# Demonstration of Noether's Theorem [closed]

So, as many, many people before me, I'm trying to get some insight on Noether's Theorem and its demonstration. As I'm in the process of self-teaching here, there are several things I'm "missing" or "not understanding". I made myself a way through Tong's notes, books like Peskin & Schroeder, Mandl & Shaw, Greiner (Field Quantization), etc., and several questions in PSE like this one and this one. Nevertheless, the more I read, the more questions I have. I will try to compile everything I've got so far, and I will write my questions bolded. I'm sorry if this post turns out to be a little too long, but I'm trying to be as accurate as possible, but without diving into the realm of group theories, that is, I'm trying so demonstrate the theorem at a pre-Ph.D. level. So, without further ado, I'll simply start.

Suppose we have an infinitesimal- transformation $$x^{\mu}\rightarrow\tilde{x}^{\mu}=x^{\mu}+\delta x^{\mu}$$, which transforms our integration region $$\Omega\rightarrow\tilde{\Omega}$$. We have that under such a transformation, the fields may have a total variation $$\Delta\phi^a$$ such that: $$$$\tag{1} \Delta\phi^a = \tilde{\phi}^a(\tilde{x})-\phi^a(x)\,.$$$$

1) Where is $$\Delta\phi^a$$ evaluated at? $$x$$ or $$\tilde{x}$$?

I will assume, without understanding why, that $$\Delta\phi^a = \Delta\phi^a(x)$$. This variation must not be confounded with the local variation showing only how the fields change:

$$$$\tag{2} \delta\phi^a(x) = \tilde{\phi}^a(x)-\phi^a(x)\,.$$$$

Nevertheless, they are related, because at least at first order (and assuming $$\delta x^{\mu}$$ and $$\delta \phi^a(x)$$ of the same order):

$$$$\tag{3} \Delta\phi^a(x) \simeq \delta\phi^a(x) + \delta x^{\mu}\frac{\partial\phi^a}{\partial x^{\mu}}(x)\,.$$$$

This can be easily proven by using the definition of derivative at a first order:

$$$$\frac{df}{dx}(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\simeq\frac{f(x+h)-f(x)}{h}\rightarrow f(x+h) \simeq f(x)+h\frac{df}{dx}(x)\,,$$$$ i.e. $$\tilde{\phi}^a(\tilde{x}) = \tilde{\phi}^a(x+\delta x) = \tilde{\phi}^a(x) + \delta x^{\mu}\frac{\partial\tilde{\phi}^a}{\partial x^{\mu}}(x)$$, and using equation (2) sometimes.

Later will also come in handy to have an expression for $$\Delta(\partial_{\mu}\phi^a(x))$$. My first assumption here is that: $$$$\tag{4} \Delta(\partial_{\mu}\phi^a) = \tilde{\partial}_{\mu}\tilde{\phi}^a(\tilde{x})-\partial_{\mu}\phi^a(x)$$$$ 2) Same question as above, where is $$\Delta(\partial_{\mu}\phi^a)$$ evaluated at?

3) Is expression (4) actually correct?

I will calculate this by starting with $$\partial_{\mu}(\Delta\phi^a)$$ :

\begin{align} \partial_{\mu}(\Delta\phi^a) &= \partial_{\mu}\tilde{\phi}^a(\tilde{x}) - \partial_{\mu}\phi^a(x)\\ &= \tilde{\partial}_{\mu}\tilde{\phi}^a(\tilde{x})- \partial_{\mu}\phi^a(x) + \partial_{\mu}\tilde{\phi}^a(\tilde{x}) - \tilde{\partial}_{\mu}\tilde{\phi}^a(\tilde{x})\\ &= \Delta(\partial_{\mu}\phi^a) + \frac{\partial\tilde{x}^{\sigma}}{\partial x^{\mu}}\frac{\partial\tilde{\phi}^a}{\partial\tilde{x}^{\sigma}}(\tilde{x})-\tilde{\partial}_{\mu}\tilde{\phi}^a(\tilde{x})\,. \end{align} To make things shorter, last two terms can be worked by using $$\tilde{x}^{\mu}=x^{\mu}+\delta x^{\mu}$$, expanding $$\tilde{\phi}^a(\tilde{x}) = \tilde{\phi}^a(x+\delta x)$$ at first order, replacing, and keeping again all terms at first order, arriving to: $$$$\tag{5} \Delta(\partial_{\mu}\phi^a) = \partial_{\mu}(\Delta\phi^a) + \partial_{\mu}(\delta x^{\sigma})\partial_{\sigma}\phi^a(x)\,.$$$$

