All Questions
15 questions
-1
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Four gradient relation
I'm doing an exercise in QFT and I have to calculate the energy-momentum tensor for the Klein-Gordon Lagrangian and by doing it I got the following term:
$$ \frac{\partial \ \partial^{\nu}\phi}{\...
0
votes
0
answers
75
views
Deriving Klein-Gordon equation from Euler-Lagrangian equation: Taking partial derivative inside [duplicate]
Lagrangian for Klein-Gordon equation is given by
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu\phi - m^2\phi^2/2.$$
To derive Klein-Gordon Equation I have to Compute derivative in Euler-Lagrange ...
1
vote
2
answers
78
views
What is the meaning of the differential in Doran and Lasenby's discussion of Noether's theorem for spacetime transformations?
Doran and Lasenby (Geometric Algebra for Physicists, pg. 450) state that if a transformation involves spacetime dependence (this brings to my mind common examples: translation and rotation), then ...
0
votes
1
answer
155
views
Finding the Euler-Lagrange equation for a scalar field
Consider a scalar field with the following lagrangian density:
$$\mathscr{L}=-\frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi-V(\phi).$$
I want to find the corresponding Euler-Lagrange equation, ...
2
votes
1
answer
159
views
Notation and Terminology Questions from Schwartz' QFT Book
I am finding some of the notation confusing in Chapter 3 on Classical Field Theory in Schwartz' QFT book a bit confusing.
First off, on page 34 he defines a translation of a field to first order as
$$...
1
vote
1
answer
246
views
Four-vector differentiation (E-M Euler-Lagrange eq.)
$$\partial_{\mu} \frac{\partial(\partial_{\alpha}A_{\alpha})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\...
0
votes
1
answer
51
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Derivative with respect a partial derivative in lagrangian density [duplicate]
I am reading about field theory and lagrangian densities, and I found the following lagrangian density in my book:
$$
\mathcal{L} = \dfrac{1}{2} (\partial_\mu \phi)^2 - \dfrac{1}{2} m^2 \phi^2\ ,
$$
...
4
votes
1
answer
401
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Calculating equation of motion in gauge theories: using ordinary derivatives or covariant derivatives?
For general gauge theories, the total Lagrangian density is given as $$L=-\frac{1}{4}F^2+L_M(\psi, D\psi)$$ where $L_M(\psi, D\psi)$ is the matter field with the ordinary derivative replaced by the ...
0
votes
2
answers
274
views
Tensor Question (Klein–Gordon equation) [closed]
I have a question following the derivation of the Klein-Gordon equation from a lagrangian. From
Eq. (13d), where does $\delta^\mu_\nu$ come from? I guess it's a conversion factor of some sort.
2
votes
2
answers
164
views
Is there a equivalent theorem for closed form/ exact form for derivative with respect to fields
For a vector (one-form) $A_\mu$, when
\begin{eqnarray}
\partial_{[\mu}A_{\nu]}=0
\end{eqnarray}
then, there exists a scalar $\phi$ such that
\begin{eqnarray}
A_\mu =\partial_\mu\phi
\end{eqnarray}
...
0
votes
2
answers
244
views
Why partial w/respect to $\phi$ of d'Alembertian of $\phi$ = 0?
Why is $\frac{\partial }{\partial \phi}(\frac{\partial^2 \phi}{\partial x^2})$ = 0?
I'm happy to read an article, but I don't know what keywords to search on.
Background:
Apologies if this is a math ...
2
votes
1
answer
388
views
How to take derivative with respect to Lagrangian of complex field?
Basics: The Lagrangian in field theory was written as
$$\frac{\partial \mathfrak{L}}{\partial \varphi}=\partial_\mu(\frac{\partial\mathfrak{L}}{\partial(\partial_\mu\varphi)})$$.
Question 1:
Is $\...
2
votes
2
answers
442
views
How do I find the function derivative $(\delta/\delta \phi) (\partial_\mu \phi)$?
The question is simple: How do I find the function derivative of $$(\delta/\delta \phi(x)) (\partial_\mu \phi(x))~?$$ As far as I can tell, I cannot use any of the standard computational rules for the ...
8
votes
3
answers
3k
views
Is the shorthand $ \partial_{\mu} $ strictly a partial derivative in field theory?
The Euler-Lagrange equation for particles is given by
$$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q},\tag{1}$$
and for fields it is
$$ \partial_{\mu} \frac{\...
6
votes
2
answers
2k
views
Variation of Lagrangian density $\mathcal{L}$ w.r.t $x^{\mu}$
If a function $f(x(t),y(t))$ has no explicit dependence on the variable $t$, then $\frac{\partial f}{\partial t}=0$.
In quantum field theory, the Lagrangian density $\mathcal{L}(\phi,\partial_\mu\phi)...