How do angular displacement and rapidity relate in special relativity?

[Logic314]

I have previously studied special relativity, but only at an introductory level. So I decided to explore the subject more in detail later by thinking and working things out on my own, in addition to doing research online. In particular, I seem to have noticed some intriguing patterns between quantities in special relativity, and based on these, have some conjectures to the mathematical relationships governing these quantities:

I know that a general infinitesimal Lorentz transformation has two components, a rotation component which rotates the inertial coordinate system by an infinitesimal angular displacement, and a boost component which boosts it by an infinitesimal relative velocity (or rapidity).

Also, I have noticed that rapidity plays a very similar role (in the context of Lorentz transformations) to angular position. When purely spatial Lorentz transformations are expressed in terms of angle, the trigonometric cosine and sine are used, whereas when space-time Lorentz boosts are expressed in terms of rapidity, the hyperbolic cosine and sine are used.

Moreover, I have noticed that in Noether's theorem, infinitesimal rotations are generated by the angular momentum pseudovector L = x × p, while infinitesimal Lorentz boosts are generated by the polar vector N = tp - Ex (where E is the energy).

I know that for any two four vectors (a_t,a_x) and (b_t,b_x), the pseudovector a_x×b_x and the vector a_tb_x-b_ta_x form an antisymmetric relativistic tensor.

Thus, the quantities L and N form an antisymmetric four-tensor. Similarly, the magnetic and electric fields B and E form an antisymmetric four-tensor.

Now, I know the infinitesimal angular displacement dθ (i.e. rotation) of an inertial frame is a pseudovector that can be expressed as ½(∇×δx) (where δx is the vector field that describes the infinitesimal linear displacement at each space-time point attached to the reference frame during an infinitesimal rotation), while the infinitesimal change in rapidity dφ describing an infinitesimal Lorentz boost is a polar vector.

Since the gradient operator ∇ acts as the spatial component of a four-vector (-∂/∂t,∇) and δx is of course the spatial component of the four-vector (δt,δx), where δt is the scalar field representing the infinitesimal change in local time at each point attached to the reference frame during an infinitesimal Lorentz boost. (Note: Because of the relativity of simultaneity, I believe δt is not the same throughout the frame: it varies with space, and thus has a nonzero gradient ∇(δt))

Thus, the bottom line of all of this is that the angular displacement dθ together with the quantity -½ ( ∂/∂t(δx) + ∇(δt) ) should form an antisymmetric four-tensor.

Based on the symmetry between angular displacement and rapidity, and between angular momentum L and rotations on the one hand, and the quantity N and Lorentz boosts on the other hand, I have reason to suspect that the quantity -½( ∂/∂t(δx) + ∇(δt) ) associated with a Lorentz boost is in fact equal to the infinitesimal change in rapidity dφ: In other words, my conjecture is that the quantities dθ and dφ describing an infinitesimal Lorentz transformation form an antisymmetric relativistic tensor, just like L and N, or like B and E.

This would certainly be a nice symmetry between the quantities. I am not sure whether all this reasoning is correct or not. I have tried to research the answer to this online, but nowhere I have seen is this relationship between angular displacement, rapidity, and angular momentum explored to this extent. So I wanted to ask if my reasoning (and my conjecture) is correct or not.

 

[vanhees71]

You get this much easier by using the covariant notation.

A Lorentz-transformation matrix must obey
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma},$$
where $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$.
For an infinitesimal transformation you have
$${\Lambda^{\mu}}_{\rho} = \delta_{\rho}^{\mu} + \delta {\Omega^{\mu}}_{\rho}.$$
Plugging this into the formula you get
$$\eta_{\mu \nu} (\delta_{\rho}^{\mu} + \delta {\Omega^{\mu}}_{\rho})(\delta_{\sigma}^{\nu} + \delta {\Omega^{\nu}}_{\sigma}) = \eta_{\rho \sigma} + \delta \Omega_{\rho \sigma} + \delta \Omega_{\sigma \rho} +\mathcal{O}(\delta^2)=\eta_{\rho \sigma},$$
i.e.,
$$\delta \Omega_{\rho \sigma}=-\delta \Omega_{\sigma \rho},$$
i.e., $\delta \Omega_{\rho \sigma}$ is antisymmetric under exchange of the indices.

The 3 generators of the boosts are given by the $\delta \Omega_{0k}=-\delta \Omega_{k0}$ with $k \in \{0,1,2 \}$ and the 3 generators of rotations by $\delta \Omega_{jk} = \delta \omega_l \epsilon_{jkl}$ ($j,k,l \in \{1,2,3 \}$).
For a more detailed discussion, see Sect. 1.7 in
https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Logic314]

I see, so the infinitesimal angular displacement dθ and the infinitesimal change in rapidity dφ associated with an infinitesimal Lorentz transformation do indeed form an antisymmetric tensor (of the same type as the angular momentum tensor). The derivation does indeed seem quite simple in covariant notation! But I'm still wondering whether the rest of the things I said above (e.g. concerning vector fields) is correct (even though it may have been needlessly complicated for the purpose of deriving the final result relating angular displacement and rapidity)? For instance, am I correct to claim that for an infinitesimal Lorentz boost the polar vector field -1/2 (d/dt(δx) + ∇(δt)) equals dφ at all points in space-time, just as for an infinitesimal rotation the infinitesimal pseudovector field 1/2 ∇×δx equals dθ at all points in space (where δx and δt are the infinitesimal spatial and temporal displacement vector and scalar fields associated with an infinitesimal Lorentz transformation of a space-time coordinate system)? I am pretty sure this is correct (especially since you have given me the thumbs-up that dθ and dφ do indeed go together as an antisymmetric four-tensor in the expected way), but I just wanted to make sure!

 

[vanhees71]

That sounds all correct. You can also check this by looking at how fields transform under Lorentz transformations. As the most simple example take a scalar field, which transforms according to
$$\Phi'(x')=\Phi(x)=\Phi(\hat{\Lambda}^{-1} x').$$
For an infinitesimal transformation,
$$x'=x+\delta \hat{\Omega} x, \quad x=x'-\delta \hat{\Omega} x' + \mathcal{O}(\delta^2)$$
you get
$$\Phi'(x')=\Phi(x')-\delta {\Omega^{\mu}}_{\nu} x^{\prime \nu} \partial_{\mu} \Phi(x').$$