| Brad
Rodgers |
In The Feynmann Lectures on Physics (page 13-9), Feynmann uses potential energy to prove that a sphere's center of gravity is it's geometric center. How is he justified, though, in using potential energy? Is it true that if the potential energy between two bodies is equivalent, the magnitude of the force between the two bodies is equivalent ? Feynmann seems to implicitly assume this, but it certainly isn't obvious... I'll post more later; I'm in a rush. Brad |
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| Andre
Rzym |
It depends what you mean by equivalent. Potential energy is a relative concept, (when is potential energy zero?), so it is not true to say that if the potential energy between two bodies in system 1 is equal to the potential energy between two bodies in system 2 then the magnitude of the force between the two bodies is the same in both systems . However, since dW=Fdx (W being work, F being force, x being distance), and dW=change in potential energy, it follows that if the derivative of potential energy between two bodies in system 1 with respect to distance is equal to the derivative of potential energy between two bodies in system 2 with respect to distance then the magnitude of the force between the two bodies is the same in both systems Is this what you wanted? Andre |
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| Brad
Rodgers |
What I was going to post earlier but didn't get a chance to: Is it true that an acceleration equal to a = B + C/R2 , where R is the distance between two bodies, and B and C are constants (unaffected by mass and R), is the only sort of acceleration (varying only with R) for which the forces at play on a sphere are equivalent to the forces at play on the sphere's center? If I understand Feynmann's argument correctly, I think this is correct -- but perhaps I just misunderstand his argument. [I'm somewhat tired, so I may be stating my initial "acceleration question" poorly] In any case, I'll have to think over your last post Andre; I'll probably post a response tomorrow. Brad |
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| Andre
Rzym |
Brad, Could you clarify what you mean by constant B? Does it represent a force independent of distance (but presumably dependent upon the product of masses)? If so, then I don't see how one could be indifferent between a sphere, and all mass at the sphere centre. The sum of the 'magnitude' of the forces resulting from all bits of mass in the sphere may add up to the magnitude of the force were the mass to be at the centre, however in the case of the sphere, we need to resolve the force (by symmetry) along the axis between the centre of the sphere and the other body. Thus the net force would necessarily smaller. Or were you thinking of constant potential energy? In which case don't you want C/R rather than C/R2 to represent the inverse square force? Andre |
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| Brad
Rodgers |
Perhaps it would be best for me to be more concrete. Feynmann uses the expression for gravitational potential energy, -Gm'm/R, to prove that the net force a hollow sphere exerts on a particle outside the sphere is the same as the force exerted on the particle were the mass of the sphere concentrated in the geometrical center of the sphere . To do so he shows that the net potential energy is the same in both cases. I was at first confused by this, and Feynmann was no help; he explains the motivation for using potential energy (it's easier to calculate), but does not further justify using the net potential energy rather than net force. It was as though he said, "Let's just deal with surface area, rather than forces, because surface area doesn't involve direction and is thus easier to use." In any case, I think I now understand the justification: Suppose we were to place the particle a distance of R away from the center of the sphere. Then if the net potential energy 'generated' by the sphere equals the potential energy 'generated' by the hypothetical center mass, then Fc - vc = Fs - vs or ||Fc || x ||vc || = ||Fs || x ||vs || where F and v are the respective forces and velocities of the center mass (subscript c) and the sphere (subscript s); the second equation because we know if we just drop the particle without previous forces, the velocity will be in the same direction as the force. And, since velocity is determined by force, we can conclude that the magnitude of the forces are the same. At least, I think. My questions: a) Is this reasoning correct? b) If so, is there a more straightforward way to deduce this? The method I've used is, of course, not too hard to see, but it is hard enough that I would expect see some mention of it in a book before the concept it derives is used. My last question (in my most recent post) is more complicated. Suppose that we had the potential energy between two bodies as m1 m2 P(R), where P is a function of distance (R) and only of distance. Is it true that, for the net force of the sphere upon the particle to be the same as the force of the mass in the center upon the paticle, we would have to have P(R) = A x R + B + C/R, or the acceleration equal to a = A - C/R2 , where as above, A and C don't vary with R? I have a lengthy mathematical 'proof' of this, but I'd like to see if it's a standard result before I post it. If it's not, or if it's untrue, I'll probably post the proof Friday; I'm not sure I'll get a chance Thursday. Brad |
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| Andre
Rzym |
Brad, a) Yes your reasoning is correct. b) You have made the justifications more complex than necessary by talking about velocity -it is more usual and more straightforward to link PE and force by talking saying: (work done)=-(change in PE)=(force)*(distance moved) -dP = F.dx or in vector notation, F =-grad(P) See my first post. c) PE is indeed sometimes easier to work with because of its directionless nature [just like it can be easier to use the Lagrangian rather than F=ma] d) As for your last point, unless I have made a mistake, I do not agree with you. I think we require P(R)=B+C/R, i.e. A=0. I'll post my proof if you wish. It is based on potential energy and fairly short, so if yours is based on force, then it would be good to compare their complexities. Andre |
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| Andre
Rzym |
