Is there an intuitive basis for the Lagrangian?


by PerpStudent
Tags: basis, intuitive, lagrangian
PerpStudent
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#1
Jul1-12, 12:30 PM
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Since it is based on the kinetic energy less the potential energy, what does the Lagrangian actually represent? Is there some intuitive way to understand why it is defined so and why it is such a fruitful concept using the principle of least action?
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vanhees71
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#2
Jul2-12, 04:03 AM
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The question, why the Lagrangian works the way it does in Hamilton's least-action principle cannot be answered satisfactorily within classical physics.

There you realize this principle the first time when you analyze Newton's equations of motion in terms of a variational principle, i.e., how you can get these equations as the Euler-Lagrange equations of an appropriate functional. The advantage of this method is more on the calculational and theoretical side than that it would provide new laws of nature. However, it is utmost important to study the foundations of the natural laws of a given theory or model.

Particularly it enables us to formulate symmetry principles in a mathematically elegant way (Lie-group theory), and in this way it reveals a one-to-one relationship between symmetries and conservation laws (Emmy Noether 1918). In this way it also enables us to calculate the fundamental quantities in any model based on the symmetries of Newtonian, Minkowski or psesuo-Riemannian spacetimes as the generators of the fundamental symmetries of the spacetime geometry, i.e., if there is translation invariance, you can define momentum as the generator of the corresponding translation transformation on the space-time variables, etc.

Anyway, all this doesn't explain why the Lagrangian looks as it does (btw. it's not always just T-V as in Newtonian mechanics). This question is answered by quantum theory. It's revealed in the most intuitive way when you think in terms of path integrals. Of course you must start with the Hamiltonian version as a functional of trajectories in phase space (parametrized by the generalized coordinates and their conjugated momenta) and then work you way through to the Lagrangian form of the path integral. In many cases it turns out that the Hamiltonian path integral, where you integrate over position and conjugated momentum coordinates, can be brought into the form of the Lagrangian path integral by integrating out the momenta exactly and also quite often you even get the same Lagrangian as you naively would start with in the classical analogy of the quantum theory in question.

Particularly that's the case for the most simple non-relativistic Hamiltonians,

[tex]\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x}).[/tex]

After you have established this relation, you may ask, under which circumstances, a quantum particle can be described approximately as behaving like a classical particle. It turns out that this is achieved whenever the path integral can be approximated with help of the stationary-phase approximation. The leading order gives the classical trajectories of the classical model of the same representation since the stationary phase of the path integral is precisely determined by the stationary point of the classical action.

In such cases, only at classical turning points, where quantum tunneling plays an important role to get the correct wave functions out, you have to do more to appropriately describe the particle's behavior, leading to the singular perturbation theory (knows as WKB method in the wave-function formuliation of quantum mechanics).
PerpStudent
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#3
Jul2-12, 01:47 PM
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Thanks for your response. I am familiar with some of the applications of the Lagrangian that you describe, but I have no experience with the quantum mechanical applications. It's quite remarkable that there is no explanation for it's usefulness and meaning in classical physics. In light of that, what might have been the historical motivation for it's development? Wikipedia describes the Lagrangian as "a function that summarizes the dynamics of the system." Why is that? In any case, the principle of least action is based on a generalized function of t, q(t) and q'(t) leaving one to speculate that not just the Lagrangian but any function of those quantities might also be the basis of finding laws of physics.

Mathematech
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#4
Dec25-12, 10:56 AM
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Is there an intuitive basis for the Lagrangian?


If you want a quick easy intuitive interpretation of the Lagrangian - energy in a mechanical system is divided into kinetic energy (T) and potential energy (V) - the Lagrangian is a measure of the unevenness of the distribution of the energy into these two types (T-V).
Vanadium 50
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#5
Dec25-12, 11:03 AM
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Apart from the fact that this thread is six months old, "unevenness" makes no sense, since I am always free to add a constant to the potential energy.
Mathematech
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#6
Dec25-12, 11:53 AM
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1) Yes you can add a constant consistently to the potential
2.) No that doesn't change the fact that T-V can be thought of as a measure of unevenness, it only changes the particular value that represents an "even division" - this value doesn't have to be 0.
3.) Yes it does shed light on why its partial derivatives the of the Lagrangian, action integral etc that are of more importances that its value.
Vanadium 50
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Dec25-12, 04:14 PM
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Quote Quote by Mathematech View Post
No that doesn't change the fact that T-V can be thought of as a measure of unevenness, it only changes the particular value that represents an "even division" - this value doesn't have to be 0.
Again, that's silly. Two people looking at the same system could then come to different conclusions as to whether over time it is becoming more even or less even. (Which is probably one reason why conventional physics has not adopted this idea of "evenness")
Mathematech
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#8
Dec25-12, 04:40 PM
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No, what _you_ are saying doesn't make sense. If two people people's description of a system differs only in a constant added to V, then they measure the same increases and decreases in the Lagrangian over time.
Jorriss
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Dec25-12, 04:59 PM
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Quote Quote by Mathematech View Post
No, what _you_ are saying doesn't make sense. If two people people's description of a system differs only in a constant added to V, then they measure the same increases and decreases in the Lagrangian over time.
I guess I just don't quite know what you mean by uneveness. Can you clarify?
Mathematech
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#10
Dec25-12, 05:32 PM
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hmm perhaps its not the best word, not sure what would be a better term.

Changes in the Lagrangian reflect how much the system is tending towards increasing or decreasing its kinetic energy in comparison to its potential. Because V and hence the Lagrangian are only defined up to the addition of an arbitrary constant, its the changes in value that matter not the actual value.

