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In famous quantum physics book of Cohen-Tannoudji, the authors write in the conclusion of the first chapter, that in classical physics, one needs both the vector $\vec{r}(t)$ and $\vec{v}(t)$ to describe a particle.

But why don't we only need $\vec{r}(t)$ since the vector of speed could be obtained by the derivation of the position ?

Mathieu Krisztian
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  • Possible duplicate: https://physics.stackexchange.com/q/885/50583 and its linked questions – ACuriousMind Mar 14 '21 at 16:05
  • ok, but the answer of the guy is not completely clear : "if you ask <<what is the kinetic energy of a particle?", then it is not enough to know the position of the particle>>" : the kinetic energy could be deduced directly from two consecutive position r(t) informations. – Mathieu Krisztian Mar 14 '21 at 16:16
  • "The guy"? There are eleven answers there. Also, no one is saying that $r(t)$ does not suffice to know everything about the particle's movement, they're just explaining why Lagrangian classical mechanics treats position $q$ and velocity $\dot{q}$ (not a solution $q(t)$ and its derivative $\dot{q}(t)$, see also this answer of mine) independently. You're likely misunderstanding/reading too much into the "needs both". – ACuriousMind Mar 14 '21 at 16:20
  • the guy is meaning the one whose answer was accepted : "Greg Graviton". ok. Thank you. Your explanation is the key point : they are discussing about Lagrangien (as does Cohen-Tannoudji probably in his conclusion, although he does not state : he says "description of the system"). So Apparently, Cohen-Tannoudji is discussion about Lagrangien (without writting it explicitely). Thank you – Mathieu Krisztian Mar 14 '21 at 16:23

3 Answers3

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In Newtonian Mechanics, we basically solve the equation: $$F(x,t)= m \frac{d^2x} {dt^2}$$ Since this is a second-order differential equation, you need two boundary conditions. In the Newtonian framework, boundary conditions are initial position and initial velocity.

So putting the values you get the trajectory of the particle.

Suppose, you get the trajectory of the particle as $y=f(x,t)$, Then you can get velocity as $v= \frac{dx}{dt}=f'(x,t)$

So knowledge of $v(t)$ is implicit in classical mechanics. without the knowledge of $v$ you don't have analytical knowledge of position, so you can't find out the velocity.

Also, In other formulations such as Lagrangian and Hamiltonian formulation position and velocity are very much intertwined. Also, there is well-built concept of phase space, that I am not discussing.

Finding the trajectory in real space is not at all easy as it seems. Thats why concept of lagrangian formulation and phase space have been developed.

crabNebula
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  • Let's ignore the problem of the initial conditions (for which, I agree, we miss the speed). You write $v=f'(x)$, so do you agree that there is no need of $v$ information : we already have it thanks to the function $f(x)$. If so, why do you write : "without the knowledge of v you don't have analytical knowledge of position" – Mathieu Krisztian Mar 14 '21 at 16:26
  • to know f(x), you should know v. Also, this is an oversimplified situation. At layman level, lets say you don't know v(t), you don't know v , you don't know f(x). – crabNebula Mar 14 '21 at 16:29
  • I'm lost : $f(t)$ is the set of the position x for various time. We know it. Thus, we just derivate it and obtain directly $f'(t)=v$. What is wrong in my sentence ? – Mathieu Krisztian Mar 14 '21 at 16:31
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    knowing velocity at initial conditon, means that you have knowledge of v at any time. – crabNebula Mar 14 '21 at 16:33
  • I do agree perfectly for the very initial condition. Apart this very initial condition, the speed could be obtained for all time. Do you agree ? – Mathieu Krisztian Mar 14 '21 at 16:34
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    Now I ask you what if force itself depends upon velocity. – crabNebula Mar 14 '21 at 16:37
  • You will not be able to find trajectory easily in real space. so you will need to use phase space and other formulations. – crabNebula Mar 14 '21 at 16:40
  • I agree with your point, but then I reply to your question with an alternative question: "what if force depends on the N-derivative of the position ? (for example N=2) " Then, the position and the speed are still not enough to describe the system... And by doing this reasoning, we could imagine N=infinity. – Mathieu Krisztian Mar 14 '21 at 16:40
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But why don't we only need $\vec{r}(t)$ since the vector of speed could be obtained by the derivation of the position ?

The way we express it is not perfectly clear, but the authors are saying that we need only $\vec{r}(t)$ and $\vec{v}(t)$ at one instant in time in order to find $\vec{r}(t)$ for all $t$.

If you already had $\vec{r}(t)$ for all $t$, then of course you could find $\vec{v}(t)$. But also, you wouldn't have any problem left to solve.

The Photon
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Of course if $\mathbf x = \mathbf f(t)$ is known, we don't need more information.

The problem is how to determine $\mathbf f(t)$ knowing $\mathbf F(\mathbf x)$ (force as a function of position)? We need $\mathbf x(t_0)$ and $\mathbf v(t_0)$ for some $t_0$.

Claudio Saspinski
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  • Your answer is almost useless without a definition of the symbols. I assume that you wanted to start saying that "if we know the position as a function of time...". But this was just a guess. – GiorgioP-DoomsdayClockIsAt-90 Mar 14 '21 at 16:59