In the literature on point particles in 2+1 gravity, I often see the identity  $$d^2 \phi = 2\pi \delta^{(2)}(r)dx\wedge dy$$  being used. Here, the particle is a puncture located at the origin and $(r,\phi)$ are polar coordinates around this puncture.  One example is in  (unfortunately it's behind a paywall), where in equation 3.4 the author uses the above identity to show that the torsion and curvature are both described by delta distributions. For example, for the curvature, since the connection is given in equation 3.3b by $\omega=M d\phi$, the curvature is  $$F=d\omega=Md^2\phi=2\pi M \delta^{(2)}(r)dx\wedge dy.$$  Other papers seem to be using this result as well, and cite the above paper. One example is , equations 4.6-4.7.  I know that $d^2$ is supposed to be always zero because the partial derivatives commute, so this has to fail somehow for the identity to be satisfied, but I don't know how that would happen in this case.
122048
So I know Stokes theorem for integrating exact differential forms over a chain: $$\int_cd\omega=\int_{\partial c}\omega.$$ But what if there are holes? More precisely, consider the following famous example:  Consider the punctured plane $U=\Bbb R^2\setminus\{(0,0)\}$ and the differential form $$\omega=\frac{1}{x^2+y^2}(-ydx+xdy)$$ on $U$. It is known that line integral of $\omega$ along a circle $c:[0,1]\to U, c(t)=(\cos 2\pi t,\sin 2\pi t)$, is $2\pi$.  Stokes theorem does not really apply here because neither is $\omega$ exact nor is $c$ the boundary of another chain. But it is possible to "define" the following (I saw similar things from physics textbook and notes):$$d\omega=2\pi\delta(x,y)dx\wedge dy$$where $\delta$ is the Dirac delta function, giving infinity when $x=y=0$, and $0$ otherwise, such that the integral of $\delta(x,y)dx\wedge dy$ over a subset $S$ of $\Bbb R^2$ is $1$ if the origin is an interior point of $S$. Regard $c$ as the boundary of the unit disk $D$ at the origin and Stokes theorem can be "applied" just fine, by$$\int_Dd\omega=\int_{c}\omega.$$  Is there a way to formalise these ideas? If possible, I would like to see some reference to formal definitions of such generalised exterior derivative and proofs of such generalised Stokes theorem.  I know some complex analysis (Cauchy's integral formula), the definition of a , and the definition of de Rham cohomology if it will help.
14321
Consider a scientist who is accelerating at a constant rate along the x direction, in flat minkovski space. The relevant coordinate transformations are  $$\begin{cases}t'=t \\ x'=x+\frac{1}{2}at^2 &amp;&amp; a=\text{const}  \\ y'=y \\ z'=z\end{cases}$$  The question is, do you expect the curvature tensor to vanish in the primed coordinates?   One argument is that the curvature tensor is indeed a tensor. Since the space is flat, the curvature tensor will vanish for a stationary observer and hence will vanish in all coordinate systems. This makes sense because the curvature of space-time is an intrinsic property of the manifold and doesn't  care about which reference frame (RF) we measure it in.  But when i try to apply the equivalence principle (EP) to answer the same question i wind up contradicting myself. Which is worrying because it likely means that my basic understanding of the concept is flawed.   The EP says that the scientist cannot decide whether his laboratory is uniformly accelerating in flat space or if his laboratory is fixed on the surface of his home planet, who's surface conveniently results in an acceleration of 'a' outward from the planet (recalling that a freely falling observer is now the definition of unaccelerated). So actually, the primed coordinates aren't those of someone who is accelerating uniformly in flat space, they're  equivalently the coordinates of someone fixed in a gravitational field, the space of which is curved and so has a non zero curvature tensor.   Please correct where my logic fails, it's something quite basic and fundamental so i would really like to straighten it out. Thanks.
