Working on the Maxwell field as a gauge theory, at some point the following derivative comes up:  $\frac{\partial(\partial_iA_0)}{\partial A_0}=0$  which must be, accordingly to the theory, zero.   My doubt is: why is that necessarily true? Is this a specific feature of the form of the fields in this case? Or, maybe, is always true because of some assumption that the field and its derivatives are independent variables?   I'm confused about this because mathematically this is trivially not true and the most obvious "one-dimensional" example is:  $f=e^x$  $\frac{\partial}{\partial f}\frac{\partial f}{\partial x}=1$
29135
In the calculus of variations, particularly Lagrangian mechanics, people often say we vary the position and the velocity independently. But velocity is the derivative of position, so how can you treat them as independent variables?
20041
A function is "differentiable" if it has a derivative. A function is "continuous" if it has no sudden jumps in it.  Until today, I thought these were merely two equivalent definitions of the same concept. But I've read some stuff today which seems to be claiming that this is not the case.  The obvious next question is "why?" Apparently somebody has already asked:    Several answers were given, but I don't understand any of them. In particular, Wikipedia and one of the replies above both claim that $|x|$ has no derivative. Can anyone explain this extremely unexpected result?  Edit: Apparently some people dislike the fact that this is non-obvious to me. To be clear: I am not saying that the result is untrue. (I'm sure many great mathematicians have analysed the question very carefuly and are quite sure of the answer.) I am saying that it is extremely perplexing. (As a general rule, mathematics has a habit of doing that. Which is one of the reasons why we demand proof of everything.)  In particular, can anyone explain precisely why the derivative of $|x|$ at zero is not simply zero? After all, the function is neither increasing nor decreasing, which ought to mean the derivative is zero. Alternatively, the expression  $$\frac{|x + a| - |x - a|}{a}$$  becomes closer and closer to zero as $a$ becomes closer to zero when $x=0$. (In fact, it is exactly zero for all $a$!) Is that not how derivatives work?  Several answers have suggested that the derivative is not defined here "because there would be a jump in the derivative at that point". This seems to assert that a continuous function must never have a discontinuous derivative; I'm not convinced that this is the case. Can anyone confirm or refuse this argument?
84280
In my calculus class, $e$ was defined to be the number such that $\frac{d}{dx}e^x = e^x$.  From the definition of the derivative, we have  \begin{align*} \frac{d}{dx}a^x &amp;= \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\ &amp;= a^x \lim_{h \to 0} \frac{a^h - 1}{h} \end{align*}  Thus $e$ is the number such that  $$ \lim_{h \to 0} \frac{e^h - 1}{h} = 1 $$  But how is it proven that there exists such number?
120385
Suppose we are given a function $f(x_{1}, x_{2})$. Does showing that $\frac{\partial f}{\partial x_{i}} = 0$ for $i = 1, 2$ imply that $f$ is a constant? Does this hold if we have $n$ variables instead?
100787
One of the most common gauges in QED computations are the $R_{\xi}$ gauges obtained by adding a term  \begin{equation} -\frac{(\partial_\mu A^{\mu})^2}{2\xi} \end{equation} to the Lagrangian. Different choices of $\xi$ correspond to different gauges ($\xi=0$ is Landau, $\xi=1$ is Feynman etc.) The propagator for the gauge field is different depending on the choice of gauge. The choice of Landau gauge forces $\partial_\mu A^\mu=0$, but I have never seen a similiar statement for the other gauges. I would like to know what constraint on the gauge field is produced by the other covariant gauges. For instance, what is the constraint on $A_\mu$ when $\xi=1,2,3,...$ etc. Is it still $\partial_\mu A^\mu=0$ or something different (it seems like it should be different)?
52484
In Heaviside-Lorentz units the Maxwell's equations are:  $$\nabla \cdot \vec{E} = \rho $$  $$ \nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$ $$ \nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$ $$ \nabla \cdot \vec{B} = 0$$  From EM Lagrangian density: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} - J^\mu A_\mu$$   I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it?
