Proving Energy Conservation in a Gravitational System with Multiple Bodies

AI Thread Summary
The discussion focuses on proving energy conservation in a gravitational system with multiple bodies, where the total energy is expressed as the sum of kinetic and potential energy. The user encounters difficulties in demonstrating that the gravitational forces between bodies lead to a zero change in energy over time. A key point raised is the need to express forces as pairwise interactions, which simplifies the analysis of energy conservation. The conversation also emphasizes the importance of correctly manipulating indices in summations and understanding the symmetry of forces to derive valid conclusions. Ultimately, the participants clarify the rules for switching indices in double sums and the implications of force symmetry in the context of energy conservation.
Zebx
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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.
 
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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
 
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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##?
 
Zebx said:
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}##
 
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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##?
 
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}##
 
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All clear, thank you very much! :smile:
 
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