# Do Galilean boosts and Lorentz boosts share the same generators?

Gottfried and Yan's Quantum Mechanics introduces a generator $N$, called the boost, which generates Galileo transformations. I think in other terminology one might say $N$ generates Galilean boosts, but I think that is just terminology. It is defined as $$\mathbf{N}=\mathbf{P}t-M\mathbf{X}.$$ This matches the definition on Wikipedia of the dynamic mass moment which is part of the same bivector as Angular momentum. That makes me think that mass moment should serve as the generator of Lorentz boosts. Indeed, this Phys.SE question and this Wikipedia page imply to me that it does, as long as $N$ is the same as $K$. So how can these two transformations have the same generator?

Galilean symmetry is a symmetry of non-relativistic physics. In reality the symmetry of physics is Lorentz invariance and non-relativistic physics arises as approximation in the limit $c\rightarrow\infty$

Acting on functions $f(x^\mu)$ the generator of Lorentz boost can be written as, \begin{equation} K_i\equiv M_{0i}=x_0 \partial_i-x_i \partial_0=ct\partial_i+\frac{x^i}{c}\partial_t \end{equation} and gives Lorentz boost by exponentiation $\Lambda=e^{K_i w}$ with argument $w=\operatorname{arctanh}\,{v/c}$ known as rapidity.

In the limit $c\rightarrow\infty$ we get, \begin{equation} w\rightarrow \frac{v}{c},\quad K_i w\rightarrow K_i^{G} v,\quad K_i^G=t\partial_i \end{equation} \begin{equation} e^{K_i^{G}v} \begin{pmatrix}t\\\vec{x}\end{pmatrix}=\begin{pmatrix}t\\x+vt\end{pmatrix} \end{equation} When we consider transformations of canonical coordinate and momenta for particle it works somewhat differently - we can rewrite in that case Lorentz generator in terms of canonical variables for relativistic particle as \begin{equation} \tilde{K}_i=-ct p^i+\frac{x^i}{c}H \end{equation} So that canonical functions transform as $f(x,p)\mapsto f(x,p)+w\{f(x,p),K_i\}+\ldots$ Note that it's not just mass but energy that is involved in terms with $x^i$.

However for particle $H=c\sqrt{m^2c^2+\vec{p}^2}\sim mc^2+\frac{p^2}{2m}+O(1/c^2)$

Therefore we get, \begin{equation} \tilde{K}_i w\sim v(-t p^i+mx^i)\equiv \tilde{K}_i^G v \end{equation} so that on non-relativistic canonical coordinates and momenta it acts as \begin{equation} \begin{pmatrix}\vec{x}\\\vec{p}\end{pmatrix}\mapsto\begin{pmatrix}x+vt\\ \vec{p}+mv\end{pmatrix} \end{equation}

So no, generators are not the same but they are connected through non-relativistic limit.

• Under Lorentz transformations, L transforms with the Galilean generator, right? (from the wikipedia article). I would expect it to be relativistically connected to relativistic boost generator. – NoethersOneRing Nov 19 '16 at 19:12
• Does the generator of rotations have a different form in relativistic physics? – NoethersOneRing Nov 19 '16 at 19:13
• @chuck.steel What's $L$? $\Lambda$? I'm not sure what you are confused about in wiki article. Rotations are same in Lorentz and Galilei – OON Nov 19 '16 at 20:53
• By L I just meant angular momentum. The wikipedia article says that the relativistic version of angular momentum is a tensor whose components are L (angular momentum) and N (dynamic mass moment). N is what you call K^G. I was surprised that the generator of rotations is relativistically connected to the generator of Galilean rather than Lorentz boosts. Does my question make any more sense? – NoethersOneRing Nov 20 '16 at 15:11
• Oh, I see now. The thing is that there's this notion of "relativistic mass", basically we can define it as $E/c$ (or $H/c$ if we treat energy as Hamiltonian) Then you can write Lorentz boost generator in a way similar to Galilean one. However I look at this as another way to confuse everyone and your question proves it. – OON Nov 20 '16 at 15:19