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Physics LibreTexts

13.5: Acceleration Components

In Section 3.4 of Chapter 3 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.

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