Below, for any function $Q$ the notation $Q_x$ means $\frac{\partial Q}{\partial x}$, and $Q_{xx}$ means $\frac{\partial^2 Q}{\partial x^2}$.

In physics, the trajectory of a particle is given by the Euler-Lagrange condition:

$$\frac{d L_{\dot{x}}}{dt}=L_x \:\:\:\:\:\:\:\ (1)$$

This condition guarantees that the particle travels along a trajectory of stationary action.

I would appreciate it if someone could help me understand the physical implications of a trajectory where (1) is satisfied but also the condition

$$L_{\dot{x}}=L_x \:\:\:\:\:\:\:\ (2)$$

is satisfied.

For example, if we have the action

$$S=f(x) \:\:\:\:; \:\:\:\:x=x(t)$$

and thus the lagrangian

$$L(x,\dot{x},t)\equiv\frac{\delta S}{\delta t}=f_x\dot{x}$$ $$S=\int_{t_0}^{t_1} L(x,\dot{x},t) \:\:dt$$ then $$L_x=f_{xx}\dot{x}$$ $$L_{\dot{x}}=f_{x}$$

and if you differentiate $L_{\dot{x}}$ w.r.t. $t$ you'll see that the the Euler Lagrange condition (1) is satisfied. Now, if condition (2) is also imposed, then the trajectory is a stationary action trajectory, but also

$$L_{\dot{x}}=L_x \Rightarrow \:\:\: f=e^{t+\ln{x}+C} \:\:\:\:\:\:\:\ (result)$$

where $C$ is the constant of integration.

Specifically, my question: I thought condition (1) already imposed a unique path upon the particle. But adding condition (2) seems to impose an even "unique-er" path upon the particle (i.e. the "result"). Does this mean that the path of stationary action is not necessarily unique? That there are many trajectories that satisfy the principle of stationary action?

And more generally, I would appreciate it if someone could help me understand the real world physical implications of a trajectory that satisfies condition (2) and of the "result."

  • $\begingroup$ Comment to the question(v2): The new condition (2) violates dimensional analysis if the variable $t$ has dimension. $\endgroup$ – Qmechanic Jan 3 '13 at 16:15
  • $\begingroup$ @Qmechanic Ok, thanks, but note that the combined conditions (2) and (3) in (v1) were mathematically the same as the new condition (2) in (v2), and yet you did not make this dimensional analysis observation. $\endgroup$ – ben Jan 3 '13 at 17:08
  • $\begingroup$ I did notice it. But at the time I saw more important issues to comment on. $\endgroup$ – Qmechanic Jan 3 '13 at 18:02

The Euler-Lagrange equations are obtained by employing the principle of least action. This means that if we make a slight change to the path of the particle (call it $\delta x$(t)), the action should not change (to first order in $\delta x$). We write $\delta S$ for the change of the action, so the Euler-Lagrange equations are obtained by requiring that $\delta S = 0$ for any slight change of the path $x(t)$ with the ends held fixed.

Let us calculate the change of the action $\delta S$ (we ommit terms of higher order in $\delta x$):

$\delta S = \int_{t_0}^{t_1}L(x + \delta x,\dot x + \delta\dot x,t)\mathrm{d}t - \int_{t_0}^{t_1}L(x,\dot x,t)\mathrm{d}t = \int_{t_0}^{t_1}(L_{\,x}\delta x+L_{\,\dot x}\delta \dot x)\mathrm{d}t = \int_{t_0}^{t_1}(L_{\,x} - \frac{\mathrm{d}}{\mathrm{d}t} L_{\,\dot x})\delta x \; \mathrm{d}t + \left.L_{\,\dot x}\delta x\right|_{t_0}^{t_1}$

In the last step we integrated by parts. Now the last term can be omitted because the ends of the path are held fixed, so $\delta x(t_0) = \delta x(t_1) = 0$. Since $\delta S = 0$ should hold for otherwise arbitrary $\delta x$ we conclude that

$L_{\,x} - \frac{\mathrm{d}}{\mathrm{d}t} L_{\,\dot x} = 0$

which is precisely the Euler-Lagrange equation.

Now the problem with your Lagrangian is that it is essentially zero. This can be seen by noting that I can add a total time derivative of an arbitrary function $\frac{\mathrm{d}}{\mathrm{d}t}g(x,t)$ to a Lagrangian without changing its physical meaning. This added term would amount to the expression $g_x\delta x|_{t_0}^{t_1}$ in the change of the action and therefore give zero, since again the ends of the path are held fixed.

