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        riddles >> general problem-solving / chatting / whatever >> motion in non-uniform electric field
        
(Message started by: jarls on May 2nd, 2009, 5:39pm)
Title: motion in non-uniform electric field
Post by jarls on May 2nd, 2009, 5:39pm
A point charge of charge q is fixed in place. a charge of mass m is placed a distance d from this charge. This mass is not fixed in place.  With respect only to time what would be the acceleration function of the charged mass?
Title: Re: motion in non-uniform electric field
Post by towr on May 4th, 2009, 1:24am
I think it comes down to solving the differential equation
x''(t) = -C/x(t)2
But the only solution I've found (x(t)=A(1-Bt)2/3)doesn't fit the curves I get from numerical approximation.
Not to mention it ought to be periodic. It should also be what you get from gravity, like a planetary orbit where the planet goes through the sun :P (i.e. the minor axis of the ellipse approaches 0). But I can't even find a parametrization for that anywhere.
Title: Re: motion in non-uniform electric field
Post by Grimbal on May 4th, 2009, 4:35am
with
  x = (C1/3)·y
you even get
  y" = -1/y2

If you consider it as a limit case of planetary motion, you should remember that the Sun is one of the focal points of the ellipse.  So the limit case would be more like bouncing off the Sun. ;D

If I remember well, there is no algebraic expression of a planet's position in terms of x(t) and y(t).  But you can compute the time t in terms of x(t) and y(t).
Title: Re: motion in non-uniform electric field
Post by jarls on May 4th, 2009, 10:40am

on 05/04/09 at 01:24:15, towr wrote:
I think it comes down to solving the differential equation
x''(t) = -C/x(t)2
But the only solution I've found (x(t)=A(1-Bt)2/3)doesn't fit the curves I get from numerical approximation.
Not to mention it ought to be periodic. It should also be what you get from gravity, like a planetary orbit where the planet goes through the sun :P (i.e. the minor axis of the ellipse approaches 0). But I can't even find a parametrization for that anywhere.


Imagine water is pumped through a hole on a flat surface such that it spreads outward on the surface radially. If this gradient is in fact perfectly radially symmetrical if there were placed  a test mass anywhere it would move outward along the velocity motion vector the water at that point had (straight line originating from the center/pump). Only when a mass placed in the water has a non-zero vectorial component of it motion which is orthogonal to the gradient at its point of placement, will there be orbital motion.


If the charged mass has no initial motion at all; if it is just placed in the radially symmetric field, why would there result any orbital motion?
Title: Re: motion in non-uniform electric field
Post by towr on May 4th, 2009, 11:09am

on 05/04/09 at 10:40:57, jarls wrote:
Imagine water is pumped through a hole on a flat surface such that it spreads outward on the surface radially. If this gradient is in fact perfectly radially symmetrical if there were placed  a test mass anywhere it would move outward along the velocity motion vector the water at that point had (straight line originating from the center/pump).
So should I take this to mean that the charged mass and fixed charge repel eachother, rather than attract, as I had assumed?



Quote:
Only when a mass placed in the water has a non-zero vectorial component of it motion which is orthogonal to the gradient at its point of placement, will there be orbital motion.
I don't see how there will be an orbital motion if it's pushed outward.


Quote:
If the charged mass has no initial motion at all; if it is just placed in the radially symmetric field, why would there result any orbital motion?
I'm not saying there would be an orbital motion. But if the mass is attracted, then due to continuity, we must be dealing with the limit case of an orbit whose minor axis is reduced to zero. You can get an arbitrarily close approximation to the case where there is no initial movement by looking at the case where the initial movement is very very small.
But if you're not talking about attracting charges, that's a moot point.
Title: Re: motion in non-uniform electric field
Post by jarls on May 4th, 2009, 11:25am

on 05/04/09 at 11:09:48, towr wrote:
So should I take this to mean that the charged mass and fixed charge repel eachother, rather than attract, as I had assumed?

Yea they're both charge q



on 05/04/09 at 11:09:48, towr wrote:
I'm not saying there would be an orbital motion. But if the mass is attracted, then due to continuity, we must be dealing with the limit case of an orbit whose minor axis is reduced to zero. You can get an arbitrarily close approximation to the case where there is no initial movement by looking at the case where the initial movement is very very small.
But if you're not talking about attracting charges, that's a moot point.


Yea I see what you're saying. Assuming they're both q though.
Title: Re: motion in non-uniform electric field
Post by Grimbal on May 4th, 2009, 3:15pm
A  solution to
   x" = C/x2
is given by
   x = -(9C/2)1/3 · t2/3
but it is not the general solution.  In fact, it doesn't cover the case where x0>0.  The general solution should have 2 arbitrary constants.
Title: Re: motion in non-uniform electric field
Post by jarls on May 4th, 2009, 10:11pm

on 05/04/09 at 15:15:50, Grimbal wrote:
A  solution to
   x" = C/x2
is given by
   x = -(9C/2)1/3 · t2/3
but it is not the general solution.  In fact, it doesn't cover the case where x0>0.  The general solution should have 2 arbitrary constants.



