Timeline for Is this a new derivation of Lorentz transformations?
Current License: CC BY-SA 4.0
5 events
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| Dec 19, 2020 at 14:33 | comment | added | robphy | It should be $\sinh w=+(v/c)\gamma=\beta\gamma$, where $\beta=\tanh w=v/c$. Note that $u= \exp w=\cosh w +\sinh w = (\cosh w)(1+\tanh w) =\frac{1+\tanh w)}{\sqrt{1-\tanh^2 w}} =\sqrt{\frac{ 1+\tanh w}{1-\tanh w}}=k$. | |
| Dec 18, 2020 at 23:09 | comment | added | ZeroTheHero | @robphy sure but you just did provide an interpretation for this eigenvalue, which the OP did not do. Moreover you have just ipso facto answered the OP’s question in the negative (and made me learn about Bondi k-calculus). | |
| Dec 18, 2020 at 23:04 | comment | added | robphy | The parameter u is the Doppler factor, sometimes called the Bondi-k factor. The Bondi k-calculus goes quite far using just k and the radar method. My “relativity on rotated graph paper” method relies heavily on it. Since it is an eigenvalue of the boost, it is a mathematically natural variable to work with. | |
| Dec 18, 2020 at 22:28 | history | edited | ZeroTheHero | CC BY-SA 4.0 |
added 141 characters in body
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| Dec 18, 2020 at 22:01 | history | answered | ZeroTheHero | CC BY-SA 4.0 |