Since any two Lagrangians are equivalent if they differ by a total time derivative $\frac{\mathrm{d}}{\mathrm{d}t}g(x,t)$ and yours is obtained from a total time derivative ($\frac{\mathrm{d}f}{\mathrm{d}t}$), your Lagrangian is equivalent to the Lagrangian $L=0$.

So it is not surprising that your Lagrangian does not give unique equations of motions, in fact it does not give any equations of motions at all.

This also happens when there is gauge symmetry in a physical system. Then the equations of motions don't uniquely determine the evolution of all variables used to describe the system. In your system the only degree of freedom is a gauge symmetry, so there is no physical content in your Lagrangian.

  • $\begingroup$ Thanks for clarifying the notation, although I think Susskind refers to the Lagrangian as Lagrangian density on several occasions in his online Stanford lectures on GR. As for your remarks about the total time derivative, I don't quite understand since (I thought) every Lagrangian is basically a total time derivative. There is partial differentiation w.r.t. position and velocity, but w.r.t. time it's always $d$ not $\partial$ (right?). $\endgroup$ – ben Jan 3 '13 at 15:43
  • $\begingroup$ Note I have edited the post in accordance with your comments, and also to modify condition (2) and remove condition (3). Do your comments still apply given these modifications? $\endgroup$ – ben Jan 3 '13 at 15:58
  • $\begingroup$ @ben: The Lagrangian density is something that arises in theories with fields. There the Lagrangian is the space integral over the Lagrangian density. $\endgroup$ – Friedrich Jan 3 '13 at 16:33
  • 1
    $\begingroup$ @ben: A Lagrangian is not supposed to be a total time derivative as in your example, because, as stated earlier, it is essentially zero then. The Lagrangian for a free particle of mass $m$ is $L=m\frac{\dot x^2}{2}$. This can not be written as a total time derivative, and it does give unique equations of motion. The equation defining your Lagrangian is not correct, because $S$ is a functional. It does neither depend on $\dot x$ nor on $t$, it depends on the whole path $x(t)$ takes between times $t_0$ and $t_1$. So you cannot write $L=\frac{\mathrm{d}S}{\mathrm{d}t}$. $\endgroup$ – Friedrich Jan 3 '13 at 16:34

Comments to the question(v2):

  1. If the Lagrangian is a total derivative $L=\frac{dF}{dt}$, the Euler-Lagrange equations are trivially satisfied, and hence provide no information at all. Concerning the role of total derivatives, see also e.g. this Phys.SE post.

  2. Euler-Lagrange equations does not uniquely determine the classical path. One have to impose boundary conditions as well, and even then there may multiple classical solutions (sometimes called instantons). Concerning the role of boundary conditions, see also e.g. this Phys.SE post.

  • $\begingroup$ Thanks for the notational clarification. Post is duly edited. (Although it seems to me that Susskind refers to the Lagrangian as the Lagrangian density at several points during his online Stanford lectures on GR.) I have also edited the post to modify condition (2) and remove condition (3), which was unnecessary to make my point. Do your comments still apply considering this modification? $\endgroup$ – ben Jan 3 '13 at 15:57
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 3 '13 at 20:01

The ''Lagrangian density'' is the spatial integral over the Lagrangian. The ''action'' is defined as $$ S \equiv \int dt \, L \quad \text{and}\quad L = \int dx \, \mathcal{L} $$ giving $$ S = \int dx \, dt \, \mathcal{L} $$ you want variations of the action to be zero to first order, $\delta S = 0$. Since we already have a Lagrangian for this problem, $L(x) = f(x) - \lambda q$, Lagrange's equation becomes $$ \delta S = \int dt \, ( \delta f(x) - \lambda \, \delta x)=\int dt \, \left( \frac{\partial f}{\partial x}-\lambda \right) \delta x =0 $$ which gives $$ \frac{\partial f}{\partial x}-\lambda = 0 \implies f(x) = \lambda \, x + c_0 $$ which doesn't contain any useful information. If $L(x,\dot{x}) = (f_x -\lambda) \dot{x} - \dot{\lambda}x$, then the EOM are $$ \frac{d}{dt}(f_x - \lambda)-\dot{\lambda} = 0 $$ Now if $\frac{\partial L}{\partial x}=0$ then we know that the conjugate momentum, $\frac{\partial L}{\partial \dot{x}}$ is conserved since it is independent of time by the EOM, however in this case, it is not zero, since $\frac{\partial L}{\partial x}=\dot{\lambda}$.

There is something called ''caustics'' is optics, where the path to get from the light source to the focal point isn't unique, in the sense that it is degenerate symmetrically through the lens. This may be something you are interested in.


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