Are the directions of charge flow (=current?) absolute?

In summary, the three observers, who are assumed to be moving inertially, have different perspectives on the motion of the two charged beams. Observer 1 sees the beams moving in the same direction, while observer 2 sees them moving in opposite directions. Observer 3 also sees the beams moving in the same direction, but at different speeds. This difference in perspective is due to the direction dependence of the magnetic and electric forces acting on the charged particles in the beams. The stronger the relative velocity between the beams and the observer, the more apparent the effects of these forces become. This phenomenon can be compared to the different perspectives of drivers on a motorway when overtaking cars in different lanes.
  • #1
kmarinas86
979
1
Let's say I have two (-) charged beams.

According to observer 1:
Beam 1 is shot at 0.998 c.
Beam 2 is shot at 0.999 c.
>> Therefore, the beams are moving in the same direction.
Observer 2 is moving at 0.9985 c.
Observer 3 is moving at 0.9995 c.

According to observer 2:
Beam 1 is shot at -0.143 c [=(0.998-0.9985)/(1+0.998*(-0.9985)) c].
Beam 2 is shot at 0.200 c [=(0.999-0.9985)/(1+0.999*(-0.9985)) c].
>> Therefore, the beams are moving in opposite directions.

According to observer 3:
Beam 1 is shot at -0.600 c [=(0.998-0.9995)/(1+0.998*(-0.9995)) c].
Beam 2 is shot at -0.333 c [=(0.999-0.9995)/(1+0.999*(-0.9995)) c].
>> Therefore, the beams are moving in the same direction.

We can assume that for the sake of argument that all three observers are moving inertially.

Does the magnetic force act against the electric repulsion, as in two currents moving in the same direction? Or does the magnetic force act together with the electric repulsion, as in two currents moving in opposite directions?

Is it really possible that two currents can really be going in the same direction in one inertial frame and opposite directions in a different inertial frame without there being two different possible futures according to each frame?

In the two attached pictures below, I present a similar example, but this time it involves currents in loop wire rather than charged beams. The second attached picture depicts the view of the system from a non-inertial perspective. This differs from my newest example above, which assumes no non-inertial motion of any of the three observers.
 

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  • #2
With the direction dependence, there's nothing mysterious going on, and you don't need any relativity to see this sort of thing.

Imagine you are driving along the motorway, in the slow lane going at 50, and see car 1 in the middle lane going at 60 and car 2 in the fast lane going at 70. They overtake you: they're both going in the same direction from your point of view
Now you accelerate to 65 so you can overtake car 1, but car 2 is still zooming off into the distance while car 2 looks like he's going backwards: they're going in opposite directions.
Finally you put your foot down and go up to 80, leaving both car 1 and 2 in your dust, looking like they're going backwards, in the same direction.

The amazing thing is that with charged particles the same force appears in different ways to different observers: sometimes it's magnetic, sometimes it's electric.

For a beam of charged particles, the apparent charge density increases the faster you move relative to it. If you travel slowly compared to both beams, there appears to be a large electric repulsive force and a smaller magnetic attractive one. If you accelerate to the same speed as beam 1, the magnetic attraction disappears but the electric repulsion decreases to compensate. If you again accelerate so that the beams appear to go in opposite directions, there must now be a repulsive magnetic force, so the repulsive electric force must decrease further to compensate: the apparent decrease in charge for beam 2 outweighs the increase for beam 1. The apparent electric force is minimal in the frame in which the beams have equal and opposite speeds.

I hope this all makes sense; if you want explicit calculation to show how it works I'll be happy to provide, but I doubt it'll be very enlightening.
 
  • #3
henry_m said:
With the direction dependence, there's nothing mysterious going on, and you don't need any relativity to see this sort of thing.

Imagine you are driving along the motorway, in the slow lane going at 50, and see car 1 in the middle lane going at 60 and car 2 in the fast lane going at 70. They overtake you: they're both going in the same direction from your point of view
Now you accelerate to 65 so you can overtake car 1, but car 2 is still zooming off into the distance while car 2 looks like he's going backwards: they're going in opposite directions.
Finally you put your foot down and go up to 80, leaving both car 1 and 2 in your dust, looking like they're going backwards, in the same direction.

The amazing thing is that with charged particles the same force appears in different ways to different observers: sometimes it's magnetic, sometimes it's electric.

For a beam of charged particles, the apparent charge density increases the faster you move relative to it. If you travel slowly compared to both beams, there appears to be a large electric repulsive force and a smaller magnetic attractive one. If you accelerate to the same speed as beam 1, the magnetic attraction disappears but the electric repulsion decreases to compensate. If you again accelerate so that the beams appear to go in opposite directions, there must now be a repulsive magnetic force, so the repulsive electric force must decrease further to compensate: the apparent decrease in charge for beam 2 outweighs the increase for beam 1. The apparent electric force is minimal in the frame in which the beams have equal and opposite speeds.

I hope this all makes sense; if you want explicit calculation to show how it works I'll be happy to provide, but I doubt it'll be very enlightening.

