2
$\begingroup$

I'm taking a course of QFT and in the notes that the professor gave us he says that for a Lorentz transformation $\Lambda^\mu_\nu$ we have $|\Lambda^0_0| \ge 1$.

He doesn't give further information about this. I understand that he refers to the diagonalized (or Jordan's) form of the matrix and in this case I understand that $\Lambda^0_0 \neq 0$ so there are two different connected components for the values of $\Lambda^0_0 \in \mathbb{R}$ so we can't reverse time continuously.

But I don't understand why the modulus has to be greater than one. Why does it have to be the case?

$\endgroup$
5
$\begingroup$

It's because the invariant squared length of the 4-vector $a_\nu=\Lambda^0_\nu$ is equal to one, and the length for the Minkowski signature is given by $$ |a|^2 = (a_0)^2-(a_1)^2-(a_2)^2-(a_3)^2 = 1 $$ Because the last three, spatial terms $-(a_i)^2$ are negative (more precisely non-positive), the first term $(a_0)^2$ has to be greater than (or equal to) the right hand side $1$.

The fact that the 4-vector $a_\nu=\Lambda^0_\nu$ is a unit time-like vector in this sense is the Minkowskian analogy of the fact that the rows (or columns) in an orthogonal matrix are unit vectors (which are also orthogonal to other, parallel rows or columns).

If someone doesn't see why it is the case, he should return to the defining condition for an orthogonal matrix. It's one that preserves the metric tensor i.e. $\Lambda$ obeying $$ \Lambda^\mu{}_\nu g^{\nu\,\kappa} \Lambda^\lambda{}_\kappa = g^{\mu\lambda}$$ If we ignored the difference between upper and lower indices and just wrote all the components of the tensors above as matrices, the equation above would say $$ \Lambda\cdot g \cdot \Lambda^T = g $$ This is a generalization of $M\cdot M^T \equiv M\cdot {\bf 1}\cdot M^T = {\bf 1} $ for the orthogonal matrix in which $1$ is replaced by the other metric tensor we want to preserve, the matrix $g$. (Mathematicians sometimes call this Lorentzian modification of the orthogonal group as the "pseudoorthogonal group" or matrix etc.) The transposition in the latest displayed equation is needed to order the indices $\kappa,\lambda$ "properly" so that the nearby indices are contracted in the usual matrix products.

Now, return to the equation $$ \Lambda^\mu{}_\nu g^{\nu\,\kappa} \Lambda^\lambda{}_\kappa = g^{\mu\lambda}$$ and set $\mu=\lambda=0$. It reduces to $$ \Lambda^0{}_\nu g^{\nu\,\kappa} \Lambda^0{}_\kappa = g^{00}=1$$ But the left hand side is simply the squared length of the vector $a_\nu =\Lambda^0_\nu$ I started with, and I can continue with the argument that was written at the beginning.

The whole derivation may be written with the upper and lower indices completely interchanged everywhere; this other convention may be more comprehensible from some viewpoints, and this exchange corresponds to our talking about rows instead of columns and/or vice versa.

$\endgroup$

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.