Deriving Casimir operator from the Lie Algebra of the Lorentz Group

[Froggeh]

Homework Statement

Show that:
$$ \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\mu \nu} M_{\rho \sigma} = - \mathbf{J} \cdot \mathbf{K}$$
Where $M_{\mu \nu}$ is an antisymmetric tensor constructed as follows: $M_{0i} = K_{i}$ and $M_{ij} = \epsilon_{ijk}J_{k}$

Relevant Equations

$J_{i}$ are the three generators of the Lorentz rotations, and $K_{i}$ are the three generators of the Lorentz boosts (my textbook calls them pure transformations, but my understanding is that they're the same thing).
The $J$ matrices are Hermitian, while the $K$ matrices are symmetric.
Maybe it's also worth mentioning the commutator
$$ [J_i, K_j] = i\epsilon_{ijk}K_k$$

Hello everyone, I am new here, so please let me know if I am doing something wrong regarding the formatting or the way I am asking for help.

I did not really know how to start off, so first I tried to just write out all the $\mu \nu \rho \sigma$ combinations for which $\epsilon \neq 0$ and assign them signs. In essence, I just used Mathematica to find these. However, when I plugged them in, and replaced each corresponding entry of $M_{\mu \nu}$, I obtained the result $2(\mathbf{K} \cdot \mathbf{J} + \mathbf{J} \cdot \mathbf{K})$, which, unless I am missing something crucial, is just plain wrong.
Since each time I attempted this was taking up a large chunk of time, I thought of changing my approach, by fixing one index and seeing what comes out:
Say $\mu = 0$, then $\nu \rho \sigma = ijk$, and this leaves us with
$$ \frac{1}{2} \epsilon^{0ijk}M_{0i}M_{jk} = \frac{1}{2} \epsilon^{0ijk}K_{i}\epsilon_{jkl}J_{l} $$
This is where I am unable to move forward with certainty. I am thinking that $\epsilon^{0ijk} = \epsilon^{ijk}$ because if the first component is zero, then we need only consider permutations of the other three components. This in turn would allow me to easily contract the two $\epsilon$ terms into a simple Kronecker-$\delta$. If I do this, I end up with
$$ \frac{1}{2} \epsilon^{ijk}\epsilon_{jkl}K_{i}J_{l} = \frac{1}{2} \epsilon^{ijk}\epsilon_{ljk}K_{i}J_{l} = \frac{1}{2} 2\delta^{i}_{l} K_{i} J_{l} = K_{l}J_{l}$$
But this seems wrong too, I am missing a negative sign. And my indices seem to be in the wrong positions.
At this point I'm unsure of whether I am on the right track. I suppose I would have to try this for each index $= 0$, but even then I cannot see the terms combining to give me a negative dot-product.

 

[jambaugh]

One trick I often use in representing these anti-symmetric rank-2 tensors in 3+1 dim, is to define the "cross product matrix" of a 3-vector. $[\mathbf{a}\times]$ is the matrix that maps a 3-vector $[\mathbf{b}]$ (as a column matrix) to the vector $\mathbf{a}\times\mathbf{b}$. (This same situation occurs with the covariant EM field tensor $F_{\mu\nu}$ which is a similar construction from the $\mathbf{E},\mathbf{B}$ 3-vectors.)

Thus for example:
$$\mathbf{u}=\left[\begin{array}{c}u\\v\\w\end{array}\right], [\mathbf{u}\times] = \left[\begin{array}{rrr} 0 & -w & v\\ w & 0 & -u\\ -v & u & 0\end{array}\right]$$
Then also note that $[\mathbf{v}]^T[\mathbf{u}\times] = (-[\mathbf{u}\times][\mathbf{v}])^T = -[\mathbf{u}\times\mathbf{v}]^T = [\mathbf{v}\times\mathbf{u}]^T$ (these are row matrices).

Then you have to work out how the $\epsilon^{ijk}$ contracts with vector components. A little work will show you that, (up to sign) contracting with anyone index with a 3-vector's components will yield this "cross product matrix". So let me check that sign:
The $[\mathbf{i}\times]$ matrix will be of the form:
$$[\mathbf{i}\times] = \left[\begin{array}{rrr} 0 & 0 & 0\\ 0 & 0 & -1\\ 1& 0& 0\end{array}\right]$$
So $\epsilon^{ijk}[\mathbf{i}]_k = \epsilon^{ij1}=\pm?[\mathbf{i}\times]^{ij}$ which gives $\pm[\mathbf{i}\times]^{23}=1, \pm[\mathbf{i}\times]^{32}=-1$ checking with the matrix form (indices are row then column) I see that the sign must be minus. (Check my work both for your understanding's sake and because I'm figuring this as I type and can easily have made a mistake.)