Back to my original example, we had observers 1, 2, and 3, and beams 1 and 2.

Observer 1 should see a strong concentration of the E-field of both beams, 1 and 2. Observer 1 also sees a strong magnetic field from both. Slightly stronger concentration of the E-field and the magnetic field is observed with beam 2. The force due to the E-field is proportional to the Lorentz factor (I think...), while the force of the latter is proportional to v (I think...). The former seems to have no limit but infinity, while the latter has a finite limit... (Either that or I am missing something there...) But either way, I think that the increase of one helps to negate the increase in the other (they are opposed forces after all).

Observer 2 should see more E-field line concentration of beam 2 than beam 1. Observer 2 sees a stronger magnetic field from beam 2 than beam 1.

Observer 3 should see more E-field line concentration of beam 1 than beam 2. Observer 3 sees a stronger magnetic field from beam 1 than beam 2.

Generally speaking, for your explanation to work, it would seem that for the force to somehow be constant, the following relation must apply:

constant = [A(v_0,v_0+Δv) * x] + [B(v,v_0+Δv) * y]

Where:
v_0 = a reference velocity
Δv = a difference in velocity
x = force due to the electric field at some arbitrary v_0
y = force due to the magnetic field at some arbitrary v_0
A(v_0,v_0+Δv) = A scaling factor for the electric field
B(v_0,v_0+Δv) = A scaling factor for the magnetic field
 
  • #4
henry_m said:
For a beam of charged particles, the apparent charge density increases the faster you move relative to it. If you travel slowly compared to both beams, there appears to be a large electric repulsive force and a smaller magnetic attractive one. If you accelerate to the same speed as beam 1, the magnetic attraction disappears but the electric repulsion decreases to compensate. If you again accelerate so that the beams appear to go in opposite directions, there must now be a repulsive magnetic force, so the repulsive electric force must decrease further to compensate: the apparent decrease in charge for beam 2 outweighs the increase for beam 1. The apparent electric force is minimal in the frame in which the beams have equal and opposite speeds.

How does the E-field concentration "decrease further to compensate"?

Obviously this can't be due to a length contraction effect. The contraction only explains an increase in the concentration at right angles.

Alternatively, if reshaping of the E-field were somehow tied to the Doppler effect (like the "focusing of a head lamp"), all that would seem to do is to make the repulsion asymmetric, but not actually decrease the repulsion overall, expect maybe at a right angle. It would however also make it asymmetric along the path of the beams themselves. Again, with this, even if we integrate this field along the length of a continuous beam of charge, it would not seem to matter which direction the beam would go. The density of the E-field line at a 90-degree angle would be the same magnitude whether the relative velocity was +0.1 c or -0.1 c (the gradient of density of those lines would simply flip), but of course we know that the magnetic field will be seen to have different signs in each case. So then, just how do the E-field lines seem to decrease in strength when the (observed) magnetic field ramps up as one accelerates the external observer?
 
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  • #5
Right, it turns out the calculations are actually very clean so they should be pretty enlightening. I'll work in units where c=1 to make things a little easier; if you aren't comfortable with this let me know and I'll help put the c's back in (though you should learn to think in terms of c=1, because it is!).

The current 4-vector is [itex]j=(\rho,\mathbf{j})[/itex], where [itex]\rho[/itex] is the charge density and [itex]\mathbf{j}[/itex] the current density. This, as the name suggests, is a four-vector, so it transforms in the same way as four-velocity, four-momentum etc. It suffices here to work in (1+1) dimensions, where [itex]\mathbf{j}=I[/itex] is just the current.

For simplicity, I'll assume that the beams are identical apart from their speed, so in the rest frame of each they have the same charge density. Now consider the 'COM' frame in which the beams appear to have equal and opposite velocities. The currents will be [itex]j_1=(\rho,-I)[/itex] and [itex]j_2=(\rho,I)[/itex] for some fixed [itex]\rho[/itex] and [itex]I[/itex].

Now boost with speed v. In this new frame, the currents will be [itex]j_1=\gamma(\rho-vI,-I-v\rho)[/itex] and [itex]j_2=\gamma(\rho+vI,I-v\rho)[/itex]. In your description, observer 1 is the case [itex]v<-I/\rho[/itex], and observer 3 is the case [itex]v>+I/\rho[/itex]. Observer 2 is anything in between.

The electric force will be proportional to the product of the charge densities (the first term), and the magnetic force the product of the currents (the second term). These are:
[tex]
F_E\propto \gamma^2(\rho^2-v^2 I^2)
[/tex]
[tex]
F_B\propto -\gamma^2(v^2\rho^2- I^2)
[/tex]
and the constants of proportionality are the same in each case. (The signs are such that positive implies repulsive). By simply adding these up, and using [itex]\gamma^2=1/(1-v^2)[/itex], the total force is then
[tex]F\propto \rho^2+I^2[/tex]
which is, thankfully, independent of [itex]v[/itex].

