Lagrangians: the obvious questions

[Rasalhague]

(1) How can a generalized velocity function, $\dot{q}$, be "independent" of the corresponding generalized position function, $q$. One is the derivative of the other.

(2) How can any Lagrangian function be "time-independent", given that its component functions are defined as functions that depend on time, $(q^1,...,q^n)$, and their time derivatives, $(\dot{q}^1,...,\dot{q}^n$? Is dependence not a transitive relation?

*

Is it that on a specific path through $\mathcal{T}(\mathcal{C})$, the tangent bundle of a configuration space, velocity can perhaps be parameterized by position, and therefore made dependent on position, somewhat as as a path in $\mathbb{R}^2$ can be parameterized by either one of its two component functions (so long as it's not parallel to the other axis for any way), or by some other parameter, call it $t$--but to name an arbitrary point in the set of all possible states of a system (before any constraints are specified, and thus before any curve is defined) requires independent coordinate values for position and velocity, since either could take any value for all we know yet (given the degrees of freedom), just as, in $\mathbb{R}^2$, it takes two real numbers to unambiguously identify a point, given all the possible points that exist in that set.

And in answer to 2, I'm guessing dependence is indeed not transitive, since the domain of a function, $f$, is not necessarily the same as the domain of the composition, $f \circ g$ of that function with another, $g$. I suppose an analogy would be containment, which is transitive in everyday English, but not in set theory: $x \in S$ and $S \subseteq T$ does not imply $x \in T$.

 

[diazona]

Is it that on a specific path through $\mathcal{T}(\mathcal{C})$, the tangent bundle of a configuration space, velocity can perhaps be parameterized by position, and therefore made dependent on position, somewhat as as a path in $\mathbb{R}^2$ can be parameterized by either one of its two component functions (so long as it's not parallel to the other axis for any way), or by some other parameter, call it $t$--but to name an arbitrary point in the set of all possible states of a system (before any constraints are specified, and thus before any curve is defined) requires independent coordinate values for position and velocity, since either could take any value for all we know yet (given the degrees of freedom), just as, in $\mathbb{R}^2$, it takes two real numbers to unambiguously identify a point, given all the possible points that exist in that set.

I think you might have it there. The point of Lagrangian mechanics (specifically action minimization and the Euler-Lagrange equation) is to find the path through phase space, but before the path is determined, we can't make any assumptions about what points in the space will be on the path or not (except for the endpoints).

On point #2, it's possible for the positions and velocities to be combined in a way that is time-independent, even if none of them individually are. Typical examples are conserved quantities like energy and momentum. In these cases, the equation expressing the conserved quantity in terms of the positions and velocities could be considered a constraint (although it's one that emerges from the Lagrangian itself, not an externally imposed constraint).

 

[K^2]

Time independence means the dynamic does not depend on whether the starting point is at t₁ or some other t₂. So long as initial position and velocity are the same, trajectory will be the same. Mathematically, it means that partial derivative of L with respect to time is zero.

 

[Rasalhague]

Thanks to you both for your replies.

K^2, the examples that I've seen so far have mostly involved functionals defined in terms of integrals with fixed limits. It's all fairly abstract to me at the moment, and so this may well be a silly question, but is there never a case where it's necessary to distinguish (even conceptually) between time-independence, as such, and dependence on the starting point (first limit)?

 

[K^2]

Sure. Imagine you are solving a problem of a ball rolling in a half-pipe. You can easily set up the Lagrangian, and for small angle, solve it approximately. Now, imagine that the half pipe itself oscillates up and down. Now your potential (or constraint, depending on how you set it up) has a time-dependent term, and the solution will depend on what the starting time is.

 

[quantum123]

Lagrangian is just a mathematical invention that treat q and q-dot as independant variables just as Hamiltonian treat q and p as independent variables.
I just wonder how did these things get invented from Newton's Laws in the first place, and how did they get so popular historically. Seems that mathematically you can invent even more of such things...

 

[K^2]

Lagrangian has deep meaning if you look at path integral formulation of quantum mechanics. But yes, in classical mechanics, it does look arbitrary.

 

[Derivator]

(1) How can a generalized velocity function, $\dot{q}$, be "independent" of the corresponding generalized position function, $q$. One is the derivative of the other.

[...]

