Tag Info

New answers tagged

2

In this answer we apply the general non-local theory developed in my Phys.SE answer here to OP's non-local example. Let us for simplicity assume that time belongs to the unit interval $[t_i,t_f]=[0,1]$. OP's non-local Lagrangian action functional reads (modulo some sign conventions$^1$) $$ \left. S[q,v]\right|_{v=\dot{q}}, \tag{A} $$ where $$ ...


0

Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the ...


0

I've found the inconsistency. What follows is a derivation of the Hamiltonian for convolutional Lagrangians. Starting with the total variation of a convolutional Lagrangian, we have: $$\delta \mathbb{L}=\int^t_0 \left(\frac{\delta \mathbb{L}}{\delta \dot{q}}\delta\dot{q}(t-\tau)+\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau ...


4

Let here consider point mechanics (as opposed to field theory) for simplicity, i.e. the generalization to field theory is left as an exercise. I) Bad news. If the Lagrangian action functional $S[q]$ is non-local, the usual definition of Lagrangian momenta as a partial derivative $$p_i~:=~\frac{\partial L}{\partial v^i} \tag{1}$$ does not apply. II) Good ...


2

The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order ...


2

I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ...


0

There is this famous reality: Addition or subtraction of any constant to potential energy doesn't change the equations of motion. In your case $U_0$ is just a constant that one can add or subtract freely. You can assume that system had an initial constant potential energy (independent of your generalized coordinates of course) just before you started to ...


1

Before going to field theory, it seems instructive to first ask the same questions in point mechanics: Can the Lagrangian $L(q,v,t)$ depend on time explicitly? Yes. The Lagrangian $L(q,v,t)$ can depend explicitly on time. E.g. there could be external sources. On the other hand, if the Lagrangian does not depend on time explicitly, then the ...


1

OP wrote (v1): Why the Lagrangian density becomes a function of the spatial derivatives? Well, one intuitive answer is, that if the theory is supposed to be relativistic, and if the Lagrangian density has temporal field derivatives, then it must also contain spatial field derivatives. Another answer is that if the theory is supposed to be local, this ...


1

To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ ...


1

What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$. and why $\partial\dot q_{j} = \partial\dot ...


1

Comments to the question (v7): The director $\vec{n}(\vec{r},t)$ is a vector-valued field. Ericksen-Leslie theory is a field theory. Before studying variational calculus in field theory, and asking which variables are independent, and which are not, it is highly recommended to understand the corresponding problem in point mechanics, see e.g. this Phys.SE ...


2

The extra derivative in Polchinski comes from the following version of the Fundamental Lemma of Calculus of Variation (FLCV): $$\tag{1} \left[ \forall g : ~~\int_a^b\! dx~ g(x) ~=~0 \quad\Rightarrow \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f^{\prime}~=~0.\quad $$ FLCV (1) states in words: If it is true that for all functions ...


0

Finding the potential radii is actually quite simple. I already have: $$V_{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2} $$ I was mistakenly graphing $1/r^2 - 1/r - 1/r^3$, when in actuality it makes more sense to take $1/r^2 - 1/r - 0.1/r^3$, since $1/r^3$ is sufficiently small. Since the orbits are circular, the potential radii ...


1

Comments to the question (v3): On one hand, traditionally, the Batalin-Vilkovisky (BV) operator $\Delta$ in Lagrangian BRST formulation encodes geometric data of the antisymplectic phase space for the model, specifically the antisymplectic structure [i.e. the so-called antibracket $(\cdot,\cdot)$, or odd Poisson bracket] and a path integral volume density ...


0

Your equation has the form $$ V_\text{eff}(r) = \frac{\alpha}{r^2} - \frac{\beta}{r} - \frac{\gamma}{r^3} $$ If you set $\alpha=\beta=\gamma=1$, then you're overestimating the $r^{-3}$ term, which is supposed to be a small correction. You will only find two extrema if the derivative has two roots: $$ V_\text{eff}'(r) = -\frac{2\alpha}{r^3} + ...


0

$$\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{q}_j}=\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{r}_i}\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}=2\dot{r}_i^T\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}$$ (A simple chain rule, while bearing in mind that $\dot{r}_i^T\dot{r}_i$ is essentially $\dot{r}_i^2$. You just need a transposed vector on the ...


2

I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums. Let \begin{equation} \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} \end{equation} be a lagrangian over $T M$, with time in ...


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...


1

We have such equation: $$H = \frac{\partial L}{\partial \dot{q}} \dot{q} - L$$ You can show by calculation, that it holds in your special case, too. Now we use chain rule, and Euler-Lagrange equation: $$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}} \ddot{q} =\frac{d}{dt}\left(\frac{\partial L}{\partial ...


