Undergrad Metric Transformation b/w Inertial Frames: Analyzing Effects

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The discussion centers on the transformation of the metric tensor in a 1-D inertial frame, specifically examining how it changes after a Lorentz boost. The initial metric tensor is defined as η = diag(-1, 1), and the transformation rule is provided. The participant calculates the new metric component g_{00} and finds it does not equal the expected value of -1, indicating a change after the boost. A key issue identified is the mixing of units where c = 1 and c ≠ 1, which leads to the discrepancy in the metric's elements. Clarification on unit consistency resolves the confusion regarding the metric transformation.
Jufa
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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?
 
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You seem to be mixing units where ##c = 1## with units where ##c \neq 1##.
 
Orodruin said:
You seem to be mixing units where ##c = 1## with units where ##c \neq 1##.
Oh yes. It is definitely that. Many thanks.
 
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