New answers tagged

1

This is easiest done not using index notation. Define the column vector $$ \vec{\phi} = \begin{pmatrix} \phi_1 \\ \phi_2 \\ \phi_3 \end{pmatrix} $$ The derivative acts on the column vector as $\mathbf{1}_3\partial_\mu$ where $\mathbf{1}_3$ is the identity matrix on the same vector space in which $\vec{\phi}$ lives. That is, $$ \partial_\mu \vec{\phi} = \...


0

This well-known problem is to find the curve joining two points, along which a particle falling from rest under the influence of gravity travels from the higher to the lower point in the least time. It's supposed that the motion on the curve is frictionless. The time to travel from point 1 to point 2 is \begin{equation} t_{12}\boldsymbol{=}\int\limits_{1}^{...


1

You are nearly finished. Consider your result $$L' = L+ \theta(2xy(a-b)-3cxy^2) + O(\theta^2).$$ In order to make $$\theta(2xy(a−b)−3cx^2) = 0 \quad \text{for all } x \text{ and } y$$ you must have the conditions $a=b$ and $c=0$.


1

Lagrangian mechanics is equivalent to Newtonian mechanics, so every problem doable with Newton you can do with Lagrange and Hamilton. Of course, each one of the types are easier to work with in different situations (dissipative forces work best with Newton, for example), but ultimately all three give the same answers, and can be used for the same systems.


1

The term you computed is identically zero, since partial derivatives commute, and the $\epsilon$ symbol is fully antisymmetric. In the language of differential forms, the extra term in the Lagrangian is $\mathrm{d}A\wedge\mathrm{d}A$, and the contribution to the equations of motion is simply $\mathrm{d}(\mathrm{d}A)=0$, since the exterior derivative is ...


2

By definition of the Levi-Civita symbol it is antisymmetric under permutations of its indices. So under permutation of $\alpha$ and $\mu$ it would obtain a minus sign. On the other hand the double derivative $\partial_\alpha\partial_\mu$ is symmetric under this permutation. The contraction of a symmetric and an antisymmetric object is 0.


3

We have $$\nabla_\mu T^{\mu\nu}=(\nabla_\mu F^\nu_{\:\,\beta})F^{\mu\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F_{\alpha\beta}\right) F^{\alpha\beta}-\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta}\tag{1}$$ Now, note the following: $$ \begin{align} -\frac{1}{4}g^{\mu\nu}\left(\nabla_\mu F^{\alpha\beta}\right) F_{\alpha\beta} &...


0

$\nabla^{α}(\nabla_{α}φ\nabla_{β}φ) -\cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ\nabla_{ν}φ) - \nabla^{α}g_{αβ}V(φ) = 0 \Rightarrow $ $\nabla^{α}\nabla_{α}φ\nabla_{β}φ +\nabla_{α}φ \nabla^{α}\nabla_{β}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ)\nabla_{ν}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla_{μ}φ\nabla^{α}\nabla_{ν}φ - \nabla^{α}g_{αβ}V(φ) = 0 \...


1

Okay here we go. $$\begin{align} \mathcal{L} & = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu}\\ &=-\frac{1}{4} g^{\mu \alpha} g^{v \beta} F_{\alpha \beta} F_{\mu \nu} \\ & = -\frac{1}{4} g^{\mu \alpha} g^{v \beta}\left(\partial_{\alpha} A_{\beta}-\partial_{\beta} A_{\alpha}\right)\left(\partial_{\mu} A_{v}-\partial_{\nu} A_{\mu}\right) \\ & = -\...


0

I think it's more straightforward to vary the action with respect to $\phi$ and cancel surface terms. Also, check this: Derivation of Klein-Gordon equation in General Relativity


2

From Ref. 1, one sees that the Euler-Lagrange equations generalize to $$\frac{\partial \mathcal{L}}{\partial\phi}=\nabla_\mu\left(\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}\right)$$ In general, when one does these things, one must make sure of two things: that the index placement is the same, and that the indexes are NOT the same in the derivative ...


