- #1
Xezlec
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So I have this book that considers the problem of a flexible vibrating string, taking [itex]\phi(x,t)[/itex] as the string's displacement from equilibrium. It then writes a Lagrangian density in terms of this [itex]\phi[/itex], takes [itex]\delta \mathcal{S} = 0[/itex], and eventually concludes that [itex]\frac{\partial}{\partial t}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) + \frac{\partial}{\partial x}(\frac{\partial \mathcal{L}}{\partial \phi'}) = 0[/itex]. Notice that the time-varying and space-varying terms have the same sign.
Two pages later, it considers a scalar field [itex]\phi(x^0,\mathbf{x})[/itex] with a Lagrangian density [itex]\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)[/itex], and concludes that [itex]\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0[/itex]. Now, unless I am having some massive brain fart on how covariant and contravariant work, the time-varying and space-varying terms have opposite signs. Right?
What gives? Why are the signs different between these two situations?
Two pages later, it considers a scalar field [itex]\phi(x^0,\mathbf{x})[/itex] with a Lagrangian density [itex]\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)[/itex], and concludes that [itex]\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0[/itex]. Now, unless I am having some massive brain fart on how covariant and contravariant work, the time-varying and space-varying terms have opposite signs. Right?
What gives? Why are the signs different between these two situations?