# 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.

• Leo Liu

Staff Emeritus
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

• Leo Liu, Zebx and (deleted member)
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##?

Staff Emeritus
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
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##?

Staff Emeritus
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
Zebx
All clear, thank you very much! 