# I Complex number 'isomorphism' error

1. Nov 20, 2016

### jk22

I felt upon a mistake I made but cannot understand. I consider the following rotation transformation inspired from special relativity :

$$\left(\begin{array}{c} x'\\ict'\end{array}\right)=\left (\begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)\left(\begin{array}{c} x'\\ict'\end{array}\right)$$

Writing the first line gives $$x'=cos(\theta)(x-i tan(\theta)ct)$$

If I want this expression to be like a Lorentz transformation I should have the form $$x'=\gamma(v)(x-vt)$$ hence $$ic tan(\theta)=v \Rightarrow tan(\theta)=-i\frac{v}{c}$$

Then $$cos(\theta)=\frac{1}{\sqrt{1+tan(\theta)^2}}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ which is the gamma factor.

This gives the usual Lorentz transformation.

The problem arises when I consider the 'isomorphism' $$i\equiv \left (\begin {array} {cc} 0 & -1\\ 1 & 0 \end {array}\right)$$

Indeed I can then write $$e^{i\theta}\equiv \left( \begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)$$

But inserting the value of theta gives : $$\left(\begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( \begin {array} {cc} 1 & i\frac{v}{c} \\ -i\frac{v}{c} & 1 \end {array}\right)\equiv \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$
Which is the Bondi k factor.

Hence using this mapping of $$i$$ to real 2x2 matrices produce an error since we get then an homotethy instead of the Lorentz transform.

Is it because the application mapping of $$i$$ is injective and not bijective ?

2. Nov 20, 2016

### Staff: Mentor

Is there a typo above? The input vector is exactly the same as the output vector. From your work below, the vector on the right side should be $\begin{bmatrix} x \\ ict\end{bmatrix}$
But you seem to be omitting that factor or $\cos(\theta)$ that multiplies $(x - i\tan(\theta)ct)$.

3. Nov 20, 2016

### jk22

Thanks, Yes there is a typo i just copied the latex formula for the vector.

But should 't this thread be moved to the relativity forum I was hesitating ?

4. Nov 20, 2016

### Staff: Mentor

I have reported this thread to see what some of the other mentors think about moving it there.

5. Nov 20, 2016

### robphy

Can you show your steps (in the equiv sign) that leads to the Bondi factor?

6. Nov 21, 2016

### jk22

Indeed it's probably there : $$\left(\begin{array}{cc} 1 & i\frac{v}{c} \\ -i\frac{v}{c} & 1\end{array}\right)\equiv 1-i\frac{v}{c} \left(\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right)\equiv 1+\frac {v}{c}$$

Last edited: Nov 21, 2016
7. Nov 21, 2016

### stevendaryl

Staff Emeritus
I think it's right. You don't actually have to use matrices to get the answer.

If $cos(\theta) = \gamma$ and $sin(\theta) = -i \frac{v}{c} \gamma$, then

$e^{i \theta} = cos(\theta) + i sin(\theta) = \gamma + \frac{v}{c} \gamma = \frac{1 + \frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}$

You can get the same result, without the $i$, by writing:

$x' = x cosh(U) - ct sinh(U)$
$t' = t cosh(U) - \frac{x}{c} sinh(U)$

where $U$ is the "rapidity", defined in terms of the usual Lorentz parameter, $v$, via:
$v = c tanh(U) \Rightarrow \gamma = cosh(U)$

Then $e^U = cosh(U) + sinh(U) = \gamma(1+\frac{v}{c}) = \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}$

The relationship between $U$ and the $\theta$ in the original post is: $U = i \theta$.

8. Nov 21, 2016

### jk22

There is obviously something wrong since :

1) the latter calculation with the contraction of the matrices to i gives $$(x',ict')=\sqrt {\frac {1+\frac {v}{c}}{1-\frac {v}{c}}}(x,ict)$$

2) whereas expanding i into matrices gives $$(x',ict')=\frac {1}{\sqrt {1-\frac {v^2}{c^2}}}(x-vt,i (ct-\frac {v}{c}x))$$

With (2) we get $$x'^2+(ict')^2=x'^2-c^2t'^2=x^2-c^2t^2$$ but with (1) this quadratic form is not conserved but it appears the factor of contraction.

I thought this comes because the function $$\mathbb {C}\rightarrow M_{2x2}(\mathbb {R})$$ is injective but not surjective ?

9. Nov 21, 2016

### stevendaryl

Staff Emeritus
I'm not exactly sure what you're trying to do. You can indeed interpret complex numbers as $2 \times 2$ real matrices. But you're trying to go the other way--interpreting a $2 \times 2$ matrix as a complex number. But then that leaves the column matrices: $\left( \begin{array} \\ x \\ ict \end{array} \right)$ with no matrix to multiply it.

10. Nov 21, 2016

### jk22

It remains a multiplication by a scalar we can insert a unit matrix if we want

11. Nov 21, 2016

### stevendaryl

Staff Emeritus
But the whole point of the Lorentz transformations is that you are "mixing up" the x coordinate and the t coordinate when you go to x' and t'. Multiplication by a scalar (or a unit matrix) isn't going to do that.

12. Nov 23, 2016

### jk22

Indeed. In fact the difference comes from the difference in the equivalence when we use a tensor product when we factorize i :

$$i\otimes i\neq-\mathbb{1}_4$$ when we see i as a 2x2 matrix.

But if we see i as a complex scalar it's -1.