I am a beginner in QFT, and  am reading it from Quantum Field Theory Demystified by David McGowan,  a Tata McGraw-Hill publication.  Here, in this book, the author at one point, while explaining Laggan giant Field Theory,  writes that the first derivative of a field  $\varphi $ is denoted by $\partial _\mu \varphi $. What does the $\mu $ mean?  A screenshot is attached:
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I've stumbled across the following notation a couple times reading physics articles on wikipedia: $$\partial_{\mu}$$ But what does it mean? They don't clarify.  Source:
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What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?    If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.  I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.  Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.   Do we know the limit for making up these objects? What is the fundamental justification for order two?
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I am going over the lectures on Machine Learning at Coursera.   I am struggling with the following. How can the partial derivative of   $$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$  where $h_{\theta}(x)$ is defined as follows  $$h_{\theta}(x)=g(\theta^{T}x)$$ $$g(z)=\frac{1}{1+e^{-z}}$$  be $$ \frac{\partial}{\partial\theta_{j}}J(\theta) =\sum_{i=1}^{m}(h_\theta(x^{i})-y^i)x_j^i$$  In other words, how would we go about calculating the partial derivative with respect to $\theta$ of the cost function (the logs are natural logarithms):  $$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$
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What is the exact use of the symbols $\partial$, $\delta$ and $\mathrm{d}$ in derivatives in physics? How are they different and when are they used? It would be nice to get that settled once and for all.  $$\frac{\partial y}{\partial x}, \frac{\delta y}{\delta x}, \frac{\mathrm{d} y}{\mathrm{d} x}$$   For what I know, $\mathrm{d}$ is used as a small infinitisemal change (and I guess the straight-up letter $\mathrm{d}$ is usual notation instead of italic $d$, simply to tell the difference from a variable). Of course we also have the big delta $\Delta$ to describe a finite (non-negligible) difference. And I have some vague idea that $\partial$ is used for partial derivatives in case of e.g. three-dimensional variables. Same goes for $\delta$, which I would have sworn was the same as $\partial$ until reading this answer on Math.SE:    Then to make the confusion total I noticed an equation like $\delta Q=\mathrm{d}U+\delta W$ and read in a physics text book that:     The fact that the amount of heat [added between two states] is dependent on the path is indicated by the symbol $\delta$...   So it seems $\delta$ means something more? The text book continues and says that:     a function [like the change in internal energy] is called a state function and its change is indicated by the symbol $\mathrm{d}$...   Here I am unsure of exactly why a $\mathrm d$ refers to a state function.   So to sum it up: down to the bone of it, what is $\delta$, $\partial$ and $\mathrm{d}$ exactly, when we are talking derivatives in physics.  Addition  Especially when reading a mathematical process on a physical equation like this procedure:  $$\delta Q=\mathrm{d}U+p\mathrm{d}V \Rightarrow\\ Q=\Delta U+\int_1^2 p \mathrm{d}V$$  It appears that $\delta$ and $\mathrm{d}$ are the same thing. An integral operation handles it the same way apparently?
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In the book of first order logic I'm reading they say that if $(\varphi_1,...,\varphi_n)$ is a proof from a set of formulae $\Sigma$, then so is $(\varphi_1,...,\varphi_n,\forall x \varphi_j)$ for all $1\leq j \leq n$. They assume that as an axiom but it doesn't make sense to me. Why if $\varphi$ is true then so is $\forall x \varphi$? For example if $\varphi$ is $x=1$ that wouldn't mean that $\forall x : x=1$ , or would it?  Maybe what I don't understand is: why $(\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)$ is true but $(\varphi \implies \forall x \varphi)$ is not true?  The book is this:
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I am trying to find a general derivative for the function: $f(x)=x^{x^{x^{...^{x}}}}$however to do that I must find $f^{\prime }$ and $f^{\prime \prime}$...etc. I am now trying to write down a general expression for $f^{\prime \prime}$ and I have stumble opon the series in the picture below, I wonder, do anyone have an idea of what sort of expantion it migth be ? I have tried breaking it down into three different sums, but I wasn't very successful!  EDIT  So since this post has gotten some attention and the title has been change I will tell you what I am on about, although I already have a post just for the question in the title from before. I am as I said earlier trying to find a general derivative for the function $f(x)=x^{x^{x^{...^{x}}}}$ where $n$ is an arbitrary number, natural number and greater or equal to 2. I have already made success when I tried to find the first derivative, and I have also found some interesting pattern emerge when the derivatives as well as n incresses. But my question here was if anyone knew what this expation might be or look similar to be, since I have yet to be lernt series etc. I did not know that the question needed such background since it's straight a forward question.