6

Comments to the question (v2): To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint $$g ~:=~ \{f,H^{\prime}\}_{PB} ...


1

The Hamiltonian is defined by $$ H = \sum_{i=1}^n \left( \frac{\partial L}{\partial \dot q_i} \dot q_i \right) - L $$ So in your case: $ H' = H - \lambda f $


1

You are overthinking, it looks like a simple inelastic collision. Before: Vertical (potential energy): $E_{plate} = 0$, $E_{object} = mgh$ Horizontal (kinetic energy): $E_{plate} = \frac{1}{2}Mv^2$, $E_{ball} = 0$ After: Vertical: The object smashes, not bounces, so its entire potential energy dissipates Horizontal: The plate now has a combined mass of ...


2

This is the Euclidean classical action $S_{cl}[\phi]=\int d^{4}x (\frac{1}{2}(\partial_{\mu}\phi)^{2}+U(\phi))$. It would be nice if somebody could explain the structure of the potential. I don't understand why $\phi$ is used instead of a position vector $\textbf{r}$. Also, how can $(\frac{1}{2}(\partial_{\mu}\phi)^{2}$ be interpreted as ...


3

In terms of a metric $g_{ab}$, the Riemann curvature tensor is given by, $$R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{c e} \Gamma^e_{d b} - \Gamma^a_{de} \Gamma^e_{c b}$$ and consequently changing $d$ does not change the formula, though of course the actual numerical measures of curvature may change. Nevertheless, there are ...


2

If you have a result stating that the non-linear completion of a free massless spin-2 field must be general covariant, then that means the action must only depend on curvature invariants like $R$, $R_{ab}R^{ab}$, $R_{abcd}R^{abcd}$, ... Because we started with the action for a free massless spin-2 field, we can narrow down the allowed dependence on the ...


3

In the action formalism a linear Euler-Lagrange (EL) equation corresponds to a quadratic action, i.e. an action which is quadratic in the dynamical field variables of the theory. On the other hand, self-coupling or interaction terms in the action correspond to cubic or higher terms. Such terms leads to non-linear EL equations. See also this related Phys.SE ...


0

The mathematical concept that I was searching for in this question is the following: http://en.wikipedia.org/wiki/Hodge_dual I will not elaborate more, but except to say that, the Hodge dual allows you to define a conserved current corresponding to any choice of the "time axis". (Sorry for the vagueness).


1

I think there is some sort of bug near the end of the first answer (joshphysics), possibly in the variational step which involves integration by parts. Unless some further assumptions were used. Otherwise the implied conclusion seems to be that the EL eqns for any functional with integrand $F(x(s),\dot{x}(s))$ are generally the same as for the functional ...


1

From the Maxwell-Dirac Lagrangian $$ \mathcal L = -\frac{1}{2}F^2 + \overline{\psi}(i\gamma^\mu D_\mu +m) \psi $$ where $D_\mu$ is the gauge covariant derivative it is clear that the 4-current that acts as the source term in Maxwell's equations is $$ j^\mu_D = q\overline{\psi} \gamma^\mu \psi. $$ Using the Dirac equation it can be shown that (see, e.g., ...


0

The answer is two, as can be seen by considering the angular positions of the pulleys: each pulley can be set independently. I think your error is that the length of rope of the "lone" mass (call it mass number 1) is not a constraint. The position of mass 1 can be set independently of masses 2 and 3. On the other hand, the position of mass 2 cannot be ...


2

I think you bring up an interesting question, I'm not sure why there were some derogatory comments to this... First off, the fermionic current doesn't couple to the gauge field due to its dimension. The Dirac field is dimension $ 3/2 $ and the current is dimension $ 3 $. Therefore, the coupling of the fermion to the gauge fields is of higher order. On the ...


1

Second possibility: $$\dfrac{\partial}{\partial u}\int u''^2 \mathrm{d}x=\dfrac{\partial}{\partial u}\int > u\,u^{(4)}\mathrm{d}x=\int u^{(4)}\mathrm{d}x$$ by double integration by part and because $\dfrac{\partial > u^{(4)}}{\partial u}=0$. I am really not sure about this latter argument either. Second possibility is closest to correct. ...


1

In the curved space of GR, the straight line path between two points is generalized to the 'shortest' path between two points in spacetime. So we want $$\int \mathrm{d}s$$ to be extremized. Or, in terms of a parameter $u$, we want $$\int \frac{\mathrm{d}s}{\mathrm{d}u}\mathrm{d}u$$ to be extremized. This means we want $$\int L\,\mathrm{d}u$$ to be ...


