Review Note

Last Update: 12/30/2024 08:30 AM

Current Deck: Physik::T2

Published

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Front
Sei 
\(x^{\mu}(\tau) \) ein Vierervektor.

Was ergibt sich dann für: \(\left\Vert\frac{dx}{d\tau}\right\Vert^2 =\;\; ?\) 
Back
\[\left\Vert\frac{dx}{d\tau}\right\Vert^2 =\ c^2\]

Tags:

Spezielle_Relativitätstheorie

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