Is it that on a specific path through $\mathcal{T}(\mathcal{C})$, the tangent bundle of a configuration space, velocity can perhaps be parameterized by position, and therefore made dependent on position, somewhat as as a path in $\mathbb{R}^2$ can be parameterized by either one of its two component functions (so long as it's not parallel to the other axis for any way), or by some other parameter, call it $t$--but to name an arbitrary point in the set of all possible states of a system (before any constraints are specified, and thus before any curve is defined) requires independent coordinate values for position and velocity, since either could take any value for all we know yet (given the degrees of freedom), just as, in $\mathbb{R}^2$, it takes two real numbers to unambiguously identify a point, given all the possible points that exist in that set.

[...]

This is a nice explanation, why to treat q and q-dot independently. But how can one tell, that q and q-dot are indeed the correct two variables needed to specify a classical system uniquely? I think, we can only say this from experience. Hence the statement 'Any classical mechanical system is uniquely specified by q and q-dot' should be treated as an axiom. Can someone comment on this, please?

 

[diazona]

There's nothing special about $q$ and $\dot q$, you can use any two (sets of) variables that fully parametrize phase space. It's not even necessary that one of them be the derivative of the other - for example, in Hamiltonian dynamics, where the conventional choice of variables is $q$ and $p$.

 

[Derivator]

yes, of course you could use any two variables. The question is, why a point in phase space uniquely describes the state of a mechanical system. I think, this is just based on experience.

 

[diazona]

If you're asking about why only two variables are enough, there's a very good (at least, I have reason to believe it's very good) resource on this that I'm unfortunately not allowed to link to, but the gist is that if you had a dependence on higher derivatives of the position, the Hamiltonian would not have a stable ground state.

 

[Derivator]

is it possible, that you give a reference to this resource, without linking? maybe i have acces to this document (via the university library)

 

[quantum123]

Just name the price.

 

[Fredrik]

(1) How can a generalized velocity function, $\dot{q}$, be "independent" of the corresponding generalized position function, $q$. One is the derivative of the other.

(2) How can any Lagrangian function be "time-independent", given that its component functions are defined as functions that depend on time, $(q^1,...,q^n)$, and their time derivatives, $(\dot{q}^1,...,\dot{q}^n$? Is dependence not a transitive relation?

The post is old, but I'll answer it anyway because so many people have difficulties with this. These questions have a really simple answer. Suppose that $f:\mathbb R^2\rightarrow\mathbb R$. For each $x,y\in\mathbb R$, $f(x,y)$ is a real number. Would you ask how x and y can be independent here? Of course you wouldn't. The question wouldn't make much sense. The variables that represent the components of a member of the domain are called "the independent variables". It's not something we prove, it's just a name we have for them. If we write z=f(x,y), we can call z "the dependent variable". (I find this terminology useless, but it seems to be standard). In question (1), $q(t)$ and $\dot q(t)$ are indepedent variables in exactly this sense. For example, in the case of a single particle in 1 dimension, the Lagrangian $L:\mathbb R^3\rightarrow\mathbb R$ is defined by $$L(a,b,c)=\frac{1}{2}mb^2-V(a)$$ for all $a,b,c\in\mathbb R$. $L(q(t),\dot q(t),t)$ is just a number in the range of that function.

How can the Lagrangian be time-independent? Because the phrase "the Lagrangian is time-independent" is defined as "$D_3L(a,b,c)=0$ for all $a,b,c\in\mathbb R$". (That's how it's defined when the domain of the Lagrangian is $\mathbb R^3$). Here $D_3$ denotes partial differentiation with respect to the third variable. Since there's no c on the right-hand side of $L(a,b,c)=\frac{1}{2}mb^2-V(a)$, we certainly have $D_3L(a,b,c)=0$ for all $a,b,c\in\mathbb R$.

I think the main reason why physics students get confused by things like this is that they've been taught to think about expressions like f(x,y) as "a function that depends on x and y". Don't do that. f is the function. f(x,y) is a member of its codomain. People often scratch their heads in confusion when they see the Euler-Lagrange equations, and in particular the expression $$\frac{\partial L}{\partial\dot q}.$$ It wouldn't cause any confusion if people could just remind themselves that a partial derivative is always with respect to variable number k, for some k, and then ask themselves what function are we dealing with here? What's its domain and codomain? Can we define it with a formula?

 

[Fredrik]

yes, of course you could use any two variables. The question is, why a point in phase space uniquely describes the state of a mechanical system. I think, this is just based on experience.

It's based on a theorem about existence and uniqueness of solutions of the relevant kind of differential equations. There's exactly one solution for each point in phase space. But the choice to only consider that kind of differential equations is of course based on experience. We don't seem to need any other kind.