3100
So I have been trying to understand why there is a negative sign in the following$\Delta U= -\int\vec{F}\cdot d\vec{l}$  I wanted to try to understand it by trying to derive the gravitational potential energy $E_p=-\dfrac{GMm}{r}$ I start by defining potential energy to be $0$ at $\infty$ and see what I get if I let gravity do work on the mass until a distance r away from a planet  $W_G=\int_{\infty}^{r} \vec{G}\cdot{d\vec{l}} = \int_{\infty}^{r} G\cos{\theta}dl = \int_{\infty}^{r} G dx$ G is the gravitational force so:  $W_G= \int_{\infty}^{r} -\dfrac{GMm}{x^2}dx = \Bigg[\dfrac{GMm}{x}\Bigg]$ applying limits of integration gives $W_G = \dfrac{GMm}{r}$  Thus I am confused why the minus sign, from the derivation I wrote it seems to have to be $\Delta U= \int\vec{F}\cdot d\vec{l}$ at least for gravity. I also did the same derivation for a mass on earth ($W_G = -mgh$ Where $\vec{G}$ is considered to be constant) and the result indeed is the correct one $\Delta U= -\int\vec{F}\cdot d\vec{l}$ so my confusion arises, why don't i get the same result for both cases.   I am also wondering if I carry out the same derivation but for elastic potential energy, what would I get... Thank you for any help!
67880
I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential.   I understand the generalized force   $$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$  that they derive from a Lagrangian of the form $L = \frac{1}{2}m|\vec v|^2 - V(\vec r,\vec v,t)$.  However, in the next (critical) step of the derivation, the author cites a theorem from Helmholtz saying      ...according to Helmholtz, for the existence of the Lagrangian, such a generalized force can be at most a linear function of acceleration, and it must satisfy the Helmholtz identities.   The three Helmholtz identities are then listed as:  $$\frac{\partial F_i}{\partial \dot{v_j}} = \frac{\partial F_j}{\partial \dot{v_i}},$$  $$\frac{\partial F_i}{\partial v_j} +  \frac{\partial F_j}{\partial v_i} = \frac{d}{dt}\left(\frac{\partial F_i}{\partial \dot{v_j}} +  \frac{\partial F_j}{\partial \dot{v_i}}\right),$$  $$\frac{\partial F_i}{\partial x_j} -  \frac{\partial F_j}{\partial x_i} = \frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial v_j} - \frac{\partial F_j}{\partial v_i}\right).$$  I'm trying to understand where this theorem comes from. Razavy cited a 1887 paper by Helmholtz. I was able to find , but it is in German, so I could not verify whether or not it proved the theorem. Additionally, I could not find it in any recent literature. I searched online and in Goldstein's Classical Mechanics.   The only similar concept that I can find is in the  where we have three equations known as Helmholtz conditions. Are these two concepts one in the same? If so, how should I interpret the function $\Phi$ and the matrix $g_{ij}$ that appear in the Helmholtz conditions I found online?   If the cited theorem from Razavy does not relate from the inverse Lagrangian problem, could I have some help finding the right direction?
88519
From the Lagrangian I've got the following equations of motion for the  in 2D. (The masses are different but the lengths of the two pendula are equal.) Let $m_2$ be the lowest-hanging mass.     $$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ -2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-(m_1+m_2)g/l\sin(\theta_1)$$   and     $$m_2\ddot{\theta_1}+2m_2\ddot\theta_2\cos(\theta_2-\theta_1)=\\ 2m_2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2)-m_2g/l\sin(\theta_1)$$   In the small angle approximation these become, respectively     $$(m_1+m_2)\ddot{\theta_1}+2m_2\ddot\theta_2= -2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1(m_1+m_2)g/l$$   and      $$m_2\ddot{\theta_1}+2m_2\ddot\theta_2= 2m_2\dot\theta_1\dot\theta_2(\theta_1-\theta_2)-\theta_1m_2g/l$$.   Most sources don't have the terms of order $\dot\theta$. This is because they apply the small angle approximation to the Lagrangian before taking the derivatives, thereby ignoring terms of order $\theta.$ What justification do we have for getting rid of these terms?