After the detour, we come back to Noether. I decided to proceed by using the invariance of the action: $$$$S[\phi^a(x),\partial_{\mu}\phi^a(x)]=\int_{\Omega}d^4x\;\mathcal{L}(\phi^a(x),\partial_{\mu}\phi^a(x))\,.$$$$

To shorten the following expressions, I'll choose to abuse the notation: $$S[\phi^a(x),\partial_{\mu}\phi^a(x)]=S[x]$$ and $$\mathcal{L}(\phi^a(x),\partial_{\mu}\phi^a(x))=\mathcal{L}(x)$$. We need our transformation to produce a variation of the action of maximum a boundary term, which is the integral of the divergence of a smooth field $$W^{\mu}$$, satisfying $$\left.W^{\mu}\right|_{\partial\Omega}=0$$, which is our condition for a well behaved variational principle. In summary, we need that:

\begin{align} \delta S &= \tilde{S}[\tilde{x}]-S[x]\\ &= \int_{\tilde{\Omega}}d^4\tilde{x}\;\tilde{\mathcal{L}}(\tilde{x}) - \int_{\Omega}d^4x\;\mathcal{L}(x)\tag{6}\\ &= \int_{\Omega}d^4x\;\partial_{\mu}W^{\mu}(x)\,. \end{align}

4) I assume that integrating $$\partial_{\mu}W^{\mu}$$ in $$\Omega$$ or $$\tilde{\Omega}$$ is exactly the same, it depends on which set of coordinates we would like to have at the end of the demonstration. Am I right?

5) In expression (6), should the first Lagrangian have a tilde or not?

If it doesn't have the tilde, and as stated by Goldstein, the underlying assumption would be that the functional form of the Lagrangian does not change with the transformation. This means, as an example, that if the Lagrangian was $$\mathcal{L}(x) = 2\phi_1(x) + \phi_2(x)$$, then after the transformation $$\tilde{\mathcal{L}}(\tilde{x}) = 2\tilde{\phi}_1(\tilde{x}) + \tilde{\phi}_2(\tilde{x})=\mathcal{L}(\tilde{x})$$. As far as I understand, this is not the general case, and there is a very good example/explanation in this post at PSE. I will therefore assume that the first term in my expression (6) must have a tilde.

In next step is where most of my headaches start. To find out what $$\delta S$$ is, I would like to turn the first term in (6) to an integral in $$\Omega$$. The story with the Jacobian for $$d^4\tilde{x}$$ is not a problem for me. My problems come with the Lagrangian. My first assumption would be, to try to go to first order in its variations, as follows:

\begin{align} \tilde{\mathcal{L}}(\tilde{x}) &= \tilde{\mathcal{L}}(\tilde{\phi}^a(\tilde{x}),\tilde{\partial}_{\mu}\tilde{\phi}^a(\tilde{x}))\\ &= \tilde{\mathcal{L}}(\phi^a(x) + \Delta \phi^a(x),\partial_{\mu}\phi^a(x)+\Delta(\partial_{\mu}\phi^a(x)))\\ &\overset{?}{\simeq} \tilde{\mathcal{L}}(x) + \Delta\phi^a(x)\frac{\partial\tilde{\mathcal{L}}^a}{\partial\phi^a}(x) + \Delta(\partial_{\mu}\phi^a(x))\frac{\partial\tilde{\mathcal{L}}}{\partial(\partial_{\mu}\phi^a)}(x)\,. \end{align}

6) Is last expression correct? If so, how do I get rid of the tilde's?

I tried to assume (I don't know why this should be true for the Lagrangian) that there is a local transformation for the Lagrangian as well, satisfying $$\tilde{\mathcal{L}}(x) = \mathcal{L}(x) + \delta\mathcal{L}(x)$$, replaced it in last expression, and then I used expressions (3) and (5), the Jacobian in $$d^4\tilde{x}$$, but I never seem to get anywhere close to the expresion I should be getting, which is expression (2.48) on the book of Greiner (note that I used here a different notation): $$$$\delta S = \int_{\Omega} d^4x\left(\delta\mathcal{L}(x) + \frac{\partial}{\partial x^{\mu}}\left(\mathcal{L}(x)\delta x^{\mu}\right)\right)\,.$$$$ The next steps after this expression is reached are not so complicated, given that Greiner expands the local transformation of the Lagrangian at first order (very much like when finding the Euler-Lagrange equations), plays with the variations a little bit, and finally gets to the conserved current.

