0
$\begingroup$

I am studyng the algebra of Poincaré group and the definition of one particle states using the Weinberg book "Quantum theory of Fields" (vol. 1), but I'm having a hard time understanding part of the book. It says that the square of 4-vector momentum $$p^{2} = \eta_{\mu \nu} p^{\mu} p^{\nu}$$ is invariant under the Lorentz transformation $\Lambda^{\mu}_{\ \ \nu}$ (it's ok, I understand this), and, for $p^{2} \leq 0$, the sign of $p^{0}$ is also invariant (this I didin't understand). And the book says that, for each value of $p^{2}$ and sign of $p^{0}$ (for $p^{2} \leq 0$) we can choose a "standard" 4-momentum $k^{\mu}$ such that $$p^{\mu} = L^{\mu}_{\ \ \nu}(p)k^{\nu},$$ where $L$ is some Lorantz transformation. Now I am really confused, I'm not understanding the idea. What would be exactly this $k$ and this $L$? And why can he express the momentum in this way?

$\endgroup$
1
1
$\begingroup$

For simplicity take 1 spatial dimension and 1 temporal dimension. As Weinberg does, we consider only proper orthochronous Lorentz transformation. All these are of the form $$ \begin{cases} p_0' = \gamma(p_0-vp_1) \\ p_1' = \gamma(p_1-vp_0) \end{cases} $$ I use the convention $p^2 = -p_0^2+p_1^2$. Also keep in mind that $|v|<1$.

For $p^2\leqslant0$ the sign of $p^0$ is invariant.

Since $p_0'=\gamma p_1(p_0/p_1-v)$ you see that if $|p_0|/|p_1|<1$ then $p_0'$ will change sign for sufficiently large values of $v$, while if $|p_0|/|p_1|\geqslant1$ it won't for any $v$. The condition $|p_0|/|p_1|<1$ is equivalent to $p^2>0$ and $|p_0|/|p_1|\leqslant1$ is equivalent to $p^2\leqslant0$, so the assertion of Weinberg follows.

For each value of $p^{2}$ and sign of $p^{0}$ (for $p^{2} \leq 0$) we can choose a "standard" 4-momentum $k^{\mu}$ such that $p^{\mu} = L^{\mu}_{\ \ \nu}(p)k^{\nu},$ where $L$ is some Lorentz transformation.

For what follows $m$ is just a fixed real number, with no physical meaning.

Any possible $p^\mu$ will be in one of these sets (I picture them as the various regions in which a Minkowski diagram is divided by the two diagonals):

  1. those with $p^2=-m^2 <0$ and $p^0>0$
  2. those with $p^2=-m^2 <0$ and $p^0<0$
  3. those with $p^2=m^2 >0$ and any $p^0$
  4. those with $p^2=0$ and $p^0>0$
  5. those with $p^2=0$ and $p^0<0$

Consider a $p^\mu$ in set 1 and the vector $k^\mu = (m,0)$. If you take a Lorentz transformation $L$ that has $\gamma = p_0/m$, you can see with some calculation that $k^\mu$ gets transformed into $p^\mu$. Therefore for all the $p^\mu$ in this set what Weinberg says is verified.

With a similar reasoning you can show that all other sets can be thought as "generated" by a particular $k^\mu$ (that will differ for each set) and various Lorentz transformations.

I hope this is clear. If not, I will be happy to provide further clarifications.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.