So if I'm correct, your 4-matrix is:
$$[M] = \left[\begin{array}{cc} 0 & [\mathbf{K}]^T \\ [\mathbf{K}] & -[\mathbf{J}\times] \end{array}\right]$$

Now the next step is harder. Working out the contraction of an anti-symmetric matrix of this form with the Levi-Civita tensor. I am not sure on the signs or any factors of 2 or 1/2 (or 4?) but know that the $\mathbf{K}$ and $\mathbf{J}$ will be swapped. You should get something like:
$$[M^\star] = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}M_{\mu\nu} = ?\cdot \left[\begin{array}{cc} 0 & \pm [\mathbf{J}]^T\\ \mp [\mathbf{J} & \pm'[\mathbf{K}\times]^{\rho\sigma}\end{array}\right]$$
I'll be lazy and leave it to your to work out the multiplier (?) and the sign choices for $\pm$ and $\pm'$.

Once that's done you'll get a double contraction between two matrices which will be the trace of their product (if adjacent indices are contracted between them) or the trace of their product with one transposed (if first contracts with first and second contracts with second.) Since they are anti-symmetric that transposed case will be yet another sign change. It looks like your problem is the latter of these cases.

$$\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}M_{\mu\nu}M_{\rho\sigma} = \mathbf{tr}\left(-[M^\star]\cdot[M]\right)$$

I know this sounds like a round-about way to get there but it's like starting welding by making your welding stand. All that effort leaves you with very useful tool. When you're done you'll be able to multiply these out and express the result in terms of cross and dot products of the 3-vectors. Specifically you'll get [\itex]\matbf{J}\times\mathbf{J}, \mathbf{K}\times\mathbf{K}[/itex] components which are, or course, zero and you'll get double cross product matrices such as $[\mathbf{J}\times]^2$ which you can resolve by applying to an arbitrary vector, $[\mathbf{J}\times]^2[\mathbf{v}] = [\mathbf{J}\times(\mathbf{J}\times \mathbf{v})]$. From there apply the double cross product identity remembering your defining an operator acting on $\mathbf{v}$ so insert the identity matrix between scalar multipliers and v, i.e. $(\mathbf{J}\bullet\mathbf{J})\mathbf{v}=(\mathbf{J}\bullet\mathbf{J})I\cdot\mathbf{v}$. The result then will be fairly easy to express in matrix form and take the trace.

 

[Froggeh]

Thanks for the help. While I had trouble understanding the cross-product matrix, the suggestion of defining $[M^{*}]$ like so actually helped me working through the algebra in a more efficient manner and I was able to reach an answer. Turns out the factor of $1/2$ was not exactly correct, and it was $1/8$ instead.

I was particularly confused with the $[\mathbf{i}\times]$ matrix. What in particular is $\mathbf{i}$?

I solved the problem, but I still have some questions about the method that you illustrated, which seems quite useful in general.

 

[vanhees71]

Homework Statement:: Show that:
$$ \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\mu \nu} M_{\rho \sigma} = - \mathbf{J} \cdot \mathbf{K}$$
Where $M_{\mu \nu}$ is an antisymmetric tensor constructed as follows: $M_{0i} = K_{i}$ and $M_{ij} = \epsilon_{ijk}J_{k}$
Relevant Equations:: $J_{i}$ are the three generators of the Lorentz rotations, and $K_{i}$ are the three generators of the Lorentz boosts (my textbook calls them pure transformations, but my understanding is that they're the same thing).
The $J$ matrices are Hermitian, while the $K$ matrices are symmetric.
Maybe it's also worth mentioning the commutator
$$ [J_i, K_j] = i\epsilon_{ijk}K_k$$

Hello everyone, I am new here, so please let me know if I am doing something wrong regarding the formatting or the way I am asking for help.