Basically if your Lagrangian increases, the system's energy is becoming more kinetic compared to potential than what it was previously, due to any increase in T that occurred being more than any increase in V that occurred. (Treat a decrease as a negative increase to avoid having to repeat the above in the context of decreases or combinations of decreases and increases.)
Jorriss
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#11
Dec25-12, 05:38 PM
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Quote Quote by Mathematech View Post
hmm perhaps its not the best word, not sure what would be a better term.

Changes in the Lagrangian reflect how much the system is tending towards increasing or decreasing its kinetic energy in comparison to its potential. Because V and hence the Lagrangian are only defined up to the addition of an arbitrary constant, its the changes in value that matter not the actual value.

Basically if your Lagrangian increases, the system's energy is becoming more kinetic compared to potential than what it was previously, due to any increase in T that occurred being more than any increase in V that occurred. (Treat a decrease as a negative increase to avoid having to repeat the above in the context of decreases or combinations of decreases and increases.)
How is that a useful way of thinking about Lagrangians though? It's pretty much just rephrasing the considerations that go into discussing turning points, classically forbidden zones, etc. It doesn't seem to add anything new and I can't imagine how that makes the euler-lagrange equations more intuitive, etc.

And also, what about lagrangians that are not T-V? Lagrangians are also not unique beyond adding a constant. Certain functions may be added that do not change the equations of motion.
Mathematech
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#12
Dec25-12, 05:51 PM
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I was talking specifically about the typical Lagrangian L = T - V and giving a simple intuitive way of understanding what quantity it represents. But no its not more helpful for understanding the Euler-Lagrange equations because Lagrangians and Hamiltonians arise from purely mathematical problem solving techniques not physical intuition.
Vanadium 50
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Dec25-12, 07:34 PM
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This is exactly why your "simple intuitive way" of "evenness" is not used in the textbooks. It provides nothing but confusion.
Mathematech
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#14
Dec26-12, 02:36 AM
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hmm, its not _my_ way, it comes from what has been said in textbooks, but it is very "rough". Explaining the Hamiltonian as "total energy" is also "rough" because its more a mathematical fluke that its typically total energy, not the reason for its use.
Cleonis
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#15
Dec28-12, 03:06 PM
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Quote Quote by Mathematech View Post
[...] Because V and hence the Lagrangian are only defined up to the addition of an arbitrary constant, its the changes in value that matter not the actual value.
Let me elaborate on that aspect of changes in value.

I take the following case: an object is thrown upward. Gravity decelerates it, the object reaches its highest point, gravity accelerates it on the way back down.

The work/energy theorem expresses that at each point in time the rate of change of kinetic energy matches the rate of change of potential energy. The two change in opposite directions, but they are at all points in time a match for each other.

The graph of the kinetic energy as a function of time (red line) is a parabola. The graph of the potential energy as a function of time (green line) is a parabola too.



Mirror the graph of the potential energy with respect to the horizontal axis, so that it is upside down. Now the two parabola's are pointing in the same direction. For the true trajectory you have the following property: the two graphs are parallel at each point in time.

We have the common Lagrangian:
L = T - V

The minus sign corresponds to the graph of the potential energy being turned upside down.



The boundary of the shaded region represents the sum of the red graph and the green graph. As we know the Action is the integral of the Lagrangian from the start point to the end point. So the surface area of the shaded region corresponds to the Action of the Principle of Least Action.

As we know, to find the true trajectory you try a range of possible trajectories, and you find the magnitude of the Action as a function of some parameter that sweeps out the range of possible trajectories.

The Action is least when at every point in time the rate of change of kinetic energy matches the rate of change of potential energy.
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kinetic_potential.png   kinetic_minuspotential.png  
Mathematech
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#16
Dec29-12, 05:19 AM
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Its more that under such conditions variation in the other parameters does not change the action, there is no "principle of _least_ action" its statitionary action.
Cleonis
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#17
Dec29-12, 07:47 AM
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Quote Quote by Mathematech View Post
Its more that under such conditions variation in the other parameters does not change the action, there is no "principle of _least_ action" its statitionary action.
It is of course well known that in its generalized form the criterium is stationary action. Stationary action subsumes least action.

Still, in a majority of cases the stationary point is a point of least action. For the purpose of explaining the intuitive basis for the Lagrangian the extra precision of 'stationary action' over 'least action' is distracting rather than helpful.


There is Edwin Taylor's Principle of Least Action website

In the article Variational mechanics in one and two dimensions Taylor and his co-authors list their reasons for using the expression 'Principle of Least Action'

We use the name least action instead of the technically correct stationary action for several reasons:
(a) Many cases involve a local minimum of the action.
(b) The value of the action is always a minimum for a sufficiently small segment of the curve.
(c) The word least is self-descriptive, but stationary requires additional explanation.
(d) The word least does not lead to the error that the value of either form of action, Eqs. (4) and (17), can be a maximum for an actual path, which it cannot.
(e) Least action is the name most often used in the historical literature on the subject.
We recommend that the term stationary action be introduced, with careful explanation, not long after the term least action itself.
Mathematech
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#18
Dec29-12, 09:03 AM
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The additional explanation required by stationary is necessary to get an accurate view of what is going on. You indeed never get a maximum w.r.t. all parameters but a saddle point is not uncommon and with more than one parameter this implies we have a maximum w.r.t to some parameters not a minimum.

Approaches to mechanics that try invoke a non-existant law that nature is magically trying to minimize action ;) just don't work in these very real saddle point cases.


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