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Lagrangian invariance, short question

  1. Apr 7, 2015 #1
    Consider a Lagrangian: ##L(x,x',t)##

    Define now: ##L'(x,x',t) = L + x ##

    We have seen that Lagrangians can differ up to a total time derivative of some function ##F(x,t)## in such cases and give the same equation. When checking explicitly these two give different equations. Why would it be wrong then to say that ##x=x(t)## and say that this function ##F## is the primitive function of ##x(t)##?

     
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  3. Apr 8, 2015 #2

    Orodruin

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    The addition to the Lagrangian must be a total time derivative, i.e., if it is the total derivative of ##F(x,t)##, you must be able to write it as
    $$
    \frac{dF}{dt} = \dot x \frac{\partial F}{\partial x} + \frac{\partial F}{\partial t}.
    $$
    Your function ##x## is not on this form.
     
  4. Apr 8, 2015 #3
    I was wondering why one can't consider ##x=x(t)## at that moment since technically ##x## is indeed a function of time. And so ##x(t)## is a total time derivative of its primitive function.
     
  5. Apr 8, 2015 #4

    Orodruin

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    This is not how things work. The variable ##x## is what you write down the EL equations for, not its derivative. The point is to use the fact that it is a total derivative to integrate it and fix it using the boundary conditions.
     
  6. Apr 9, 2015 #5
    Do you have any suggestions on some text that gives a very thorough explanation on exactly the subtle treatment of variables in Lagrangian mechanics? Most things I've found really skim over it very fast and I think it would benefit me because the treatment of ##q## and ##q'## always confused me when I tried to think about it on a little deeper level than just solving problems in class.
     
  7. Apr 9, 2015 #6

    Orodruin

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    I do not think this is a subtle issue to be honest. It is more a question of going back to the derivation of the fact that having a total derivative does not change the EoM. Assume that you add a function ##\Delta\mathcal L = dF(x,t)/dt = \dot x \partial_x F + \partial_t F## to the Lagrangian. Your EoM will change by
    $$
    \frac{\partial \Delta\mathcal L}{\partial x} - \frac{d}{dt}\frac{\partial \Delta \mathcal L}{\partial \dot x} =
    \dot x \partial_x^2 F + \partial_t \partial_x F - \frac{d(\partial_x F)}{dt}
    = \dot x \partial_x^2 F + \partial_t \partial_x F - \dot x \partial_x^2 F - \partial_t \partial_x F = 0.
    $$

    Alternatively just observe that
    $$
    S'[x] = \int_{t_0}^{t_1} (\mathcal L + \frac{dF}{dt}) dt = \int_{t_0}^{t_1}\mathcal L dt + F(x(t_1),t_1) - F(x(t_0),t_0).
    $$
    The boundary conditions depend only on the function $x$ on the boundary and cannot affect the equations of motion. You cannot do the same procedure if you insert ##x## instead of ##dF/dt##.
     
  8. Apr 9, 2015 #7

    vanhees71

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    ...and this brings us to the most general definition of a symmetry (within classical mechanics): A transformation that leaves the variation of the action invariant is a symmetry transformation, because then and only than it doesn't change the equations of motion.
     
  9. Apr 9, 2015 #8

    PeroK

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    When you talk about subtleties, do you mean this sort of thing?

    If we define ##\mathcal{L} = \mathcal{L}(x, y, z, \dot{x}, \dot{y}, \dot{z})##

    Then the Lagrangian is a time-independent function of 6 variables, as are all its partial derivatives wrt these variables. So, how can you take ##\frac{d}{dt}## of these partial derivatives?

    What does ##\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}}## mean?

    Is that the sort of subtlety that mechanics texts tend to skim over?
     
  10. Apr 9, 2015 #9
    Yeah I think this is one of these things. The fact that ##L(x,v,t)## is a function where the coordinates are independent however once we pick our path of minimal action they become dependent. Nowhere I've seen any elaboration on this.
     
  11. Apr 10, 2015 #10

    PeroK

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    I suspect most mechanics texts will skip merrily over this. If you really want to understand things, you will probably have to strengthen your maths (multi-variable calculus and real analysis). The key is often establishing exactly what is a function of what.

    To take the example I gave. The way I see it is, there are several important points:

    1) First, you have the concept of a Lagrangian for the system. This is a static (time-independent) real-valued function of 6 variables. The variables are independent and can take any value.

    2) Second, you can take the partial derivative of this "system" Lagrangian wrt each of the 6 variables. That gives you 6 more real-valued functions - technically, still of 6 variables.

