- #1

Doug007

- 6

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Doug007
- Start date

- #1

Doug007

- 6

- 0

- #2

rbj

- 2,227

- 9

Doug007 said:

but your premise that the movement of a uniformly smooth charged rod is indistiguishable from a stationary one, that premise is unfounded. there is a difference between an infinite uniform colomn of water moving past you and an identical one that is not moving.

and an observer traveling alongside the uniform line of charge will see it differently than the observer "at rest" who sees the uniform line of charge moving past at high speed. the first observer sees no magnetic field since the charge is static relative to him. the second sees a magnetic field.

- #3

Antiphon

- 1,683

- 3

Once you pass from point charges to massless charge density, how do you assign

the charge density a velocity?

You do it in the same way you would with any continuous field. The problem you have

is not unlike what first year calculus students experience when they have the vansishingly

small dx adding up to a finite value.

You simply associate a velocty distribution function V(r) to the continuous charge

density. It's velocity is an axiom of your classical EM problem.

- #4

Doug007

- 6

- 0

rbj said:but your premise that the movement of a uniformly smooth charged rod is indistiguishable from a stationary one, that premise is unfounded. there is a difference between an infinite uniform colomn of water moving past you and an identical one that is not moving.

It's not the quality of the rod being infinite that makes the moving rod indistiguishable from a stationary one (in my mind) but the uniform charge density - I introduce infinity just to remove the ends that would give away the motion of the rod, as clearly as individual point charges (electrons) would. If water had no structure to it but was instead a quality that was uniform across the column you mention then how would the motion of the water be distinguishable from a stationary one - without touching it?

and an observer traveling alongside the uniform line of charge will see it differently than the observer "at rest" who sees the uniform line of charge moving past at high speed. the first observer sees no magnetic field since the charge is static relative to him. the second sees a magnetic field.

How do your two observers "see" the rod differently? It's uniformly charged without ends so it's motion (in my mind) is not obvious.

This is a tricky issue to describe. I think it may come down to the problem of assigning a velocity to a infinite uniform field as hinted at by the other reponse.

- #5

Doug007

- 6

- 0

Antiphon said:

Once you pass from point charges to massless charge density, how do you assign

the charge density a velocity?

You do it in the same way you would with any continuous field. The problem you have

is not unlike what first year calculus students experience when they have the vansishingly

small dx adding up to a finite value.

You simply associate a velocty distribution function V(r) to the continuous charge

density. It's velocity is an axiom of your classical EM problem.

Thanks for replying. I do understand your reply but I still have the problem. I've tried to reformulate it as...

If I'm sitting above an infinite continuous 2D plane how can I tell if I'm moving relative to the plane or not (not nearer or closer to, just parallel to)? Presumably I must know this before I can assign a velocity distribution to it and presumably to know this I would need to observe some fixed point on the plane moving relative to me. No point exists for the above empty plane nor would it - as far as i can see - if the plane was uniformly charged so it's motion, or mine relative to it would be impossible to determine. The generation of a magnetic field would seem to assign points to the plane - but there arn't any?

- #6

krab

Science Advisor

- 893

- 3

- #7

Doug007

- 6

- 0

krab said:

But how can a reference frame for the charges be constructed because there arn't any? It's a uniform continuous distribution of charge with no discrete points of any form - charge or otherwise - to refer to!

- #8

rbj

- 2,227

- 9

Doug007 said:But how can a reference frame for the charges be constructed because there arn't any? It's a uniform continuous distribution of charge with no discrete points of any form - charge or otherwise - to refer to!

but the "soup" is moving in reference to the "stationary" observer. it matters not that it is discrete little chunks of soup moving past or one continuous glob of it. if the observer pokes his finger into the stream, he'll know whether or not it is moving.

FYI: here is a reposting of a thought experiment you can do to see how this magnetic field comes from the electrostatic field with special relativity taken into consideration. the magnetic force is actually not a different force than the electrostatic force but is a manifestation of it in the context of moving charges in a relativistic reality.

The classical electromagnetic effect is perfectly consistent with the lone

electrostatic effect but with special relativity taken into consideration.

The simplest hypothetical experiment would be two identical parallel

infinite lines of charge (with charge per unit length of [tex] \lambda \ [/tex]

and some non-zero mass per unit length of [tex] \rho \ [/tex] separated

by some distance [tex] R \ [/tex]. If the lineal mass density is small enough

that gravitational forces can be neglected in comparison to the electrostatic

forces, the static non-relativistic repulsive (outward) acceleration (at the instance

of time that the lines of charge are separated by distance [tex] R \ [/tex])

for each infinite parallel line of charge would be:

[tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]

If the lines of charge are moving together past the observer at some

velocity, [tex] v \ [/tex], the non-relativistic electrostatic force would appear to be

unchanged and that

with the lines of charge would observe.

