Complex Isomorphism Error in Lorentz Transform

• I
• jk22
In summary: But then that leaves the column matrices: \left( \begin{array} \\ x \\ ict \end{array} \right) with no matrix to multiply...No I'm doing the other way, from the complex number i, to the matrix transformation, by making the exponential of the matrix of i. Not the complex multiplication by the matrix of i. I don't understand what you are trying to do. The way you are using the matrix multiplication, it is not defined.I could be wrong but i think you are not understanding what i am trying to do. The goal is to find a connection between the complex number i and the matrix transformation. I am trying to do this by making the exponential of the matrix of i.
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 ?

jk22 said:
I felt upon a mistake I made but cannot understand. I consider the following rotation transformation inspired from special relativity :

$$\left(\begin{array}{c} x'\\ict'\end{array}\right)=\left (\begin {array} {cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end {array}\right)\left(\begin{array}{c} x'\\ict'\end{array}\right)$$
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}##
jk22 said:
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 $$ic tan(\theta)=v \Rightarrow tan(\theta)=-i\frac{v}{c}$$
But you seem to be omitting that factor or ##\cos(\theta)## that multiplies ##(x - i\tan(\theta)ct)##.
jk22 said:
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 ?

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 ?

jk22 said:
But should 't this thread be moved to the relativity forum I was hesitating ?
I have reported this thread to see what some of the other mentors think about moving it there.

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

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

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

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: $U = i \theta$.

vanhees71
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 ?

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.

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

jk22 said:
It remains a multiplication by a scalar we can insert a unit matrix if we want

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.

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.

1. What is complex isomorphism error in Lorentz transform?

Complex isomorphism error in Lorentz transform is a phenomenon in physics where the equations used to describe the transformation of coordinates between two reference frames in special relativity lead to complex solutions. This error occurs when the relative velocity between the two frames approaches the speed of light.

2. How does complex isomorphism error affect the calculations in Lorentz transform?

Complex isomorphism error can cause discrepancies in the calculations of time dilation, length contraction, and other physical quantities between two reference frames in special relativity. This can lead to incorrect predictions and interpretations of experimental data.

3. What causes complex isomorphism error in Lorentz transform?

Complex isomorphism error is caused by the use of Lorentz transformations, which are based on the assumption that the relative velocity between two frames is less than the speed of light. When this assumption is violated, the resulting equations lead to complex solutions.

4. How can complex isomorphism error be avoided?

Complex isomorphism error can be avoided by using alternative equations, such as the Wigner rotation, to describe the transformation between reference frames in special relativity. These equations do not suffer from the same issues as the Lorentz transformations and provide accurate results even at high relative velocities.

5. Is complex isomorphism error a major problem in the field of physics?

Complex isomorphism error is a relatively minor issue in the field of physics, as it only occurs at extreme relative velocities that are not typically encountered in everyday situations. However, it is important for scientists to be aware of this error and to use appropriate equations to avoid any potential inaccuracies in their calculations.

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