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I Complex number 'isomorphism' error

  1. Nov 20, 2016 #1
    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. jcsd
  3. Nov 20, 2016 #2

    Mark44

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    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)##.
     
  4. Nov 20, 2016 #3
    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 ?
     
  5. Nov 20, 2016 #4

    Mark44

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    I have reported this thread to see what some of the other mentors think about moving it there.
     
  6. Nov 20, 2016 #5

    robphy

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    Can you show your steps (in the equiv sign) that leads to the Bondi factor?
     
  7. Nov 21, 2016 #6
    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
  8. Nov 21, 2016 #7

    stevendaryl

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

    If [itex]cos(\theta) = \gamma[/itex] and [itex]sin(\theta) = -i \frac{v}{c} \gamma[/itex], then

    [itex]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}}}[/itex]

    You can get the same result, without the [itex]i[/itex], by writing:

    [itex]x' = x cosh(U) - ct sinh(U)[/itex]
    [itex]t' = t cosh(U) - \frac{x}{c} sinh(U)[/itex]

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

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

    The relationship between [itex]U[/itex] and the [itex]\theta[/itex] in the original post is: [itex]U = i \theta[/itex].
     
  9. Nov 21, 2016 #8
    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 ?
     
  10. Nov 21, 2016 #9

    stevendaryl

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    I'm not exactly sure what you're trying to do. You can indeed interpret complex numbers as [itex]2 \times 2[/itex] real matrices. But you're trying to go the other way--interpreting a [itex]2 \times 2[/itex] matrix as a complex number. But then that leaves the column matrices: [itex]\left( \begin{array} \\ x \\ ict \end{array} \right)[/itex] with no matrix to multiply it.
     
  11. Nov 21, 2016 #10
    It remains a multiplication by a scalar we can insert a unit matrix if we want
     
  12. Nov 21, 2016 #11

    stevendaryl

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    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.
     
  13. Nov 23, 2016 #12
    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.
     
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