New answers tagged variational-calculus
1
$\epsilon$ is a real, positive, small number that modulates the strength of the perturbation $\delta$ (remember that $\delta$ is a function $\delta:[t_1,t_2] \rightarrow R^n$ in this context). When $\epsilon$ goes to zero, the total perturbation goes to zero, no matter what $\delta$ is.
This variable is a way to translate the "magnitude" of a ...
0
The answer to your first question is no, not all field variations are given by coordinate transformations. For example, if you have a complex scalar field, then performing a $U(1)$ transformation (multiplying by a phase) is certainly not a coordinate transformation.
For your second point, I'm not sure I see the question as both the statements you've made are,...
0
Will the first variation of the action functional be zero if we plug the solution that we found from the ICP?
Yes, if we [besides assuming the appropriate initial boundary condition (IBC)] also assume an appropriate final boundary condition (FBC).
Concerning consistent choices of boundary conditions (BCs), see also e.g. this Math.SE post.
2
Sure, one can for starters use an infinitesimal variation
$$ \delta x~=~\epsilon f(x,\dot{x},t) $$
with an arbitrary function $f$. Repeating dfan's & BebopButUnsteady's argument from this Phys.SE post then leads to infinitely many virial theorems
$$ \langle m\dot{x}\cdot \frac{df}{dt}\rangle~=~\langle f\cdot \frac{\partial V}{\partial x}\rangle$$
for the ...
3
There are a number of different approaches one can find to the calculus of variations, so here I will only mention the approach which make the most sense to me, which goes by the name "covariant phase space." The idea is to call the space of classical solutions to the equations of motion the phase space (up to subtleties involving gauge ...
0
Assuming real fields (the complexification is an easy exercise), @Chandrahas's point is that$$\int\tfrac12\nabla\phi\nabla\phi d^3x=-\int\tfrac12\phi\nabla^2\phi d^3x$$up to a boundary term. This is the $3$-dimensional equivalent of how we use $4$-dimensional integration by parts in Lagrangian field theory, when deriving an Euler-Lagrange equation from the ...
1
It was because in my second line of working I did not contract the first and third indices in my equation for the variation of the Riemann tensor.
It should look like this
$$\implies g^{\mu \nu}\delta R_{\mu \nu} = g^{\mu \nu} R^{\lambda}_{\mu \lambda \nu} = g^{\mu \nu} \nabla_{\lambda}\delta \Gamma^{\lambda}_{\nu \mu} - g^{\mu \nu} \nabla_{\nu} \delta \...
Top 50 recent answers are included
