Generators of Lorentz transformations

In Chapter 3 of Peskin and Schroeder's Introduction to Quantum Field Theory they write

For the rotation group, one can work out the commutation relations by writing the generators as differential operators; from the expression $$J=x\times p = x \times (-i\nabla),$$ the angular momentum commutation relations follow straightforwardly. The use of the cross product is special to the case of three dimensions. However, we can also write the operators as an antisymmetric tensor, $$J^{ij} = -i(x^{i}\nabla^j - x^{j}\nabla^i),$$ so that $$J^3 = J^{12}$$ and so on. The generalization to four-dimensional Lorentz transformations is now quite natural: $$J^{\mu \nu} = -i(x^{\mu}\partial^\nu - x^{\nu}\partial^\mu).$$

It is not clear to my why the generalization should take this form. If we wanted to represent rotations/angular momentum in 4 dimensional space, i.e. $$SO(4)$$, then this generalization would seem appropriate. But boosts are similar to but not identical with a 4d rotation. Why does the generalization seem to look like space and time are treated exactly equally?

• Generators are basically elements of a Lie algebra. The Lorentz group can be differentiated at the identity to yield it's Lie algebra, the Lorentz algebra. The elements of this algebra is traditionally known as generators in the physics literature. Oct 16, 2021 at 4:17
• @MoziburUllah Yes, my question is more that it seems to me like the Lorentz algebra doesn't seem like it should take this form. Oct 16, 2021 at 4:19
• Well, there are lots of different representations of Lie groups and Lie algebras. It all depends how you decide to build your representing vector space. Boosts, by the way are 4d rotations in Lorentz signatures. Oct 16, 2021 at 6:03
• Have you tried checking the commutation relations and just seeing whether or not it is, in fact, the Lorentz algebra? Oct 16, 2021 at 9:43

The generalization to four-dimensional Lorentz transformations is now quite natural: $$J^{\mu \nu} = -i(x^{\mu}\partial^\nu - x^{\nu}\partial^\mu).$$

You need to be very careful about upper and lower indices here.
Spacetime position is defined as the 4-vector $$x^\mu=(ct,x,y,z). \tag{1}$$ These are contravariant components, i.e. with an upper index.

Likewise the gradient is defined as the 4-vector $$\partial_\mu=\frac{\partial}{\partial x^\mu}= \left(\frac{\partial}{c\ \partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \tag{2}$$ Notice, these are its covariant components (with lower index).

But the contravariant components (with upper index) of the gradient are slightly different: $$\partial^\mu== \left(\frac{\partial}{c\ \partial t}, -\frac{\partial}{\partial x}, -\frac{\partial}{\partial y}, -\frac{\partial}{\partial z}\right) \tag{3}$$

This follows from raising and lowering indices $$A^\mu=\eta^{\mu\nu}A_\nu$$ with the Minkowski metric. I'm using the $$(+,-,-,-)$$ metric sign convention here.

If we wanted to represent rotations/angular momentum in 4 dimensional space, i.e. $$SO(4)$$, then this generalization would seem appropriate.

You are right. The Lorentz group is not $$SO(4)$$, but it is $$SO(1,3)$$.

But boosts are similar to but not identical with a 4d rotation. Why does the generalization seem to look like space and time are treated exactly equally?

Take for example the $$^{01}$$ component of Peskin's formula $$J^{\mu\nu} = -i(x^{\mu}\partial^\nu - x^{\nu}\partial^\mu).$$ Using (1) and (3) we get the boost in $$x$$ direction $$J^{01} = -i(x^0\partial^1 - x^1\partial^0) =-i\left(-ct\frac{\partial}{\partial x} - x\frac{\partial}{c\ \partial t}\right).$$

Notice that both terms got the same sign, as it should be for a Lorentz boost.
This is slightly different from a Lorentz rotation, for example the rotation around the $$z$$ axis $$J^{12} = -i(x^1\partial^2 - x^2\partial^1) =-i\left(-x\frac{\partial}{\partial y} + y\frac{\partial}{\partial x}\right)$$ where both terms have opposite signs.

So when only looking at the formula for $$J^{\mu\nu}$$ it may seem, time and space dimensions are treated exactly equally. But actually there is a subtle difference between time and space dimensions because of the minus signs in formula (3) for $$\partial^\mu$$, which in turn come from the the minus signs in the Minkowski metric $$\eta^{\mu\nu}$$.

For $$n\in\Bbb N$$ and $$g\in\Bbb R^{n\times n}$$, one can ask which $$\Lambda\in\Bbb R^{n\times n}$$ satisfy $$\Lambda^Tg\Lambda=g$$. If $$g=I_n$$, the $$\Lambda$$ are rotations; if $$g=\eta$$, they are Lorentz transformations. So from a spacetime-is-one-thing perspective, Lorentz transformations are just rotations. Regardless of $$g$$, an infinitesimal transformation is $$\Lambda=I_n+\epsilon$$, so at first-order$$0=(I_n+\epsilon^T)g(I_n+\epsilon)-g=\epsilon^Tg+g\epsilon\implies\epsilon^T=-g\epsilon g^{-1}.$$But that's with matrices. Since multiplying by $$g^{\pm1}$$ means nothing to a tensor, we're basically interested in which $$\epsilon$$ are antisymmetric; all $$g$$ changes is the meaning of lowering/raising indices. Since we need to make a rank-$$2$$ dimensionless tensor from vectors such as $$x^\mu,\,\partial^\mu$$, dimensional analysis tells us the infinitesimal transformations must have generators $$\propto i(x^\nu\partial^\mu-x^\mu\partial^\nu)$$, where the $$i$$ addresses Hermiticity concerns. This argument determines the $$J^{\mu\nu}$$ up to a $$^{\mu\nu}$$- and $$g$$-independent sign, which Peskin and Schroeder fixed from the case $$g=I_n$$. Therefore, we needn't worry we might be missing a $$-$$ sign hiding in $$\propto$$. (Actually, the same easy-case-first logic also tells us we're not missing any other proportionality constant.)

• That's more clearer. Let me ask you additional follow question. Does your reasoning (viz-a-viz dimensional analysis), and your conclusion i.e. terms of $\partial \,x$ apply equally well to generators of translation and rotation? Further, while we are discussing this: Does it apply to any other generators of Lorentz transformation; and if so, which? and if not, which not? Jan 15 at 23:58
• @MichaelLevy Generators made from coordinates and derivatives are determined, to within scaling factors you can deduce from specific examples, by dimensional analysis and (anti)symmetry principles. This is true whether they're to do with translations, boosts or rotations, whether they're to do with Galilean or Lorentz transformations etc. If you read up on it, many sources focus on commutation relations rather than differential operator expressions for generators. But you can obtain such expressions from those commutators anyway.
– J.G.
Jan 16 at 8:41