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13.5: Acceleration Components

In Section 3.4 of the Celestial Mechanics “book”, I derived the radial and transverse components of velocity and acceleration in two-dimensional coordinates. The radial and transverse velocity components are fairly obvious and scarcely need derivation; they are just $$\dot{\rho}$$ and $$\rho\dot{\phi}$$. For the acceleration components I reproduce here an extract from that chapter:

“The radial and transverse components of acceleration are therefore $$(\ddot{\rho}-\rho\dot{\phi}^{2})$$ and $$(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi})$$ respectively.”

I also derived the radial, meridional and azimuthal components of velocity and acceleration in three-dimensional spherical coordinates. Again the velocity components are rather obvious; they are $$\dot{r},r\dot{\theta}$$ and $$r\sin\theta\dot{\phi}$$ while for the acceleration components I reproduce here the relevant extract from that chapter.

“On gathering together the coefficients of $$\bf{\hat{r},\hat{\theta},\hat{\phi}}$$ we find that the components of acceleration 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: $$2\dot{r}\dot{\phi}\sin\theta+2r\dot{\theta}\dot{\phi}\cos\theta+r\sin\theta\ddot{\phi}.$$ "

You might like to look back at these derivations now. However, I am now going to derive them by a different method, using Lagrange’s equation of motion. You can decide for yourself which you prefer.

We’ll start in two dimensions. 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 don’t 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.

• If $$r$$ 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.