# Orthochronal subgroup of the Lorentz group

1. Feb 18, 2009

### martin_blckrs

This is probably very trivial, but I can't find an argument, why the orthochronal transformations (i.e. those for which $\Lambda^0{}_0 \geq 1$) form a subgroup of the Lorentz group, i.e. why the product of two orthochronal transformations is again orthochronal?

Since when you multiply two orthochronal matrices $\Lambda$ and $\overline{\Lambda}$ the term $$\Lambda\overline{\Lambda})^0{}_0 = \Lambda^0{}_0\overline{\Lambda^0{}_0} + \Lambda^0{}_1\bar{\Lambda^1{}_0} + \Lambda^0{}_1\overline{\Lambda^2{}_0} + \Lambda^0{}_3\overline{\Lambda^3{}_0},$$ where I can only say that $\Lambda^0{}_0\overline{\Lambda^0{}_0} \geq 1$ but I don't really know much about the other terms.

Last edited by a moderator: Jan 9, 2013
2. Feb 18, 2009

### sirchasm

What generates the initial terms in the LHS?

3. Feb 18, 2009

### martin_blckrs

What do you mean?

4. Feb 18, 2009

### sirchasm

You say: "when you multiply two orthochronal matrices", where are they multiplied? How do they arrive on the scene as it were?

5. Feb 18, 2009

### martin_blckrs

$(\Lambda\bar{\Lambda})^0{}_0$ denotes the 00 element of the product of $\Lambda$ and $\bar{\Lambda}$

Last edited by a moderator: Jan 9, 2013
6. Feb 18, 2009

### sirchasm

Yes, but where do you derive the need to find the product, so you have the 00 element?

Why do you start with a multiplication, and what do you multiply, if that's any clearer?
Obviously you're finding a product of a matrix and its conjugate. What is the matrix, where did it come from, or what did it drop out of?

7. Feb 18, 2009

### martin_blckrs

I'm sorry that the notation confused you, but I thought it is obvious.
Well $\bar{\Lambda}$ is not the conjugate of the matrix $\Lambda$, but another orthochronal matrix. Now I want to prove that orthochronal matrices form a subgroup of the Lorentz group, so what I wish to show is that if I multiply two orthochronous matrices, I again obtain an orthochronous matrix. Thus I multiply $\Lambda$ and $\bar{\Lambda}$, and look at the 00 element of the product. The question is: Why is it true that the 00 element of the product is also bigger than 1?
I hope it is now clear.

Last edited by a moderator: Jan 9, 2013
8. Feb 18, 2009

### sirchasm

You're deriving maybe a relativistic operator, in the Lorentz time domain?; 00 can be >1 because of the Lorentz effect..?
The RHS terms are operators on the LHS?

9. Feb 18, 2009

### martin_blckrs

Ok, I try it once again.
Do you know what an orthochronous Lorentz transformation is?
If so, then my question is: How can you prove that orthochronous Lorentz transformations form a subgroup in the group of Lorentz transformations?

10. Feb 18, 2009

### Fredrik

Staff Emeritus
I find tensor notation (some indices upstairs, some downstairs) pretty awkward in this context so I'll just write all indices as subscripts. In particular, what you write as $\Lambda^\mu{}_\nu$, I write as $\Lambda_{\mu\nu}$. Just think of it as row $\mu$, column $\nu$ of the matrix $\Lambda$ and forget that you've ever even heard about tensors. My metric signature is -+++.

First note that there's a velocity associated with each Lorentz transformation, given by the formula

$$v_i=-\frac{\Lambda_{i0}}{\Lambda_{00}}$$

For example, if $\Lambda$ represents a transformation from frame F to frame F', then $\vec v$ is the velocity of F' in F. (This observation simplifies the notation a bit).

The 00 component of the condition $\Lambda^T\eta\Lambda=\eta$ is

$$-1=\eta_{00}=(\Lambda^T\eta\Lambda)_{00}=(\Lambda^T)_{0\mu}\eta_{\mu\nu}\Lambda_{\nu 0}=\eta_{\mu\nu}\Lambda_{\mu 0}\Lambda_{\nu 0}$$

$$=-(\Lambda_{00})^2+(\Lambda_{10})^2+(\Lambda_{20})^2+(\Lambda_{30})^2=-(\Lambda_{00})^2(1-\vec{v}^2)$$

Note that this implies that $|\vec{v}|<1$. It also implies that

$$\Lambda_{00}=\pm\frac{1}{\sqrt{1-\vec{v}^2}}$$

which is either ≥1 or ≤-1.

The only difficult step is to prove that the orthochronous subset is closed under matrix multiplication, i.e. that

$$(\bar\Lambda\Lambda)_{00}\geq 1$$

where $\bar\Lambda$ and $\Lambda$ are both orthochronous Lorentz transformations. Because of the result we just obtained it's actually sufficient to prove that the left-hand side is ≥0.

I'm getting database errors, so I'll end this post here and start a new one.

Last edited by a moderator: Jan 9, 2013
11. Feb 18, 2009

### Fredrik

Staff Emeritus
Define

$$u_i=\frac{\Lambda_{0i}}{\bar\Lambda_{00}}$$

The identity $\Lambda^{-1}=\eta\Lambda^T\eta$ implies that $\vec{u}$ is the velocity that corresponds to $\bar\Lambda^{-1}$ the same way $\vec{v}$ corresponds to $\Lambda$.

$$(\bar\Lambda\Lambda)_{00} =\bar\Lambda_{0\mu}\Lambda_{\mu 0} =\bar\Lambda_{00}\Lambda_{00} +\bar\Lambda_{01}\Lambda_{10} +\bar\Lambda_{02}\Lambda_{20} +\bar\Lambda_{03}\Lambda_{30} =\underbrace{\bar\Lambda_{00}\Lambda_{00}}_{\geq 1}(1-\vec{u}\cdot\vec{v})$$

$$\geq \bar\Lambda_{00}\Lambda_{00}(1-\underbrace{|\vec{u}| |\vec{v}|}_{\leq 1}) \geq 0 \implies (\bar\Lambda\Lambda)_{00} \geq 1$$

Last edited by a moderator: Jan 9, 2013
12. Feb 19, 2009

### sirchasm

OK, I think "orthochronous" means time-independent, or a spatial transform at t = t', or maybe I still don't know...

yes, I have encountered Lorentz before, but the terminology I'm little rusty on.

13. Feb 19, 2009

### martin_blckrs

Wow, Frederik, Thank You soo much for the effort! You made it very clear! Thank you very much for your time.

to sirchasm:
Why do you react on posts, where you don't have the slightest idea of what we're talking about?

14. Feb 19, 2009

### sirchasm

Because I do have the slightest idea.
See how initially I tried to prompt you to explain = think about what it was you were starting with?

When or where does an orthochronous transform appear? What are you actually deriving? What does it tell you? Stuff like that. Are you sure I don't understand the word "orthochoronous"?
You said: "I'm deriving a term based on a formula"
Where does it comes from? "It comes from a formula"...

P.S. why do you assume I don't know? Does it make you feel better?
(seriously, I'd like to know what your reaction is)

15. Feb 19, 2009

### Fredrik

Staff Emeritus
I fixed a mistake in #11. I had written

$$u_i=\frac{\bar\Lambda_{i0}}{\bar\Lambda_{00}}$$

when it should have been

$$u_i=\frac{\bar\Lambda_{0i}}{\bar\Lambda_{00}}$$

Last edited by a moderator: Jan 9, 2013