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[SOLVED] Lagrange coefficients f and g

dwsmith

Well-known member
Feb 1, 2012
1,673
I am not sure how to do this one. Nothing I try goes anywhere.

Consider the two-body equation of motion in vector form
$$
\ddot{\mathbf{r}} = -\mu\frac{\mathbf{r}}{r^3}.
$$
Show that the $f$ and $g$ functions defined by
$$
\mathbf{r} = f\mathbf{r}_0 + g\mathbf{v}_0
$$
satisfy
$$
\ddot{f} = -\mu\frac{f}{r^3},\quad \ddot{g} = -\mu\frac{g}{r^3}
$$
for arbitrary $\mathbf{r}_0$ and $\mathbf{v}_0$.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
What are the conditions on $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$? Are they fixed in time? Moving? Linearly dependent? Linearly independent?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
What are the conditions on $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$? Are they fixed in time? Moving? Linearly dependent? Linearly independent?
The are arbitrary position and velocity vectors.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,778
I am not sure how to do this one. Nothing I try goes anywhere.

Consider the two-body equation of motion in vector form
$$
\ddot{\mathbf{r}} = -\mu\frac{\mathbf{r}}{r^3} \qquad (1)
$$
Show that the $f$ and $g$ functions defined by
$$
\mathbf{r} = f\mathbf{r}_0 + g\mathbf{v}_0 \qquad (2)
$$
satisfy
$$
\ddot{f} = -\mu\frac{f}{r^3},\quad \ddot{g} = -\mu\frac{g}{r^3} \qquad (3)
$$
for arbitrary $\mathbf{r}_0$ and $\mathbf{v}_0$.
I've taken the liberty to number your equations.
Can you take the second derivative of (2)?
And substitute that together with (2) in (1)?
If so, what do you get?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The are arbitrary position and velocity vectors.
Are they initial position and velocity vectors? So they are not moving in time?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I've taken the liberty to number your equations.
Can you take the second derivative of (2)?
And substitute that together with (2) in (1)?
If so, what do you get?
$$
\ddot{\mathbf{r}} = \ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0
$$
This leads to the identity
$$
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 = -\mu\frac{\mathbf{r}}{r^3}.
$$
Next, we can plug $\mathbf{r}$ back into our new equation.
\begin{alignat*}{3}
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0 + g\mathbf{v}_0}{r^3}\\
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0}{r^3} - \mu\frac{g\mathbf{v}_0}{r^3}
\end{alignat*}
Can we just split them up at the end? If so, what is the reason for it?
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
$$
\ddot{\mathbf{r}} = \ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0
$$
This leads to the identity
$$
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 = -\mu\frac{\mathbf{r}}{r^3}.
$$
Next, we can plug $\mathbf{r}$ back into our new equation.
\begin{alignat*}{3}
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0 + g\mathbf{v}_0}{r^3}\\
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0}{r^3} - \mu\frac{g\mathbf{v}_0}{r^3}
\end{alignat*}
Can we just split them up at the end? If so, what is the reason for it?
This is precisely why I asked the questions I did. If $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$ are linearly independent, then this sort of breaking-up is possible. You could try writing them in terms of the regular basis vector and see what happens.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
This is precisely why I asked the questions I did. If $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$ are linearly independent, then this sort of breaking-up is possible. You could try writing them in terms of the regular basis vector and see what happens.
I can't tell you anymore about r0 and v0 because that is the whole question.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,778
$$
\ddot{\mathbf{r}} = \ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0
$$
This leads to the identity
$$
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 = -\mu\frac{\mathbf{r}}{r^3}.
$$
Next, we can plug $\mathbf{r}$ back into our new equation.
\begin{alignat*}{3}
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0 + g\mathbf{v}_0}{r^3}\\
\ddot{f}\mathbf{r}_0 + \ddot{g}\mathbf{v}_0 & = & -\mu\frac{f\mathbf{r}_0}{r^3} - \mu\frac{g\mathbf{v}_0}{r^3}
\end{alignat*}
Can we just split them up at the end? If so, what is the reason for it?
This is precisely why I asked the questions I did. If $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$ are linearly independent, then this sort of breaking-up is possible. You could try writing them in terms of the regular basis vector and see what happens.
There are 2 cases: either they are independent or they are not.

If $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$ are independent you can break them up.
In that case they form a so called basis of the vector space that the object moves around in.
Any vector can be written uniquely as a linear combination of the basis vectors.
On the LHS you have a linear combination of the basis vectors and on the RHS you also have a linear combination of the basis vectors.
In other words, their linear factors must be the same.

If $\mathbf{r}_{0}$ and $\mathbf{v}_{0}$ are dependent, that means they are parallel.
In that case there is no clear distinction between f and g.
You might as well set g for instance to zero and treat $\mathbf{r}_{0}$ as your basis.
The movement of the object is on a line: it moves up and down along the line defined by $\mathbf{r}_{0}$ which is parallel to $\mathbf{v}_{0}$.
An alternative for the choice of f and g is to use the same solution you have for the independent case.
To keep things simple you might as well do that.
In other words, you can also break up the expression.

So in the first case you are forced to, while in the second case you have a free choice to arrive at (3). $\qquad \blacksquare$