Can equation of motion depend on the choice of metric?

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Discussion Overview

The discussion centers on whether the equations of motion in field theory are dependent on the choice of metric signature. Participants explore implications of different metric conventions in the context of Lagrangian formulations, particularly in relation to scalar fields in 1+1 dimensions.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the equation of motion appears to depend on the choice of metric, presenting two cases with different signatures leading to different forms of the equation.
  • Another participant suggests that to achieve a unique equation of motion, the signs of the kinetic term in the Lagrangian density must be opposite for different metric conventions.
  • A third participant explains that changing the overall sign of the Lagrangian results in a corresponding change in the sign of the potential, indicating a symmetry in the Lagrangian formulation.
  • Further clarification is provided that the same sign change occurs when expressing scalar proper time in terms of metrics with different conventions.

Areas of Agreement / Disagreement

Participants express differing views on the implications of metric choice for the equations of motion, with no consensus reached on whether the dependence is problematic or how to resolve it.

Contextual Notes

Participants highlight the need to consider the implications of metric choice on the Lagrangian and the resulting equations of motion, but do not resolve the underlying assumptions or dependencies involved.

arroy_0205
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It seems strange to me that in such a simple case as the following the equation of motion depends on the choice of metric! In flat space we have say:
[tex] \partial^{\mu}\partial_{\mu}\phi=\frac{dV}{d\phi}[/tex]
Suppose \phi depends on space alone and we work in 1+1 dimension. then
[tex] \eta_{\mu \nu}=\left (-1,1 \right ) \rightarrow \frac{d^2\phi}{dx^2}=\frac{dV}{d\phi}[/tex]
and
[tex] \eta_{\mu \nu}=\left ( +1,-1 \right ) \rightarrow -\frac{d^2\phi}{dx^2}=\frac{dV}{d\phi}[/tex]
What is wrong here? Obviously the field configuration can not depend on the metric choice! Generally particle physicists and cosmologists use opposite signatures. Can anybody please point out?
 
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I want to add: It seems the only way to get unique equation of motion is to choose signs of kinetic term in the Lagrangian density opposite in these two cases (that is for two different metric conventions). Is that true?
[tex] L=\pm \frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V(\phi)[/tex]
 
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You have to change by an overall sign all the Lagrangian. In your example this turns out a change into the sign of V. Indeed, this is a symmetry of the Lagrangian as you can change all by an overall sign and you always get an extremum giving the right equation of motion (you have not to care about this being a minimum or a maximum). This means that when you choose a metric or the other a change into the sign of the potential is required.

Jon
 
arroy_0205 said:
I want to add: It seems the only way to get unique equation of motion is to choose signs of kinetic term in the Lagrangian density opposite in these two cases (that is for two different metric conventions). Is that true?
[tex] L=\pm \frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V(\phi)[/tex]

Yes, and this makes sense.

If [itex]\eta = - \eta'[/itex], and

[tex]L = \frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V \left(\phi\right),[/tex]

then

[tex]L = - \frac{1}{2}\eta'{}^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V \left(\phi\right).[/tex]

The same change of sign occurs when the scalar proper time is expressed in terms of spacetime metrics that have different sign conventions.
 
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