27781
I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$.  One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th edi, p442) is as following: $$ \langle x|[x,p]|y \rangle = \langle x|xp-px|y \rangle = (x-y) \langle x|p|y \rangle. $$ On the other hand,  $ \langle x|[x,p]|y \rangle = i \langle x|y \rangle = i \delta (x-y)$. Thus $$ (x-y) \langle x|p|y \rangle = i \delta(x-y). \tag{1} $$  We use $(x-y) \delta(x-y) = 0$. Take the derivative with respect to $x$; we have $\delta(x-y) + (x-y) \delta'(x-y) = 0$. Thus $$ (x-y) \delta'(x-y) = - \delta(x-y). \tag{2} $$  Comparing Eqs. (1) and (2), we identify $$ \langle x|p|y \rangle = -i \delta'(x-y). \tag{3} $$  In addition, we can add $\alpha \delta(x-y)$ on the right-hand side of Eq. (3), i.e. $$ \langle x|p|y \rangle = -i \delta'(x-y)  + \alpha \delta(x-y), $$ and $[x,p] = i$ is still satisfied. We can also add $$ \frac{\beta}{\sqrt{|x-y|}}\delta(x-y)$$ on the RHS of Eq. (3).  Here $\alpha$ and $\beta$ are two real numbers.  My question is, what is the most general expression of $\langle x|p|y \rangle$? Can we always absorb the additional term into a phase factor like Dirac's quantum mechanics book did?  Thank you very much in advance.
43500
The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.  Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?
58552
In classical General Relativity (meaning not modified) one can think of geodesics in two ways.    One way is to say that a geodesic is the curve which is the straightest (in analogy with the flat case) among all curves. More or less the story goes like this (correct me if I'm wrong): in the flat case geodesics are of the form $x^\mu(s)=st^\mu +b^\mu$ where $t$ and $b$ are constant vectors and $s$ is the curve parameter. The curve tangent vector is $\frac{dx^\mu}{ds}=t=const$ so $$\frac{dt^\mu}{ds}=t^\nu\partial_\nu t^\mu=0.$$ This is a tensor equation and by general covariance it  holds true in a general spacetime, mutatis mutandis: $$t^\nu\nabla_\nu t^\mu=0.$$ Another way is to obtain the equation by finding the minimum of the length functional $$L=\int \sqrt{g(t,t)} \,\, d\tau,$$ we get $$\frac{dt^\mu}{d\tau}+\Gamma^\mu_{\alpha\beta}t^\alpha t^\beta=0.$$ The above equation turns out to be just $t^\nu\nabla_\nu t^\mu=0$, with a certain parametrization choice.    The two formulations are thus equivalent. My question is when and why is it so? Is there a deep reason?  Looking up Wald's book I found the following, unclear to me, argument:     "On a manifold with a Riemannian metric, one can always find curves of arbitrarily long length connecting any two points. However, the length will be bounded from below, and the curve of shortest length connecting two points (assuming the lower bound in length is attained) is necessarily an extremum of length and thus a geodesic. Thus, the shortest path between two points is always a straightest possible path."    Later in the book he also says something about conjugate points.  I always thought it had to do with torsion: If we relax the condition $\Gamma^\mu_{\alpha\beta}-\Gamma^\mu_{\beta\alpha}=0$ then the connection is not Levi-Civita, hence $\delta_g L=0$ remains unchanged but $t^\nu\nabla_\nu t^\mu=0$ does and gives a different geodesic equation.. so Straightest $\neq$ Shortest anymore.  Can anyone clarify things to me?
61285
I'm reading the derivation of the QED generating functional on Mandl&amp;Shaw, "Quantum Field Theory" 2nd ed., 12.5.2..  The authors start from the expression (schematically): $$Z[j]=\frac{\langle 0\vert S[j]\vert 0\rangle}{\langle 0 \vert S\vert 0\rangle}$$and derive the formula: $$Z[j] = \frac{e^{V}Z_0[j]}{e^{V}Z_0[0]},\\V=i\intop \text d^4 x \mathscr L _{\text {int}}(\frac{i\delta}{\delta j(x)}),$$which is precisely what one obtains from the path integral expression: $$Z[j]=\frac{\intop \text D[\phi]e^{iS[\phi]-i(j,\phi)}} {\intop \text D[\phi]e^{iS[\phi]}}.$$ Here $Z_0$ is the generating functional for the free theory. Let me know if my notation is not clear.  In doing so, they completely disregard the normal ordering prescription in the interaction density $$\mathscr L _{\text {int}}\to \colon \mathscr L _{\text {int}}\colon.$$ Now, omitting the normal ordering amounts to include equal time contractions in the application of Wick's theorem. At the end of page 271, the authors state (and prove) that the only diagram which arises from this omission in QED is the tadpole diagram:    which, in QED, is identically zero. Also, in exercise 12.7, they state the in scalar QED, the tadpole diagram leads (in second order) to a simple correction $\delta m^2$ to the self-energy of the charged scalar.  I'm wondering, how does this work in general, that is, what is the effect of including equal time contraction in the perturbative expansion of Green functions? Do they always simply lead to constant (i.e. momentum-independent) corrections to the self-energy, which are incorporated in the physical masses through renormalization?