 

[diazona]

is it possible, that you give a reference to this resource, without linking? maybe i have acces to this document (via the university library)

Sorry I didn't get to reply earlier; the document is not available offline. In retrospect I probably shouldn't have brought it up. (If I have time I'll see if I can explain the argument in some more detail, but I can't make any promises.)

 

[Rasalhague]

In question (1), $q(t)$ and $\dot q(t)$ are indepedent variables in exactly this sense.

This notation makes them look as if they're the values of two functions $\mathbb{R}\rightarrow \mathbb{R}$, one called $q$, and the other $\dot{q}$. If I was trying to interpret $q(t)$ and $\dot q(t)$ as independent variables in exactly the same sense as x and y above, I might conclude that $q$ and $\dot{q}$ are function-valued variables, labelled according to which argument slot of $L$ their values are going to be inserted into - supposing we want to compose them with $L$ - each one representing an arbitrary member of the set of all functions from $\mathbb{R}\rightarrow \mathbb{R}$, not necessarily related in any way. A problem with this interpretation is that when the letter $q$ appears in an equation, representing a specific (albeit arbitrary) function of this kind, the context (e.g. the Euler-Lagrange equation) does tend to require $\dot{q}$ to be the derivative of that same specific function $q$, rather than just another arbitrary function, doesn't it? If so, then they don't represent independent variables.

In the Euler-Lagrange equation, the Lagrangian,

$$L:\mathbb{R}^3\rightarrow \mathbb{R} \; | \; (a,b,c) \mapsto L(a,b,c),$$

is composed with another function to give

$$L\circ(f,f',Id):\mathbb{R}\rightarrow \mathbb{R}\; | \; t\mapsto L(f(t),f'(t),t)$$

where $f$ is a differentiable function, and $f'$ its derivative function. Alright, $L(f(t),f'(t),t)$ is a number in the range of $L$, but so is

$$L(f(t),g'(t),t)$$

and

$$L(f(t),f'(u),v)$$

and

$$L(f''(t),t,f'(t)).$$

It seems the notation says more than that $L(f(t),f'(t),t)$ is just a number in the range of $L$.

Does $L(q,\dot q, t)$ means something like $L(f(t),f'(t),t)$ or something like $L(a,b,c)$? I get the impression it could mean either.

 

[Rasalhague]

My impression is that there are two conventions here, and they tend - confusingly - to be used both at the same time. According to one convention, $q$ and $\dot q$ are simply labels for independent variables, so that a point in $\mathbb{R}^3$ can be labelled $(q,\dot q,t)$, just as it could be labelled $(x,y,z)$. In this sense, $\dot q$ doesn't depend on $q$, and $q$ and $\dot q$ don't depend on $t$.

The only significance of this special choice of labels is that $q$ labels points in the conguration space of the dynamical system being modeled with $\mathbb{R}^3$, while $\dot q$ labels possible values of a certain set of functions $A_2=\left \{ \alpha_2, \beta_2, \gamma_2, ... \right \}$, defined, in relation to certain other functions $A=\left \{ \alpha, \beta, \gamma, ... \right \}$, by a differential equation specific to the dynamical system. Let the functions in $A$ be of the form

$$\gamma:\mathbb{R}\rightarrow \mathbb{R}^3 \; | \; \gamma(t) = (\gamma_1(t),\gamma_2(t),t)$$

with$\gamma_2 = \gamma_1'$. Then $\dot q$ labels values in the range of $\gamma_2$ for all elements, $\alpha, \beta, \gamma, ...$ of $A$, for all possible inputs. And the differential equation imposes a condition on the elements of $A$ which specifies their ranges, the trajectories. Initial conditions then select particular element of $A$ from those sharing some particular range.

But according to the other convention, $q$ actually denotes the component function $\gamma_1$, for an arbitrary $\gamma \in A$, and $\dot q$ denotes the function $\gamma_2 = \gamma_1'$, for the same $\gamma \in A$. Or, in a variant of this convention, $q$ and $\dot q$ denote values of these functions, for the same $\gamma$, evaluated at the same input $t$. When used in this sense, $\dot q = \gamma_2(t) = \gamma_1'(t)$ depends both on $q = \gamma_1$ and $t$.

 

[Fredrik]

It seems the notation says more than that $L(f(t),f'(t),t)$ is just a number in the range of $L$.