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The following Lagrangian density describes a charged scalar field. Determine the global U(1) symmetry. Make this global symmetry local and give the Lagrangian density. The Lagrangian density is given by: $$L = \partial_\mu \psi \partial^\mu \psi^* - \frac{1}{2}m^2|\psi|^2 - \frac{1}{4}\lambda|\psi|^2.$$  My first idea was to perform the following transformation (because it is mentioned we are dealing with U(1) invariance): $$\psi \rightarrow e^{i\theta}\psi$$ $$\psi^* \rightarrow e^{-i\theta}\psi^*$$ $$\partial_\mu \psi \rightarrow D_\mu \psi = (\partial_\mu + iqA_\mu)\psi$$ and $$\partial_\mu \psi^* \rightarrow D_\mu \psi^* = (\partial_\mu - iqA_\mu)\psi^*$$ However I think this is what you have to do to obtain a local symmetry out of the global one. I have no idea how to find the global one. In short my questions are:  How do you find the global U(1) symmetry for a given Lagrangian density? Is the way to determine the local symmetry out of the global U(1) symmetry achieved by transforming in the following way: $\psi \rightarrow e^{i\theta}\psi$, $\psi^* \rightarrow e^{-i\theta}\psi^*$, $\partial_\mu \psi \rightarrow D_\mu \psi = (\partial_\mu + iqA_\mu)\psi$ and $\partial_\mu \psi^* \rightarrow D_\mu \psi^* = (\partial_\mu - iqA_\mu)\psi^*$?  Thanks in advance!
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My question is: Why do we consider Lagrangian densities in  field theory (as opposed to Lagrangians as in point mechanics)? Is it simply because of the following? We wish the theories to be Lorentz invariant and as in Special Relativity space and time are no longer independent of one another we must consider the physics at each spacetime point $x^{\mu}$, implying that the theory should be locally described by a density. Additionally, to obey Poincare invariance the Lagrangian density should not be explicitly dependent on $x^{\mu}$, and for it to be Lorentz invariant it must be dependent on spatial derivatives $\nabla\phi$ of the fields as well as temporal derivatives $\partial_{t}\phi$ (as time derivatives in one reference frame will correspond to a combination of time and spatial derivatives in another frame). (Physically does the fact that $\mathscr{L}$ depends on $\nabla\phi$ as well as $\partial_{t}\phi$ because the field $\phi$ is defined at each spacetime point and so a fluctuation in $\phi$ at a point $x^{\mu}$ will produce temporal and spatial derivatives at that point which will affect fields in its immediate neighbourhood (i.e. infinitesimally close to it)?)
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Weinberg at page 300 of The Quantum Theory of Fields - Volume I says:  $L$ itself should be a space integral of an ordinary scalar function of $\Psi(x)$ and $\partial \Psi(x)/\partial x^\mu \,$, known as the Lagrangian density $\mathscr{L}$: $$ L[\Psi(t), \dot{\Psi}(t)]= \int d^3x \, \mathscr{L}\bigr(\Psi({\bf x},t), \nabla \Psi({\bf x},t), \dot{\Psi}({\bf x},t)\bigl) $$  So he says that $\mathscr{L} \, $ is a function. But Gelfand and Formin at page one of their book Calculus of variations say:  By a functional we mean a correspondence which assigns a definite (real) number to each function (or curve) belonging to some class.  So from that I'd say it is a functional. The notes of quantum field theory of my professor stay on this side, explicitly calling the lagrangian density a functional. I'm very confused at the moment. I always used this latter way of defining functionals (the Gelfand way) so Weinberg saying that $\mathscr{L}$ is a function confuses me. Can someone makes some clarity about this?
