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Derivation in classical physics

  1. Aug 29, 2012 #1
    Hi, Im getting started on analytical mechanics but need some help understanding some of the things that is said in my book. Therefore I have attached a section of the derivation and for that I have some questions:

    1) First of all the author says that in cartesian coordinates, i.e. (x,y,z), the derivative of T=kinetic energy with respect to the generalized coordinate q (which I assume is now taken to be a coordinate in a cartesian coordinate system, right?) vanishes. On the other hand it doesn't for angular coordinates. Can someone explain this? Why can't the kinetic energy depend on the position of the object? Certainly it can, so I must be misunderstanding something.

    2) Secondly, he later says that it is possible to find a set of generalized coordinates (q1,q2...qn) which are independent of each other such that any virtual displacement, δj, is independent of any other δk. I don't understand this, does this not only hold for orthogonal coordinates? Or is it not that property which assures that the above can be assumed?

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  2. jcsd
  3. Aug 29, 2012 #2

    Simon Bridge

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    I'll do the easy one first:
    2. you missed out a bit - what he's said is that if you have a system with some sort of additional constraints on it, you can always find a coordinate system where the constraints are built in.

    eg. if your system is constrained to move on a circle, then in Cartesian coordinates, a small displacement in one coordinate requires a displacement in another one. But you can choose polar coordinates ... in which case the motion is entirely angular: and does not affect any other.

    The author is saying that it doesn't matter how complicated the constraints are, you can always define a coordinate system like this.

    The other one is harder - but I suspect the confusion arises from a similar source: you have sort-of glossed over an important part of the description. I'm not sure I have enough information to pinpoint it though.

    For 1. it looks like he's talking about some sort of circular motion: spot the mention of "centripetal acceleration". The vanishing or not of partials will be a characteristic of the system under consideration. He's saying if {qi} are chartesian, the the partials vanish but if they are something else they may not - in the example of polar coordinates, you get a centripetal acceleration term arising from the angular coordinate.
  4. Aug 29, 2012 #3


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    The statement isn't about how the object is moving. It's about the coordiate system used to describe the motion.

    Take the simple example of a point mass. moving in a straight line at constant speed.

    In cartesian coordiates the KE is just ##T = m(\dot x^2 + \dot y^2)/2##. Clearly ##\partial T / \partial x = \partial T / \partial y = 0## because ##x## amd ##y## don't appear in the expression for ##T##.

    In polar coordiates, the radial and tangential velocities are ##\dot r## and ##r\dot \theta##, so ##T = m (\dot r^2 + r^2 \dot \theta^2)/2## and ##\partial T / \partial r## is not zero. ##\partial T / \partial \theta## is zero. The difference between ##r## and ##\theta## is that constant values of ##r## are "curved" (geometrically) but constant values of ##\theta## are not. In general, constant values of all of the generalised coordinates can be "curved".
    Last edited: Aug 29, 2012
  5. Aug 29, 2012 #4

    Simon Bridge

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    Ah - that's better. Thanks.
    I should have realized that :(
  6. Aug 29, 2012 #5
    The kinetic energy is defined as a function of velocity (Cartesian). So it cannot have any explicit dependence on position (Cartesian).
  7. Aug 29, 2012 #6
    But why can't velocity depend explicitly on position? What's the difference between being able to write it as a function of position for a system (because that should be possible for a lot of system) and it not depending explicitly on it? Sorry if it's a stupid question, but I want to get this completely straight :)
  8. Aug 29, 2012 #7

    Simon Bridge

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    You can write velocity as a function of position - but you lose the time axis when you do that. eg. in kinematics you can write [itex]v^2(x)=2ax[/itex] (acceleration from rest).

    Gives [itex]T=max[/itex] so [itex]\frac{dT}{dx} = ma[/itex] ... look familiar?

    T=max is just the expression for work. However, isn't eq 1.52 constrained to be workless?
  9. Aug 29, 2012 #8
    Well I guess since it only considers virtual displacements. But still not sure why eliminating time violates anything of such. That is, why is it that velocity is said to be independent of position - independent in what sense?
  10. Aug 29, 2012 #9


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    It can. For example, motion in a conservative force foild (like gravity, as in Simon Bridge's example)

    It's not a stupid question. I think the key point here (and it applies to more than just that specific question) is this:

    The physical system doesn't know or care what coordinate system you use to model it. It just does what it does.