101346
The definition of a partial derivative is the "derivative of a multi-variable function relative to a single variable when all other variables are held constant".  But isn't the regular derivative (for one-variable functions) just a trivial case of this, where there are no other variables to hold constant? Why do we need the separate notation for partial derivatives (that is, writing $\displaystyle \frac{\partial f}{\partial x}$ rather than just $\displaystyle \frac{df}{dx}$)?
59245
The electromagnetic field tensor is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, and it appears in the Lagrangian as  $$L = -\frac{1}{4}F_{\mu\nu}^2 - A_{\mu}J_{\mu}.$$ Schwartz's QFT textbook says that (chapter 8, page 116) $$F_{\mu\nu}^2 = 2(\partial_{\mu}A_{\nu})^2 - 2(\partial_{\mu}A_{\mu})^2,$$ but I do not see how that is correct.   The first step of the expansion gives $F_{\mu\nu}^2 = (\partial_{\mu}A_{\nu})^2 - \partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu} - \partial_{\nu}A_{\mu}\partial_{\mu}A_{\nu} + (\partial_{\nu}A_{\mu})^2$, and I see that the first and last term add up to produce $2(\partial_{\mu}A_{\nu})^2$. However, the second term $\partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu}$ when expanded has a term like $\partial_{0}A_{1}\partial_{1}A_{0}$, and I don't see how this is present in $(\partial_{\mu}A_{\mu})^2 = (\partial_{0}A_{0} + \partial_{1}A_{1} + \partial_{2}A_{2} + \partial_{3}A_{3})^2$.
110881
Taking the curl of the electric field must be possible, because Faraday's law involves it: $$\nabla \times \mathbf{E} = - \partial \mathbf{B} / \partial t$$ But I've just , where it says  The curl of the gradient of any twice-differentiable scalar field $\phi$ is always the zero vector: $$\nabla \times (\nabla \phi)=\mathbf{0}$$  Seeing as $\mathbf{E} = - \nabla V$, where $V$ is the electric potential, this would suggest $\nabla \times \mathbf{E} = \mathbf{0}$. What presumably monumentally obvious thing am I missing?
55850
It is well-known that Maxwell added the displacement current term to Ampère's Law to make electrodynamics whole. As it is taught in the modern context (I am currently reading Griffiths's text, Introduction to Electrodynamics), we can motivate the addition of the displacement current term by noting that its addition to Maxwell's equations means that Maxwell's equations imply the continuity equation. However, as Griffiths remarks, this nicety (the fact that the continuity equation falls out of Maxwell's equations) is not incontrovertible evidence that the addition of the specific form of the displacement current term is necessarily correct. Indeed, he says that there &quot;might, after all, be other ways to doctor up Ampère's Law&quot;. My question is, therefore, twofold: (1) Is it true, as Griffiths says, that there are conceivably other ways to &quot;fix&quot; Ampere's Law? That is, can we let $$\nabla \times \mathbf{B}=\mu_{0}\mathbf{J}+\mathbf{v}$$ for some arbitrary vector function $\mathbf{v}$ and still develop a consistent theory? I'm not sure how to define &quot;a consistent theory&quot; here but, perhaps, we can roughly say that a consistent theory would mean no contradictions with the other three Maxwell equations (mathematically-speaking). At least to me, I would suspect that the answer is &quot;yes&quot; since the problem (at least as it is understood in the more modern language of vector calculus, as compared to what Maxwell was doing) with Ampere's Law without Maxwell's correction is that the divergence of the right-hand side does not in general vanish, as it must. Thus we would be requiring that (using continuity and Gauss's Law) $$\nabla \cdot \mathbf{v}=-\nabla \cdot(\mu_{0}\mathbf{J})=\mu_{0}\frac{\partial\rho}{\partial t}=\mu_{0}\nabla \cdot(\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t})$$ but, of course, the divergence of a vector function does not fully specify that vector function. However, assuming we choose $\mathbf{v}$ to satisfy the above, and putting aside experimental verification for the moment, would choosing something else for $\mathbf{v}$ break the structure of Maxwell's theory somewhere else? (2) Moving now to consider experimental verification, Griffiths says that Hertz's discovery of EM waves confirmed Maxwell's choice for the displacement current term. I understand that Maxwell's equations imply wave solutions that were observed experimentally, but perhaps someone can (at a high-level, even) explain why any other choice of the displacement current term would have yielded inconsistencies with experiment (assuming that my attempt at answering (1) above was correct for, if there are mathematical inconsistencies, then we are done).
