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I am a Physics undergraduate, so provide references with your responses.

Landau & Lifshitz write in page one of their mechanics textbook:

If all the co-ordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. Mathematically this means that, if all the co-ordinates $q$ and $\dot{q}$ are given at some instant, the accelerations $\ddot{q}$ at that instant are uniquely defined.

They justify this as being "known from experience", which is not entirely satisfactory. What is the basis for their assertion?

Similar: Why are there only derivatives to the first order in the Lagrangian?

Is his question equivalent to mine, even though his solely refers to Lagrangian Mechanics?

Furthermore, this might just indicate how mathematically crude my mind is, but why is it not sufficient simply to give the coordinates $q$, and determine $\dot{q}$ from that, i.e. if $q$ is given by some smooth function, can we not determine all further derivatives from that alone?

Community
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Mark Allen
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  • Related: http://physics.stackexchange.com/q/18588/2451 , http://physics.stackexchange.com/q/4102/2451 and links therein. – Qmechanic Feb 23 '16 at 20:53

5 Answers5

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You should think of this by timestepping Newton's laws--- if you know the positions and velocity and one instant, you know the force, and the force determines the acceleration. This allows you to determine the velocity and an infinitesimal time in the future by

$$ v(t+dt) = v(t) + dt F/m $$ $$ x(t+dt) = x(t) + dt v $$

You then find the position and velocity at the next time step, and you find the new force, and continue forever. This is an algorithm to solve Newton's laws, and all that LL are saying is that Newton's laws are known from experience with objects, they are inducted from observations.

Ron Maimon
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  • +1 good premise, be careful not to imply observational results alone are enough. – Argus Jun 09 '12 at 04:47
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    You seem to be saying that you can calculate the force, rather than the force being specified for you. – Larry Harson Jun 10 '12 at 13:58
  • @Argus: "Induction" means experimental results plus whatever it is that allows us to do induction. – Ron Maimon Jun 11 '12 at 04:13
  • @LarryHarson: You calculate the force from the law of force, and the positions and velocities of the particles. – Ron Maimon Jun 11 '12 at 04:14
  • The law of force is $F =m\frac{dv}{dt}$ and you're given $F$, whereas you calculate $dv$ from $F$ and $dt$. Your second equation should have a $vdt$ instead of $dtx$ – Larry Harson Jun 11 '12 at 08:56
  • @LarryHarson: No, no, no. You are not given F as a function of time, you are given F as a function of position. The position is unknown until you do the simulation, so to find the value of the force at time t, you need to calculate it from the position at time t. The law of force is not F=ma, it's F=GMm/r^2, or F= -a/r^6 + b/r^12, or something like that--- it's a law that tells you the force between particles as a function of their positions. – Ron Maimon Jun 11 '12 at 12:45
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"known from experience" in here means "known from experience that first order derivatives in the Lagrangian or second order in the equations of motion are enough". I think their basis for this assertion is very Occamian (but who could know with certainty what L&L were thinking about?) The non-Occamian approach to this answer is given in the post you quoted in your question.

For the last question indeed you can determine $\dot{q},\ddot{q},\ldots,$ from $q$ alone if you already know $q$. But wait! wasn't finding $q$ the problem? and, how are you suppose to determine $q$? Solving a second order equation, for which you need initial conditions ($q_0,\dot{q}_0$).

c.p.
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    Sorry, the second questions seems comically obvious now, I don't how I managed to get confused, I somehow managed to interpret the statement "if all the coordinates $q$ and $\dot{q}$ are given at some instant" as the statement "if the coordinates $q$ [and $\dot{q}$] are given at all instances" – Mark Allen Jun 09 '12 at 07:51
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Given a second order differential equation, the solution to it will be uniquely determined by two sets of data.

yca
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In order to know about a system, we need to know all the forces acting on that system at some instant. Classically, we always solve Newton's equation of motion i.e conservation of momentum. Again, momentum is a function of velocity and velocity is function of coordinates.

So, if the velocity and coordinates are known, we solve Newton's equation of motion and hence will able to define the system. However, determining all the forces in a general system may not be easy. So, generalized coordinates and velocities are introduced. Using generalized coordinates and velocities, the solution of the equation of motion becomes easier and more efficient.

tpg2114
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sushmita
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i think it's becoz all forces we observe experimentaly are functions of r and v only like the Lorentz force or the gravitational force etc

baz
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