- #1

asdf1

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where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?

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- Thread starter asdf1
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- #1

asdf1

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where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?

- #2

selfAdjoint

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asdf1 said:where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?

From

[tex] \left( \begin{array}{c} x' \\ t' \end{array} \right) = \left( \begin{array}{cc} cosh(\theta) & -sinh(\theta) \\ sinh(\theta) & cosh(\theta) \end{array} \right) \left( \begin{array}{c} x \\ t \end{array} \right) [/tex]

where [tex]v = tanh(\theta)[/tex] and c is taken to be 1.

- #3

Perspicacious

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There are many elegant ways to derive the Lorentz transformation:asdf1 said:where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?

http://www.everythingimportant.org/relativity/

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000043000005000434000001 [Broken]

http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf [Broken]

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- #4

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It should be [tex]+\sinh(\theta)[/tex].

(The determinant has to be 1.)

(The determinant has to be 1.)

- #5

learningphysics

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robphy said:It should be [tex]+\sinh(\theta)[/tex].

(The determinant has to be 1.)

Or rather they should both be [tex]-\sinh(\theta)[/tex] I believe.

- #6

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learningphysics said:Or rather they should both be [tex]-\sinh(\theta)[/tex] I believe.

Yes, of course :tongue2: , considering the original post. Thanks.

- #7

asdf1

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thanks! :)

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