Once again, I'm sorry for the extension of the post, I have invested weeks already on this, and I do not seem to be able to advance somehow. Any inputs would be very much appreciated!

• Check section 2.4 in A First Book of Quantum Field Theory Second Edition by Lahiri and Pal. It might help. Jan 30 at 11:20
• @KasiReddySreemanReddy Thank you for the recommendation! I checked it and it confirmed my suspicions, that the answers to my questions 1) and 2) is "evaluated at $x$", but I still don't know why. And in equation (2.38) the book assumes $\tilde{\mathcal{L}}(\tilde{x})=\mathcal{L}(\tilde{x})$ also without explaining why. Jan 30 at 11:59
• @Condereal for 1) and 2) there is no special reason why $\Delta\phi^a$ is defined as a variable of $x$. $x$ and $\tilde{x}$ have a one-to-one correspondence. So we could have defined it using $\tilde{x}$, but it will be simple if we express it in $x$ then the entire right-hand side will be expressed in $x$. Jan 30 at 12:21
• are you comfortable with some basic differential geometry? differential forms, pullbacks, Lie derivatives? Feb 1 at 3:20
• @peek-a-boo I have acquired some of the basic skills during the years, I know the terminology and the theory, but maybe cannot apply them directly to a case like this one. Feb 1 at 18:53

These are the same issues I struggled with, and the resolution is obviously just to have clear definitions, not hidden behind (what I find to be) the mysterious tilde/ primed notation. I'll restrict attention just to the case of a scalar field. I'll first talk about the different variations of a scalar field, and then talk about variations of the Lagrangian then the action.

1. Changes to Scalar fields.

Fix a smooth oriented $$n$$-dimensional manifold $$M$$ (we interpret this as spacetime with $$n=4$$). For simplicity, let us just stick to real scalar fields, meaning smooth maps $$\phi:M\to\Bbb{R}$$. Before we get to the actual details, some preliminaries:

• Consider a smooth deformation $$\Phi$$ of $$\phi$$. This means we consider a smooth map $$\Phi:I\times M\to \Bbb{R}$$, where $$I\subset \Bbb{R}$$ is some open interval containing the origin, such that $$\Phi(0,\cdot)=\phi(\cdot)$$. It is of course tradition to write the values $$\Phi(\epsilon, p)$$ as $$\Phi_{\epsilon}(p)$$ for $$(\epsilon,p)\in I\times M$$.

• Consider a smooth vector field $$X$$ on $$M$$, and let $$\Psi$$ denote its flow (this is 'basic' ODE theory). The name "flow" really is apt here, because if you imagine a little marble being dropped in a river, it will follow a trajectory naturally as described by the flow of the water. More mathematically, the interpretation of the flow map $$\Psi$$ is that given a point $$p\in M$$ and sufficiently small $$|\epsilon|$$, the quantity $$\Psi_{\epsilon}(p)$$ can be thought of as "where the point $$p$$ ends up $$\epsilon$$ units of time later, if it is left under the influence of the vector field $$X$$".

It is standard terminology to call the vector field $$X$$ the "infinitesimal generator of $$\Psi$$". Here the adjective "infinitesimal" refers to the fact that this is at the level of tangent spaces. In a local coordinate chart $$(U, (x^1,\dots, x^n))$$, we can write the vector field as $$X=X^{\mu}\frac{\partial}{\partial x^{\mu}}$$. These $$X^{\mu}$$ are what you've written as $$\delta x^{\mu}$$.

Now that we have the concept of a deformation and a flow, we can start talking precisely about the various ways things change.

1. The first way to change a scalar field, is to consider a deformation of it. At an "infinitesimal level" this gives rise to a variation $$\delta \phi$$. By definition $$\delta\phi:M\to\Bbb{R}$$ is a smooth map defined for each $$p\in M$$ as $$\delta\phi(p):=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Phi_{\epsilon}(p)$$.

2. The second way to investigate a change in a scalar field arises due to the effect of the vector field $$X$$ itself, which moves points in the manifold around. More precisely, this is the idea of the Lie derivative $$L_X\phi$$. By definition, this is a smooth map $$M\to\Bbb{R}$$, defined for each $$p\in M$$ as $$(L_X\phi)(p):=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}(\Psi_{\epsilon}^*\phi)(p)=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\phi(\Psi_{\epsilon}(p))$$. In a local coordinate chart we can write $$L_X\phi=X^{\mu}\frac{\partial \phi}{\partial x^{\mu}}$$ (traditional physics texts write this term as $$\delta x^{\mu}\partial_{\mu}\phi$$).

