Conservation of energy proof

[Zebx]

Hi all. I'm trying to prove energy conservation in a (maybe) uncommon way. I know there are different ways to do this, but it is asked me to prove it this way and I'm stucked at the end of the proof. I'm considering $N$ bodies moving in a gravitational potential, such that the energy is $E = K + V$, with $K$ kinetic energy, $V = Gm_im_j/r_{ij}$ the potential energy ($i \neq j$) and $r_{ij} = \sqrt{(x_i - x_j)^2 + (y_i - y_j)^2 + (z_i - z_j)^2}$ the distance between the bodies. The complete expression for the energy is
$$
E = \frac{1}{2} \sum_{i=1}^{N} m_i \dot{\vec{r}}_i \cdot \dot{\vec{r}}_i - \frac{1}{2} \sum_{i,j=1}^{N} \frac{Gm_im_j}{r_{ij}},
\tag{1}
$$
with dotted variables representing the derivative with respect to time and the $1/2$ term before the second summation is there to avoid to consider the same values of $V$ two times (the term with $(i,j) = (a,b)$ are the same as the one with $(i,j) = (b,a)$, with $a,b$ from $1$ to $N$). If $E$ is conserved, then $\dot{E} = 0$:
$$
\dot{E} = \frac{1}{2} \sum_{i=1}^{N} 2m_i\dot{\vec{r}}_i \cdot\ddot{\vec{r}}_i + \frac{1}{2} \sum_{i,j=1}^{N} \frac{Gm_im_j}{r_{ij}^3}(\vec{r}_i - \vec{r}_j) \cdot (\dot{\vec{r}}_i - \dot{\vec{r}}_j),
\tag{2}
$$
with $(\vec{r}_i - \vec{r}_j)(\dot{\vec{r}}_i - \dot{\vec{r}}_j)/r_{ij} \equiv \dot{r}_{ij}$. What I do then is
$$
\begin{align}
\dot{E} & = \sum_{i=1}^{N} \vec{F}_i \cdot \dot{\vec{r}}_i + \frac{1}{2} \sum_{i=1}^{N} \vec{F}_i \cdot \dot{\vec{r}}_i - \frac{1}{2} \sum_{i,j=1}^{N} \vec{F}_i \cdot \dot{\vec{r}}_j \nonumber \\
& = \sum_{i=1}^{N} \vec{F}_i \cdot \dot{\vec{r}}_i - \frac{1}{2} \sum_{i=1}^{N} \vec{F}_j \cdot \dot{\vec{r}}_i - \frac{1}{2} \sum_{i,j=1}^{N} \vec{F}_i \cdot \dot{\vec{r}}_j \nonumber
\end{align}
\tag{3}
$$
with $\vec{F}_i = m_i \ddot{\vec{r}}_i$ being the gravitational force experienced by the mass $i$ from the $j$ other bodies, so it is also $\vec{F}_i = \sum_{j=1}^{N}Gm_im_j(\vec{r}_i - \vec{r}_j)/r_{ij}^3$. This is the point where I'm stucked. If everything's correct, I should prove that $\vec{F}_j \cdot \dot{\vec{r}}_i = \vec{F}_i \cdot \dot{\vec{r}}_j$ but I don't see any chance for this to happen unless I impose $\dot{\vec{r}}_i + \dot{\vec{r}}_j = 0$, but of course it can't be done so I don't know how could I proceed.

 

[stevendaryl]

I think you need to be more careful about forces. Let $F_{ij}$ be the force on particle $i$ due to particle $j$. Then you will find

$\dot{E} = \sum_i F_i \cdot \dot{r_i} - \frac{1}{2}\sum_{i\neq j} F_{ij} \cdot (\dot{r_i} -\dot{r_j})$

Then if you split up the second part, it becomes
$\dot{E} = \sum_i F_i \cdot \dot{r_i} - \frac{1}{2}\sum_{i\neq j} F_{ij} \cdot \dot{r_i} + \frac{1}{2}\sum_{i\neq j} F_{ij}\cdot \dot{r_j}$

Since $F_{ij} = - F_{ji}$, we can combine the two sums on the right to get:
$\dot{E} = \sum_i F_i \cdot \dot{r_i} - \sum_{i\neq j} F_{ij} \cdot \dot{r_i}$

I’m using $\sum_{i\neq j}$ to mean $\sum_i \sum_j$, but skipping the case of $i=j$.

