Metric for non-inertial coordinate system

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SUMMARY

The discussion centers on transforming the metric of a 2D flat spacetime in Cartesian coordinates to a non-inertial coordinate system defined by \( t' = t \) and \( x' = x - vt - \frac{1}{2}at^2 \). The participants debate two potential line elements: OPTION 1 suggests \( ds^{2} = -dt^{2} + dx'^{2} \) with \( dx' = dx - (v + at)dt \), while OPTION 2 proposes \( ds^{2} = -dt^{2} + dx^{2} \) with \( dx = dx' + (v + at)dt \). The correct approach requires calculating \( dx \) and \( dt \) in terms of \( (t', x') \) and substituting them into the original metric equation.

PREREQUISITES
  • Understanding of 2D flat spacetime metrics
  • Familiarity with non-inertial coordinate transformations
  • Knowledge of differential geometry concepts
  • Proficiency in calculus, particularly in derivatives and integrals
NEXT STEPS
  • Study the derivation of metrics in non-inertial frames using general relativity principles
  • Learn about the implications of coordinate transformations on physical laws
  • Explore the concept of proper time and its relation to non-inertial frames
  • Investigate the effects of acceleration on spacetime metrics
USEFUL FOR

Students and researchers in physics, particularly those focusing on general relativity, differential geometry, and the study of non-inertial reference frames.

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Homework Statement


Hey guys.

So here's the problem:

Consider an ordinary 2D flat spacetime in Cartesian coordinates with the line element
ds^{2}=-dt^{2}+dx^{2}

Now consider a non-inertial coordinate system (t',x'), given by

t'=t, x'=x-vt-\frac{1}{2}at^{2}

(1) What is the metric in these coordinates?

There are some more questions apart from this but I think I can do those if I know how to do this part.

Homework Equations



None


The Attempt at a Solution



Okay so here's why I'm confused. How do I get the line element in these coordinates? Here are the two options in my mind...which one is correct?

OPTION 1
The line element they are looking for is ds^{2}=-dt^{2}+dx'^{2} where dx'=dx-(v+at)dt

OPTION 2
The line element they are looking for is ds^{2}=-dt^{2}+dx^{2}, where dx=dx'+(v+at)dt

Both of these options give different metrics...so which one (if any) is the way to go?

Thanks guys!
 
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All you have to do is calculate ##dx## and ##dt## in terms of the ##(t',x')## coordinates and plug them into ##ds^2 = -dt^2 + dx^2## to get the metric in ##(t',x')## coordinates.
 

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