82001
We have $\Omega=\mathbb{R^3}\backslash \left\{ (0,0,z):z\in \mathbb{R}\right\}$ and $\omega$ the differential form: $$\omega :=\left(\frac{4x^2+2zx}{x^2+y^2}+2A(x,y)\right)dx+\left(\frac{2y}{x^2+y^2}(2x+z)\right)dy+A(x,y)\,dz,$$ where $A\in C^1(\mathbb{R^2}\backslash(0,0); \mathbb{R})$  Find a function $A$ that makes the differential form exact.  My teacher advised to use the homotopy of the curves of which I make the integral. Namely find two curves on which the integral is $0$, one that surrounds the origin and the other doesn't, and hence obtaining all the other curves through continuous deformation. For the curves that don't include the $z$-axis I can just take a ball that contains the curve, and since the ball is convex, I know that a closed differential form in a open convex set is exact (I'm pretty sure it is closed). I'm stuck on the curves that surround the $z$-axis, any ideas?
43754
Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined? For example, given the  for a  $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian? As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a  equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?  [1]: From Matthew D. Schwartz's , p.114:  A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.
51362
Let $\Omega\subset\mathbb R^n$ and consider an arbitrary functional $$ S\colon C^k(\Omega)\to\mathbb R $$ that is local, $$ \phi\overset S\mapsto \int_\Omega L(\phi,D\phi,\dots,D^k\phi) $$ for $L$ some $n$-form taking values in the $k$-jet space $J^k\phi$. We may call $S$ the action functional, and $L$ its associated Lagrangian (although we do not assume that $\delta S[\phi]=0$, i.e., we work off-shell).  There are two notions of a symmetry that are typically discussed:   Lagrangian symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ L-L'=\mathrm d\Theta $$ for some $(n-1)$-form $\Theta$, and where $L'(J^k\phi)=L(J^k\phi')$. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ S-S'=0 $$ where $S'[\phi]=S[\phi']$.   If we assume that the boundary conditions are such that $\Theta|_{\partial\Omega}=0$, then it is clear that Lagrangian symmetries are also action symmetries.  If we assume that $\Omega$ is "nice-enough" (e.g. star-shaped) then I believe that any action symmetry is also a Lagrangian symmetry, although I would like to have a more precise statement about this fact. In other words, my question is     Are these two notions of symmetry equivalent? Under what conditions?   Moreover, how much can we weaken the assumptions? For example, can we take $\Omega$ to be simply-connected instead of star-shaped? Can we take $\phi$ to only have weak derivatives instead of strong derivatives?
104664
Out of all of my time learning General relativity, this is the one identity that I cannot get around.   $$ \Gamma_{\alpha \beta}^{\alpha} = \partial_{\beta}\ln\sqrt{-g} \tag{1}$$   where $g$ is the determinant of the metric tensor $g_{\alpha \beta}$.   With the Christoffel symbol, we start by contracting  $$ \begin{align}  \Gamma_{\alpha \beta}^{\alpha} &amp;= \frac{1}{2} g^{\alpha\gamma} (\partial_{\alpha} g_{\beta\gamma}  + \partial_{\beta} g_{\alpha\gamma}  - \partial_{\gamma} g_{\alpha\beta} ) \\ &amp;= \frac{1}{2} g^{\alpha\alpha} ( \partial_{\beta} g_{\alpha\alpha}) \\ &amp;= \frac{1}{2g_{\alpha\alpha}} ( \partial_{\beta} g_{\alpha\alpha})  \end{align}\tag{2}$$   where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g_{\alpha\alpha}$.  The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,  $$\det A = \det||A^{\lambda}_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}_{\ \ \alpha}A^{1}_{\ \ \beta}A^{2}_{\ \ \gamma}A^{3}_{\ \ \delta} $$  $$(A^{-1})^{\mu}_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}_{\ \ \nu} A^{\gamma}_{\ \ \rho}A^{\delta}_{\ \ \sigma} $$  $$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$ where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}_{\ \ \mu}||$ whose entries are one-forms.  I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.
