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0

The numerator doesn't depend on the frequency, so you just want to minimize the denominator. That's the same as minimizing the square of the denominator. Now it should be simple!


5

The places in physics where commutation of partial derivatives tends to be important are in the identities of vector calculus. The situations where these identities might seem to break down is when there is some kind of topological winding. Then the partial derivatives commute at almost all points except some small set where they are undefined but still can ...


1

Singularities in functions often lead to non commuting second derivatives. As for a Physical interpretation I think the following exercise may help: The partial derivative can be from First Principles can be written as df(x,y)/dx = (f(x+h,y)-f(x,y))/h i.e the function is incremented by h and then the derivative is found. (x,y+h). .(x+h,y+h) (x,y). ...


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The easiest trick is to write (say, in three dimensions. For higher dimensions, add indices to the Levi-Civita symbol, and factors of g): $$g = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}g_{ax}g_{by}g_{cz}$$ Then, the variation is easy. I'll leave it as an excersise to work out the variation, and how to translate the result into factors of $g_{ab}$ and $g$


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Use the identity that if $M$ is invertible and $\delta M$ is "small" compared to $M$, then we have $$ \det (M + \delta M) = \det(M) \det( 1 + M^{-1} \delta M) \approx \det(M) \left[ 1 + \text{tr} (M^{-1} \delta M) \right]. $$ In the case of the metric, this implies that $$ -\det(g + \delta g) \approx -\det(g) \left[ 1 + g^{ab} \delta g_{ab} \right] $$ and ...


2

A discontinuity in the flow of water could be a wall, or a clog that water is still getting around, but not flowing directly through. In finite electric current it could be a substance with a different conductivity, notably zero or ∞. In theory, the mixed partial second derivatives would not be generally equal, just on the cusp of a boundary such as these. ...


2

Formally, the meaning you assign is just the usual meaning of the derivative. $$\partial_\mu \psi(x^\nu) = \lim_{h \to 0} \frac{\psi(x^\nu + h\delta^\nu_\mu) - \psi(x^\nu)}{h}$$ You can indeed compute it componentwise, because you can subtract two spinors, as in the equation above, just by subtracting their components. The object you get has sixteen ...


0

Basically your confusion is caused by a bias about what a vector is. Everyone agrees that you can add two vector and get another. Everyone agrees you can scale a vector and get another vector. Sometimes we square a vector and get a scalar, but some people say that is "merely" an abuse of notation. But that is just a special case of ...


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I would avoid using a spline in data analysis in general. The spline draws smooth curves through arbitrary sets of points but destroys a lot of the information in the points and adds extraneous, meaningless "information". Use it for making things pretty, not for analysis, unless you really think your data should follow a power series (which would be rather ...


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Model two consecutive measurements as the real values plus some noise. Call the first measured temperature $T_1$ and the second $T_2$. Call the measured noises $\gamma_1$ and $\gamma_2$, and suppose that they are drawn from a distribution $\Gamma(\gamma)$ and are uncorrelated. The (approximation to the) derivative is $$\text{Derivative} \approx \frac{(T_2 ...


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The spin connection has a geometric meaning - it is a connection associated to a particular non-coordinate frame, which diagonalizes the metric. Here's how: Let $M$ be our spacetime. The tangent bundle $TM$ may be thought of as the associated bundle to an $\mathrm{SO}(n)$-principal bundle, where the $\mathrm{SO}(n)$ matrices represent ordered orthonormal ...


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I think the answer is neither one sadly. If you do a taylor expansion you find \begin{equation} \partial_\mu e^{i t^a \phi_a} = \sum_n \left(\frac{i^n}{n!}\right) \partial_\mu (\phi^a t_a)^n = i \partial_\mu (\phi^a t_a) - \frac{1}{2}\left( \partial_\mu (\phi^a t_a) \phi^b t_b + \phi^a t_a \partial_\mu (\phi^b t_b) \right) + \cdots \end{equation} The second ...


1

In an ideal, theoretical world, we would only have functions from $m$ arguments to some space, and we'd say "the partial derivative with respect to the $n^{\text{th}}$ argument", with an implicit statement "holding all the other $m - 1$ arguments constant." Call this derivative of a function $f(x_1, \dots, x_m)$ the derivative $f_{n}$. But, we live in a ...


2

If you have a function $f(x_e,x_p)$ then it has the partial derivative $\partial f/\partial x_e$ (that holds $x_p$ constant and treats the function like a 1d function of $x_e$ alone) and similarly has the partial derivatives $\partial f/\partial x_p$ (that holds $x_e$ constant and treats the function like a 1d function of $x_p$ alone). But you can think of ...


2

If you want to be physical, you'd have to have a physical interpretation of the derivatives. If you've already taken two derivatives you can ask yourself whether it is possible to take the gradient of those second derivatives. If so, then the second derivatives commuted, if not then the second derivatives are weird (if something wasn't weird you could take ...


5

The general requirement you are looking for is that the particular function be of class $C^1$, where ...if all order $p$ partial derivatives evaluated at a point $\mathbf a$: $$\frac{\partial^p}{\partial x_1^{p1}\partial x_1^{p2}\cdots\partial x_n^{pn}}f\left(\mathbf x\right)\vert_{\mathbf x=\mathbf a}$$ exist and are continuous, where $p1,\,p2, ..., ...


1

If $$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$ then $$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$ and $$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$ on the other hand $$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$ and $$ \left | d\overrightarrow{r} ...


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As shown in the diagram $|dr|$ represents the magnitude of the vector difference(that involves the laws of vector addition/subtraction) between $\vec{r_2}\quad \& \quad \vec{r_1}$ while $d|r|$ represents the difference between magnitudes of two vectors which is simply the difference in their lengths.


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Using polar coordinates it holds $ |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}$. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the $d|{\bf r}|$ should be moreover pointing ...


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In general you are allowed to use any parameter $q$ to describe the motion of an object as $\vec{r}(q)$ where the parameter changes with time $q=q(t)$. The parameter can be an angle or a distance or any combination that best makes sense. So now you have expressions for velocity and acceleration defined from the chain rule $$ \vec{v} = \frac{{\rm d} ...


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In the case of a circular motion, the tangential velocity $v$ can be expressed in terms of $\omega$ and $r$, i.e. $ v= \omega r$. The formula in your textbook $$a=r\frac{d\omega}{dt} $$is obtained by taking the derivative of the previous equation, while taking into account that $r$ is constant. Of course, you could consider $v$ as a function of $\theta$, ...


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What I don't understand is why we have to apply the rule in this specific way? How do I know how the chain rule must be applied? We don't have to. You don't know. Somebody just found out that by using that specific method, the result ended up neat and simple. Nothing is wrong with another method. You get the same thing in another expression. Let's ...



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