Note that each pair has to be taken into account twice.
Once for each member of the pair = twice
Then you were calculating the gravitational effect of an infinite number of external particles on a finite number of particles, which makes things worse.
No, I calculated it for every particle, and therefore I calculated it on an infinite number of particles. Obviously my calculations only considered one particle at any time, but the same could be said of calculations in problems involving finitely many particles.
You calculate the gravitational force on the first particle. There are N forces on it; you now have N unpaired forces.
Now you calculate the gravitational force on the second particle. There are N forces on it; you now have 2N-2 unpaired forces (you added N more forces, but one of the new ones and one of the old ones are paired).
Then you calculate the gravitational force on the third particle. There are N forces on it; you now have 3N-6 unpaired forces (you added N more forces, and 3 choose 2 are paired).
The number of unpaired forces will grow until it peaks and then shrinks down to zero (as the number of forces that pair up will eventually overcome the added forces).
What happens if you take this approach with an infinite number of particles? Well, at each moment you add an infinite number of new forces, and only a finite number pair up. Thus, as soon as you start the calculations you have an infinite number of unpaired forces and that number only grows.
The right way to do this is the buddy system: never let a force enter the calculations without its reverse force. To do this, you add particles to your consideration one by one. The number of considered forces is N choose 2, and you never have unpaired forces.You let the number of considered particles grow to infinity, and it still works.
What happens if you take this approach with an infinite number of particles? Well, at each moment you add an infinite number of new forces, and only a finite number pair up. Thus, as soon as you start the calculations you have an infinite number of unpaired forces and that number only grows.
It certainly doesn’t grow, it becomes infinity and stays there.
Okay, this is getting into set theory, but its actually the case that adding infinitely many things and then taking finitely many away can eventually leave you with zero.
Suppose at step n you add the following numbers, 2n-1, 2(2n-1), 4(2n-1)… (2^k)(2n-1) for all k. Then you take away the number n.
This procedure adds infinitely many at every step, and only takes one away, so by your argument it should only ever grow and end up as infinity. However, its pretty clear that every number gets added once and then taken away once, so we end up with nothing.
I think its pretty clear that a similar thing is going on here.
Again, I agree that the limits are being done wrong, I just disagree with your description of the N vs N+1 thing.
It certainly doesn’t grow, it becomes infinity and stays there.
Okay, this is getting into set theory, but its actually the case that adding infinitely many things and then taking finitely many away can eventually leave you with zero.
It’s not clear to me what you mean by “actually the case.” I agree that it could be the convention among set theorists that things work that way, and that there may even be benefits from taking that approach. In my experience, it only serves to foster confusions about infinity.
Consider a simple version of your scenario: start off with a ball labeled 1 in an urn of infinite capacity. Add two balls labeled with the two integers immediately above the largest label in the urn, then remove the ball with the smallest label from the urn. Repeat those two steps an infinite number of times. It is clear that every time you perform the two steps, the number of balls in the urn increases by one, but also that for any individual ball, at some point it will get removed from the urn. I disagree with the conclusion that the set of balls in the urn is empty after an infinite number of steps; I would rather call it ill-defined, as it should have both no members and an infinite size. (If you don’t paint numbers on the balls, and just add two balls then remove one, it’s obvious the urn contains an infinite number of balls after an infinite number of steps. Why would painting numbers on balls make them disappear from the urn? Because you run out of numbers?)
I just disagree with your description of the N vs N+1 thing.
I would call it the mistake of taking a theorem about Newtonian physics that was proven by assuming finitely many objects, and generalising it to infinitely many (which would be allowed by limits, but this isn’t using limits.
(If you don’t paint numbers on the balls, and just add two balls then remove one, it’s obvious the urn contains an infinite number of balls after an infinite number of steps. Why would painting numbers on balls make them disappear from the urn? Because you run out of numbers?)
I would say that if you don’t paint numbers on the balls you could end up with 0, or infinity, or something else, depending on how you do it. If you do it randomly you do have a probability 1 of ending up with infinity, but if you’re careful you can make it less.
I would call it the mistake of taking a theorem about Newtonian physics that was proven by assuming finitely many objects, and generalising it to infinitely many (which would be allowed by limits, but this isn’t using limits.
I’m confused. Are you saying that the CoM would accelerate in this situation, and thus the proof that CoMs only accelerate under external forces does not apply to this situation, or are you saying that this counterexample is misconstructed, and the CoM would not accelerate?