    Going back to the mass moving with constant speed, in a polar cooirdinate system the equation for the velocity happens to contain ##r##, but the reason for that is because of what polar coordiates are, not because of what the particle is doing.

    On the other hand if the particle was moving under a gravitational potential field, the gravitational force doing work on the particle and changing its KE is part of the physical system, and the math that describes the system has to include it in some way whatever coordinate system you use.

    In general, if there is an explicit relationship between say velocity and position, there will be some force acting on the system to maintain that relationship, and the force will appear in the Lagrangian formulation as a constraint on the behaviour of the system.
  11. Aug 29, 2012 #10

    Simon Bridge

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    It doesn't. The derivation you are interested in explicitly states that no work is being done (I'm inferring that from the passage you showed us) - ergo, no unbalanced forces. From the above dT=ma.dx if the force=ma=0, then dT/dx=0.
    In the sense that the space derivative of the kinetic energy is zero. It's a tautology: saying there is "no work involved" is the same thing as saying that the kinetic energy does not depend on position.

    Can you find a system of zero work where this is not true?

    But what the author wants you to realize is that he could have used a different coordinate system and got a different result ... the statement is about the coordinate system used to represent the motion, not about the physics of the motion. i.e.
    ... the map is not the territory.

    That's the first part of the lesson.

    There's a problem: if these things are statements about coordinate systems - how can we do any physics at all? It's all arbitrary right?

    The coordinates are independent of the physics, but we need our models to bear some relation to the physics in order for them to be useful. The second part of the lesson is that, although an arbitrary coordinate system need not tell us anything helpful, we can always choose one which has a special relationship to the physics in question.

    The special relationship is that the coordinate axes encode the constraints on the system. This is why we change coordinate systems with the symmetry of the system we study.

    In a way this is the power of the Lagrangian mechanics - it formalizes (and generalizes) the process of changing coordinates to make the math easier. This makes complicated systems easier to analyse - you do the math in whatever coordinate system makes the math easiest and then transform the answer into something your brain finds easy to think about.

    We can go further: by generalizing the coordinates, you can identify those parts of the physics that do not depend on what representation we choose - things like that. A lot of the rest of the course will have you doing just that.

    (caveat: it's been a while so I'm a bit rusty - I'm sure AlephZero will redirect where I am not careful enough with my wording and emphasis.)
  12. Aug 30, 2012 #11
    So am I to understand it like this. Say we had some cartesian frame in which we represented our system. Shifting that frame to another origin would not alter the values of the kinetic energy of any of our particles. On the other hand: If we did the same to a particle moving in a system described by polar coordinates, then we would alter the value of its KE by shifting the origin?
  13. Aug 30, 2012 #12

    Simon Bridge

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    No - the derivative of the KE with respect to a position coordinate can differ with the coordinate system. dT/dx=0 but dT/dθ may not be.

    Of course you realize that a coordinate system moving wrt the first one will yield a different value for KE right?

    What the author has shown you is that some force components can get introduced by some coordinates. Remember - this is a statement about the coordinates not a statement about the physics. The introduced terms are characteristic of the "curvature" of the coordinate. The key lesson is that any generalized coordinate can be chosen to have any curvature. The choice of coordinate system is arbitrary.

    Note: you already know, right, that in a workless system - the statement "kinetic energy is a constant" is just a statement of the conservation of energy? If there is work as well you have to say that the total energy is a constant.

    Have a look at what happens the other way - start out with polar coordinates and an object with a constant angular velocity. Is that a workless system? Does kinetic energy depend on position? In cartesian coordinates - the velocity changes with position. What happens with the representation above?

    Try not to get too caught up in particular coordinate systems: let yourself get used to generalized systems. Also, try not to draw conclusions about the physics from the behavior of the coordinates - how you get to the physics from the model is later in the course - let yourself get used to treating mathematical models entirely abstractly.