I would say that the case in which displacement from the point charge is zero would be erroneous because the placement of a charge zero distance from a another would require infinite force (or energy depending on how you approach the problem).
Title: Re: motion in non-uniform electric field
Post by towr on May 5th, 2009, 12:24am

on 05/04/09 at 15:15:50, Grimbal wrote:
A  solution to
   x" = C/x2
is given by
   x = -(9C/2)1/3 · t2/3
but it is not the general solution.  In fact, it doesn't cover the case where x0>0.  The general solution should have 2 arbitrary constants.

x = -(9C/2)1/3 · (L+t)2/3 satisfies the differential equation, but it still doesn't work as a solution. x should approach a line (as v approaches a constant)
Title: Re: motion in non-uniform electric field
Post by Grimbal on May 5th, 2009, 6:04am
The solution to the differential equation is correct.
The problem is that for negative values of x, we must in fact satisfy
   x" = -C/x2
That's why it doesn't give a solution to the 2-particle problem.  We need a solution that works with x>0.
Title: Re: motion in non-uniform electric field
Post by Grimbal on May 5th, 2009, 6:59am
The expression
   x" = C/x2  [1]
Can be rewritten as
   x' = y  [2]
   y' = C/x2  [3]
[2] => dx/dt = y
   =>  dt = dx/y  [4]
[3] => dy/dt = C/x2
   => dt = dy·x2/C  [5]
equating [4] and [5] gives
   dx/y = dt = dy·x2/C
   => C·dx/x2 = dy·y  [6]
integrating [6]
   C·(-1)/x = y2/2 - D2/2  (an arbitrary constant.  The square is OK if we restrict to x>0)
   => x = 2C/(D2 - y2)  [7]
plugging [7] into [2]
   y = x' = 2C·-1/(D2-y2)2·-2y·y'
   => y' = (D2-y2)2/4C  [8]
The solution I gave is for the special case D=0.  But we want D>y to have x>0.
   dy/dt = (D2-y2)2/4C
   => dy/(D2-y2)2 = dt/4C [9]
Let's integrate.
According to Wolfram, the left integrates to
   1/(2D3)·( Dy/(D2-y2) + tanh-1(y/D) )
For the right side, let's write the constant as E/4C.  [9] integrates to
   1/(2D3) · ( Dy/(D2-y2) + tanh-1(y/D) ) = 1/4C·t + E/4C
   => 4C/(2D3)·( Dy/(D2-y2) + tanh-1(y/D) ) = t + E
   => 2C/D3·( (y/D)/(1-(y/D)2) + tanh-1(y/D) ) = t + E
Substituting  y/D = sqrt(1-2C/(D2x)) and 1/(1-(y/D)2) = D2x/2C gives
   => 2C/D3·( sqrt(1-2C/(D2x))·D2x/2C + tanh-1(sqrt(1-2C/(D2x))) ) = t + E
   => x/D·sqrt(1-2C/(D2x)) + 2C/D3·tanh-1(sqrt(1-2C/(D2x))) = t + E
Now we just need to express x in terms of t...

[edit] corrected missing parentheses [/edit]
Title: Re: motion in non-uniform electric field
Post by towr on May 8th, 2009, 11:35am

on 05/05/09 at 06:59:48, Grimbal wrote:
 => x/D·sqrt(1-2C/D2x) + 2C/D3·tanh-1(sqrt(1-2C/D2x)) = t + E
Now we just need to express x in terms of t...
That looks to be the right curve, compared to what I got from simulating an example.  
Although, it confused me a moment that 2C/D2x was meant as 2C/(D2x)

Is there a way to get x''(t) (or t(x'') ) without expressing x in terms of t first?
Title: Re: motion in non-uniform electric field
Post by Grimbal on May 9th, 2009, 5:24pm
I don't know.  I guess if the relation is between x and x" you can hardly get rid of x.

but I found a faster way.
  x" = C/x2
Let's compute (basically the conservation of energy)
  (x'2+2C/x)'
  = 2x'x" + (-1)2C/x2·x'
  = 2x'·(x" - C/x2) = 0
Therefore
  x'2 + 2C/x = D2 (an arbitrary constant)
  => x' = sqrt(D2-2C/x)
  => dx/dt = D·sqrt(1-2C/(D2x))
  => dx/sqrt(1-2C/(D2x)) = D·dt
We integrate (using Wolfram's online integrator)
  x·sqrt(1-2C/(D2x)) + C/D2·log[C-D2x·(sqrt(1-2C/(D2x))+1)] = D·(t + E)
  x/D·sqrt(1-2C/(D2x)) + C/D3·log[C-D2x·(sqrt(1-2C/(D2x))+1)] = t + E

The funny thing is that I don't get the same solution...


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