1

As far as the single Gaussian $x_1$ integration goes, OP is doing the correct thing (up to possible typos) in eq. (7). However since the time-increment $$\epsilon~\ll~ \omega^{-1}\tag{i}$$ is supposed to be small (in order for Feynman's fudge factor $1/A$ to be valid), we have under the square root $$ \frac{\sin (\omega \epsilon)}{2\epsilon^2\omega\cos (\...


2

The action that goes in the path integral is a functional of the path $x(t)$. Namely, $$S[x(t)]=\int_0^T dt \frac{m}{2}\left(\dot{x}^2-\omega^2 x^2\right)$$ which can be discretized very simply $$S=\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}-\epsilon\omega^2 x_i^2$$ The $x$ in this action is not necessarily a classical trajectory, but this is ...


0

When we do a variation $\delta S$ of the action, we must state for which variables the variation is done: a variation $\delta x^a$ on the particle's trajectory or a variation $\delta A^a$ of the field. Not both at the same time. Then, the variation $\delta S$ separates the various lagrangians: the variation on the particle's path drops the field action, ...


1

You are just missing something very simple. There are two appearances of the variable $\phi_i$ in that sum $$\frac{d}{d\phi_i}\sum_j(\phi_{j+1}-\phi_j)^2=\frac{d}{d\phi_i}\left((\phi_{i+1}-\phi_i)^2+(\phi_{i}-\phi_{i-1})^2\right)=-2(\phi_{i+1}-\phi_i)+2(\phi_{i}-\phi_{i-1})$$


2

Hints: Recall that the action $S=\int \! \mathbb{L}$ is an integral over the Lagrangian 4-form $\mathbb{L}~=~ d^4x~ {\cal L}$, where ${\cal L}$ is the Lagrangian density. It's a density in the sense that under a coordinate transformation $x\to x^{\prime}$, it transforms as ${\cal L}^{\prime}={\cal L}/J$ with the inverse Jacobian $J:=\det\frac{\partial x^{\...


3

Is the answer as simple as taking the time derivative of the path to calculate velocity (which will allow a calculation of KE), and obtaining PE using mgh? Yes, it is. But that doesn't seem reasonable, since conservation of energy can easily be violated here. It is true that a path drawn down at random easily violates energy conservation, but this is ...


2

Yes it is that simple. For any particular trajectory $\vec{r}(t)$ you can compute the velocity $\vec{v}(t)$ and thus the kinetic and potential energies. More generally, the Lagrangian can be an arbitrary function $L=L(\vec{r},\vec{v},t)$ but these are still all things you know if you know the time-parametrised trajectory. The only 'rule' when calculating the ...


0

The $(\partial\phi)^2$ will give you the momentum dependence in the propagator, $\sim 1/(p^2+m^2)$. Just as you can expand in $g$ you can as well expand in the other parameters $\lambda, \alpha$ and $\beta$. But in all honesty, if these are really questions you have, then trying to learn QFT seems a bit ambitious to say the least and I would suggest you to ...


1

I figured this out a long time ago, should probably have deleted this or just added my own answer. Anyway, it is really the angular momentum and I should have never passed to coordinates. Also, I should have kept the mass $m$, for psychological reasons. The point is that $m \det(\gamma(t),\dot{\gamma}(t), v) = \langle m\gamma(t)\times \dot{\gamma}(t), v\...


1

The type of non-holonomic constraints, that Ref. 1 is discussing at this point, are so-called semi-holonomic constraints $$ a_{\ell}(q,\dot{q},t)~\equiv~ \sum_{j=1}^n a_{\ell j}(q,t)\dot{q}^j+a_{\ell t}(q,t)~=~0, \qquad \ell~\in \{1,\ldots, m\}. \tag{1}$$ The upshot is the following: On one hand, it is possible to use d'Alembert's variational principle to ...


