don't see how this two terms equal to zero

Assuming you meant to write $$A_{\alpha \beta} = i \left( M^{\nu}{}_{\alpha} \ M_{\beta \nu} - M^{\nu}{}_{\beta} \ M_{\alpha \nu} \right) ,$$ then by lowering the index $\nu$ $$A_{\alpha \beta} = i \left( \eta^{\nu \rho} M_{\rho \alpha} \ M_{\beta \nu} - \eta^{\nu \rho} M_{\rho \beta} \ M_{\alpha \nu} \right) .$$ In the second term, change the dummy indices $\nu \leftrightarrow \rho$ and find $$A_{\alpha \beta} = - i \eta^{\nu \rho} [ M_{\alpha \rho} , M_{\beta \nu} ] .$$ Now, using the Lorentz algebra $$i [M_{\alpha \rho} , M_{\beta \nu} ] = \eta_{\alpha \beta} M_{\rho \nu} - \eta_{\rho \beta} M_{\alpha \nu} + \eta_{\alpha \nu} M_{\beta \rho} - \eta_{\rho \nu} M_{\beta \alpha} ,$$ you find $$A_{\alpha \beta} = ( 2 - n ) M_{\alpha \beta} ,$$ where $n$ is the dimension of space-time.
Clearly, $A_{\alpha \beta} = 0$, only for $n = 2$, i.e., in 2-dimensional space-time.
However, if by $A$ you meant the trace $A = \eta^{\alpha \beta} A_{\alpha \beta}$, then yes $A = 0$ identically.

Sam

 

[Mentz114]

if $M$ is antisymmetric then ( I believe) $M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}$ is symmetric in $\alpha,\ \beta$

 

[Petraa]

What I am trying to prove is that the commutator of the operator W^2 constructed from the Pauli-Lubanski pseudo vector is a Casimir operator for the poincare group.

The whole commutator [W^2,M_ij] splits up into two parts, my textbook says that the first part should be zero and I've revised multiple times my calculations and I always get what I have written in my first post...

@Mentz114 if you are correct then of course is 0 but I can not prove it.

 

[Orodruin]

@Mentz114 if you are correct then of course is 0 but I can not prove it.

It is really only a matter of doing what I proposed you to do. I suggest first switching the indices of the M with both indices down, then you switch which index is up and which is down, then you switch the indices of the other. You can never have $M^\mu_{\phantom\mu \nu} = -M^{\phantom \mu \nu}_\mu$, the different sides have different indices contra/covariant.

 

[Petraa]

It is really only a matter of doing what I proposed you to do. I suggest first switching the indices of the M with both indices down, then you switch which index is up and which is down, then you switch the indices of the other. You can never have $M^\mu_{\phantom\mu \nu} = -M^{\phantom \mu \nu}_\mu$, the different sides have different indices contra/covariant.

If I play with the indexs, lowering them etc. I just walk in circles. For example I get

$$i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-g^{\nu\xi}M_{\xi\beta}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\xi\beta}M_{\alpha}^{\thinspace\thinspace\xi}\right)$$

 

[Orodruin]

If I play with the indexs, lowering them etc. I just walk in circles. For example I get

$$i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-g^{\nu\xi}M_{\xi\beta}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\xi\beta}M_{\alpha}^{\thinspace\thinspace\xi}\right)$$

I also told you to switch the names of the summation indices in one of the terms ... Do that and factorise out $g^{\nu\xi}$ from the expression ...

Edit: And use that the metric is symmetric ...

