Orthochronous subspace of Lorentz group.

1. Apr 14, 2014

LayMuon

In a Lorentz group we say there is a proper orthochronous subspace. How can I prove that the product of two orthchronous Lorentz matrices is orthochronous? Thanks. Would appreciate clear proofs.

2. Apr 14, 2014

robphy

Sounds like homework...

As a start, what is your definition of orthochronous?

3. Apr 14, 2014

LayMuon

$\Lambda^0_0 \geq 1$

4. Apr 14, 2014

samalkhaiat

Consider two such matrices, $A$ and $B$, call their product $C$. Now take the $(0,0)$ entries of the following matrix equations
$$A B = C , \ \ \ A \ \eta \ A^{ T } = B^{ T } \eta \ B = \eta .$$
Now, if you use the Schwarz inequality
$$- \sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ i }{}_{ 0 } )^{ 2 } } \leq A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 } ,$$
and
$$A^{ 0 }{}_{ 0 } \geq 1 , \ \ B^{ 0 }{}_{ 0 } \geq 1 ,$$
you should be able to show that $C^{ 0 }{}_{ 0 } \geq 1$

5. Apr 14, 2014

LayMuon

Thanks for reply. Why is there a minus sign in Schwartz inequality? Isn't it $\sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ i }{}_{ 0 } )^{ 2 } } \geq |A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 }|$?

$$C^0 {}_0 - A^0 {}_0 B^0 {}_0 = A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 }$$

How to proceed?

6. Apr 15, 2014

Fredrik

Staff Emeritus
I don't (yet) see what Sam is doing there. But what you wrote isn't the standard Cauchy-Schwarz inequality either. I think what you're looking for is
$$\left|\sum_i A^0{}_i B^i{}_0\right|\leq \sqrt{\sum_i(A^0{}_i)^2}\sqrt{\sum_j(B^j{}_0)^2}.$$ Do you know how to prove that if $\Lambda$ is orthochronous, its 00 component is either $\geq 1$ or $\leq -1$? If you have proved that, it will be sufficient to prove that the 00 component of the product transformation is $\geq 0$. I think that makes the problem much easier. One approach is to try to show that (in a notation with all indices of matrix components written downstairs) $(\bar\Lambda\Lambda)_{00}=\bar\Lambda_{00} \Lambda_{00} X,$ where X is a non-negative real number.

Last edited: Apr 15, 2014
7. Apr 15, 2014

LayMuon

So how does it work out?

8. Apr 15, 2014

Fredrik

Staff Emeritus
Have you tried this approach? (I have, so I know that it works). What do you get?

9. Apr 15, 2014

micromass

Staff Emeritus
Moved to homework since it is a homework type question.

Please provide an attempt at the solution.

10. Apr 15, 2014

LayMuon

$$|C^0 {}_0 - A^0 {}_0 B^0 {}_0 | \leq \sqrt{\sum_i(1-B^0{}_0{}^2)}\sqrt{\sum_j(A^0{}_i)^2}$$. I don't know how to express $$A^0{}_i^2$$ in terms of A00

Last edited: Apr 15, 2014
11. Apr 15, 2014

Fredrik

Staff Emeritus
I don't see what you're doing there.

I worked through the solution by my method again. I had to use Cauchy-Schwartz, the familiar version for vectors on $\mathbb R^3$, but not until the very end.

You didn't answer if you have already proved that the 00 component of a Lorentz transformation can't be in the interval (-1,1). You should start with that, because it makes your task easier. You will find that result if you look at the 00 component of the matrix equation $\Lambda^T\eta\Lambda=\eta$. (Use the definition of matrix multiplication). The next step is to use the definition of matrix multiplication on the 00 component of $\bar\Lambda\Lambda$. Feel free to use the notation AB instead of $\bar\Lambda\Lambda$ if you prefer.

12. Apr 15, 2014

LayMuon

Yes, that's clear.

13. Apr 15, 2014

Fredrik

Staff Emeritus
OK, I should also mention that I find it useful to know that if $\Lambda$ is the transformation from the coordinate system S to the coordinate system S', then (in my notation, where all matrix indices are downstairs), $\Lambda_{i0}/\Lambda_{00}$ is the velocity of S in S'. It's not absolutely necessary to know that, but it makes some steps easier to see.

14. Apr 15, 2014

samalkhaiat

Ok, doesn’t this imply that
$$- \sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ j }{}_{ 0 } )^{ 2 } } \leq A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 } .$$

Now, from the other two equations, you have (sums over indices are implies)
$$( A^{ 0 }{}_{ i } )^{ 2 } = ( A^{ 0 }{}_{ 0 } )^{ 2 } - 1 ,$$
$$( B^{ j }{}_{ 0 } )^{ 2 } = ( B^{ 0 }{}_{ 0 } )^{ 2 } - 1 .$$
Using these, the Schwarz inequality becomes equivalent to
$$0 \leq | A^{ 0 }{}_{ 0 } | \ | B^{ 0 }{}_{ 0 } | \left( 1 - \sqrt{ ( 1 - | A^{ 0 }{}_{ 0 } |^{ - 2 } ) ( 1 - | B^{ 0 }{}_{ 0 } |^{ - 2 } ) } \right) \leq C^{ 0 }{}_{ 0 } .$$
Now you can check that $C^{ 0 }{}_{ 0 } \geq 1$.

15. Apr 15, 2014

LayMuon

How did you get this: $( A^{ 0 }{}_{ i } )^{ 2 } = ( A^{ 0 }{}_{ 0 } )^{ 2 } - 1$? I understand it for B. how are $A^0{}_j$ and $A^j{}_0$ related?

16. Apr 15, 2014

samalkhaiat

Lower indices number the rows in the matrix. Choose any convention you like and stick with it.

17. Apr 15, 2014

LayMuon

I still don't understand. They are different parts of lorentz matrix. One can only use the definition of lorentz group, I.e. orthogonality.

18. Apr 15, 2014

samalkhaiat

I don't realy do homework. Any way, orthogonality is what I used in writing the matrix equations
$$\eta = A \eta A^{ T }, \ \ \eta = B^{ T } \eta B$$
Now, forget about upper and lower indices, Can you find the first matrix element, i.e. the $(0,0)$ entry in each of the above equations?

19. Apr 16, 2014

Fredrik

Staff Emeritus
OK, I see now that I got confused by the notation. The inequality that samalkhaiat wrote down is an almost immediate consequence of Cauchy-Schwartz for vectors in $\mathbb R^3$. If we use your inequality from the quote above, and then use that $|x|\geq -x$ for all real numbers x,...

Regarding notational conventions: Some people would write $x_i x_i$ rather than $(x_i)^2$ to ensure that the index to be summed over explicitly appears twice, but I suppose that can be weird too when we use a convention that typically has one of the indices upstairs and the other downstairs. I would denote the component on row $\mu$, column $\nu$ of an arbitrary matrix X by $X_{\mu\nu}$, $X^\mu_\nu$ or $X^\mu{}_\nu$. Samalkhaiat apparently uses $X^\nu{}_\mu$. That's fine too. As he said, "choose any convention you like and stick with it".