    3) Now imagine a particle taking a path over time through this 6-dimensional space. Note that here "path" is not just the physical path, but includes the velocity components. The book I have doesn't emphasise this point at all. We're really talking about a path through a 6-dimensional position/velocity space.

    For example, a starting point of (0, 0, 0, v, 0 , 0) (starting at the origin with velocity v in the x-direction) is a different starting point from (0, 0, 0, 0, 0, v).

    4) At every time t, the particle is at a point ##(x(t), y(t), z(t), \dot{x}(t), \dot{y}(t), \dot{z}(t))##. Note that most paths are physically impossible and, in fact, the whole point is that there is only one valid path given any starting point.

    5) Now you can take your 7 functions - the "system" Lagrangian and its 6 partial derivatives - and define 7 time-dependent, real-valued functions of one variable (t) for your particle's path. It is these functions that can be time differentiated to get the E-L equations. (This also explains why you have a total derivative wrt t and not a partial derivative). You might call these time-dependent functions the "specific-path" Lagrangians.

    6) Specifically, here's an explanation of one of the E-L equations:

    ##\frac{\partial \mathcal{L}}{\partial x} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}}##

    The LHS is the partial derivative of the "system" Lagrangian wrt x, evaluated using the particle's position/velocity at time t. The LHS is a function of t.
    The RHS is the time derivative of the following function: the partial derivative of the "system" Lagrangian wrt ##\dot{x}##, evaluated using the particle's position/velocity at time t. The RHS is a function of t.

    For an arbitrary (physically impossible) path, you will not have equality here. But, for a physically valid path, you will have equality for all t. And this is equivalent to Newton's laws etc. And again, "path" is a 6-dimensional path through position/velocity space.

    I believe you'll have to do this sort of "filling in the mathematical gaps" if you want to see what's happening "under the covers". The key point is probably to establish clearly how you are defining your functions and how one function is derived from another. Try not to get bogged down in the mathematical background, though.
     
    Last edited: Apr 10, 2015
  12. Apr 10, 2015 #11
    This was EXACTLY what is was looking for. I can't believe I havent found anything like this before. It's almost enitrely clear to me now. The only last question I have is how variations come into this picture with relationship to this path you speak of in the derivation of the E-L equations. If you look at the derivation, no where do we ask that ##v(t)=\dot{x(t)}## , we do ask that ##\delta{v} = \dot{\delta{x}}## though, while it seems similar I'm not sure about it. To me this seems that well at first you can pick ANY path you like without regards to velocity being the time derivative of position, and only then you consider these restricted variations on the path.

    I guess I could summarize my final confusion in this: what if we pick an unphysical path where ##v(t)## is not the derivative of ##x(t)## but this turns out to have ##\delta{S}=0## under these restricted variations ##\delta{v} = \dot{\delta{x}}##?

    Anyway, I'm very content with your answer above and this is just a small detail.
     
  13. Apr 10, 2015 #12

    PeroK

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    The quick answer is that in the 6-dimensional space, there is no relationship between the variables. Some paths will be physically impossible because, say, ##x## will be increasing and ##\dot{x}## will be negative. It's physically absurd when you map these variables to the real system and interpret ##x## as the position of a particle and ##\dot{x}## as the velocity. But, it's still a path in your abstract 6-D space (where you can study any path you like).

    Such paths will simply not satisfy the E-L equations.

    One of the things the E-L equations must reveal is precisely this sort of relationship: if a path satisfies the E-L equations, then ##x## must be increasing when ##\dot{x}## is postive and decreasing when it's negative.
     
  14. Apr 10, 2015 #13
    Yeah where in the derivation do we ask this, namely that the path is physical? The only constaint we ask is that ##\delta{v}=\delta{\dot{x}}## but not that ##v(t)=\dot{x(t)}##, is this necessarily the same? The reason I doubt is as explained because one could start from an unphysical path and only then remember the fact to use these physical variations.
     
  15. Apr 10, 2015 #14

    Orodruin

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    This is not correct, we do require this. There are three EL equations if you have a three dimensional configuration space, resulting in three differential equations for fixing your three functions. If you want x and v to be independent, you need to look to the Hamilton formulation of mechanics using the phase space and first order differential equations rather than second order. The path is fixed once you know x(t).