3. The third way is to combine both effects. We define $$\Delta \phi:M\to\Bbb{R}$$ as $$(\Delta \phi)(p):=\frac{d}{d\epsilon}\bigg|_{\epsilon}(\Psi_{\epsilon}^*\Phi_{\epsilon})(p)=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Phi_{\epsilon}(\Psi_{\epsilon}(p))$$. Using the chain rule, you can show $$\Delta \phi=\delta \phi+L_X\phi$$ (as a hint for how to efficiently prove this, fix a point $$p\in M$$ and consider the function $$h(s,t)=(\Psi_s^*\Phi_t)(p)=\Phi_t(\Psi_s(p))$$. This is a smooth function of two real variables, defined on a small open set around the origin. The goal is to calculate $$\frac{d}{d\epsilon}\bigg|_{\epsilon=0}h(\epsilon,\epsilon)$$, which by the chain rule is just $$\frac{\partial h}{\partial s}(0,0)+\frac{\partial h}{\partial t}(0,0)=\frac{d}{ds}\bigg|_{s=0}h(s,0)+\frac{d}{dt}\bigg|_{t=0}h(0,t)$$, and recall $$\Psi_0=\text{id}_M$$ and $$\Phi_0=\phi$$).

Hopefully this answers your question of where things are evaluated. We have $$\Delta \phi=\delta\phi+L_X\phi$$; this is an equality of smooth functions on $$M$$. You evaluate everything at the same point $$p$$. This equation is just telling you that the total change in the scalar field $$\phi$$ arising from the deformation and the vector field's flow moving points around is imply the sum of the two separate effects (by chain rule).

You next ask about things like $$\Delta (\partial_{\mu}\phi)$$ and so on, but if you just stick to the $$\Psi_{\epsilon}$$ and its pullback, and to the deformation $$\Phi_{\epsilon}$$, this makes things much clearer (for me anyway), and you'll never even have to write things like $$\Delta(\partial_{\mu}\phi)$$.

2. Changes to Lagrangian.

One technical detail is that rather than considering the Lagrangian as a real-valued function, it is much more convenient to consider it as being $$n$$-form valued, i.e to include $$dx^1\wedge \cdots \wedge dx^n$$ as part of its definition. So, let $$\Lambda$$ denote this object; it eats scalar fields and their derivatives and outputs $$n$$-forms. So, in a local coordinate chart $$(U,(x^1,\dots, x^n))$$ and points $$p\in U$$, we can write \begin{align} \Lambda_{\phi}(p)&:=\Lambda(\phi(p),\partial_{\mu}\phi(p)):=\mathcal{L}(\phi(p),\partial_{\mu}\phi(p))\,(dx^1\wedge \cdots \wedge dx^n)(p) \end{align} So, $$\Lambda_{\phi}$$ is a perfectly good object to integrate on the oriented manifold $$n$$ (remember that $$n$$-forms should be integrated on oriented $$n$$-manifolds).

Again, there are three ways to change the Lagrangian:

1. Simply plug in the deformed field, i.e consider $$\Lambda_{\Phi_{\epsilon}}$$.
2. Plug in the original field $$\phi$$ but consider pullback by the flow map $$\Psi_{\epsilon}$$, i.e $$\Psi_{\epsilon}^*(\Lambda_{\phi})$$.
3. Do both of the above steps by considering $$\Psi_{\epsilon}^*(\Lambda_{\Phi_{\epsilon}})$$.

Again, for each sufficiently small $$\epsilon$$, the three procedures above give us a new $$n$$-form on $$M$$. We can differentiate at $$\epsilon=0$$ to obtain three variations of the lagrangian. THe first we can denote $$\delta(\Lambda_{\phi})$$ (variation due to plugging in deformed field), the second is by definition the Lie derivative $$L_X(\Lambda_{\phi})$$ (the variation due to the vector field's flow moving points around), and the final we can denote $$\Delta(\Lambda_{\phi})$$, and it equals the sum $$\Delta(\Lambda_{\phi})=\delta(\Lambda_{\phi})+L_X(\Lambda_{\phi})$$ (the reason for the sum is same chain rule reasoning as above).