Since I have a little time, I'll post my proof. 1) Consider a shell (not a solid sphere) of mass M, radius r, centre O, and a particle outside the shell mass m at point P, where P O=R. 2) We will assume a power law for potential energy of the form m M/Rn and answer the following question: ''For which value(s) of n is the potential energy of the system (as a function of R) independent of r (to within a constant)?''. Hopefully you agree that proving this is sufficient to answer your question. 3) The density (mass/unit area) of the shell is r = M/4pr2 4) Consider a 'band' of mass on the shell, such that every point, B, on the band forms an angle B O P in the range [q,q+dq] The radius of the band is r sin(q). The width of the band is r.dq. The area of the band is 2pr2sin(q).dq. The mass of the band is M sin(q)/2.dq 5) All points on the band are equidistant from P, a distance x where x2=R2+r2-2R r cos(q) Therefore from our assumed power law for potential energy, the system has a potential energy dPE=m M sin(q)/2.dq/(R2+r2-2R r cos(q)) n/2 Integrating, the total PE PE=ò0p m M sin(q)/2/(R2+r2-2R r cos(q)) n/2.dq 5)This, if I am not mistaken, integrates to m M /(4r Rn-1).[1/(-n/2+1)].[(1+r/R)-n+2-(1-r/R) -n+2] When is this independent of r? Expand the two terms at the end as power series. We require that there is no power of r/R higher than 1, implying n=1 or 0 (we ignore n=2 since that integrates to a logarithm which clearly has an r sensitivity). So the power law of potential energy must be proportional to 1/R or a constant (or linear combinations thereof). The latter, of course, gives rise to no force. Andre |
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| Brad
Rodgers |
I follow your argument - it's roughly the same as mine (or, more accurately, Feynman's). Actually, looking back at my working, A×R cannot be part of the potential energy; I proved that U(r)=m1 m2 g(r)/r iff g(R)=1/2 òR+1R-1g(x)dx, for all R, where U is the potential energy. For analytic g, this is satisfied only for g(x) = A x+B, which is what you've proved. I included an x2 term before because I differentiated the above condition without fully justifying doing so. I can post my proof if anyone would like, but it's nothing you can't devise fairly easily from looking at Feynman's proof. In essence, I integrate with respect to Andre's x, rather than q. Leaving out the latter makes the equations slightly neater. Brad |
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| Andre
Rzym |
Brad, If you are interested (and have not already done so), there are a couple of things you may wish to look at: i) Show that with an inverse square (force) law, a particle inside a shell experiences no force from the mass of the shell. ii) Show for what other power laws (if any) this result holds. iii) Show that if you drill a hole (any direction) in the earth (assumed to have uniform density) until you come back out again, a particle dropped into the (smooth) hole will execute SHM. Andre |
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| Brad
Rodgers |
Hi, i) is in Feynmann. ii) Is it only for the inverse square force law? This is what I get, though if I'm wrong (or if anyone's interested) I can post my work. iii) I'm not sure about this one. I have a method to prove it, and with sufficient computation, I'm sure I could in fact use this method to prove it -- but "sufficient computation" could take some time. I'm guessing one has to show that a shell adequately models the sphere of which it is the shell, but, at present, I'm not sure how to justify this... In any case, this problem seems interesting, and I'd like to try it on my own; if possible, could you just give a hint or two? Thanks, Brad |
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| Andre
Rzym |
Brad, ii) Yes, I think it's only for inverse square force laws. The algebra for this case should be pretty much the same as when you are outside the sphere. iii) I'm not sure what's obvious and what's not, so apologies in advance: a) The hole can be characterised by how close it ever gets to the centre of the sphere. b) Wherever the particle is, the earths mass between the particle and the centre of the sphere can be regarded as being at the centre. The remaining mass can be ignored . Let me know if you need more ... Andre |
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| Colin
Prue |
Almost exactly what Andre said came up on BPhO2 on monday, with the exception of the SHM part. However, then they stared asking about gravitational forces inside a ring! Anyway, but how would you tackle the following question: An estimate of the mass of the Coma cluster can be made from the velocity of the galaxies relative to the centre of the cluster. the velocities are around 2000kmh-1 . Assuming steady state, predict the mass of the cluster, using simple asumptions. The radius of the cluster is about 1021 sorry if you think i'm hijacking the conversation, feel free to ignore me |
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| Andre
Rzym |
You are definitely not hijacking the conversation. This is what I think is going on: I think a cluster is a collection of galaxies. I think we are to assume that the galaxies are ''orbiting'' a centre in the formation of a ring. The motion is as though you glue galaxies to the tyre of a bicycle wheel, and the wheel is rotating. For numerous enough galaxies, we can regard the mass as being uniform per unit length of the circle. Assume mass per unit length of r, then total mass is 2prr. Now consider a small section (length dx, mass rdx) of the ring. You know the total mass of the ring, and it sounds like (from the previous part of the question) that you can calculate the gravitational attraction of this mass towards the centre. This must be balanced by the centripetal force on this mass. This should allow you to solve for r and hence the total mass. Andre |
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| Colin
Prue |
I guess so, yes, but then how does that allow that all of the galaxies travel at 2000kms-1 (this is what it should have been), as if we model as a bycicle tyre, then the tangential velocities would be variable (linear with distance from centre). But i wrote something similar to what you have given above Andre, so that's reassured me that i'll get SOME marks for trying at least thanks, colin |
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| Andre
Rzym |
The hardest part of this question is trying to understand what the examiners were on about. I think the galaxies are all 'on the bicycle tyre', not in the region between the tyre and the hub. In other words, they are not distributed evenly on a disc , but rather evenly on a ring . Just like a Polo mint. That way, all galaxy speeds are the same. Andre |