13.5: Componentes de aceleración

En la Sección 3.4 del “libro” de Mecánica Celestial, derivé los componentes radial y transversal de velocidad y aceleración en coordenadas bidimensionales. Los componentes de velocidad radial y transversal son bastante obvios y apenas necesitan derivación; son justos$\dot{\rho}$ y$\rho\dot{\phi}$. Para los componentes de aceleración reproduzco aquí un extracto de ese capítulo:

“Los componentes radial y transversal de la aceleración son, por lo tanto$(\ddot{\rho}-\rho\dot{\phi}^{2})$ and $(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi})$ respectively.”

También derivé los componentes radial, meridional y acimutal de velocidad y aceleración en coordenadas esféricas tridimensionales. Nuevamente los componentes de velocidad son bastante obvios; son$\dot{r},r\dot{\theta}$ and $r\sin\theta\dot{\phi}$ while for the acceleration components I reproduce here the relevant extract from that chapter.

“Al reunir los coeficientes de$\bf{\hat{r},\hat{\theta},\hat{\phi}}$ we find that the components of acceleration are:

• Radiales:$\ddot{r}-r\dot{\theta}^{2}-r\sin^{2}\theta\dot{\phi}^{2}$
• Meridional:$r\ddot{\theta}+2\dot{r}\dot{\theta}-r\sin\theta\cos\theta\dot{\phi}^{2}$
• Azimutal:$2\dot{r}\dot{\phi}\sin\theta+2r\dot{\theta}\dot{\phi}\cos\theta+r\sin\theta\ddot{\phi}.$ "

Quizás te gustaría mirar hacia atrás a estas derivaciones ahora. Sin embargo, ahora voy a derivarlos por un método diferente, usando la ecuación de movimiento de Lagrange. Puedes decidir por ti mismo cuál prefieres.

Empezaremos en dos dimensiones. Let$R$ and $S$ be the radial and transverse components of a force acting on a particle. (“Radial” means in the direction of increasing $\rho$; “transverse” means in the direction of increasing $\phi$.) If the radial coordinate were to increase by $\delta\rho$, the work done by the force would be just $R \delta\rho$. Thus the generalized force associated with the coordinate $\rho$ is just $P_{\rho}=R$. If the azimuthal angle were to increase by $\delta\phi$, the work done by the force would be $S\rho\delta\phi$. Thus the generalized force associated with the coordinate $\phi$ is $P_{\phi}=S\rho$. Now we do not have to think about how to start; in Lagrangian mechanics, the first line is always “$T$= ...”, and I hope you’ll agree that

$$T=\frac{1}{2}m(\dot{\rho}^{2}+\rho^{2}\dot{\phi}^{2}). \label{13.5.1}$$

If you now apply Equation 13.4.12 in turn to the coordinates $\rho$ and $\phi$, you obtain

$$P_{\rho}=m(\ddot{\rho}-\rho\dot{\phi}^{2}) \quad and \quad P_{\phi}=m\rho(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi}), \label{13.5.2a,b}\tag{13.5.2a,b}$$

and so

$$R=m(\ddot{\rho}-\rho\dot{\phi}^{2}) \quad and \quad S=m(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi}). \label{13.5.3a,b}\tag{13.5.3a,b}$$

Therefore the radial and transverse components of the acceleration are $(\ddot{\rho}-\rho\dot{\phi}^{2})$ and $(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi})$ respectively.

We can do exactly the same thing to find the acceleration components in three-dimensional spherical coordinates. Let $R$, $S$ and $F$ be the radial, meridional and azimuthal (i.e. in direction of increasing $r$, $\theta$ and $\phi$) components of a force on a particle.

• Si$r$ increases by $\delta r$, the work on the particle done is $R \delta r$.
• If $\theta$ increases by $\delta\theta$, the work done on the particle is $Sr \delta \theta$.
• If $\phi$ increases by $\delta\phi$, the work done on the particle is $Fr\sin\theta\delta\phi$.

Therefore $P_{r}=R,\quad P_{\theta}=Sr$ and $P_{\phi}=Fr\sin\theta$.

Start:

$$T=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+r^{2}\sin^{2}\theta\dot{\phi}^{2}) \label{13.5.4}\tag{13.5.4}$$

If you now apply Equation 13.4.12 in turn to the coordinates $r, \theta$ and $\phi$, you obtain

$$P_{r}=m(\ddot{r}-r\dot{\theta}^{2}-r^{2}\sin^{2}\theta\dot{\phi}^{2}), \label{13.5.5}\tag{13.5.5}$$

$$P_{\theta}=m(r^{2}\ddot{\theta}+2r\dot{r}\dot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}) \label{13.5.6}\tag{13.5.6}$$

and

$$P_{\phi}=m(r^{2}\sin^{2}\theta\ddot{\phi}+2r^{2}\dot{\theta}\dot{\phi}\sin\theta\cos\theta+2r\dot{r}\dot{\phi}\sin^{2}\theta). \label{13.5.7}\tag{13.5.7}$$

Therefore

$$R=m(\ddot{r}-r\dot{\theta}^{2}-r\sin\theta\dot{\phi}^{2}), \label{13.5.8}\tag{13.5.8}$$

$$S=m(r\ddot{\theta}+2\dot{r}\dot{\theta}-r\sin\theta\cos\theta\dot{\phi}^{2}) \label{13.5.9}\tag{13.5.9}$$

and

$$F=m(r\sin\theta\ddot{\phi}+2r\dot{\theta}\dot{\phi}\cos\theta+2\dot{r}\dot{\phi}\sin\theta). \label{13.5.10}\tag{13.5.10}$$

Thus the acceleration components are

• Radial: $\ddot{r}-r\dot{\theta}^{2}-r\sin^{2}\theta\dot{\phi}^{2}$
• Meridional: $r\ddot{\theta}-2\dot{r}\dot{\theta}-r\sin\theta\cos\theta\dot{\phi}^{2}$
• Azimuthal: $2\dot{r}\dot{\phi}\sin\theta-2r\dot{\theta}\dot{\phi}\cos\theta+r\sin\theta\ddot{\phi}$.

Be sure to check the dimensions. Since dot has dimension T^-1, and these expressions must have the dimensions of acceleration, there must be an $r$ and two dots in each term.