So my question is, when we write the Lorentz transformation in the following form

\begin{equation} \Lambda = e^{- \frac{\mathrm{i}}{2} \omega ^{\rho \sigma} M_{\rho \sigma}} \end{equation}

Since the argument in the exponent is a scalar, do we get $\Lambda$ as a scalar? If not, it should be a tensor, but in what configuration of its indices? Is it $\Lambda ^{\mu \nu}$ or $\Lambda ^{\mu}_{~~\nu}$? And if we were to convert, say, $\Lambda ^{\mu \rho}$ to $\Lambda ^{\mu}_{~~\nu}$, do we multiply $g_{\rho \nu}$ to all terms in the expansion of the exponential or do we keep the first term in the expansion, which is the identity, invariant? Why?

Short, easy-to-reference answer

You need to understand that the expression $M_{\rho\sigma}$ in $$ \begin{equation} \Lambda = e^{- \frac{\mathrm{i}}{2} \omega ^{\rho \sigma} M_{\rho \sigma}} \end{equation} $$ is but a shorthand for $(M_{\rho\sigma})^\mu_{\ \nu}$. Apparently, you use a convention that lowers the $\mu$ index, which makes it more confusing. The expression $x_{\rho} p_{\sigma} - x_{\sigma} p_{\rho}$ should stand for $x_{\rho} p_{\sigma} \hat \theta^{(\rho)} \otimes\hat \theta^{(\sigma)} - x_{\sigma} p_{\rho}\hat\theta^{(\sigma)} \otimes\hat \theta^{(\rho)}$, where $\hat \theta^{(\sigma)}$ are the basis covectors.

Long, easy-to-understand answer

We shall formulate this in terms of generalization of complex numbers. If we define $$ i\equiv \left(\begin{matrix}0&1\\-1&0\end{matrix}\right), $$ then $i^2=-1$ and all the complex number rules apply. Suppose we want to define a 2D-rotation, according to Euler's identity, $$ R=e^{i\theta} $$ where $R$ is the rotation matrix.

Similarly, we can try to define a matrix $i_{\mu\nu}$ for each rotation plane in four dimensions (rotations in 4 dimensions has a plane as an 'axis'), then when we are considering rotations purely around one plane, we can just apply the formula for $R$ above.

However, when we try to rotate along a 'slant' plane (with direction described as $\omega^{\rho\sigma}$, because a plane in four dimensions has six antisymmetric components), the problem arises: since rotations does not commute, we cannot simply decompose the tensor $\omega^{\rho\sigma}$, and rotate them separately. Instead, we consider rotating an infinitesimal amount along each component each time, and 'mixing' the rotations. This is exactly what $\exp()$ does. We can express 4-dimensional Euclidean rotation as $$ {R^\mu}_\nu=e^{\omega^{\rho\sigma}i_{\rho\sigma}}. $$ Remember that each $i_{\rho\sigma}$ is a matrix! After contracting, we are still left with a matrix. In fact, this is called the exponential map, which maps the $\omega$'s to the $R$'s. We can modify that to get the Lorentzian version, which is up to you and your book.

  • Can I understand $\hat{\theta}^{\left( \sigma \right)} \otimes \hat{\theta}^{\left( \rho \right)}$ as the metric $g_{\mu \nu}$? – zyy Sep 14 at 15:34
  • If I’m remembering correctly, it’s actually $(M_{\rho\sigma})^\mu_\nu$.. I understand that it’s just convention, but the standard is to put the indices that label the matrix on the inside of the parentheses – InertialObserver Sep 14 at 15:53
  • why do you need basis covectors $\theta$? – OkThen Sep 15 at 3:18
  • @zyy Not really, because they are covectors not components, whereas $g_{\mu\nu}$ is actually 16 components of a tensor. – Trebor Sep 15 at 11:40
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    The form $\hat \theta^{(\sigma)}$ is the most commonly used form. They satisfy $\hat \theta^{(\sigma)}\hat e_{(\mu)}=\delta^\sigma_\mu$. – Trebor Sep 15 at 15:45

The argument in the exponent is not a scalar, $M_{\rho\sigma}$ is not a component of a matrix/tensor, but a full matrix/tensor, namely the generator of the Lorentz algebra labeled by $\rho\sigma$.

$\Lambda$ is the same type of tensor as $M_{\rho\sigma}$ is, i.e. whatever index position for all the $M_{\rho\sigma}$ you choose, $\Lambda$ inherits.

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    Is $M_{\rho \sigma}$ not $x_{\rho} p_{\sigma} - x_{\sigma} p_{\rho}$? – zyy Sep 14 at 13:41

As Trebor correctly points out, $M_{\rho\sigma}$ is indeed an abbreviated notation for something more than a $4\times4$ collection of scalars. But I think for the purposes of general physics education, it would be helpful if you really understand what they are, which the other answers don't get into very much.

