Re: Force between 2 charged particles

On Thu, 06 Oct 2005 11:46:19 +0200, Julian Becker wrote:

> I got a little question, being the following: Imagine two equally
> charged particles with a certain distance between them, being
> connected by a rigid force measuring instrument. Both particles be
> resting in the system K. The instrument shows us a certain force
> corresponding the coulomb-force. But when I look at those two
> particles in a system K' which moves inertially in the plan to which
> the distance of the particles is perpendicular, the charges are
> moving and could be thought of as two currents having the same
> direction. And normally two parrallel currents attract each
> other. This attraction thus counteracts the electrostatic
> repulsion. Consequently, the same force-meter should give diferent
> values as seen from two different inertial systems, which is
> impossible. Please would someone show me where the error is!?

You've already gotten a bunch of answers but I'm not sure how helpful
any of them have been, so I'll throw my 2 cents in as well.

First, note that pointing out the error is easier than providing the
correct solution!

There are two errors: You've assumed that a single point charge,
moving along by itself, can be treated as a current, which is the
limiting case of a line of arbitrarily closely spaced point charges
moving in tandem. That's not right: a single moving point charge is
rather different.

The second is that you're mixing the forces as measured in the two
frames. Your force meter, as someone (Bill Hobba?) already pointed
out, is moving with the charges, and so is affected by all the same
things that affect the charges. In other words, the value on the
force meter's face is the force _in_ _frame_ K. It is _not_ the same
as the force as measured in frame K'.

With that said, note that actually _measuring_ the force on a moving
object is not simple! You need to use a stationary force-meter, and
coupling the meter to the moving object is likely to be difficult.

* * *

Now let's see if I can say anything useful about the _correct_
conclusion.

I won't say anything more about the force transformations just now.
Instead, let's go back and look at the E and B fields. As a rule, if
you can determine the electric and magnetic fields exactly in any
particular inertial frame, then you can just transform them to any
other frame. In this case, in frame K we know that each particle
"sees" an E field but no B field. The E field points directly away
from the other particle (assuming they're positively charged).

Let's put particle P0 at the origin, and particle P1 on the Y axis at
some positive index. Assume that motion of frame K' is along the X
axis. Let's see what fields will be "seen" by particle P1 in frame
K'.

K' moves at velocity "v", along the X axis, relative to frame K.
Then, plugging into the transforms for the E and B fields (which are
in many textbooks as well as being scattered around the web), we have

g = gamma = 1/sqrt(1 - v^2) (for convenience)

E' = g*E (in the Y direction, in this case)

The only nonzero component of the B' field will be along the Z
direction (perpendicular to v as well as to the electric field in
frame K) and it's given by

B' = -v*g*E

If the original value for E is positive (pointing "up" the Y axis),
and "v" is positive, then B' must be negative: it points along the Z
axis, in the -z direction. In frame K' (which is moving to the
right), our charge P1 is moving TO THE LEFT, so it's traveling at
velocity -v along the X axis. Using the handy right-hand-rule, we see
that it feels a force from the B' field which will be directed
downward, along the Y axis. There's your force pulling two parallel
current-carrying wires together, of course.

More precisely, using the Lorentz force law in frame K', if particle
P1 has charge "q", with B' perpendicular to P1's velocity, after
expanding out the cross product "vxB" and plugging in "-v" for the x
component of the velocity, we'll have

F' = q*(E' + vB')

= q*g*E*(1 - v^2)

= q*E*(1 - v^2)/sqrt(1 - v^2)

= q*E/g

On the other hand, the force on P1 in frame K, in which both particles
are at rest, is just

F = q*E

Now, the "inertial mass" of particle P1 is not the same in the two
frames. In frame K it's just the rest mass of P1, which we'll call
"m"; in frame K', where P1 is moving, it's given as

inertial mass = g*m

Since the force is perpendicular to the line of motion, gamma will be
constant, and so in this particular case, acceleration in frame K'
will be force divided by mass, as you might expect. So, if we let
particle P1 go, its acceleration, observed in frame K', will be

a' = F'/(inertial mass)

= (q*E/g)/(g*m)

= q*E/m*g^2

By comparison, the same acceleration in frame K will be observed as

a = q*E/m

These are not the same! Why is that?

They're different because the clocks aren't running at the same rates
in the two frames, and so the very same motion of particle P1 will
seem to go more slowly in one frame than the other.

In English, we can explain it more or less as follows: In frame K,
where clocks are ticking g times slower, a transverse velocity will
appear g times faster (more distance is covered per tick). The time
it takes to accelerate to that same velocity will again appear g times
shorter in frame K (fewer ticks elapse during acceleration). So,
overall, a transverse acceleration will be measured as g^2 times
larger in frame K than in frame K'. (In German, I couldn't explain it
at all, sorry!)

The English explanation is somewhat misleading, unfortunately, because
it conceals the fact that the situation is symmetric. Whose clock
appears to be running slower depends on exactly how you measure the
clock rates; neither is "really" running slower.

I don't know if any of this helped, but I hope, at least, it didn't
add too much confusion! :-)


--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org

.

Relevant Pages

  • Re: could this be possible
    ... >> existence of magnetic fields for continuous charges. ... >> single point particle as moving then it is well known that magnetic ... > the ring remains the same? ... > the x-axis of some inertial frame. ...
    (sci.physics)
  • Re: could this be possible
    ... >> existence of magnetic fields for continuous charges. ... >> single point particle as moving then it is well known that magnetic ... > the ring remains the same? ... > the x-axis of some inertial frame. ...
    (sci.physics.relativity)
  • Re: Force between 2 charged particles
    ... But when I look at those two particles in a system K' which moves inertially in the plan to which the distance of the particles is perpendicular, the charges are moving and could be thought of as two currents having the same direction. ... as the force as measured in frame K'. ... from the other particle. ...
    (sci.physics.relativity)
  • Re: Force between 2 charged particles
    ... > moving along by itself, can be treated as a current, which is the ... > limiting case of a line of arbitrarily closely spaced point charges ... > as the force as measured in frame K'. ... > from the other particle. ...
    (sci.physics.relativity)
  • Re: Proper quantities in SR
    ... measured by a clock attached to the particle (hence, ... inertial frame in which the particle is moving with velocity u. ...
    (sci.physics.relativity)