Now, if special relativity is considered, the in-motion observer's clock

would be ticking at a relative

from the point-of-view of the stationary observer because of time dilation. Since

acceleration is proportional to (1/time)

an acceleration scaled by the square of that rate, or by [tex] {1 - v^2/c^2} \ [/tex],

compared to what the moving observer sees. Then the observed outward

acceleration of the two infinite lines as viewed by the stationary observer would be:

[tex] a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]

or

[tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} [/tex]

The first term in the numerator, [tex] F_e \ [/tex], is the electrostatic force (per unit length) outward and is

reduced by the second term, [tex] F_m \ [/tex], which with a little manipulation, can be shown

to be the classical magnetic force between two lines of charge (or conductors).

The electric current, [tex] i_0 \ [/tex], in each conductor is

[tex] i_0 = v \lambda \ [/tex]

and [tex] \frac{1}{\epsilon_0 c^2} [/tex] is the magnetic permeability

[tex] \mu_0 = \frac{1}{\epsilon_0 c^2} [/tex]

because [tex] c^2 = \frac{1}{ \mu_0 \epsilon_0 } [/tex]

so you get for the 2

[tex] F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} [/tex]

which is precisely what the classical E&M textbooks say is the magnetic force (per unit length)

between two parallel conductors, separated by [tex] R \ [/tex], with identical current [tex] i_0 \ [/tex].

Last edited:

- #9

Doug007

- 6

- 0

"If I'm sitting above an infinite continuous 2D plane how can I tell if I'm moving relative to the plane or not (not nearer or closer to, just parallel to)? Presumably I must know this before I can assign a velocity distribution to it and presumably to know this I would need to observe some fixed point on the plane moving relative to me. No point exists for the above empty plane nor would it - as far as i can see - if the plane was uniformly charged so it's motion, or mine relative to it would be impossible to determine. The generation of a magnetic field would seem to assign points to the plane - but there arn't any?"

As for sticking your finger into a plane of this nature no movement would be detected as it has no mass or any other tangible quality apart from it's smoth uniform charge distribution - In Antiphons repsonse, a continuous field. The generation of a magnetic field would bely this motion and in doing so allocate points of reference to a plane that has none!

- #10

rbj

- 2,227

- 9

Doug007 said:The first paragraph of your reply is relevant to my question, but I'm afraid the rest isn't

that's okay, i was just picking up on krab's point and i had the text with equations already written (hanging out in wikipedia's magnetic field talk page).

- I understand the generation of magnetic field via the equations you qoute. I have reformulated the OP as...

"If I'm sitting above an infinite continuous 2D plane how can I tell if I'm moving relative to the plane or not (not nearer or closer to, just parallel to)? Presumably I must know this before I can assign a velocity distribution to it and presumably to know this I would need to observe some fixed point on the plane moving relative to me. No point exists for the above empty plane nor would it - as far as i can see - if the plane was uniformly charged so it's motion, or mine relative to it would be impossible to determine. The generation of a magnetic field would seem to assign points to the plane - but there arn't any?"

i guess no points that you can see. a geometric point is not a concept that requires matter or a pixel or a

As for sticking your finger into a plane of this nature no movement would be detected as it has no mass or any other tangible quality apart from it's smooth uniform charge distribution

which is moving (relative to you, the observer).

- In Antiphons repsonse, a continuous field. The generation of a magnetic field would bely this motion and in doing so allocate points of reference to a plane that has none!

sure there are points of reference (whether you see them or not). they are points of reference around which no movement of this uniform charge exist.

your infinite plane example

in our 3D classical existence, moving away from a point charge or even a little sphere of charge is an inverse-square effect on field strength (the sphere gets smaller in 2 visible dimensions). moving away from an infinite line of charge is an inverse-proportional effect on field strength (the line gets smaller in 1 dimension). moving away from an infinite plane of charge is one power less in the denominator, a constant field strength because that infinite plane is not getting any smaller (in some hypothetical field of vision), no matter how far you get from it.

- #11

Doug007

- 6

- 0

i guess no points that you can see. a geometric point is not a concept that requires matter or a pixel or a thing to define it.

But a charge would need to be attached to a geometric point in order for movement to be defined and this would necessitate the introduction of non uniform charge distribution. The attachment of a uniform plane of charge would not distinguish it - in terms of movement - from an unattached one!

sure there are points of reference (whether you see them or not). they are points of reference around which no movement of this uniform charge exist.

But the quality of being stationary to the plane cannot be determined (apart from the debatable generation of a magnetic field). If magnetism did not exist how would you determine your motion relative to the uniformly charged plane as you move parallel to it - as distinct from the underlying geometry that makes the math work.

You're right about the inverse square, inverse and constant electric fields that apply to points, lines and planes and it raises an analogues situation: how would an observer situated within such a constant electric field no he was moving relative to it without sighting the plane (which would be a definite point of reference)? Ignore magnetic field generation in this example for the sake of the underlying geometric point I'm making.

Last edited:

Share:

- Replies
- 11

- Views
- 178

- Replies
- 9

- Views
- 415

- Replies
- 4

- Views
- 257

- Last Post

- Replies
- 1

- Views
- 312

- Replies
- 16

- Views
- 360

- Replies
- 7

- Views
- 376

- Last Post

- Replies
- 1

- Views
- 135

- Last Post

- Replies
- 30

- Views
- 1K

- Replies
- 41

- Views
- 1K

- Replies
- 9

- Views
- 458