Can equation of motion depend on the choice of metric?

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SUMMARY

The equation of motion in field theory is directly influenced by the choice of metric signature, as demonstrated in the discussion regarding the Lagrangian density L = ±(1/2)ημνμφ∂νφ - V(φ). In 1+1 dimensions, using the flat metric ημν = (-1, 1) leads to the equation d²φ/dx² = dV/dφ, while using ημν = (1, -1) results in -d²φ/dx² = dV/dφ. This discrepancy necessitates a change in the sign of the potential V(φ) to maintain consistency across different metric conventions. Thus, the unique equation of motion requires careful consideration of the kinetic term's sign in the Lagrangian.

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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:
<br /> \partial^{\mu}\partial_{\mu}\phi=\frac{dV}{d\phi}<br />
Suppose \phi depends on space alone and we work in 1+1 dimension. then
<br /> \eta_{\mu \nu}=\left (-1,1 \right ) \rightarrow \frac{d^2\phi}{dx^2}=\frac{dV}{d\phi}<br />
and
<br /> \eta_{\mu \nu}=\left ( +1,-1 \right ) \rightarrow -\frac{d^2\phi}{dx^2}=\frac{dV}{d\phi}<br />
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?
<br /> L=\pm \frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V(\phi)<br />
 
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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?
<br /> L=\pm \frac{1}{2}\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V(\phi)<br />

Yes, and this makes sense.

If \eta = - \eta&#039;, and

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

then

L = - \frac{1}{2}\eta&#039;{}^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi -V \left(\phi\right).

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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