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Fourier transform of a functional derivative

A more direct way to derive the result is with the aid of the chain rule. So, consider a function $E[f]$, where $f(x)$ is a function of the one dimensional variable $x$. It is related to a spectrum ...
flippiefanus's user avatar
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Variation of the Einstein-Hilbert action to derive the metricity condition

To see that this implies the covariant conservation of the metric, we simply look at some of the traces and contractions of the equations presented here. For instance, tracing over $\mu$ and $\rho$ ...
Eletie's user avatar
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2 votes

Difference in definition of conserved current in Quantum Field Theory

Comparing the two resulting currents you see that they only differ by the addition of the $-F^\mu$ term. (The replacement of $X_a $ by $\delta \phi / \delta \alpha$ is only due to a different ...
Thomas Tappeiner's user avatar
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Taking functional derivatives of generating functional

Be careful. Your first line is correct, but the statement in the second line is not. You can only use it inside the integration to get rid of the delta function. As I will show you how to compute the ...
Everlin Martins's user avatar
1 vote

Taking functional derivatives of generating functional

My main confusion is about how to take derivatives of function like $J(y_1)$ wrt to $J(x)$. For example what is the value of $$\frac{\delta J(y_1)}{\delta J(x)}$$. Is that equal some kind of delta ...
hft's user avatar
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1 vote
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How did the boundary term vanish in deriving equation of motion from Lagrangian?

The Lagrangian $L$ is a smooth (say atleast of class $C^2$) function. So the boundary terms are \begin{align} \left[\frac{\partial L}{\partial\dot{q}}\delta q\right]_{t_1}^{t_2}&= \frac{\partial L}...
peek-a-boo's user avatar
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How did the boundary term vanish in deriving equation of motion from Lagrangian?

That's not how it works: We either impose essential/Dirichlet BC or natural BC at each endpoint $t_i$ and $t_f$, but not both. And that's enough to make the boundary terms vanish. Both BC would lead ...
Qmechanic's user avatar
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1 vote

Calculus of variations -- how does it make sense to vary the position and the velocity independently?

I want to start with asserting that nowhere in the evaluation there is a stage where the two factors $y$ and $\tfrac{dy}{dx}$ are treated as independent. The purpose of this answer is to address the ...
Cleonis's user avatar
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Variation of the kinetic term wrt the metric in scalar field theory

As mentioned in the comments, the first method is incorrect. $\partial_\mu \phi$ is something that is always well-defined in a manifold (with respect to some coordinate system) and does not depend on ...
Níckolas Alves's user avatar
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Equation of Motion for Polyakov Action with a mass term

Your problem is in your matter action you forgot a 1/2 factor in the tension, the action should be $$S = - \frac{T}{2} \int \mathrm{d}^2 \xi \sqrt{- \gamma} \left( \gamma ^{a b} \partial _a x^{\mu} \...
Luigi Teixeira de Sousa's user avatar
3 votes
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Computing variation of triple wedge term in Chern-Simons action

The Lie-bracket $[\cdot,\cdot]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$ and the Killing form $\kappa:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C}$ are extended to Lie algebra-valued ...
Qmechanic's user avatar
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3 votes
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Free scalar field deriving Ehrenfest using the path integral

Dozens of questions on this site remind you that $$ \frac{\delta}{\delta X(\tau)}~\int\!\! d^2\sigma ~ \partial_\alpha X(\sigma) \partial^\alpha X(\sigma) =-2\partial^2 X(\tau) .$$ Crudely, you may ...
Cosmas Zachos's user avatar

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