Questions tagged [exterior-derivative]

For questions related to exterior derivative. The exterior derivative of a function $f$ is the one-form $df=\sum_i\frac{\partial f}{\partial x_i}dx_i$ written in a coordinate chart $(x_1,\dots,x_n)$.

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159 questions
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Applying the exterior differential to the first law of thermodynamics

I'm working on an exercise for an advanced statistical physics course. The question I'm struggling with is this: $$TdS=dE+PdV-\mu dN\tag{1}$$ Write the first law $(1)$ as $dS=....$ Applying the exterior differential $d$ to the resulting equation…
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How can we write Leibniz rule in differential forms notation?

In the proof of Faraday's law, the following identity is used: $$ \frac{d}{dt} \left[ \int_{\sum(t)} B(t) \cdot dA \right]|_{t_o}= \int_{\sum(t_o)} \partial_t B|_{t_o} \cdot dA + \frac{d}{dt} \int_{\sum(t)} B(t_o) \cdot dA $$ How would I wrote the…
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Proving that $d\theta_{fX}=df.\theta_X+\theta_X.df+d(f\theta_X)$

Let $\theta_X$ be a $1$-form. "Riemannian Geometry" by Peter Petersen, on pg 25, has the following formula: $$d\theta_{fX}=df.\theta_X+\theta_X.df+d(f\theta_X)$$ What does the dot product of two $1$-forms mean? What does $df.\theta_X$ mean? Where…
user67803
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multiple-indices in proving Invariant formula of exterior derivative

I'm trying to prove the invariant formula by induction $$d\omega(X_0,\cdots,X_p)=\sum (-1)^i X_i(\omega(\cdots,\hat{X}_i,\cdots))+\sum_{i
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Exterior derivative of a vector field

The exterior derivative of a scalar function is $d f(x,y,z) = ( \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz )$ Am I correct in assuming then that $d\left( F_x(x,y,z) e_x + F_y(x,y,z) e_y +…
R. Emery
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Exterior derivative of a $1$-form

Let the $1$-form $\theta_X$ be defined as $\theta_X(Y)=g(X,Y)$. Then my textbook says $$(d\theta_X)(\partial_k,\partial_l)=\partial_k g(X,\partial_l)-\partial_l g(X,\partial_k)-g(X,[\partial_k,\partial_l])$$ I tried looking up exterior…
user67803
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3 dimensional Exterior Derivative

I have $dx=x^{1/2}\omega_1$, (so $\omega_1=x^{-1/2}dx$), and $\star dx=x^{1/2}\omega_2 \wedge \omega_3$ (where $\star$ is the Star Hodge Operator). Then I have $d(x^{1/2}\omega_2 \wedge \omega_3)= -p \omega_1 \wedge \omega_2 \wedge \omega_3$, where…
user333046