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- Summary:
- Found something weird when calculating the transformation due to a boost.

The metric tensor in an inertial frame is ## \eta = diag(-1, 1)##. Where I amb dealing with only 1-D space. The metric tranformation rule after a crtain coordinate chane is the following:

$$ g_{\mu \nu} = \frac{\partial x^\alpha}{\partial x'^{\mu }} \frac{\partial x^\beta}{\partial x'\nu } \eta_{\alpha \beta} $$

with ##x^0 = t## and ##x^1= x##

Given the particular form of ## \eta ## we obtain for ## \mu = \nu = 0 ## :

$$ g_{00} = -\Big(\frac{\partial t}{\partial t' }\Big)^2 + \Big(\frac{\partial x}{\partial t' }\Big)^2 = -\gamma ^2 + v^2\gamma^2 = \frac{v^2-1}{1-v^2/c^2} \neq -1 = \eta_{00}$$

So I get that after a Lorentz boost one of the metric's elements has changed.

Where am I wrong?

$$ g_{\mu \nu} = \frac{\partial x^\alpha}{\partial x'^{\mu }} \frac{\partial x^\beta}{\partial x'\nu } \eta_{\alpha \beta} $$

with ##x^0 = t## and ##x^1= x##

Given the particular form of ## \eta ## we obtain for ## \mu = \nu = 0 ## :

$$ g_{00} = -\Big(\frac{\partial t}{\partial t' }\Big)^2 + \Big(\frac{\partial x}{\partial t' }\Big)^2 = -\gamma ^2 + v^2\gamma^2 = \frac{v^2-1}{1-v^2/c^2} \neq -1 = \eta_{00}$$

So I get that after a Lorentz boost one of the metric's elements has changed.

Where am I wrong?

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