 

[samalkhaiat]

if $M$ is antisymmetric then ( I believe) $M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}$ is symmetric in $\alpha,\ \beta$

No, you are very wrong. I will try to repeat what I have already done in my previous post. $$M^{\nu}{}_{\alpha} M_{\beta \nu} = g^{\nu \mu} M_{\mu \alpha} M_{\beta \nu} = g^{\nu \mu} ( - M_{\alpha \mu} ) ( - M_{\nu \beta} ) = g^{\nu \mu} M_{\alpha \mu} M_{\nu \beta} . \ \ (1)$$ $$M^{\nu}{}_{\beta} M_{\alpha \nu} = g^{\rho \sigma} M_{\rho \beta} M_{\alpha \sigma} = g^{\nu \mu} M_{\nu \beta} M_{\alpha \mu} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ Now, subtract (2) from (1) $$M^{\nu}{}_{\alpha} M_{\beta \nu} - M^{\nu}{}_{\beta} M_{\alpha \nu} = g^{\nu \mu} \left( M_{\alpha \mu} M_{\nu \beta} - M_{\nu \beta} M_{\alpha \mu} \right) = g^{\nu \mu} [ M_{\alpha \mu} , M_{\nu \beta} ] \neq 0 .$$

 

[samalkhaiat]

If I play with the indexs, lowering them etc. I just walk in circles. For example I get

$$i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-g^{\nu\xi}M_{\xi\beta}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\xi\beta}M_{\alpha}^{\thinspace\thinspace\xi}\right)$$

Did you read my first post? The expression you wrote IS NOT zero.

 

[samalkhaiat]

What I am trying to prove is that the commutator of the operator W^2 constructed from the Pauli-Lubanski pseudo vector is a Casimir operator for the poincare group.

First, you need to show that $W_{\mu}$ is a Lorentz vector, i.e. you need to prove that $$[ M_{\mu \nu} , W_{\rho} ] = i \left( g_{\nu \rho} W_{\mu} - g_{\mu \rho} W_{\nu} \right) . \ \ \ (1)$$ Once you have done that, the rest is easy. Use $$[ M_{\mu \nu} , W^{2} ] = g^{\rho \sigma} [ M_{\mu \nu} , W_{\rho} ] W_{\sigma} + g^{\rho \sigma} W_{\rho} [ M_{\mu \nu} , W_{\sigma} ] . \ \ (2)$$ Now, insert (1) in (2) and do the trivial algebra, you find $$[ M_{\mu \nu} , W^{2} ] = 0 .$$ Proving $[ W^{2} , P_{\mu} ] = 0$ is even easier, because $$[ W_{\rho} , P_{\mu} ] = \frac{1}{2} \epsilon_{\rho \sigma \alpha \beta} g_{\mu \nu} P^{\sigma} [ M^{\alpha \beta} , P^{\nu} ] = \frac{i}{2} \epsilon_{\rho \sigma \alpha \mu} [ P^{\sigma} , P^{\alpha} ] = 0 .$$

 

[Mentz114]

No, you are very wrong. I will try to repeat what I have already done in my previous post. $$M^{\nu}{}_{\alpha} M_{\beta \nu} = g^{\nu \mu} M_{\mu \alpha} M_{\beta \nu} = g^{\nu \mu} ( - M_{\alpha \mu} ) ( - M_{\nu \beta} ) = g^{\nu \mu} M_{\alpha \mu} M_{\nu \beta} . \ \ (1)$$ $$M^{\nu}{}_{\beta} M_{\alpha \nu} = g^{\rho \sigma} M_{\rho \beta} M_{\alpha \sigma} = g^{\nu \mu} M_{\nu \beta} M_{\alpha \mu} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ Now, subtract (2) from (1) $$M^{\nu}{}_{\alpha} M_{\beta \nu} - M^{\nu}{}_{\beta} M_{\alpha \nu} = g^{\nu \mu} \left( M_{\alpha \mu} M_{\nu \beta} - M_{\nu \beta} M_{\alpha \mu} \right) = g^{\nu \mu} [ M_{\alpha \mu} , M_{\nu \beta} ] \neq 0 .$$

What about the definition of the EM energy-momentun tensor $T^{\mu\nu}={F^\mu}_\rho F^{\rho\nu}$ which alway gives a symmetric T ?
I checked this and it is definitely symmetric in $\nu\mu$ but not if $\nu$ is interchanged with another index (obviously) .