You've probably heard of Lie groups, and you may know that the Lorentz group is one particular Lie group. Specifically, your $\Lambda$ is an element of the Lie group. Every Lie group also has an associated Lie algebra, which is basically the differential behavior of the group near the identity element (and is closely related to the differential behavior near any element). Like any algebra, the Lie algebra is also a vector space. In this case, the quantities represented by $M_{\rho\sigma}$ are basis elements of that vector space, so the sum $\omega^{\rho\sigma} M_{\rho\sigma}$ is a general element of the Lie algebra. One important way to remember the difference is that in the group you should only multiply elements together (sometimes also called "composing" transformations), whereas in the algebra you can multiply (possibly via the Lie bracket; possibly in other ways) as well as add elements together.

But while the group and the algebra are different, they are also related quite closely. A fundamental part of Lie theory is the exponential map, which maps elements of the Lie algebra into the Lie group. This is really quite general for all Lie algebras, but you've probably seen this before in the special case of rotations, where the group is $SO(3)$ and the algebra is $\mathfrak{so}(3)$. The algebra has a basis $J_i$ (the "generators of rotation") and a general rotation is given as something like $e^{- \mathrm{i}\, \theta^k J_k}$ — possibly with an $\hbar$ thrown in for some reason in quantum mechanics. Of course, $J_k$ is just the "dual" to $J_{ij}$, which actually spans a subspace of your $M_{\rho\sigma}$ in some sense. And indeed, you will frequently see expressions like $J_{ij} = x_i p_j - x_j p_i$, especially in the quantum context, in which case you need to consider the $p_j$ to be derivatives with respect to $x_j$ — which then extends naturally to $M_{\rho\sigma}$ as you suggested in the comment to ACuriousMind's answer. But the point there is that you can't consider $p_j$ to be a scalar; it's really an operator which happens to have a nice representation in matrix form.

So what do you actually do with an expression like $e^{-\frac{\mathrm{i}}{2}\omega^{\rho\sigma} M_{\rho\sigma}}$? Well, first of all, you need to realize that $M_{\rho\sigma}$ for any particular values of $\rho$ and $\sigma$ is an element of the algebra, mapping to an element of the group. So if you want your group to be a linear transformation taking vectors to vectors, you might want to write your group element as ${\Lambda^\alpha}_\beta$ — which means that you might want to write your algebra basis elements as ${\left(M_{\rho\sigma}\right)^\alpha}_\beta$. Second, you probably want to actually evaluate that exponential. The standard way to define it is to just use the standard power series: \begin{equation} e^{-\frac{\mathrm{i}}{2}\omega^{\rho\sigma} M_{\rho\sigma}} = 1 + \left(-\frac{\mathrm{i}}{2}\omega^{\rho\sigma} M_{\rho\sigma}\right) +\frac{1}{2} \left(-\frac{\mathrm{i}}{2}\omega^{\rho\sigma} M_{\rho\sigma} \right)^2 + \ldots \end{equation} The $1$ in the first term really represents the identity matrix, and the terms in parentheses evaluate to matrices, so when you see one of those squared for example, you just multiply the matrix by itself as usual. A particular case of interest is $\omega^{\rho\sigma} = 0$, in which case we have $e^{-\frac{\mathrm{i}}{2}\omega^{\rho\sigma} M_{\rho\sigma}} = 1$, which is the identity element of the group (so you might be more comfortable writing it as an identity matrix).

This is really a part of Lie theory.

All Lie groups determine a Lie algebra, for example, the 3d rotation group $SO(3)$ determines the Lie algebra, $so(3)$.

We can reverse this procedure and determine a Lie group given a Lie algebra and this is what exponentiation here is doing. The argument of the exponential is an element of the Lie algebra and it's value is an element here of the Lie group.

In fact, the relationship between Lie groups and Lie algebras isn't one-to-one - unfortunately. But it can be made to be so by using local Lie groups - this is the germ of the Lie group at the the identity. Then we get a functorial association between local Lie groups and Lie algebra, that is a functor $LocLieGrp \rightarrow LieAlg$.

In your specific example, the argument is an element of the Lie algebra of the Lorentz group, $Ltz =~ SO(1,3)$ and so is $ltz =~ so(1,3)$ and exponentiating gives back an element of the Lorentz group itself.

Generally, in physics, when we do calculations we don't work with the abstract groups themselves but with a representation. This just means we work with matrices. For example, the Pauli matrices span the Lie algebra of the Lorentz group in its fundamental representation - ie it's smallest matrix representation.

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