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2

This is the purpose of the Legendre transform, which lies at the core of hamiltonian mechanics. You have a function $E(V,S)$, which is convex for certain interval of $V$, and want to find $E'(P,S)$, where $P \equiv -\left.\frac{\partial E}{\partial V}\right|_S$. This function is given by the Legendre transform of $E$: $$E'(P,S) = PV + E(V,S)$$ for the ...


2

when you put the Maxwell equation in the vlasov equation, you calculate the averages and that is how the terms $\left\langle \frac{\partial f}{\partial t}\right\rangle =0 $ since the distribution is not dependent on time and $\left\langle v.\nabla f\right\rangle =0$ because distribution is uniform on an average. similarly if you differentiate the third ...


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df/dt is 0 for stationary condition. but that distribution function is for molecules. Without any charge or potential. However, in your third term you have dependence on intensity and charge, therefore, I suppose, your MB equation should have term for some potential in exponent. Electric or electrostatic. In case of electrostatic, then df/dt = 0. maxwell-...


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A property of the Vlasov equation is that any distribution that is only a function of constants of motion is its solution. So if the velocity of the case you present is not a function of time, the distribution would trivially be a solution of Vlasov.


2

For the purposes of dimensions (units), you can treat a derivative like a division. So when you apply $\frac{{\rm d}}{{\rm d}t}$ to a function you divide the dimensions of the function by a unit of time. In your example I get: $$\frac{{\rm d}S}{{\rm d}t}\left[{\rm m}\,{\rm s}^{-1}\right] = v \left[{\rm m}\,{\rm s}^{-1}\right] + a\,\left[{\rm m}\,{\rm s}^{-2}...


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There is no relation between the area being two-dimensional in your graph and what it means. For example, consider you make the $x$-axis as an indicator of the temperature, so what is the difference between $x= 5 K$ and $x = 8 K$? of course it's $ \Delta x = 3K $. Now, isn't $x$ a one-dimensional quantity? So the graph gains its meaning from you not from ...


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I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement , which are 1 dimensional quantities. You are right. Area is a 2 dimensional quantity but what you have missed out is that you didn't use dimensional analysis properly. The dimensions of velocity is $[LT^{-1}]$. So if you multiply velocity with time, you get the ...


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If we consider the velocity-time graph area under the whole line is the distance. For example, Now look at the second part of the diagram (rectangular, dark blue) $$v=\frac{dx}{dt}\rightarrow\int{vdt}=\int{dx}\rightarrow x=vt$$ But the first part (triangle), we should consider acceleration $$a=\frac{d^2x}{d^2t}=\frac{dv}{dt}\rightarrow\int{adt}=\int{dv}\...


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The interpretation of the area under a curve, depending upon the curve, will vary. If it is a Velocity v. Time Graph, the area from a given time to another time, will be the distance traveled between those times. If it is an Acceleration v. Time Graph, the area from one time to another, will be the change in velocity of the object between those two times. ...


1

Consider some path $\gamma^{\mu}(\tau)$, and some vector $x^{\mu}$. Parallel transport is the condition when $\gamma^{a}\nabla_{a}x^{b} = 0$ Torsion is present if, for two paths $\gamma^{a}$ and $\delta^{a}$ that satisfy $\partial_{a}\gamma^{b} = \partial_{a}\delta^{b} = 0$, it is the case that parallel transport of $\gamma$ along $\delta$ produces a ...


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This is mere math. In the maximum or minimum points, the slope of the curve is zero. And the differentiation (the derivative) is the slope. So, set the derivative equal to zero, and the solutions are the peaks and valleys on the curve. http://mathsfirst.massey.ac.nz/Calculus/SignsOfDer/MaxMin.htm


1

Physicists do you the rules of calculus but can be sloppy in their notation. The symbols $\delta, \Delta$ and $d$ often seem to be used interchangeably to mean a (small or infinitesimal) change in something or better still a final value minus an initial value. So in your equation $dU$ is the change in internal energy of a system or final internal energy ...


5

Note that, for example, \begin{align} [A_\mu,\partial_\nu]f&=A_\mu\partial_\nu f-\partial_\nu(A_\mu f)\\ &=A_\mu\partial_\nu f-\partial_\nu(A_\mu)f-A_\mu\partial_\nu f\\ &=-f\partial_\nu A_\mu\,. \end{align} So you don't get terms like $A_\mu\partial_\nu$.


0

I misunderstood what Menzel had intended by "covariant nature of the differential operator". He did not mean that the differential operation is synonymous with covariant differentiation. As Menzel is wont to do, he proceeded to expose a series of non-trivial equivalences, and then tacked on a final equivalence which does not obviously follow from the ...


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I think your mistake is in assuming that there are other terms. The way I see it, your only vector quantity is $w$, $u_{c} \nabla^{c} $ for example is a sum/scalar and doesn't have any vector character or components, thus does not need to be expanded into base vectors.


1

Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then $$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)\nabla_l(\delta^i{}_k\delta^m{}_j)=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$ where $C(k,m)$ is the contraction of the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.


2

The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero). One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because ...


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You can write the Kronecker delta tensor as a product of the metric tensor $$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} + g^{ac}\nabla_a g_{bc} $$ As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity. Version without using the metricity of the connection : ...


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$$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v $$ The physics notation makes the truth of the derivation above clear because $dx/dx$ cancels. In mathematicians-style presentation of the rule, it's the chain rule for the derivative of a composite function, $v'(t(x)) =\dots$



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