# Orthochronal subgroup of the Lorentz group

• martin_blckrs
I hope that caused no confusion. In summary, the question being discussed is how to prove that orthochronous Lorentz transformations form a subgroup in the group of Lorentz transformations. The key is to show that the product of two orthochronous matrices is again orthochronous. This is done by considering the 00 element of the product and using the fact that only the 00 element needs to be greater than 1. The velocity associated with each transformation is used to simplify the notation, and a proof is provided using the formula for the inverse of a Lorentz transformation.
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.

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What generates the initial terms in the LHS?

What generates the initial terms in the LHS?

What do you mean?

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

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

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

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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?

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.

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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?

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?

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.

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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 $\Lambda$.

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

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

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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.

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?

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)

I fixed a mistake in #11. I had written

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

when it should have been

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

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## 1. What is an orthochronal subgroup of the Lorentz group?

An orthochronal subgroup of the Lorentz group is a subset of the larger Lorentz group, which is a mathematical representation of the symmetries of special relativity. This subgroup only includes transformations that preserve the direction of time, also known as proper orthochronous Lorentz transformations.

## 2. What are the properties of an orthochronal subgroup?

An orthochronal subgroup must have the properties of a subgroup, meaning it must be a subset of the larger group and must itself be a group. Additionally, it must only include transformations that preserve the direction of time, and must be closed under composition and inversion operations.

## 3. How is an orthochronal subgroup different from the full Lorentz group?

An orthochronal subgroup is a subset of the full Lorentz group, which includes all possible Lorentz transformations. The main difference is that the orthochronal subgroup only includes transformations that preserve the direction of time, while the full Lorentz group includes both proper and improper transformations.

## 4. What is the significance of the orthochronal subgroup in physics?

The orthochronal subgroup is significant in physics because it represents the symmetries of special relativity, which is a fundamental theory in modern physics. This subgroup helps us understand the behavior of objects moving at high speeds and the concept of time dilation.

## 5. Can the orthochronal subgroup be extended to include other symmetries?

Yes, the orthochronal subgroup can be extended to include additional symmetries, such as spatial rotations, to form the full Lorentz group. This extension is important in understanding the symmetries of spacetime and plays a crucial role in theories such as quantum field theory.

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