29985
Specifically, I'm asking this because I'm taking a class on General Relativity, and in Hartle's book , in Ch. 12, after having spent some time using Schwarzschild coordinates, we are introduced to two new coordinate systems, the  coordinates and the  coordinates.  The book claims that the three of these coordinates are just different coordinate representations of the same spacetime geometry.  How does one prove this claim, particularly since the coordinates do not even remotely behave identically (e.g.: when $r=2M$ in Schwarzschild coordinates, the system appears to blow up, but really the issue is with its representation in Schwarzschild coordinates)?  Obviously, there would be no point having 3 different coordinate systems if they all behave identically everywhere, but I'm having difficulty reconciling these different behaviours as all being different descriptions of the same spacetime geometry.
4081
I've started today the book of Landau and Lifshitz Vol.2: The Classical Theory of Fields $\S 2$. They start from the invariance of the speed of light, express it as the fact that $$c^2(\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2=0$$ is preserved when we change inertial frame, so they consider $$ds^2=c^2dt^2-dx^2-dy^2-dz^2,$$ and say     We have observed that if $ds=0$ in one frame then $ds'=0$ in another frame. But $ds$ and $ds'$ are infinitesimal of the same order. So it follows that $ds^2$ and $ds'^2$ have to be proportional that is $ds^2=ads'^2$...    and he goes on to prove that $a=1$.  How to translate this argument in a rigorous one? I'm really interested in this, both to understand this deduction and also to be able in future to make similar ones.
61647
I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}z^2\phi^4. $$  It was my understanding that terms of power 2 in the fields always give Feynman rules that are propagators. However, it appears that the counterterms with power 2 in the fields give a Feynman rule looking like $$i(p^2\delta_z-\delta_m),$$ instead of something with a denominator that would be more familiar. Like $\frac{i}{p^2-m^2+i\epsilon}$, from the first terms. Why is this the case? Is the idea that any term with power 2 in the fields gives a propagator wrong?
16952
I have a question about the following passage on pg. 89 of Zee's QFT in a nutshell:     At first sight, what Dirac wanted does not make sense. The equation is supposed to have the form "some linear combination of $\partial_\mu$ acting on some field $\psi$ is equal to some constant times the field." Denote the linear combination by $c^\mu\partial_\mu$. If the $c^\mu$'s are four ordinary numbers, then the four-vector $c^\mu$ defines some direction and the equation cannot be Lorentz invariant.   The simplest equation I can construct that he is referring to is  $$(c^{\mu}\partial_{\mu}-m)\phi(x)=0, $$  for $m&gt;0$. Taking $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$, and $c^{\mu}\rightarrow\Lambda^{\mu}_{\ \ \nu}c^{\nu}$. Then, the left term transforms like   \begin{align*} \Lambda^{\mu}_{\ \ \nu}c^{\nu}(\Lambda^{-1})^{\sigma}_{\ \ \mu}\partial_{\sigma}\phi(\Lambda^{-1}x) &amp;= \delta^{\sigma}_{\ \ \nu}c^{\nu}\partial_{\sigma}\phi(\Lambda^{-1}x)\\ &amp;=c^{\sigma}\partial_{\sigma}\phi(\Lambda^{-1}x). \end{align*}  Then, we have that   $$ (c^{\sigma}\partial_{\sigma}-m)\phi(\Lambda^{-1}x)=0.$$  But doesn't this show that the equation is Lorentz-invariant?