Then by definition, $F_i = \sum_{j} F_{ij}$. So summing over $j$ gives 0

 

[Zebx]

Thank you for your answer. I actually already tried to write everything using $F_{ij}$ as you did, but once I reached the second equation you wrote I had problem with $\dot{r}_j$, for instance if I used $F_{ij} = -F_{ji}$ I didn't turn also the index of $\dot{r}_j$ in $\dot{r}_i$. So I don't understand your last equation: how could you swap the indeces of $F$ and also the index of $\dot{r}_i$?

 

[stevendaryl]

Thank you for your answer. I actually already tried to write everything using $F_{ij}$ as you did, but once I reached the second equation you wrote I had problem with $\dot{r}_j$, for instance if I used $F_{ij} = -F_{ji}$ I didn't turn also the index of $\dot{r}_j$ in $\dot{r}_i$. So I don't understand your last equation: how could you swap the indeces of $F$ and also the index of $\dot{r}_i$?

So you have
$- \frac{1}{2} \sum_{i,j} F_{ij} \cdot (\dot{r_i} - \dot{r_j})$
$= - \frac{1}{2} \sum_{i,j} F_{ij} \cdot \dot{r_i} + \frac{1}{2} \sum_{i,j} F_{ij} \cdot \dot{r_j}$

On the second sum, swap the names $i$ and $j$. That gives:
$- \frac{1}{2} \sum_{i,j} F_{ij} \cdot \dot{r_i} + \frac{1}{2} \sum_{i,j} F_{ji} \cdot \dot{r_i}$

Now, in the second sum, you use the fact that $F_{ji} = - F_{ij}$ to get

$- \frac{1}{2} \sum_{i,j} F_{ij} \cdot \dot{r_i} - \frac{1}{2} \sum_{i,j} F_{ij} \cdot \dot{r_i}$

 

[Zebx]

Ok, I was not sure that in this case I could exchange indeces that way. Is there some sort of "rule of thumb" which one can refer to when it comes to swap indeces? I mean, in this case I'm sure I can use $F_{ij}$ simmetry, but how can I know I will not "ruin" the general expression by changing also the $\dot{r}_i$?

 

[stevendaryl]

Well, if a double sum is over a finite number of terms, you can always switch the order of summation:

$\sum_{i=1}^N \sum_{j=1}^N F_{ij} \cdot \dot{r_j}= \sum_{j=1}^N \sum_{i=1}^N F_{ij} \cdot \dot{r_j}$

That doesn’t have anything to do with any symmetry of the problem. It’s always valid (for finite sums, anyway).

The second thing that’s always valid is renaming dummy variables. I used $i$ and $j$, but I could have used $m$ and $n$, or anything. So swapping names of dummy variables doesn’t do anything. In particular, I can rename $i$ by $j$ and vice-versa. So

$\sum_{j=1}^N \sum_{i=1}^N F_{ij} \cdot \dot{r_j} = \sum_{i=1}^N \sum_{j=1}^N F_{ji} \cdot \dot{r_i}$

Again, this doesn’t have anything to do with symmetry of $F_{ij}$. It’s always valid.

But now, if I do know that, for example, $F_{ji} = - F_{ij}$, then I can rewrite it again.

$\sum_{i=1}^N \sum_{j=1}^N F_{ji} \cdot \dot{r_i}= -\sum_{i=1}^N \sum_{j=1}^N F_{ij} \cdot \dot{r_i}$

 

[Zebx]

All clear, thank you very much!