Lorentz algebra elements in an operator representation

In summary, the conversation discussed the confusion regarding Einstein summation and the notation used for Lorentz transformations. The conversation also touched upon the representation of the Lorentz group and how it acts on different spaces of functions/fields. Finally, the conversation addressed the derivation of the Lorentz generators and their role in transforming fields under the action of the Lorentz group.
  • #1
Jason Bennett
49
3
Homework Statement
see below
Relevant Equations
see below
1) Likely an Einstein summation confusion.
Consider Lorentz transformation's defined in the following matter:

Please see image [2] below.

I aim to consider the product [tex]L^0{}_0(\Lambda_1\Lambda_2).[/tex] Consider the following notation [tex]L^\mu{}_\nu(\Lambda_i) = L_i{}^\mu{}_\nu.[/tex] How then, does [tex]L^0{}_0(\Lambda_1\Lambda_2) = L_1{}^0{}_\mu L_2{}^\mu{}_0?[/tex]

2) To right the J and K generators of the Lorentz group in a compact way, one can write [tex](M^{lm})^j{}_k=i (g^
{lj}g^m{}_k - g^{mj}g^l{}_k)[/tex] where on the left hand side, it is helpful to think of l and m as indices/labels, and the j and k as rows/columns for the whole matrix.
(Apparently) One can write this in an operator representation as [tex]M^{\mu\nu} = i(x^\mu \partial^\nu-x^\nu\partial^\mu).[/tex]

a) where does this come from?

b) why and how is it used? What is it operating on, [tex]x^\mu?[/tex]

c) how does [tex]\partial^\nu x^\sigma= \frac{\partial}{\partial x_\nu} x^\sigma[/tex] equal [tex]g^{\nu\sigma}?[/tex]

[2]: https://i.stack.imgur.com/uPsLc.png
 
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  • #2
Regarding your first question, calculating the element of product Lorentz transformation is given by the formula you assumed. It follows from the Lorentz group multiplication rule ##L(\Lambda_1)L(\Lambda_2) = L(\Lambda_1\Lambda_2)##, so you have to sum over ##\mu## index in order to get the component you're looking for.

As for the angular momentum generator, generators generate Lorentz tranformations on spaces of functions/fields. We know how Lorentz group acts on coordinates. It's a simple coordinate transformation:
$$x'^\mu = \Lambda^\mu_{\hphantom{\mu}\nu}x^\nu$$
Where ##\Lambda## is defined as matrix that preserves the Minkowski metric, as you've defined it in the picture you've attached. We now ask ourselves how this transformation is going to act on functions of these coordinate. Let's take the example of scalar field, since it is the simplest one(it has zero spin, so there is no additional spin transformation). In order to see how these transformations act on fields, we need to create a mapping that maps our group elements(Lorentz tranformations) into some operators on the space of functions/fields we want to act on. This mapping is called the representation of the group. Let's take scalar field ##\phi(x)##. Acting on it with Lorentz tranformation, we get the transformed field ##\phi'(x')##, where prime above field indicates the change of form of the function, and prime above x is just the change of coordinates. We wish to investigate how this field differs from the field we started with when the tranformation is infinitesimal. So we define total variation of the field
$$\delta\phi = \phi'(x') - \phi(x) = \phi'(x') - \phi'(x) + \phi'(x) - \phi(x) = \phi'(x') - \phi'(x) + \delta_0\phi(x)$$
We call ##\delta_0\phi(x)## the form variation of the field, that is, the change of function irrespectible of coordinate change. To further clarify the expression, we can use Taylor series to simplify the expression:
$$\delta\phi = \frac{\partial \phi}{\partial x^\mu}(x'^\mu -x^\mu) + \delta_0\phi(x)$$
However, if we look now at the definition of a scalar, scalar is a function that stays invariant under changes of coordinates, that is for scalar fields we have ##\phi'(x') = \phi(x)##. So the variation of a scalar field is zero by definition, but that doesn't mean the form variation is zero, too. So, we will now investigate the form variation. We define the difference between old and new coordinates by ##\xi^\mu = x'^\mu - x^\mu##. We get:
$$\delta_0\phi(x) = -\xi^\mu\partial_\mu\phi(x)$$
Now if we look what infinitesimal transformation does to coordinates, we get:
$$x'^\mu = x'^\mu(x) = x^\mu + \frac{\partial x'^\mu}{\partial x^\nu}x^\nu = x^\mu + \omega^\mu_{\hphantom{\mu}\nu}x^\nu$$
Where ##\omega^\mu_{\hphantom{\mu}\nu}## is infinitesimal matrix of Lorentz transformation(You get from the law of how coordinates transform that the derivative of new coordinates with respect to old ones equals the Lorentz matrix). Now we rewrite the form variation.
$$\delta_0\phi(x) = -\omega^\mu_{\hphantom{\mu}\nu}x^\nu\partial_\mu\phi(x) = -\frac{1}{2}\omega^{\mu\nu}M_{\mu\nu}\phi(x)$$
We have rewritten it in the form where ##\omega^{\mu\nu}## is an antisymmetric matrix of parameters and ##M_{\mu\nu}## are some operators which now act on ##\phi(x)##. These operators are what we call the generators of Lorentz group. Direct comparison gives:
$$M_{\mu\nu} = x_\mu\partial_\nu - x_\nu\partial_\mu$$
The factor of ##i## that appears in your formula is there for Hermiticity reasons and is not the important part, it depends on how you define the representation of the group. So we have here derived the Lorentz generators as operators which change the form of scalar field under the action of Lorentz group. It is simple since scalar field doesn't change under those transformation so total variation is zero. If we do the same with vector and tensor fields, we will see that they transform as some matrix, which is function of these generators, that is, the representation of the group is different. But in general generators generate infinitesimal transformations on spaces of fields.

As for your last question, it is trivial, since you define ##\partial^\mu = g^{\mu\nu}\partial_{\nu}## Acting with that on coordinate ##{x^\mu}## and using ##\tfrac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu## gives the answer to your question.

I hope I didn't expand on this too far, I tried to be comprehensive so that you understand the point of generators. If you have more questions, feel free to ask.
 

Related to Lorentz algebra elements in an operator representation

1. What is the Lorentz algebra?

The Lorentz algebra is a mathematical structure that describes the symmetries of space and time in special relativity. It is a Lie algebra, meaning it consists of a set of elements (operators) that satisfy certain algebraic properties.

2. What are Lorentz algebra elements?

Lorentz algebra elements are operators that represent the generators of the Lorentz group, which is a set of transformations that preserve the laws of physics in special relativity. These elements can be represented as matrices or other mathematical objects.

3. How are Lorentz algebra elements used in physics?

Lorentz algebra elements are used to describe the behavior of particles and fields in special relativity. They are essential in understanding the symmetries and transformations that occur in physical systems, and are crucial in formulating theories such as the Standard Model of particle physics.

4. What is an operator representation of the Lorentz algebra?

An operator representation of the Lorentz algebra is a way of expressing the elements of the algebra as mathematical operators acting on a vector space. This allows for calculations and manipulations to be performed using these operators, making it a useful tool in theoretical physics.

5. How are Lorentz algebra elements related to other mathematical structures?

The Lorentz algebra is a subalgebra of the Poincaré algebra, which describes the symmetries of spacetime in both special and general relativity. It is also related to other mathematical structures such as Lie groups and Lie algebras, which have wide applications in physics and mathematics.

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