0

Comments to the question (v2): It is often is possible to formulate (discrete) time-evolution equations/equations of motion (eoms) in a discretized theory. This is of course useful in computational physics. However OP asks for a variational action principle for a fully discretized theory. Hence we will not discuss further the case where eoms (without a ...


0

This is frequently good enough, but in your specific case its actually much easier to show this holds exactly in the finite case. I won't do this for you, but note that this is a global symmetry, i.e. alpha has no dependence on x. At most, you might also need to use the Baker-Campbell-Hausdorff formula.


0

You say but position and velocity are independent along any trajectory too. No, a trajectory is defined as a given vector-function $\vec r(t)$ from which follows $\vec v(t) = d\vec r(t)/dt$. So, given a trajectory, the vector-function $\vec v(t)$ depends clearly on $\vec r(t)$ as being its derivative. Now, forget the trajectories. You have an ...


0

Doesn't the result follow from a simple diagram? If you change the X,Y position by a small amount dx, dy, the change in length of the hypotenuse is simply $$\begin{align} &= \sqrt{(a+dx)^2 + (a+dy)^2} - \sqrt{a^2 + a^2}\\ &= \sqrt{2a^2 + 2a dx + 2a dy + ...} - \sqrt{2a^2}\\ &= a\sqrt{2}\left(1 + \frac{dx}{a} + \frac{dy}{a} - 1\right)\\ &= ...


4

Comments to the question (v8) concerning numerical integration: On one hand, to solve a Hamiltonian system numerically, there exist the numerical integration schemes of symplectic integrators (SI), where each (finite) numerical iteration step is a canonical transformation/symplectomorphism, which preserves certain properties, such as, e.g., energy, and ...


3

In general, when working out the Lagrangian, start in coordinates that you know and then rewrite in generalized coordinates. Kinetic energy in this case is proportional to $v^2 = \dot x^2 + \dot y^2 + \dot v_z^2$. In your spherical coordinates $$x = r \sin \alpha \cos \theta;~ y = r \sin \alpha \sin \theta;~ z = r \cos \alpha.$$ Take full time derivatives ...


1

So, the nice thing about Lagrangian mechanics is generalized coordinates. The crummy thing about it is that you at least in principle need to know about variational calculus to understand why it works and how to get the equations of motion. Finally, Emmy Noether's theorem gives you a really generic sense for conservation laws. The nice thing about ...


0

Not all conserved charges are obtained by integrating the time component of some conserved current. For example, momentum and angular momentum are conserved charges and are obtained by integrating a spatial component of a conserved current. So the equations and interpretation for conserved charges in a Euclidean theory are the same as in the ...


1

The kinetic energy of the particle depends just on the path it is following - if you imagine the cone is suddenly invisible, the particle continues to go around in a circle. That means that there is no reason to add $\sin\theta$ in your expression for the kinetic energy if you used $r$ to mean (as drawn) the distance from the axis of rotation. Note - your ...


0

After some research, it seems a similar problem has been explored in [Vainberg, 1956], wherein he formulates a variational principle as: $$ F\left[\mathbf{x}\right]=\int^{\lambda= 1}_{\lambda= 0} \left\langle \frac{\partial \eta}{\partial \lambda},N\!\left(\eta\right)\right\rangle $$ Where $\eta$ is some curve in phase space which goes from the initial ...


2

Assuming no quantum gravity, $\eta^{\mu\nu}$ is a constant and can be pulled out of the derivative and what remains looks like a $\delta^k_{\mu}$ or $\delta^k_{\nu}$-type expression (in the sense of a Kronecker $\delta$), pulling the $k$ into the $\partial^\mu$ or $\partial^\nu$ respectively. If you are confused about where the minus sign comes from, I ...


2

Comments to the question (v1): I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$ However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path ...


1

Think about an action principle as an abstract mapping of trajectories $\mathbf r = (\vec r(t), t_0, t_1)$ to some number $S(\mathbf r)$ which no longer is explicitly time-dependent. Now, sometimes, the action can be computed from another function. The other function just has a bunch of parameters, which we can call $\{\alpha_i\}, ~ \{\beta_i\}, ~ \tau$. ...


0

You say " When we calculate the action as the integral of the Lagrangian for a wiggly trajectory, the velocity is obviously dependent on time and so is the Lagrangian". How exactly is the velocity dependent on time? Before applying the least action principle and find the trajectory of the object, we have a Lagrangian dependent on velocity (through the ...



Top 50 recent answers are included