Conservation of energy proof

In summary, the conversation discusses the process of proving energy conservation in a unique way by considering N bodies moving in a gravitational potential. The energy is expressed as a sum of kinetic and potential energy, and it is stated that if the energy is conserved, then the derivative of the energy with respect to time is equal to zero. The conversation then delves into the use of forces and the use of Fij to represent the force on particle i due to particle j. It is suggested to use the symmetry of Fij to rewrite the equation and prove that Fj⋅ri=Fj⋅ri. The conversation concludes with a discussion on the rules of swapping indices and renaming dummy variables. It is stated that swapping indices and renaming
  • #1
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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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  • #2
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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  • #3
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##?
 
  • #4
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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  • #5
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##?
 
  • #6
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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  • #7
All clear, thank you very much! :smile:
 

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