- #1
binbagsss
- 1,266
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Okay the question is, after just attaining an expression for a second-order Taylor series expansion of the Coulomb potential Vc about an arbitrary value r = a, call this *,. to use this to attain an expression for V(x1,x2,x3) with the values of a determined by the equillibrium values of x1 , x2 and x3?
Where vc=k(1/r), where r is a constant
Background:
x1,x2 and x3 are the distances to 3 point charges respectively.
The equilibrium position of the whole system has been attained. The system is these three charges followed by a plate of equal charge to the 3 charges which is immovable, and the spheres slide on a friction less rod of length 4l, all charges may be treated as point-wise. Equillibrium: x1=l, x2=2l, x3=3l.
This was done via attaining a basic expression for the potential considering only nearest neighbor interactions, and differentiated with respect to xi, were i=1,2,3 , in turn. And then solving simultaneously.
My main question:
The solution is given by determining a by the equiibrium of x1,x2,x3 IN TURN, and SUMMING UP these potentials.
I was, however, tempted to do this via 3 taylor expansions and then add them together , where v(x1) is done by knowing that x1=l is the equilibrium position, so via * I attain c(x1-l)+k(x1-l)^2, where c and k are the relevant contants, for the second and third term. similar for x2, for these two terms I would get c(x2-2l) + k(x2-2l)^2, where c and k are two more, different , constants. I would do the same for x3, about 3l, then simply sum up.
Looking at the solution, this is not the case, but instead the '(r-a)' terms correspond to x1-l, x2-x1-l, x3-x2-l and 4l-x3-l.
I believe my main issue lies in the lack of understand of what it means to do a taylor expansion about equilibrium, tied with, the dependence /Independence of the variables involved. My first method corresponds only to the value of x1, x2 and x3 , when the WHOLE SYSTEM is in equilibrium.
The hint is when we Taylor expand around xi we are only fixing xi to its equilibrium value, not necessarily any other variables.
So I believe this means that we need to cover all points of equilibrium , and not just the one corresponding to the whole system equilibrium? (Which explains why we end up with the variables as functions of each other - e.g x3 of x2.)
But from this I have three questions:
1) I am struggling to see how we deduce the correct 'r-a' for each xi. Solving simultaneously we attain x1=l, x2=2l, x3=3l, so these correspond to an equilibrium of the whole system. But in the taylor expansion, we are only after fixing each xi to be in equilibrium in turn. So I am struggling to see how the l ties in/ how we deduce the 'r-a' for each xi. Surely x2-xl=l etc,is based on solving all of these equations simultaneously,but for the expansion we only require the xi to be in equilibrium independently in turn, not the particular equilibrium case when all other xi are in equilibrium. ( From the symmetry I can see that in equilibrium the charges must be equally spaced, which seems to be along to right lines of getting to the correct 'r-a', but , to me, the question of distinguishing between whole system equilibrium and independent xi equilibrium still arises, as surely this equally spacing again corresponds to whole system equilibrium.
(I'm guessing the conclusion yielded from these independent equations will be needed to answer my question : dv/dx1 yielded 2x1=x2, dv/dx2 that 2x2=x3+x1, and dv/dx3 that 2x3=4l+x2)
2) For v(xi) ,before adding all i to attain v(x1,x2,x3), would it be correct to say that we treat the other xi as a constant? For example when 'r-a'=x3-x2-l, is it correct to say that x2 is treated as a constant? My justification being that, if not, wouldn't we need to turn to a multi-variable taylor expansion, rather than adding together the independent xi's?
3) Also does the fact that we don't do a multi-variable Taylor expansion imply that xi are independent? Then how is it that we attain them as a function of each other , when considering the equilibrium positions of them? Or would this not be defined as a function of each other as we treat the variable we are not fixing to equilibrium as a constant?
Many many thanks to anyone who can shed some light on any of this :) !
Where vc=k(1/r), where r is a constant
Background:
x1,x2 and x3 are the distances to 3 point charges respectively.
The equilibrium position of the whole system has been attained. The system is these three charges followed by a plate of equal charge to the 3 charges which is immovable, and the spheres slide on a friction less rod of length 4l, all charges may be treated as point-wise. Equillibrium: x1=l, x2=2l, x3=3l.
This was done via attaining a basic expression for the potential considering only nearest neighbor interactions, and differentiated with respect to xi, were i=1,2,3 , in turn. And then solving simultaneously.
My main question:
The solution is given by determining a by the equiibrium of x1,x2,x3 IN TURN, and SUMMING UP these potentials.
I was, however, tempted to do this via 3 taylor expansions and then add them together , where v(x1) is done by knowing that x1=l is the equilibrium position, so via * I attain c(x1-l)+k(x1-l)^2, where c and k are the relevant contants, for the second and third term. similar for x2, for these two terms I would get c(x2-2l) + k(x2-2l)^2, where c and k are two more, different , constants. I would do the same for x3, about 3l, then simply sum up.
Looking at the solution, this is not the case, but instead the '(r-a)' terms correspond to x1-l, x2-x1-l, x3-x2-l and 4l-x3-l.
I believe my main issue lies in the lack of understand of what it means to do a taylor expansion about equilibrium, tied with, the dependence /Independence of the variables involved. My first method corresponds only to the value of x1, x2 and x3 , when the WHOLE SYSTEM is in equilibrium.
The hint is when we Taylor expand around xi we are only fixing xi to its equilibrium value, not necessarily any other variables.
So I believe this means that we need to cover all points of equilibrium , and not just the one corresponding to the whole system equilibrium? (Which explains why we end up with the variables as functions of each other - e.g x3 of x2.)
But from this I have three questions:
1) I am struggling to see how we deduce the correct 'r-a' for each xi. Solving simultaneously we attain x1=l, x2=2l, x3=3l, so these correspond to an equilibrium of the whole system. But in the taylor expansion, we are only after fixing each xi to be in equilibrium in turn. So I am struggling to see how the l ties in/ how we deduce the 'r-a' for each xi. Surely x2-xl=l etc,is based on solving all of these equations simultaneously,but for the expansion we only require the xi to be in equilibrium independently in turn, not the particular equilibrium case when all other xi are in equilibrium. ( From the symmetry I can see that in equilibrium the charges must be equally spaced, which seems to be along to right lines of getting to the correct 'r-a', but , to me, the question of distinguishing between whole system equilibrium and independent xi equilibrium still arises, as surely this equally spacing again corresponds to whole system equilibrium.
(I'm guessing the conclusion yielded from these independent equations will be needed to answer my question : dv/dx1 yielded 2x1=x2, dv/dx2 that 2x2=x3+x1, and dv/dx3 that 2x3=4l+x2)
2) For v(xi) ,before adding all i to attain v(x1,x2,x3), would it be correct to say that we treat the other xi as a constant? For example when 'r-a'=x3-x2-l, is it correct to say that x2 is treated as a constant? My justification being that, if not, wouldn't we need to turn to a multi-variable taylor expansion, rather than adding together the independent xi's?
3) Also does the fact that we don't do a multi-variable Taylor expansion imply that xi are independent? Then how is it that we attain them as a function of each other , when considering the equilibrium positions of them? Or would this not be defined as a function of each other as we treat the variable we are not fixing to equilibrium as a constant?
Many many thanks to anyone who can shed some light on any of this :) !
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