I did not really know how to start off, so first I tried to just write out all the $\mu \nu \rho \sigma$ combinations for which $\epsilon \neq 0$ and assign them signs. In essence, I just used Mathematica to find these. However, when I plugged them in, and replaced each corresponding entry of $M_{\mu \nu}$, I obtained the result $2(\mathbf{K} \cdot \mathbf{J} + \mathbf{J} \cdot \mathbf{K})$, which, unless I am missing something crucial, is just plain wrong.
Since each time I attempted this was taking up a large chunk of time, I thought of changing my approach, by fixing one index and seeing what comes out:
Say $\mu = 0$, then $\nu \rho \sigma = ijk$, and this leaves us with
$$ \frac{1}{2} \epsilon^{0ijk}M_{0i}M_{jk} = \frac{1}{2} \epsilon^{0ijk}K_{i}\epsilon_{jkl}J_{l} $$
This is where I am unable to move forward with certainty. I am thinking that $\epsilon^{0ijk} = \epsilon^{ijk}$ because if the first component is zero, then we need only consider permutations of the other three components. This in turn would allow me to easily contract the two $\epsilon$ terms into a simple Kronecker-$\delta$. If I do this, I end up with
$$ \frac{1}{2} \epsilon^{ijk}\epsilon_{jkl}K_{i}J_{l} = \frac{1}{2} \epsilon^{ijk}\epsilon_{ljk}K_{i}J_{l} = \frac{1}{2} 2\delta^{i}_{l} K_{i} J_{l} = K_{l}J_{l}$$
But this seems wrong too, I am missing a negative sign. And my indices seem to be in the wrong positions.
At this point I'm unsure of whether I am on the right track. I suppose I would have to try this for each index $= 0$, but even then I cannot see the terms combining to give me a negative dot-product.

I've not followed your calculation in detail, but if only the sign is different it may be simply a question of convention. The subtlety is that in SR
$$\epsilon_{\mu \nu \rho \sigma} =-\epsilon^{\mu \nu \rho \sigma},$$
and the absolute sign depends on which of the symbols, i.e., whether the covariant or contravariant tensor components of the Levi-Civita (pseudo) tensor you define in the usual way. In the HEP community usually one defined $\epsilon^{0123}=+1$ and then has consequently $\epsilon_{0123}$, but it depends on the textbook/paper which convention is followed. So it may well be that you use another convention than used in the problem.

 

[jambaugh]

I was using $[\mathbf{i}\times]$ where $\mathbf{i}$ is the unit vector in the x - direction. (Yes, that ole i).
Thus $[\mathbf{i}\times] \mathbf{j} = [\mathbf{k}]$ and $[\mathbf{i}\times]\mathbf{k}=-\mathbf{j}$. My apologies for not being more detailed but I was already long in the post. I also have gotten so used to this technique that I don't as easily appreciate the need to set the context more clearly.

I picked this up from a short book from Springer: A Brief on Tensor Analysis by James Simmonds.

Here's an example calculation which is handy in this notation:
The triple vector product (in 3 dim) identity is:
$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\bullet \mathbf{c})\mathbf{b} - (\mathbf{a}\bullet\mathbf{b})\mathbf{c}$$
Viewing it as the composition of two "cross product operators" $[\mathbf{a}\times]\circ[\mathbf{b}\times]$ acting on $\mathbf{c}$ you can re-write the r.h.s. thusly:
$$[\mathbf{a}\times][\mathbf{b}\times]\mathbf{c} = \left( \mathbf{b}[\mathbf{a}\bullet] - (\mathbf{a}\bullet\mathbf{b})\boldsymbol{1}\right)\mathbf{c}$$
(Here I also use $[\mathbf{a}\bullet]$ as the "row matrix" orthogonal transpose of $\mathbf{a}$ as a column matrix, i.e. the linear functional acting on a 3-vector by dotting it with $\mathbf{a}$.)

Now you can drop the third vector and write it as an operator identity:
$$[\mathbf{a}\times][\mathbf{b}\times] = \mathbf{b}[\mathbf{a}\bullet] -(\mathbf{a}\bullet \mathbf{b})\boldsymbol{1}$$
or in terms of matrix representation:
$$[\mathbf{a}\times][\mathbf{b}\times] = [\mathbf{b}][\mathbf{a}]^T - [\mathbf{a}]^T[\mathbf{b}]\mathbf{I}$$

Well anyway, I'm glad my exposition did lead you to a useful path even if I got overly exuberant about this notation. Do pick up the Simmonds book. It's not extensive (It sticks to tensors over Cartesian spaces) but it establishes a very practical foundation on working with tensors and this sort of handy notation. He, for example, works with dyad notation too. I found what I learned from this text very useful when parsing some of the less often cited vector derivative identities such as curl of curl, curl of cross products and such.

 

[jambaugh]

As to the sign convention on $\epsilon^{\alpha\beta\rho\sigma}$ you are correct. I did not incorporate the sign change convention of typical SR.

Lately in my own casual research I actually been using two sets of coordinates $x^\mu$ and $t_\mu$ and treating $g^{\mu\nu}$ as the components of a "speed of light" tensor. I then absorb the metric sign convention into distinct definitions of proper distance vs proper duration. I'll do something similar with the Levi-Civita tensor.