# Can someone please explain to me what a dynamical trajectory means?

1. Jan 25, 2014

### richyw

Can someone please explain to me what a "dynamical trajectory" means?

I am reviewing the first chapter of my classical mechanics book (Fetter & Walecka), and in the very first section he uses the term "dynamical trajectory". I have seen this derivation many times, but I just don't understand exactly what that term means.

Here is the page (he goes from 1.7a to 1.7b using this principle)

Last edited by a moderator: May 6, 2017
2. Jan 25, 2014

### vanhees71

I've never heard this expression before, but obviously what Fetter and Walecka (btw. well known for their classic book on quantum many-body theory) mean a trajectory a particle takes according to the equations of motion (Newton's Law),
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}).$$
Then calculating the work done on the particle when moving along that line leads to the "work-energy theorem" (1.7a). It's important to remember that this law holds generally only for the work done when the particle is moving along the so defined "dynamical trajectory", i.e., the trajectory followed by the particle when solving the equation of motion.

The more special very important case for a conservative force, i.e., a force which has a potential,
$$\vec{F}=-\vec{\nabla}U$$
the work does not depend on the path along which the line integral, defining work is taken, but only on the start and end point of the path, as is proven in the next paragraph of the book.

Then, for any motion of the particle, i.e., "along any dynamical trajectory", the total energy of the particle is conserved:
$$\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 + U(\vec{x} \right)=m \dot{\vec{x}} \cdot \ddot{\vec{x}} + \dot{\vec{x}} \cdot \vec{\nabla} U=\dot{\vec{x}}(m\ddot{\vec{x}}-\vec{F})=0,$$
where in the last step we've made use of Newton's equation of motion.

3. Jan 25, 2014

### tiny-tim

hi richyw!
it seems to mean a trajectory in which the only force on it is the force from the field, F

then v (= ds/dt) is the same v as in the force equation F = mdv/dt

(personally, i would call that a projectile trajectory, ie a hit-and-forget trajectory of an object that, after it starts, has no propulsion other than from the field)

4. Feb 13, 2014

### mpresic

I think it means in plainest English that the particle moves. The trajectory describes the change in location as time progresses, or occasionally we may be interested in examining the past and then we consider time moving backwards.

I think the authors should omit the word dynamical which refers to systems evolving in time (nothing more). The term trajectory covers it all. Fetter and Walecka is a good theoretical mechanics textbook although I like Goldstein as well.

I think most authors occasionally put in superfluous words in their explanations occasionally which confuse an alert reader. Good catch and question

5. Feb 13, 2014

### olivermsun

I would read it this way as well. The alternative might be a particle moving along a path which is determined by kinematic constraints, e.g., moving along a curved track, attached by a string, etc.

6. Feb 13, 2014

### voko

The important part here is that the integral $\int\limits_1^2 d \boldsymbol s \cdot \boldsymbol F (\boldsymbol r)$ can in principle be taken along any curve between (1) and (2). "Dynamical", however, means that we require that Newton's second law be satisfied everywhere on the curve, which is why we can replace the force with the derivative of the momentum and obtain kinetic energy. The word trajectory alone may not be sufficient to ensure that, but that depends on how it is defined in the book (and whether it is even defined).

7. Feb 13, 2014

### mpresic

I may have been too dismissive of the term dynamical. The other posters provide good points. Hard to know the author's intent without asking them.