Connecting back to more classical notation, writing $$\tilde{\mathcal{L}}(\tilde{x})$$ means you're considering the third type of change $$\Psi_{\epsilon}^*(\Lambda_{\Phi_{\epsilon}})$$. Why? The tilde in $$\tilde{\mathcal{L}}$$ denotes that we're plugging in the deformed fields into the Lagrangian (so something like $$\tilde{\mathcal{L}}(p)$$ is shorthand for $$\mathcal{L}(\Phi_{\epsilon}(p),\partial\Phi_{\epsilon}(p))$$) and the final $$(\tilde{x})$$ indicates that we should do a pullback $$\Psi_{\epsilon}^*$$ to capture the effect of the vector field moving things around. The expression $$\tilde{\mathcal{L}}(x)$$ is taken to mean you plug in just the deformed fields, i.e what I have denoted as $$\Lambda_{\Phi_{\epsilon}}$$ (the $$(x)$$ rather than $$(\tilde{x})$$ tells us we don't do any pullback via the vector field's flow).

3. Changes to Action

I'm sure you can guess what I'm about to say here. For each small enough $$\epsilon$$, and "nice" open set $$\Omega\subset M$$ (say for example having compact closure with smooth boundary), we can consider three types of integrals: \begin{align} \int_{\Omega}\Lambda_{\Phi_{\epsilon}}\quad\text{or}\quad\int_{\Omega}\Psi_{\epsilon}^*(\Lambda_{\phi})\quad \text{or}\quad \int_{\Omega}\Psi_{\epsilon}^*(\Lambda_{\Phi_{\epsilon}}). \end{align} In words, we take the three types of varied Lagrangians (either by virute of the scalar field being deformed, or by the vector field moving points around or both), and integrate the resulting $$n$$-form over a given set $$\Omega$$. I hope you realize now that there's no need to go down to the level of coordinates and look at the images of the set $$x[\Omega]$$ or $$x'[\Omega]$$ or whatever.

The general calculation of course takes both effects into account, so we focus on the third integral. The resulting total variation in the action is \begin{align} (\Delta S)(\phi; \Omega)&:=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\int_{\Omega}\Psi_{\epsilon}^*(\Lambda_{\Phi_{\epsilon}})\\ &=\int_{\Omega}\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Psi_{\epsilon}^*(\Lambda_{\Phi_{\epsilon}})\\ &=\int_{\Omega}\Delta (\Lambda_{\phi})\\ &=\int_{\Omega}\delta(\Lambda_{\phi})+L_X(\Lambda_{\phi}) \end{align} These are exactly the two terms which one arrives at. At this stage calculating $$\delta(\Lambda_{\phi})$$ is the usual Euler-Lagrange type calculation. The term $$L_X(\Lambda_{\phi})$$ can be massaged slightly if you invoke Cartan's magic formula that Lie derivative on differential forms is given by $$L_X=d\iota_X+\iota_Xd$$, where $$\iota_X$$ denotes interior product and $$d$$ is exterior derivative. Now, since $$\Lambda_{\phi}$$ is an $$n$$-form on an $$n$$-manifold, its exterior derivarive vanishes, so we just get $$L_X(\Lambda_{\phi})=d(\iota_X \Lambda_{\phi})\equiv d(X\,\,\lrcorner \,\,\Lambda_{\phi})$$, where $$\lrcorner$$ is just another symbol for the interior product. A straightforward coordinate calculation then shows that this term is exactly $$\frac{\partial (\mathcal{L}(x)X^{\mu})}{\partial x^{\mu}}\, dx^1\wedge \cdots \wedge dx^n$$; maybe you'll find this answer of mine helpful in carrying out the coordinate calculations with exterior derivatives and interior products.

