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.