Vg vf pbeerpg gung ng g=0, rnpu cnegvpyr unf n yrsgjneqf
irybpvgl. Ohg vg vf abg pbeerpg gung gur prager bs tenivgl ng g=0
unf n yrsgjneqf irybpvgl. Guvf vf orpnhfr gur flfgrz orunirf
qvfpbagvahbhfyl ng g=0, naq fb gur irybpvgl bs gur
prager bs tenivgl pnaabg or qrgrezvarq ol gnxvat n jrvtugrq nirentr
bs gur vavgvny irybpvgvrf bs gur pbzcbaragf.
Va zber qrgnvy:
Yrg P(g) or gur cbfvgvba bs gur prager bs tenivgl ng gvzr g.
Gur irybpvgl bs gur prager bs tenivgl ng g=0 vf gur yvzvg nf g → 0
bs (P(g)-P(0))/g.
Ng nal gvzr g, P(g) vf gur fhz bire a bs Z(a)C(a,g), jurer Z(a) vf
gur znff bs gur agu cbvag naq C(a,g) vgf cbfvgvba.
Gur nethzrag eryvrf ba pbzchgvat qP/qg nf gur fhz bire a bs Z(a)qC(a,g)/qg.
Guvf fjncf gur beqre bs gjb yvzvgvat cebprffrf, fhzzngvba naq qvssreragvngvba, juvpu vf
inyvq sbe jryy-orunirq fhzf, ohg abg sbe guvf bar.
For intuition, imagine that there were just the first three particles in the sequence, and that the number 10 is replaced by something much larger. The two leftmost particles will rapidly oscillate about one another, while drifting slowly towards the rightmost. Now imagine the first four particles are present. The two leftmost will oscillate madly about each other, while that system as a whole oscillates rapidly about the third from the left (I’m conveniently ignoring what happens at the instants when particles meet and pass through each other), and the ensemble of the three leftmost drifts towards the rightmost. Repeat.
It’s acceleration that matters, not velocity (the initial velocity of all points is zero, or at least that’s how I thought of it). However, your argument does generalize nicely to acceleration, and could possibly be the correct resolution.
Yes, stupid mistake of mine. Replace velocity by acceleration and position by velocity and the same argument works. The fallacy is in the vagrepunatr bs yvzvgf.
Vg vf pbeerpg gung ng g=0, rnpu cnegvpyr unf n yrsgjneqf irybpvgl. Ohg vg vf abg pbeerpg gung gur prager bs tenivgl ng g=0 unf n yrsgjneqf irybpvgl. Guvf vf orpnhfr gur flfgrz orunirf qvfpbagvahbhfyl ng g=0, naq fb gur irybpvgl bs gur prager bs tenivgl pnaabg or qrgrezvarq ol gnxvat n jrvtugrq nirentr bs gur vavgvny irybpvgvrf bs gur pbzcbaragf.
Va zber qrgnvy:
Yrg P(g) or gur cbfvgvba bs gur prager bs tenivgl ng gvzr g.
Gur irybpvgl bs gur prager bs tenivgl ng g=0 vf gur yvzvg nf g → 0 bs (P(g)-P(0))/g.
Ng nal gvzr g, P(g) vf gur fhz bire a bs Z(a)C(a,g), jurer Z(a) vf gur znff bs gur agu cbvag naq C(a,g) vgf cbfvgvba.
Gur nethzrag eryvrf ba pbzchgvat qP/qg nf gur fhz bire a bs Z(a)qC(a,g)/qg. Guvf fjncf gur beqre bs gjb yvzvgvat cebprffrf, fhzzngvba naq qvssreragvngvba, juvpu vf inyvq sbe jryy-orunirq fhzf, ohg abg sbe guvf bar.
For intuition, imagine that there were just the first three particles in the sequence, and that the number 10 is replaced by something much larger. The two leftmost particles will rapidly oscillate about one another, while drifting slowly towards the rightmost. Now imagine the first four particles are present. The two leftmost will oscillate madly about each other, while that system as a whole oscillates rapidly about the third from the left (I’m conveniently ignoring what happens at the instants when particles meet and pass through each other), and the ensemble of the three leftmost drifts towards the rightmost. Repeat.
It’s acceleration that matters, not velocity (the initial velocity of all points is zero, or at least that’s how I thought of it). However, your argument does generalize nicely to acceleration, and could possibly be the correct resolution.
Yes, stupid mistake of mine. Replace velocity by acceleration and position by velocity and the same argument works. The fallacy is in the vagrepunatr bs yvzvgf.