0

I'll put my thoughts on this question as an answer to reduce the length of the comments. However, this answer should not(!!) be interpreted as a formal answer, but merely as a lengthy comment (there are undoubtedly more qualified people on this site to give a full answer). First of all, in the case of Chern-Simons theory with a gauge fields in any other (...


1

How about this. We wish to show $$\frac{\partial}{\partial r_{i}}\left(\frac{d \phi}{d t}\right)=\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d t}\right).$$ Consider the right hand side. $$\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d t}\right)=\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\...


2

The new action is $$ S' = \int_{t_1}^{t_2}L' \text{d}t = \int_{t_1}^{t_2}\left(L + e\frac{\text{d}\phi}{\text{d}t}\right)\text{d}t = S + e\left[\vphantom{1_1^1} \phi\right]_{t_1}^{t_2}. $$ Since the extra term is constant, the variation of the action hasn't changed: $\delta S' = \delta S$, and therefore the EL equations remain the same.


0

My bad, I see now what happened, and as @Qmechanic noted above, Lagrangians are invariant under the addition of total derivatives.


2

There are at least 2 issues with OP's discussion (v2): One should properly distinguish between total and explicit spacetime derivatives, cf. e.g. my Phys.SE answer here. In particular, an infinitesimal quasisymmetry of the Lagrangian (density), means by definition that the infinitesimal variation is a total (space)time divergence. Note that not all terms ...


1

If the oscillation is small, then $\eta(t)=A\cdot f(t)$, where $A$ is the (small) amplitude of oscillation and $f(t)$ is a sinusoidal function of some finite frequency $\omega$. The products $\eta\eta$, $\dot{\eta}\eta$, $\ddot{\eta}\ddot{\eta}$, $\ldots$ are all of the form $A^2\cdot\omega^n\cdot g(t)$, where $\omega^n$ is finite and $g(t)$ is sinusoidal. ...


1

This is necessary because the action integral should vanish for arbitrary variations.


1

Diffeomorphism invariance implies (via Noether's 2nd theorem) that $$\nabla_{\mu} T^{\mu\nu}~\approx~0.\tag{1}$$ Perhaps surprisingly, there are no conserved quantities associated with the identity (1) per se for generic spacetime metric $g_{\mu\nu}$. However, if the metric $g_{\mu\nu}$ has a Killing vector field $K^{\mu}$, then the $4$-current $$J^{\mu}~:=...


1

This isn't a fact about general relativity, it's a fact about special relativity. Based on hundreds of years worth of experiments, we know that the energy-momentum four-vector is conserved locally. This means that the stress-energy tensor has zero divergence. (The stress-energy tensor isn't conserved. What's conserved is the energy-momentum four-vector.) If ...


0

You can more or less obtain it from the definition of the expression. It is really easy to get confused here so I would encourage you to take this whole thing one big conceptual step backwards. There are two things to emphasize. The realm of mathematical purity in multivariate calculus The first is about functions. In the realm of pure mathematics, a ...


1

This is just the Chain Rule (from multivariable calculus). You have then the function (the lagrangian): $L(q_{i},\dot{q}_{i},t)$. Then applying the differentiation (total derivative) with respect to time you have: $$\frac{\mathrm{d}}{\mathrm{d}t}\Big[L(q_{i},\dot{q}_{i},t)\Big] = \frac{\partial L}{\partial q_{i}}\frac{dq_{i}}{dt}+\frac{\partial L }{\...


2

That's just the regular chain rule from multivariable calculus, recall that the Lagrangian is a function of positions, velocities and time $$\mathcal{L}\left(q,\dot{q},t\right)$$ The first two terms of your equation come from the derivatives of $\mathcal{L}$ respect to $q$ and $\dot{q}$ while the third term comes from the derivative of $\mathcal{L}$ respect ...