 

[Orodruin]

What about the definition of the EM energy-momentun tensor $T^{\mu\nu}={F^\mu}_\rho F^{\rho\nu}$ which alway gives a symmetric T ?
I checked this and it is definitely symmetric in $\nu\mu$ but not if $\nu$ is interchanged with another index (obviously) .

This is true when $F^{\mu\nu}$ is a number, meaning that the $F$s commute. It will no longer be true if it is a group element as given by the last equation of post #22 (note that if the $M$ were numbers, then the commutator would be zero).

 

[Mentz114]

What I am trying to prove is that the commutator of the operator W^2 constructed from the Pauli-Lubanski pseudo vector is a Casimir operator for the poincare group.

The whole commutator [W^2,M_ij] splits up into two parts, my textbook says that the first part should be zero and I've revised multiple times my calculations and I always get what I have written in my first post...

@Mentz114 if you are correct then of course is 0 but I can not prove it.

Right now I cannot prove it is always true.

 

[Mentz114]

This is true when $F^{\mu\nu}$ is a number, meaning that the $F$s commute. It will no longer be true if it is a group element as given by the last equation of post #22 (note that if the $M$ were numbers, then the commutator would be zero).

OK, thanks. I forget that those commutators have implied indexes.

 

[samalkhaiat]

First, you need to show that $W_{\mu}$ is a Lorentz vector, i.e. you need to prove that $$[ M_{\mu \nu} , W_{\rho} ] = i \left( g_{\nu \rho} W_{\mu} - g_{\mu \rho} W_{\nu} \right) . \ \ \ (1)$$

The proof rests on the following facts:
(1) $P^{\mu}$ is a vector operator, i.e. it transforms according to $$U^{\dagger} ( \Lambda ) P^{\mu} U ( \Lambda ) = \Lambda^{\mu}{}_{\nu} \ P^{\nu} , \ \ \ \ (1)$$ where $$U ( \Lambda ) = e^{- \frac{i}{2} \omega_{\rho \sigma} M^{\rho \sigma}} ,$$ is (infinite-dimensional) unitary representation of the Lorentz group on Hilbert space.
(2) The Lorentz’s generators $M^{\mu \nu}$ transform like tensor operators $$U^{\dagger} ( \Lambda ) M^{\mu \nu} U ( \Lambda ) = \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ M^{\rho \sigma} . \ \ \ \ (2)$$
(3) $\epsilon$ is an invariant tensor: $$\epsilon_{\mu \nu \rho \sigma} \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} = \epsilon_{\alpha \beta \gamma \tau} \ \det | \Lambda | = \epsilon_{\alpha \beta \gamma \tau} ,$$ or $$\epsilon_{\lambda \nu \rho \sigma} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} = \Lambda^{\alpha}{}_{\lambda} \epsilon_{\alpha \beta \gamma \tau} . \ \ \ (3)$$
So, let us write $$U^{\dagger} ( \Lambda ) W_{\mu} U ( \Lambda ) = \frac{1}{2} \epsilon_{\mu \nu \rho \sigma} U^{\dagger} P^{\nu} U U^{\dagger} M^{\rho \sigma} U .$$ Using (1), (2) and (3) we find $$U^{\dagger} W_{\mu} U = \Lambda^{\nu}{}_{\mu} W_{\nu} .$$ This is the finite version of the wanted commutation relation. You can obtain it in the first order approximation $$U = 1 - \frac{i}{2} \omega_{\mu \nu} M^{\mu \nu} , \ \ \ \ \Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + \omega^{\mu}{}_{\nu} .$$ One last important commutator that you can prove is $$[W_{\mu} , W_{\nu}] = i \epsilon_{\mu \nu \rho}{}^{\sigma} P^{\rho} W_{\sigma} .$$ From this commutator, you can show that the vector $W_{\mu}$ determines an intrinsic angular momentum, i.e. spin of the representation.
Sam