I don't understand your conclusions in this post. The fact that we have f(t) in the first slot and f'(t) in the second only means that we're evaluating the function L at the specific point (f(t),f'(t),t)) rather than, say, (f(t),g(t),t). We're still talking about the value of L at a point in its domain. The symbols used tell us which point that is (and absolutely nothing else), but that's also the case when we denote the point by (a,b,c).

Does $L(q,\dot q, t)$ means something like $L(f(t),f'(t),t)$ or something like $L(a,b,c)$? I get the impression it could mean either.

There's no difference. (a,b,c) is a point in the domain of L, and so is (f(t),f'(t),t). Keep in mind that f(t) isn't a function. It's a point in the codomain of the function f. So it's a number, just like a.

$(q,\dot q,t)$ is a sloppy (and misleading) notation for $(q(t),\dot q(t),t)$. It's a member of $\mathbb R^3$. Which member of $\mathbb R^3$ it is depends on the function q of course, and on the number t. The function $\dot q$ is of course completely determined by $q$.

My impression is that there are two conventions here, and they tend - confusingly - to be used both at the same time.

What's confusing is to denote the number q(t) by the symbol q which is already reserved for a function.

According to one convention, $q$ and $\dot q$ are simply labels for independent variables, so that a point in $\mathbb{R}^3$ can be labelled $(q,\dot q,t)$, just as it could be labelled $(x,y,z)$. In this sense, $\dot q$ doesn't depend on $q$, and $q$ and $\dot q$ don't depend on $t$.

That last part isn't true, since in this context q is a sloppy notation for q(t). Edit: q(t) and t should be viewed as the values of two independent variables of the function L, but obviously you can't change the value of the variable "t" without changing the value of the expression "q(t)".

The only significance of this special choice of labels is that $q$ labels points in the conguration space of the dynamical system being modeled with $\mathbb{R}^3$, while $\dot q$ labels possible values of a certain set of functions

I don't understand what you're saying here and after this. If $q$ is an n-tuple of functions, then so is $\dot q$, and each of its components is the derivative of the corresponding component of q. But in the simple case we're discussing, we have n=1, so $q$ and $\dot q$ are both functions, and $\dot q$ is the derivative of $q$.

I think you're making this much more complicated than it needs to be. In the case of a single particle of mass $m\in\mathbb R$ moving under the influence of a potential $V:\mathbb R\rightarrow\mathbb R$, the Lagrangian is defined as the function $L:\mathbb R^3\rightarrow\mathbb R$ that satisfies $$L(a,b,c)=\frac{1}{2}mb^2-V(a)$$ for all $(a,b,c)\in\mathbb R^3$. It's of course entirely trivial to compute the partial derivatives of this function at all points in its domain (at least if we are going to express them in terms of F=-V'). The action is defined as the function $S:\mathbb R^\mathbb R\rightarrow\mathbb R$ that satisfies $$S[q]=\int_{\mathbb R} L(q(t),\dot q(t),t) dt$$ for all $q:\mathbb R\rightarrow\mathbb R$.

When we're dealing with a system with n generalized coordinates, we define \begin{align}
q &=(q_1,\dots,q_n)\\
\dot q &=(\dot q_1,\dots,\dot q_n)
\end{align} and take the Lagrangian to be a function $L:\mathbb R^n\times\mathbb R^n\times\mathbb R\rightarrow\mathbb R$. The action is defined by $$S[q]=\int_{\mathbb R} L(q(t),\dot q(t),t) dt$$ for all $q_1,\dots,q_n\in\mathbb R^\mathbb R$.

If the generalized coordinates take values in something other than $\mathbb R$ we will of course have to change the domain of the Lagrangian, integrate over some other set, etc., but $L(q(t),\dot q(t),t)$ is always just a number in the range of L.

 

[Rasalhague]

What's confusing is to denote the number q(t) by the symbol q which is already reserved for a function.

Well, it's certainly been confusing me. Thanks for your help. How about this one?

$$\frac{\partial L}{\partial q}=\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot q}$$

Using round brackets for tuples and to clarify assossiation, and square brackets to enclose the argument of a function, does this mean:

$$((D_1[L])\circ \gamma_q)[t]=(D[D_2[L]\circ \gamma_q])[t]$$

where $\gamma_q:\mathbb{R}\rightarrow \mathbb{R}^3$ such that $\gamma_q[t]=(q(t),\dot q(t),t)$.

 

[Fredrik]

Yes, that's exactly what it means.