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I am a bit confused by the concepts of . In all the courses I am doing at the moment we do transformations of the form: $$ \phi(x) \rightarrow\phi'(x') = \phi(x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial' _{\nu}\phi(x') = \frac{\partial x^{\alpha}}{\partial x' ^{\nu}}\partial_{\alpha}\phi(x) $$ This is all perfectly clear to me. However, I am reading Peskin and Schoder at the moment, and they adapt an "active" point of view (their words), such that the above transformations are: $$ \phi(x) \rightarrow\phi'(x) = \phi(\Lambda^{-1}x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial _{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x). $$ I don't understand how to interpret this and especially how to derive the second equation.
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If $Z[J]$ is the generating functional for the path-integral, could any prove (or more reasonably, refer me to a proof) that $$W[J]\equiv\frac{\hbar}{i}\log\left(Z[J]\right)$$ "generates" only ?   So far I've only seen theory-dependent "examples" (basically showing how in $\phi^4$ theory the two-point function from $W$ gives only connected contributions).  I'm looking for a generic systematic proof for a general field theory.
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So I was reading this:  and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.  So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?   EDIT Ok, I changed my question a little bit:  Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: , equation 14.)  Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&amp;Lifshitz's mechanics book only consider this kind of change in the Lagrangian?
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in the book Quantum Field Theory by Itzykson and Zuber the following derivation for the stress-energy tensor is proposed (p.22):  Assume a Lagrangian density depending on the spacetime coordinates $x$ only through fields and their gradients. Under a translation we have $$\mathcal L (x+a)=\mathcal L[\phi_i(x+a),\partial_\mu\phi_i(x+a)].$$  Consider an infinitesimal $x$-dependent transformation $$\delta\phi_i=\delta a^\mu(x)\partial_\mu\phi_i(x),$$ $$\delta\partial_u\phi_i(x)=\delta a^\nu \partial_\nu\partial_\mu\phi_i(x)+\partial_\mu[\delta a^\nu (x)]\partial_\nu \phi_i(x).$$  The proof then proceeds with variation of the action and integration by parts. But why do we consider an $x$-dependent local transformation instead of a global transformation?
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I am reading about Lagrangian mechanics.  At some point the difference between the temporal derivative of a variation and variation of the temporal derivative is discussed.   The fact that the two are the same is presented in the book I am reading as a rule, commutativity, and possible non-commutative rules are mentioned too.  I do not get it: given a path $q(t)$ and its variation $\delta q(t)$, the equivalence between the variation of the derivative $\delta \dot{q}$ and derivative of the variation $\dot{\delta q}$ seems to me a fact descending straight from calculus, not an arbitrary choice.  References:   B.D. Vujanovic and T.M. Atanackovic,, (2004); p.12. C. Lanczos, The Variational principles of Mechanics.
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I am new to QFT, so I may have some of the terminology incorrect.  Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field: $$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$ The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT )  What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.  Is it simply a shortcut to obtaining the equations of motion $$(\square +m^2)\phi=0\\ (\square + m^2)\phi^*=0~?$$  I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, ; perhaps this addresses my question in a way that I don't understand.
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I am reading now E. Soper, Classical Theory Of Fields, now and sometimes it is very hard to follow the equations. So I need a side book on  to read it comfortably. Landau &amp; Lifshitz's book is not helping as its content and topics are very much different.
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I am currently in the process of teaching myself QFT.  It is not an easy task.  I have armed myself with many of the standard textbooks.  However, I am slow learner.   I get stuck on a thousand different terms which I don't understand, and a style of 'mathematics speak'  which loses me 95% of the time.   There are so many gaps in my knowledge that if I were a ship I would be at the bottom of the ocean.  What I really need is a book called 'QFT for Morons'. That's a bit harsh but 'QFT for dummies' doesn't quite express it clearly enough and besides there isn't such a book as far as I can find.  Any suggestions of resources or books that might help?   What I am really looking for is the one that explains QFT like you were talking to a five year old.  Perhaps I am asking the impossible.  I will add a bit more of a specific question here.    I have learnt that there are groups and non-abelian Lie groups like SU(2) etc., and then there a lagrangians, as well as a few other things like operators and states.   So a really dumb question is how are the lagrangians derived?  Are they determined by the group?  I am also confused because I understand some of the groups like SU(3) are used for different things in QFT.   I feel compelled to appologize for my ignorance but I am determined to make so more progress with this.   The problem when one is teaching oneself is you don't know which is the most fruitful path to follow.