Suppose at step n you add the following numbers, 2n-1, 2(2n-1), 4(2n-1)… (2^k)(2n-1) for all k. Then you take away the number n.
This procedure adds infinitely many at every step, and only takes one away, so by your argument it should only ever grow and end up as infinity. However, its pretty clear that every number gets added once and then taken away once, so we end up with nothing.
Technically speaking its a slight modification, since usually the paradox involves adding a finite number than taking away 1, rather than infinitely many.
Once for each member of the pair = twice
No, I calculated it for every particle, and therefore I calculated it on an infinite number of particles. Obviously my calculations only considered one particle at any time, but the same could be said of calculations in problems involving finitely many particles.
Suppose you have a system of N+1 particles.
You calculate the gravitational force on the first particle. There are N forces on it; you now have N unpaired forces.
Now you calculate the gravitational force on the second particle. There are N forces on it; you now have 2N-2 unpaired forces (you added N more forces, but one of the new ones and one of the old ones are paired).
Then you calculate the gravitational force on the third particle. There are N forces on it; you now have 3N-6 unpaired forces (you added N more forces, and 3 choose 2 are paired).
The number of unpaired forces will grow until it peaks and then shrinks down to zero (as the number of forces that pair up will eventually overcome the added forces).
What happens if you take this approach with an infinite number of particles? Well, at each moment you add an infinite number of new forces, and only a finite number pair up. Thus, as soon as you start the calculations you have an infinite number of unpaired forces and that number only grows.
The right way to do this is the buddy system: never let a force enter the calculations without its reverse force. To do this, you add particles to your consideration one by one. The number of considered forces is N choose 2, and you never have unpaired forces.You let the number of considered particles grow to infinity, and it still works.
It certainly doesn’t grow, it becomes infinity and stays there.
Okay, this is getting into set theory, but its actually the case that adding infinitely many things and then taking finitely many away can eventually leave you with zero.
Suppose at step n you add the following numbers, 2n-1, 2(2n-1), 4(2n-1)… (2^k)(2n-1) for all k. Then you take away the number n.
This procedure adds infinitely many at every step, and only takes one away, so by your argument it should only ever grow and end up as infinity. However, its pretty clear that every number gets added once and then taken away once, so we end up with nothing.
I think its pretty clear that a similar thing is going on here.
Again, I agree that the limits are being done wrong, I just disagree with your description of the N vs N+1 thing.
It’s not clear to me what you mean by “actually the case.” I agree that it could be the convention among set theorists that things work that way, and that there may even be benefits from taking that approach. In my experience, it only serves to foster confusions about infinity.
Consider a simple version of your scenario: start off with a ball labeled 1 in an urn of infinite capacity. Add two balls labeled with the two integers immediately above the largest label in the urn, then remove the ball with the smallest label from the urn. Repeat those two steps an infinite number of times. It is clear that every time you perform the two steps, the number of balls in the urn increases by one, but also that for any individual ball, at some point it will get removed from the urn. I disagree with the conclusion that the set of balls in the urn is empty after an infinite number of steps; I would rather call it ill-defined, as it should have both no members and an infinite size. (If you don’t paint numbers on the balls, and just add two balls then remove one, it’s obvious the urn contains an infinite number of balls after an infinite number of steps. Why would painting numbers on balls make them disappear from the urn? Because you run out of numbers?)
How would you describe it?
I would call it the mistake of taking a theorem about Newtonian physics that was proven by assuming finitely many objects, and generalising it to infinitely many (which would be allowed by limits, but this isn’t using limits.
I would say that if you don’t paint numbers on the balls you could end up with 0, or infinity, or something else, depending on how you do it. If you do it randomly you do have a probability 1 of ending up with infinity, but if you’re careful you can make it less.
I’m confused. Are you saying that the CoM would accelerate in this situation, and thus the proof that CoMs only accelerate under external forces does not apply to this situation, or are you saying that this counterexample is misconstructed, and the CoM would not accelerate?
I thought I was saying the former, although I’m no longer actually sure.
Perhaps someone should start a discussion thread in which any mention of infinity or its synonyms is forbidden.
Why would you want to?
Minor note: this is the Ross-Littlewood paradox.
Technically speaking its a slight modification, since usually the paradox involves adding a finite number than taking away 1, rather than infinitely many.