    However, there is this subtlety: The Lagrangian is a function of ##x##, ##\dot x## and ##t##. A priori, you could write those arguments as ##q##, ##v## and ##t## instead. The point is that when you vary the action, you end up with
    $$
    \delta \mathcal L = \frac{\partial \mathcal L}{\partial q} \delta q + \frac{\partial \mathcal L}{\partial v} \delta v + \frac{\partial\mathcal L}{\partial t} \delta t
    $$
    simply by the chain rule. Now, this needs to be evaluated for the variation along a path ##x(t)##, where ##v = \dot x## and you have variations ##\delta q = \delta x## and ##\delta v = \delta\dot x##. As long as you are only varying the path, ##\delta t = 0## and you end up with the Euler-Lagrange equations. It is only a matter of checking how the Lagrangian itself varies with the variation of its parameters and then inserting what these parameters are along a path. Why ##\partial \dot x /\partial x = 0## is because we are simply considering the Lagrangian as a function of the parameters it takes at this stage and checking how it varies with those parameters.

    This is only true in Hamilton's formulation of mechanics. In the Lagrange formulation, the derivatives are dependent on the coordinates. This is reflected in the fact that the arbitrary coordinate transformations ##q_i \to Q_i(q,t)## you can perform in Lagrange mechanics only form a subset of the more general canonical transformations you can do in Hamilton's version.

    You do not have that freedom in Lagrange mechanics. In Hamiltonian mechanics, unphysical paths will simply not fulfil the equations of motion, i.e., they will not be Hamiltonian flows through phase space.
     
  16. Apr 10, 2015 #15
    So to summarize if I got your points correctly:

    Lagrangian mechanics:

    1) The Lagrangian is a function of three random coordinates ##L=L(x,v,t)## where the coordinates are fully independent

    2) When considering the action ##S## one must consider a certain path along which we will integrate to find this value. It is possible to parametrize the first two coordinates as ##x(t)## and ##v(t)##.

    3) Now we artificially introduce the restriction of the path in the form of ##\delta{v} = \dot{\delta{x}}## - (**1**) this means that we only consider the paths where ##v(t)=\dot{x(t)}##. There are still infinite amount of possible paths depending on what we pick for ##x(t)## - but only for one of those we will have ##\delta{S}=0##. This is the path for which E-L holds. And so when using the E-L equations we already assumed that ##v=\dot{x}## and so that remains after using them and finding the motion equations.

    Hamiltonian mechanics:

    1) The Hamiltonian is a function of three random coordinates ##H=H(x,p,t)## where the coordinates are fully independent

    2) When considering the action ##S## we again parametrize ##x(t)## and ##p(t)## - this time we integrate over ##\dot{x}p-H(x,p,t)##

    3) However now- we do NOT introduce any restriction on the variations. We still consider them as fully independent and now arrive at two equations instead of one. (**2**) For paths where both equations hold, we have ##\delta{S}=0##. Here we made no assumptions about ##x## and ##p## and so when solving the equations of motions we can treat them as functions of time of which we have no previous knowledge of any connection.


    NOTE : (1) This is the step I don't understand - why introducing restrictions on the variations automatically induces restrictions on the path. In the E-L derivations I don't see this happening.

    (2) I know how to do this derivation but this really feels like magic.
     
    Last edited: Apr 10, 2015
  17. Apr 10, 2015 #16

    Orodruin

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    Yes, apart from that I would not call this artificial. It is just the result of looking at paths such that the velocity is the time derivative of the position. And I would say it is the other way around, you impose ##v = \dot x## for the set of paths you are considering. This also fixes ##\delta v = \delta\dot x##. If you considered ##x## and ##v## as independent you would get six pretty boring equations to fulfil.

    It really is not different from how you approach the Lagrangian. The point is that the EL equation for ##p## is trivial, since you do not have any dependence on ##\dot p## it simply becomes ##\dot x = \frac{\partial H}{\partial p}##. At the same time you have for the ##x## variable
    $$
    - \frac{\partial H}{\partial x} = \frac{d}{dt}p = \dot p.
    $$
    These are just the Hamiltonian equations of motion. It is simply rewriting one second order equation as two equivalent first order equations. In some sense, you are looking at a larger set of paths (i.e., the dimensionality is doubled!) but you also have twice the number of equations to restrict your paths. One of the equations is simply the restriction to the set of paths you were looking at from the beginning (the equation for ##\dot x##) when you were considering Lagrange mechanics.
     
  18. Apr 10, 2015 #17
    I want to thank everyone helping out with this confusion. I think I understand it much better now.
     
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