Lorentz Contraction from Minkowski Space

Asked Oct 05 '21 at 15:46

Active Oct 05 '21 at 17:51

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I looked at other similar questions, and couldn't find a fit.

I was reading this resource, which claimed you could use Minkowski space :

$\delta{s}^2 = c^2 \delta{t}^2 - \delta{x}^2$

To calculate the Lorentz contraction :

$\gamma = {{1} \over {\sqrt{1 - {{v^2}\over{c^2}}}}}$

Unfortunately, the resource doesn't give a complete example of how this is done.

I tried a few other resources, and the best I could get was this:

$\delta{s}^2 = c^2 \rightarrow c^2 = c^2v_t^2 - v_x^2$

$\rightarrow 1 = v_t^2 - {{v_x^2} \over {c^2}}$

$\rightarrow v_t^2 = 1 + {{v_x^2} \over {c^2}}$

$\rightarrow v_t = \sqrt{1 + {{v_x^2} \over {c^2}}}$

However, these two terms $\gamma$ and $v_t$ do not look the same to me. What am I missing? How is this done?

edited Oct 05 '21 at 17:51

Qmechanic

asked Oct 05 '21 at 15:46

James McLellan

1

Does this answer your question? How do I derive the Lorentz contraction from the invariant interval? – John Rennie Oct 05 '21 at 15:48
You’re right. Contraction. I’ll fix it. – James McLellan Oct 05 '21 at 16:48
The answer proposed is not correct – James McLellan Oct 05 '21 at 17:34
If dTau = dt sqrt( 1 - v^2 / c^2 ), v is close to c, and dt is 1; the answer proposed returns a dTau of near zero. Infinity is the expected result. – James McLellan Oct 05 '21 at 17:37

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