Proving Energy Conservation in a Gravitational System with Multiple Bodies

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
Zebx
12
3
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:
 

Related to Proving Energy Conservation in a Gravitational System with Multiple Bodies

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another.

2. How is the conservation of energy proven?

The conservation of energy is proven through various experiments and observations, such as the conservation of mechanical energy in a closed system or the conservation of energy in chemical reactions.

3. Can the conservation of energy be violated?

No, the conservation of energy is a fundamental law of physics and has been proven to hold true in all observed cases. Any apparent violation is due to incomplete understanding or measurement errors.

4. What are some examples of the conservation of energy in everyday life?

Examples of the conservation of energy in everyday life include the conversion of potential energy to kinetic energy in a falling object, the conversion of chemical energy to heat and light in a fire, and the conversion of electrical energy to light and sound in a lightbulb.

5. How does the conservation of energy relate to the first law of thermodynamics?

The first law of thermodynamics is a restatement of the law of conservation of energy in the context of thermodynamic systems. It states that the total energy of a closed system remains constant, and can only be transferred or transformed but not created or destroyed.

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