4. Computations in Local Coordinates.

First, I'll write out $$\delta(\Lambda_{\phi})$$ in local coordinates $$(U,(x^1,\dots, x^n))$$. Fix a point $$p\in U$$, then we get: \begin{align} \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Lambda_{\Phi_{\epsilon}}(p)&=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\mathcal{L}\left(\Phi_{\epsilon}(p), \partial_{\mu}\Phi_{\epsilon}(p)\right)\, (dx^1\wedge \cdots \wedge dx^n)(p)\\ &=\left[\frac{\partial \mathcal{L}}{\partial \phi}\bigg|_{(\phi(p),\partial\phi(p))}\cdot \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Phi_{\epsilon}(p) + \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\bigg|_{(\phi(p),\partial\phi(p))}\cdot \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\partial_{\mu}\Phi_{\epsilon}(p)\right](dx^1\wedge\cdots \wedge dx^n)(p)\\ &=\left[\frac{\partial \mathcal{L}}{\partial \phi}\bigg|_{(\phi(p),\partial\phi(p))}\cdot\delta\phi(p)+\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\bigg|_{(\phi(p),\partial\phi(p))}\cdot \partial_{\mu}(\delta\phi)(p)\right]\,(dx^1\wedge\cdots \wedge dx^n)(p). \end{align} The first line is just the chain rule, and the second line used the definition of $$\delta\phi$$ in the first term, and in the second term I used the fact that for smooth enough functions, derivatives commute which is why I could swap the $$\partial_{\mu}$$ with $$\frac{d}{d\epsilon}\bigg|_{\epsilon=0}$$.

Now, we can do the second term $$L_X(\Lambda_{\phi})$$. Here, I won't assume any familiarty with Cartan's formula or anything. We'll just work as much as possible from the definitions. One thing you should note however is that pullback of differential $$n$$-forms is different from pullback of functions, in the sense that we don't just do composition. We also have to deal with the $$dx^1\wedge \cdots\wedge dx^n$$ term; but this term transforms with the Jacobian determinant term, as you've probably seen in other courses. Let us write \begin{align} \Lambda_{\phi}=\mathcal{L}_{\phi}\,dx^1\wedge \cdots \wedge dx^n \end{align} so $$\mathcal{L}_{\phi}:U\to\Bbb{R}$$ is defined by plugging in the field: $$\mathcal{L}_{\phi}(p)=\mathcal{L}(\phi(p),\partial_{\mu}\phi(p))$$. Now, pullback behaves as follows: \begin{align} \Psi_{\epsilon}^*(\Lambda_{\phi})&=\Psi_{\epsilon}^*(\mathcal{L}_{\phi}\,dx^1\wedge \cdots \wedge dx^n)\\ &=(\Psi_{\epsilon}^*\mathcal{L}_{\phi})\cdot \Psi_{\epsilon}^*(dx^1\wedge \cdots \wedge dx^n)\\ &=(\mathcal{L}_{\phi}\circ \Psi_{\epsilon}) \cdot \det\left(\frac{\partial (x^i\circ\Psi_{\epsilon})}{\partial x^j}\right)\, dx^1\wedge \cdots \wedge dx^n \end{align} So, if we now want to differentiate with respect to $$\epsilon$$ at $$\epsilon=0$$, we apply the product rule (we have a product of functions of $$\epsilon$$). THe first term's derivative is as I discussed in section 1, the definition of the Lie derivative $$L_X(\mathcal{L}_{\phi})$$, which simplifies to $$X^{\mu}\frac{\partial \mathcal{L}_{\phi}}{\partial x^{\mu}}$$. At $$\epsilon=0$$, the second term is the determinant of identity matrix so $$1$$. Next, the derivative of the second term is \begin{align} \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\det\left(\frac{\partial (x^i\circ\Psi_{\epsilon})}{\partial x^j}\right) &=\text{trace}\left(\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\frac{\partial (x^i\circ \Psi_{\epsilon})}{\partial x^j}\right)=\text{trace}\left(\frac{\partial X^i}{\partial x^j}\right)=\frac{\partial X^{\mu}}{\partial x^{\mu}} \end{align} where in the middle, I again used commutativity of derivatives for smooth functions to swap $$\frac{d}{d\epsilon}$$ with $$\frac{\partial}{\partial x^{j}}$$. So, putting all of this together, we get \begin{align} L_X(\Lambda_{\phi})=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Psi_{\epsilon}^*(\Lambda_{\phi})&=\left[X^{\mu}\frac{\partial \mathcal{L}_{\phi}}{\partial x^{\mu}}\cdot 1+ \mathcal{L}_{\phi}\cdot \frac{\partial X^{\mu}}{\partial x^{\mu}}\right]\,dx^1\wedge \cdots \wedge dx^n\\ &=\frac{\partial (\mathcal{L}_{\phi} X^{\mu})}{\partial x^{\mu}}\,dx^1\wedge \cdots \wedge dx^n. \end{align}

• wow, now this is a detailed answer! Thank you very much! I think I got it, but I'll have to dust off my differential geometry books to be sure, and then I'll come back here. Hopefully I'll have the time for it this weekend. Once again, thank you for your effort! Feb 5 at 9:11