I was also very confused by these things at first. I think I stared at these equations for hours on many different occasions until I finally figured it out. I think some kind of mental block sets in when we see something like $$\partial_\mu \left(\frac{\partial\mathcal L}{\partial( \partial_\mu\phi)}\right)$$ where $$\mathcal L=\frac{1}{2}(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2).$$ (I'm not going to bother to think about if the signs are right). I eventually got past it by asking myself questions like "what function are we dealing with" and being extremely careful with the notation. I was actually quite surprised when I realized that $\mathcal L$ is just a polynomial in 5 variables (6 if we include t) and that the strange-looking expression in the big parentheses is nothing more than the $\mu+2$ partial derivative of that polynomial, evalutated at the point $(\phi(x),\partial_0\phi(x),\dots,\partial_3\phi(x))$ in its domain. Why couldn't someone just have told me that? It would have saved me a lot of time.

It doesn't make me as upset as when I think about how they tried to explain tensors to us, but I get a little annoyed every time I think about it. The fact that I've seen other posts about this suggests that it's being taught the same way (without any explanations of what functions we're actually dealing with) at other universities too. An even stronger indication of that is that I've never seen anyone but me explain it this way.

 

[Rasalhague]

Oh wow, at long last, progress on Lagrangians! Thank you so much, Fredrik. Let's see if I can read Hamilton's equations now. First, to get the basic idea, with one degree of freedom. The definition of momentum:

$$p[t]=((D_2[L])\circ\gamma_q)[t]$$

Wikipedia says then "find the velocities in terms of the momenta by inverting [this] expression." Not sure how to do that, but, let $\delta:\mathbb{R}\rightarrow \mathbb{R}^3$ such that $\delta_q[t]=(q[t],p[t],t)$, then perhaps the definition of the Hamiltonian function

$$H=\dot{q} p-L$$

means

$$(H\circ\delta_q)[t]=(D[q])\cdot ((D_2[L])\circ \gamma_q)[t]-L\circ\gamma_q[t].$$

And Hamilton's equations themselves,

$$\frac{\partial H}{\partial q}=-\dot p \enspace\enspace\enspace \frac{\partial H}{\partial p}=\dot q \enspace\enspace\enspace \frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t},$$

perhaps mean

$$((D_1[H])\circ\delta_q)[t]=-(D[p])[t],$$

$$((D_2[H])\circ\delta_q)[t]=(D[q])[t],$$

where, by the definition of momentum, $D[p]=D[((D_2[L])\circ\gamma_q)]$.

And

$$((D_3[H])\circ\delta_q)[t]=-((D_3[L])\circ\gamma_q)[t].$$

Expressing these three equations as one matrix equation, with a row vector on the right:

$$(H'\circ\delta_q)[t]=\begin{pmatrix}-p'[t] & q'[t] & -((D_3[L])\circ\gamma_q)[t]\end{pmatrix}$$

 

[Rasalhague]

That last part isn't true, since in this context q is a sloppy notation for q(t). Edit: q(t) and t should be viewed as the values of two independent variables of the function L, but obviously you can't change the value of the variable "t" without changing the value of the expression "q(t)".

It's statements like the following, by Roger Penrose, that made me think some people might define $q$ and $\dot q$ as independent variables, rather than as functions, using them to mean exactly what you meant by $a$ and $b$:

$$\cal{L}\times\mathbb{R} = \cal{L} (q^1,...,q^N;\dot q^1,...,\dot q^N)$$

Each $\dot q^r$ has to be treated as an independent variable (independent of $q^r$ in particular) in this expression.

- Penrose: The Road to Reality, Vintage 2005, p. 473.

I suppose people who use these terms would consider $q[t]$ to be the dependent variable of one function, namely $q$, while also considering it to be the independent variable of another $L$. But then it seems meaningless to say that some symbol represents an "independent variable" (in an apparently absolute sense) without saying with respect to which function, especially when there's another function being discussed of which it's the dependent variable.

Another alternative that occurred to me was that he might be thinking of them as a set of natural coordinate functions on the state space, the state space here being the tangent bundle $TC$ of the configuration space $C$ - and then identifying the coordinate representation of the Lagrangian with the Lagrangian as a function on $TC$, and then, in case that's too easy to follow, identifying all of these functions with their values. (And then not even mentioning $\gamma_f$, and its coordinate representation, with which these functions are implicitly composed.)

 

[Rasalhague]

I'd be very interested to hear if anyone knows of any books or online sources which deal with Lagrangian or Hamiltonian mechanics at an introductory level, explicitly defining what functions are involved, what their domains and codomains are, and avoiding the ambiguities of the Leibniz notation for derivatives.