105809
The second derivative of $f$ can be written as   $$f''(x) = \lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$$  while it can also be written as (in fact, I believe this is the definition of $f''(x)$):  $$f''(x) = \lim_{h \to 0}\frac{f'(x + h) - f'(x)}{h}$$  From the second expression, it seems clear that if $f'(x)$ does not exist, then $f''(x)$ cannot exist as well.  However, the first expression does not involve $f'(x)$, so this got me wondering: If $f'(x)$ does not exist at $x$ , can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ at $x$  exist?  How would one prove/disprove this? If this were to be true, what would be some examples?
119977
If $f$ is a differentiable real function in an open set $E \subset \Bbb R^n$ and $f$ has a local maximum at a point $\textbf{x} = (x_1, x_2, \cdots , x_n) \in E$, show $f'(\textbf{x}) = 0$.   I think it would be nice if I could show that each component is $0$, which I think means showing the partial derivative at each component is $0$, but I don't know how to show this.
12260
I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).  Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$.  I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.  I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why.  Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?
76635
I'm having a bit of trouble understanding how the gauge condition is found.  Consider the potentials $V$ and $\vec A$ and $V'$ and $\vec A'$ so that $$\vec E = -\vec\nabla V -\frac{\partial\vec A}{\partial t} = -\vec\nabla V' -\frac{\partial\vec A'}{\partial t}$$ $$\vec B = \vec\nabla \times \vec A=\vec\nabla \times \vec A'$$  The 2 sets of potentials are then related like this:  $$\vec A' = \vec A+\vec\nabla\theta$$ $$V' = V-\frac{\partial\theta}{\partial t}$$ For a function $\theta$.  The equations for the potentials are the following:  $$\vec\nabla^2\vec A-\varepsilon\mu\frac{\partial^2 \vec A}{\partial t^2}=-\mu\vec J+\vec \nabla(\vec \nabla.\vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$  $$\vec\nabla^2V -\varepsilon\mu\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\varepsilon} -  \frac{\partial}{\partial t}(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$  They can be decoupled by choosing the right potentials so that $(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})=0$  The above part is the theory in general.  Now to find the correct potentials where this is the case, let's say that this last part is true for potentials $\vec A'$ and $V'$, i.e.   $$\vec\nabla . \vec A' + \varepsilon\mu\frac{\partial V'}{\partial t}=0$$  Filling in the relations between the 2 sets of potentials from earlier gives:  $$\vec\nabla . \vec A + \nabla^2\theta + \varepsilon\mu\frac{\partial V}{\partial t} -\varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$  Yet my book tells me that this means that  $$\nabla^2\theta - \varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$  I don't see why this is the case.
23051
$$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$
33120
The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^
101641
Consider the differential form:  $$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$$  I believe that this form is not exact. i.e: There does not exist a $\psi$ such that $d\psi=\omega$.  In particular, this is supposed to be true since the line integral along a loop, $$\int_{C}\omega = 2\pi \neq 0$$  But, what about $$\psi=-\arctan(\frac x{y})$$  According to Maple,    As you can see, Maple says that $d\psi=\omega$. So what's the deal here? Is $\omega$ exact or not?  Thank You!