    I'm wondering if perhaps you could use a bit of a run-up before diving into the math?
    ... this is a lecture series: a bit flawed (imo. lecturer falls into a few pedagogical traps) but it does provide a useful, long, run at the concepts before tackling Lagrangians.
  14. Aug 30, 2012 #13
    1) Yes it's a workless system. Kinetic energy does not depend on the position. But for constant angular velocity changing r means a change in the kinetic energy, since v=wr. For a cartesian coordinate system you cannot say the same since there is not bond for the velocity between a change in x and y. I think this is also the same principle that explains why the unit vectors in a cartesian coordinate system are fixed while they change all the time in polar coordinates. Is that about right? :(
    I am little confused on this term workless system, but I guess it is just another way of viewing mechanics. I.e. the lagrangian formalism comes from d'Alemberts principle which studies how a system should move under the condition that the total work done is zero.
    Hope some of the above sounds about right.
  15. Aug 30, 2012 #14

    Simon Bridge

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    Careful - that is the tangential velocity - if the radius is constant (circular motion) this is also a constant. In the linear motion of AlephZero's example, there is also a radial component to the velocity.
    You need to work it through like the book did.
    The unit vectors are fixed in their respective coordinate systems - but are not fixed with respect to the other coordinate system.

    That's the idea - you saw that with a linear kinematic system, the kinetic energy turned into the expression for work when it was expressed in terms of position?

    When you do external work on the system it is no longer closed: you are adding (or removing) energy and there is an applied force somewhere. This is why AlephZero's example was for constant speed.

    You don't have to use workless constraints - these are just the simple practise examples while you get used to the formalism. You can have any constraints you like ... and they can get very complicated ... as you will discover ;)

    One of the things they should show you later on is that total energy and momentum should be conserved. It just "drops out" of the math after a very general assumption. iirc it is a consequence of forward and backwards determinism. But it's 1am so don't hold me to that.
  16. Aug 30, 2012 #15
    oh wait.. I understand it better now. As you say it is just simply a property that has to do with the mathematics of the coordinates in which we represent our system. In a cartesian coordinate system kinetic energy is completely independent of the coordinates since it is just the sum of the squares of the velocity in the independent direction. But in polar coordinates there is an explicit mathematical dependence between our coordinates, that is v is the product of our two variables.
    Really I don't think there is anything more to it, and I shouldn't confuse myself thinking more about it.
  17. Aug 30, 2012 #16

    Simon Bridge

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    And that is, pretty much, the idea that the author wanted you to get.
    A whole lot of things should click into place in a bit.
    Now you probably need a drink.

    Bonus marks: see if you can explain it to someone else :)
  18. Aug 30, 2012 #17
    Okay, so all in all the difference between eliminating time and then getting KE as a function of (x,y) and the above, is that the kinetic energy is mathematically dependent on the r even though v and r are viewed as entirely dependent quantities. But what is, in the language of lagrangian mechanics, the physical interpretation or significance of the fact that dT/dr is not equal to zero? Right now it just stands as a mathematical statement of the nature of the coordinates for me.

    Also, all my confusion seemed to arise from the fact that I could not distinguish between the case where you view r,v,t as explicitly independent quantites and the case where you eliminate one to get them as a function of each other. So far in all of mechanics it has always been a huge benefit to do so, but now I am being represented with a formalism that explicitly wants to view these variables as independent - what is the advantage of doing so?
  19. Aug 30, 2012 #18
    The whole point of Lagrangian mechanics is that you compose the kinetic and potential energies in a simple form via the generalized coordinates and velocities and then you get nice equations of motion.

    I know this is not obvious when you are given the material in the axiomatic way, when before you use the equations for practical work you derive them theoretically, and the derivation is quite hairy.

    What I suggest is just take the equations for granted for some time, and apply them to some problems. Take a pendulum. It can be described by one generalized coordinate: the angle from the vertical. See how simple it is to get a nice equation for the angle. A few more such examples and you will see why those equations are useful. Then you can look again at the derivation.
  20. Aug 30, 2012 #19

    Simon Bridge

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    That's what I'd suggest too - the early stage of the course is just the "getting to know the tools" bit. You already have very good tools and you don't see why you should have to bother with these funny-looking ones that seem counter-intuitive.

    We use them because they are better, more powerful, but you won't see that until you start using them so do so.

    ... that's the main lesson - it is a mathematical statement about the coordinates.

    Have you actually tried applying the lagrangian formalism with the circular motion with constant angular velocity constraint?

    The pendulum is a good one to try - also constrained to a circle, but this time with an extra force.

    A harder one, where the radius and angular velocity both get to change, is free-fall in a conservative, central, force. That usually gets introduced quite a bit later.

    I also still think the linked lecture series will help you handle it conceptually. However - I'm pretty sure the original questions have now been answered :)
  21. Aug 30, 2012 #20
    I would also add that after you play with the equation in problems, the derivation becomes easier to comprehend, because you have some intuition developed by then.
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