Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm reading Sternberg's Group Theory and Physics. I have a question about chapter 1.2 Homeomorphisms.


Background:

A Lorentz Metric is defined as $||{\bf x}||^2=x_0^2-x_1^2-x_2^2-x_3^2$

And a Lorentz Transformation $B$ as one which satisfies $||B{\bf x}||^2=||{\bf x}||^2$

Identify ${\bf x}$ with $x=x_0I+x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$ where $\sigma_i$ are the Pauli matrices.

We have $x=x^*$ ($x$ is self-adjoint) and $\det x=||{\bf x}||^2$

Let $A$ be any 2x2 matrix and define the action of $A$ on $x$ as $x \to AxA^*$ and denote the corresponding action on the vector ${\bf x}$ as $\phi(A){\bf x}$

$AxA^*$ is also self-adjoint and $\det AxA^*=(\det A)^2\det x$

Therefore if $A \in SL(2,\mathbb{C})$ then $\phi(A)$ is a Lorentz transformation and also a homeomorphism.


Ok so far, here begins my question.

The next bit of the text says: suppose $A \in SU(2)$, then $AIA^*=I$

so if ${\bf e}_0 = \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$ then $\phi(A){\bf e}_0 = {\bf e}_0$

I see why this is so. Now here's the bit I don't understand.

If a Lorentz Transformation $C$ satisfies $C{\bf e}_0 = {\bf e}_0$ then $C$ also carries the space of vectors $\left( \begin{array}{c} 0 \\ x_1 \\ x_2 \\ x_3 \end{array} \right)$ onto itself.

I don't see why this is so. Can someone please explain the logical steps which I'm missing? Thanks.

share|improve this question
    
Wow... three answers, all different but equivalent, in the space of less than three minutes! –  Michael Brown Feb 20 '13 at 13:28
    
If you like this question you may also enjoy reading this post. –  Qmechanic Feb 20 '13 at 13:40

3 Answers 3

up vote 0 down vote accepted

The space of vectors $(0,x_1,x_2,x_3)^T$ may also be defined as the space of 4-vectors $\vec v$ that satisfy $\vec v\cdot \vec e_0 = 0$. The inner product is Lorentzian. If all vector $\vec u$ are transformed to $C\vec u$, then this space of vectors is – by definition – transformed to the space of vectors that obey $$ (C\vec v)\cdot \vec e_0=0 $$ for any allowed transformation $C$. However, the condition above is equivalent to $$ \vec v\cdot (\tilde C\vec e_0) =0$$ because the inner product may also be written as $\vec u\cdot \vec v = u^T \cdot \eta\cdot v$ where $\eta$ is the usual metric tensor. I used $(CD)^T = D^TC^T$ etc. Here, $\tilde C = \eta C^T \eta^{-1}$. However, $\tilde C \vec e_0$ may be easily seen to be $\vec e_0$ as well, so the condition in the displayed formula is equivalent to $$\vec v \cdot \vec e_0$$ so the whole space (as a set) is invariant under the transformation by $C$.

The text above is an unnecessary formalism that may be reproduced by self-evident words. The space of vectors given by the three coordinates is the space of all vectors normal to $\vec e_0$, and because both $\vec e_0$ and the inner product is conserved by the transformations given by $C$ and the definition of the 3D space depends on these invariant things only, this 3D space as a set is conserved as well. $C$ must be just a rotation of the spatial coordinates into each other.

share|improve this answer

Hints:

  1. Note that the time coordinate $x^0=\frac{1}{2} {\rm tr}(x)$ is given by the trace.

  2. Note that the trace ${\rm tr}(x)$ is invariant under the $SU(2)$ action.

  3. The subspace with $x^0=0$ is invariant under $SU(2)$.

  4. Note that the stabilizer/isotropy subgroup of $e_0$ is $SU(2)$.

share|improve this answer

If $C$ doesn't change the time component then it is just a spatial rotation!

To show this mathematically we can use $\boldsymbol{e}_{0}$ to construct a projection operator that projects down to the space orthogonal to $\boldsymbol{e}_{0}$:

$$ \Pi = \mathbb{I} - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0} $$

Then the statement is that

$$ C \boldsymbol{x} = \Pi C \boldsymbol{x}$$

for spatial $\boldsymbol{x}$, that is $\boldsymbol{x}$ obeying

$$ \boldsymbol{x} = \Pi \boldsymbol{x} $$

Applying this above we get

$$ C \Pi \boldsymbol{x} = \Pi C \boldsymbol{x}$$

So if $ C\Pi - \Pi C = 0$ then the result holds. Calculating this:

$$ C \Pi = C (\mathbb{I} - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0}) = C - (C\boldsymbol{e}_{0})\otimes \boldsymbol{e}_{0} = C - \boldsymbol{e}_{0} \otimes \boldsymbol{e}_{0} $$

and similarly for $\Pi C$ after using orthogonality of $C$. Subtracting gives $C\Pi - \Pi C = 0$.

share|improve this answer
    
To the person downvoting all my old answers lately: if you care at all about the quality of this site then leave a reason for your downvote! I love constructive criticism. If I've said something wrong, I want to be corrected. But anonymous downvoting achieves nothing. –  Michael Brown Aug 13 '13 at 1:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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