3

Physicists conventionally normalize the square root Lagrangian a bit differently, namely as$^1$ $$L_0~:=~ -m\sqrt{-\dot{x}^2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\tag{1}$$ where $m>0$ is the mass. (OP assumes that $m=1$.) It is important to note that the Legendre transformation of the square root Lagrangian (1) is ...


1

Prof. Zachos gave a nice and clean argument, however since there is a comment that his reply is not yet very clear let me venture a different approach. If my answer is not satisfactory, let me know and I will delete it. What I understand from the comments is that you need some clarification for the minimum. Consider the function $E(R)$ rather than its ...


0

The boundary conditions do not affect the Lagrangian of the problem. You may use von-Karman strains to take the large rotations into account, which is simplified form of Green's strains. First, define displacement fields: $$ u_1(x,t) = u(x,t) - z \frac{\partial v(x,t)}{\partial x}$$ $$ u_2(x,t) = v(x,t) $$ I consider this problem as two dimensional ...


1

Adding minimally coupled scalar matter $\phi$ does not remove the gauge symmetry. In particular, the Legendre transformation is still singular. The Faddeev-Popov method (or one of its equivalent formulations) should still be used.


2

Rather than asking why we should impose this invariance, I think it would make more sense to ask why we should relax it. The Lagrangian is a relativistic scalar, which means that it has to be invariant under any coordinate transformation. An infinitesimal change of coordinates is just one type of coordinate transformation. The physical interpretation is ...


-2

Invariance under certain transformations mean conserved physical properties.


2

Hamilton's first principal function(al) is nowadays usually called the action (functional). What Maupertuis historically in 1744 called the action (functional) is nowadays usually called the abbreviated action (functional). See also this related Phys.SE post. References: QFT lectures of Sidney Coleman, edited by B.G. Chen et. al., 2019; footnote 2 on p. ...


0

I'm pretty sure this is the same kind of problem as that which led to covariant differentiation. In generalized coordinates our coordinate transformations are position dependent. Assuming our constraints are time-independent, we can think of the (differential) coordinate transformations as persisting in time. At each point along the configuration path, we ...


1

If $Q$ is a configuration manifold and $G$ is a Lie group acting on $Q$ (usually $G = \Bbb R$ with a $1$-parameter group of diffeomorphisms of $Q$) leaving a Lagrangian $L:TQ\to \Bbb R$ invariant, for every $X\in \mathfrak{g}$ we have the Noether charge generated by $X$, the map $\mathscr{J}^X\colon TQ\to \Bbb R$ given by $$\mathscr{J}^X(x,v) =\mathbb{F}L(x,...


2

You can use the lagrangian approach to solve it. You know that it is fixed to the surface so you can use $x, y$ as the independent coordinates. So the lagrangian is: $$L = T-V =\frac{1}{2}m\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)-mgz$$ Where $T$ is kinetic energy and $V$ is potential energy. You now use the fact that $z$ can be expressed in terms of ...


0

You might be misunderstanding the notation. $$ \delta \Phi = \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi} -\Phi +O(\alpha^2)= \frac{i}{2}\alpha_j\sigma^j \Phi, $$ so that $$ \delta \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} = \frac{i}{2} \begin{pmatrix} \alpha_3&\alpha_1-i\alpha_2\\ \alpha_1+i\alpha_2&-\alpha_3 ...


1

Correct me if I'm wrong on this, guys, and take this answer with a grain of salt until verified by someone more experienced. For a scalar field $\phi$ in flat space, Question 1: The derivative of a scalar field is a vector, which we can see because we pick up an index $\mu$ when we operate $\partial_\mu$ on $\phi$. This is like the gradient of a scalar ...


0

In case someone else bumps into the same question, I think I found the answer: The Euler's theorem states that if $f$ is a homogeneous function of degree $n$ in the variables $x_i$, then $$\sum_i x_i\frac{\partial f}{\partial x_i}=nf.$$ So for example, if $f$ is a function of two variables $x_1, x_2$ and it is homogeneous, say to 3rd degree, in ...


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