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In beginner classical mechanics, which I've just started learning, a particle with coordinates $\mathbf q\in\mathbb R^n$ has its equation of motion specified by $F(\mathbf q,\dot{\mathbf q},t)=m\ddot{\mathbf q}$. Force is a function of all the coordinates necessary to describe the (rigid) body, and should cover all the degrees of freedom in the system. However, it seems to me that since $\dot{\mathbf q}=\frac{d\mathbf q}{dt}$, it's only necessary to specify $F(\mathbf q,t)$ for a complete description. I'm not sure I understand why this isn't the case, but my best guess is as follows.

If $\mathbf q,\dot{\mathbf q}$ are given in a differential equation, such as $\dot{\mathbf q}=t^\mathbf q\mathbf q^\dot{\mathbf q}$, then it's necessary to specify all of $\mathbf q,\dot{\mathbf q},t$ in order to locate its position and velocity at any given time, unless the differential equation has a solution and we use it.

But this explanation is strange to me. Since we can have a $k$th-order differential equation which specifies $\mathbf q,\frac{d}{dt}\mathbf q,\ldots,\frac{d^k}{dt^k}\mathbf q$ with no obvious solution, wouldn't that mean our equation of motion is actually $F(\mathbf q,\frac{d}{dt}\mathbf q,\ldots,\frac{d^k}{dt^k}\mathbf q,t)=m\frac{d^2}{dt^2}\mathbf q$?

Edit The differential equation above which has a vector exponentiated by a vector is just a poorly thought-out attempt at an example of a diff eq with no obvious solution to me, it doesn't really matter what it is. Or if you want to consider that case, treat it as a 1D system, I guess.

Andrews
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ok_
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    this properly belongs to physics.SE – Mister Benjamin Dover Jan 18 '15 at 21:05
  • Oh sorry, how do I move? – ok_ Jan 18 '15 at 21:05
  • What do you mean when you write, e.g., $\mathbf{q}^{\dot{\mathbf{q}}}$? What does it mean to raise a vector to the power of a vector? (Or to perform any exponentiation with vectors, for that matter?) – wchargin Jan 18 '15 at 21:10
  • @WChargin You're right, I should restrict this to the 1D case. I was just supplying a crazy diff eq example. – ok_ Jan 18 '15 at 21:11
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    @side1066 A force might depend explicitly on the velocity, think of friction. – johndoe Jan 18 '15 at 21:14
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    @johndoe And we know that it doesn't explicitly depend on acceleration, jerk, and all the others? – ok_ Jan 18 '15 at 21:15
  • If $F$ takes $q$ as a functional argument, then sure, $F(q,t)$ can involve $q'$. But here it doesn't (if it did, the analysis would be drastically more complicated). It takes it as a fixed vector $q(t)$ for each $t$. Knowing that vector doesn't tell you the velocity. – Ian Jan 18 '15 at 21:25
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    The force depending on higher derivatives would seem rather nonphysical, because such a system would not be causal. The $n$th derivative could depend on the $0$th, $1$st, ..., $n-1$th derivatives, though. But this can be put into the same form by appropriately choosing the vector $q$. – Ian Jan 18 '15 at 21:28
  • @Ian "because such a system would not be causal" Could you clarify? I'm not too great at physics yet. – ok_ Jan 18 '15 at 21:30
  • @side1066 Basically, force is supposed to cause acceleration, not the other way around. I'm not sure of a more precise answer here, though. – Ian Jan 18 '15 at 21:34
  • OK, so this is essentially a part of our model that it doesn't extend past $\dot{\mathbf q}$, because we observe that force causes acceleration. I think I get it. – ok_ Jan 18 '15 at 21:35
  • @side1066 Sorry to jump in. Yes, independence from higher derivatives is part of the content of Newton's law $F(\mathbf q,\dot{\mathbf q},t)=m\ddot{\mathbf q}$. – johndoe Jan 18 '15 at 21:39
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    It's been thirty+ years since I studied Lagrangian/Hamiltonian mechanics, but the reason to include $\dot{q}$ in the list of variables is that, when studying generalized momenta, you want calculate partial derivatives like $\partial F/\partial \dot{q}$. Including it serves as a reminder (or rather we take those partial derivative of the Lagrangian, but anyway, $q$, and $\dot q$ are in a sense independent variables.) – Jyrki Lahtonen Jan 18 '15 at 21:50
  • Related: http://math.stackexchange.com/q/580858/11127 and http://physics.stackexchange.com/q/885/2451 – Qmechanic Feb 01 '15 at 18:47

2 Answers2

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However, it seems to me that since $\dot{\mathbf q}=\frac{d\mathbf q}{dt}$, it's only necessary to specify $F(\mathbf q,t)$ for a complete description.

While it's true that the function $q$ determines its derivative $\frac{dq}{dt}$, it's not true that the value of $q$ at a particular value $t_0$ of $t$ determines the value of the derivative $\frac{dq}{dt}$ at that same value. The Lagrangian has a value at a particular time $t_0$ which is a function of the three numbers $q(t_0), q'(t_0)$, and $t_0$. In particular, it depends on more information than $q(t_0)$, but on less information than all of the higher derivatives of $q$ at $t_0$.

Qiaochu Yuan
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    This makes a lot of sense. $F$ is a function on certain values, since it needs to be a function of all the coordinates which correspond to the system's degrees of freedom.

    But say $q\in A$ and $\dot q\in B$ and the function $\dot q(q(t))=\frac{dq}{dt}$ (evaluated at given $t$) is a bijection between $A$ and $B$. Then the value of $q$ at $t_0$ gives me the value of $\dot q$ at $t_0$, so it would be fine in this special case to consider $F(q,t)=m\ddot q$?

     – ok_ Jan 19 '15 at 03:31
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Lagrangian mechanics is based on Newtonian mechanics (specifically d'Alembert's principle of virtual works), i.e. sets of second order ODEs. Under some regularity conditions, these second order ODEs can be turned into a system of first order ODEs, where the $q$s and the $\dot q$s are independent variables. Hence if you want to retrieve the equations of motion out of a Lagrangian (or an action principle), this must depend on the $q$s and the $\dot q$s. Observe that the action integral itself is assumed to depend on the trajectory alone, i.e.

$$S[q] = \int_a^b L(q(t),\dot q(t), t)\text dt.$$

Hamilton noticed that there is a standard procedure for passing from Euler-Lagrange equations, which are the second order ODEs of Newtonian mechanics, to the equivalent system of first order ODEs. For this to work, the condition is that the Hessian of $L$ w.r.t. the $\dot q$ should not vanish (so it must be positive definite on the domain of interest, i.e $L$ is concave w.r.t. the $\dot q$s). Such equations are the Hamilton equation coming from the Hamiltonian, which is the Legendre transform of $L$ w.r.t. the generalised velocities.

Phoenix87
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