Re: [Phys-l] drop a metal cylinder through a solenoid

[John Denker <jsd@av8n.com>]

On 03/25/2012 07:02 PM, Carl Mungan wrote:
> I'm surprised. It's completely negligible. Obviously this wasn't a
> surprise to you. I was willing to believe 1 uJ. But 1e-39 J: that's
> super tiny. Can you shed any further intuitive way to see that it's
> basically zero for all practical purposes?

There are a couple of ways to make this less mysterious.
See below.

>  Maybe part of the answer
> would be: are there other systems for which the current goes like V
> dot?

1) Yes, indeed there are.  I addressed this in another thread.

At one point in my life I did literally thousands of calculations 
like this, where the power went like V dot -- or more specifically 
like I^2 R -- which was much less than V^ / R.  This was in connection 
with the low-power electronics business.  So I could set up this 
calculation literally without thinking.

2) Now for the main point of this message:  Here is another way
of attacking the problem:  Non-dimensional scaling!

Scaling laws are very powerful.

The methods we call "dimensional analysis" are a simplified form
of scaling.  For the task at hand, we will start with that but
quickly move beyond it.

Since we want to calculate the power, let's find something plausible
that has dimensions of energy per unit time.  The induced electric
field E has electrostatic field energy density ε0 E^2.  So the total 
energy will be ε0 E^2 d^3.  (You could use this to calculate the 
capacitance of a capacitor from first principles, but we don't need
to go there.)  Now we need something with dimensions of time.  The
obvious thing is the electrical RC time, which is dimensionally equal
to (and indeed numerically equal to) ε0 ρ, where ρ is the resistivity
of the metal.  Hint: ε0 is equal to 8.85 picofarads per meter, while
ρ is 28e-9 ohm-meters, so ε0 ρ has the same dimensions as RC.

So at this point we have E^2 d^3 / ρ ... which has dimensions of power.
This would give us a nonzero power for a bob sitting in a constant
E field, which we know can't be right.  We need something that depends
on the time-derivative, i.e. E dot ... or equivalently ω E.  We can't
just replace E by ω E, because that would break the dimensions.  So
instead we replace E by ω R C E, or equivalently ω ε0 ρ E.  That works
because ω ε0 ρ is dimensionless.

Because we are multiplying by something that is dimensionless, dimensional
analysis won't tell us how many powers of ω ε0 ρ we need.  However, as
the saying goes, there is more to physics than dimensional analysis.  We
know that the dissipation should to like I^2 R, which depends on V dot
and on the capacitance and on the resistance, so ω ε0 ρ E is looking
pretty good.

So at this point we have ω^2 ε0^2 ρ^2 E^2 d^3 / ρ.  So far so good.

We can plug in E = v B and v = ω L (times some dimensionless amplitude
θmax).  That gives us 
      ω^4 ε0^2 ρ L^2 θmax^2 B^2 d^3          [10]
which is the right answer AFAICT.

The ω^4 gets your attention.  It has significant consequences.  Its
origin is, however, not unduly mysterious.  One factor of ω is needed 
to turn B into E, and another is needed to turn E into E dot.  And then 
both factors get squared when E dot gets squared.

You can simplify equation [10] a little bit by dropping the
θmax factor, since θmax was specified to be of order unity in this
example.  If the amplitude ever changes you can easily stick the θmax
back in.  You can also let ω = sqrt(g/L) in which case we are left
with 
     g^2 ε0^2 ρ B^2 d^3                     [11]

It is interesting that L has dropped out.  This can be understood
since the dissipation depends on E dot which depends on v dot i.e.
acceleration, which scales like g ... independent of L.

So ... the next time you see this problem, you can do it on the 
back of an envelope, in less time than it takes to talk about it.