Now here's the critical paragraph, and this is the kind of argument that will make relativity easy. Notice that the force is in fact the Lorentz inner product of the current four-vectors, which must be scalar. This generalises to any parallel charge and current distributions (current-carrying wires and beams of particles). If you can notice this from the very beginning then you don't have to do any work at all! Though it is nice to see how the force splits into electric and magnetic components in different frames.

Worth drawing a graph to see how this works. I should mention that my earlier comment that the electric force is minimal in the COM frame is wrong; it's actually a local maximum, and there will be minima elsewhere.

I haven't addressed your concerns directly, but hopefully this should clear most of them up. If you're still worried about anything specific, post again and I'll do my best to answer.
 
  • #6
henry_m said:
Right, it turns out the calculations are actually very clean so they should be pretty enlightening. I'll work in units where c=1 to make things a little easier; if you aren't comfortable with this let me know and I'll help put the c's back in (though you should learn to think in terms of c=1, because it is!).

The current 4-vector is [itex]j=(\rho,\mathbf{j})[/itex], where [itex]\rho[/itex] is the charge density and [itex]\mathbf{j}[/itex] the current density. This, as the name suggests, is a four-vector, so it transforms in the same way as four-velocity, four-momentum etc. It suffices here to work in (1+1) dimensions, where [itex]\mathbf{j}=I[/itex] is just the current.

For simplicity, I'll assume that the beams are identical apart from their speed, so in the rest frame of each they have the same charge density. Now consider the 'COM' frame in which the beams appear to have equal and opposite velocities. The currents will be [itex]j_1=(\rho,-I)[/itex] and [itex]j_2=(\rho,I)[/itex] for some fixed [itex]\rho[/itex] and [itex]I[/itex].

Now boost with speed v. In this new frame, the currents will be [itex]j_1=\gamma(\rho-vI,-I-v\rho)[/itex] and [itex]j_2=\gamma(\rho+vI,I-v\rho)[/itex]. In your description, observer 1 is the case [itex]v<-I/\rho[/itex], and observer 3 is the case [itex]v>+I/\rho[/itex]. Observer 2 is anything in between.

The electric force will be proportional to the product of the charge densities (the first term), and the magnetic force the product of the currents (the second term). These are:
[tex]
F_E\propto \gamma^2(\rho^2-v^2 I^2)
[/tex]
[tex]
F_B\propto -\gamma^2(v^2\rho^2- I^2)
[/tex]
and the constants of proportionality are the same in each case. (The signs are such that positive implies repulsive). By simply adding these up, and using [itex]\gamma^2=1/(1-v^2)[/itex], the total force is then
[tex]F\propto \rho^2+I^2[/tex]
which is, thankfully, independent of [itex]v[/itex].

Now here's the critical paragraph, and this is the kind of argument that will make relativity easy. Notice that the force is in fact the Lorentz inner product of the current four-vectors, which must be scalar. This generalises to any parallel charge and current distributions (current-carrying wires and beams of particles). If you can notice this from the very beginning then you don't have to do any work at all! Though it is nice to see how the force splits into electric and magnetic components in different frames.

Worth drawing a graph to see how this works. I should mention that my earlier comment that the electric force is minimal in the COM frame is wrong; it's actually a local maximum, and there will be minima elsewhere.

I haven't addressed your concerns directly, but hopefully this should clear most of them up. If you're still worried about anything specific, post again and I'll do my best to answer.

The formula:

[tex]F\propto \rho^2+I^2[/tex]

...only allows the magnetic field to add to the repulsion. It does not allow one to calculate an [tex]F[/tex] less than proportional to [tex]\rho^2[/tex].

Also, the current [tex]I[/tex] must be defined. If we defined [tex]\rho[/tex] as the linear charge density, then we can have two equations for the two currents in question:

[tex]I_1 = \rho_1 * v_1[/tex]

[tex]I_2 = \rho_2 * v_2[/tex]

But what are [tex]v_1[/tex] and [tex]v_2[/tex]? It seems that from your example, [tex]v_1[/tex] and [tex]v_2[/tex] are the velocities of the currents relative to your COM (center of momentum) frame. However, this actually creates a constraint, where we end up with only currents going in opposite directions. This would be consistent with the fact that the equation you provided assumes that [tex]F[/tex] is not less than proportional to [tex]\rho^2[/tex].

However, what if we really have two currents moving in the same direction? The decrease of the Lorentz force must somehow be captured in every inertial reference frame. [tex]v_1[/tex] and [tex]v_2[/tex] must be allowed to both be positive.

So perhaps we can use this instead:

[tex]F_E\propto \gamma^2(\rho^2+v^2 \left(I_1*I_2\right))[/tex]

[tex]F_B\propto -\gamma^2(v^2\rho^2+ \left(I_1*I_2\right))[/tex]

[tex]F\propto \rho^2-\left(I_1*I_2\right)[/tex]

[tex]F\propto \left(\rho_1*\rho_2\right)-\left(\rho_1*v_1*\rho_2*v_2\right)[/tex]

[tex]F\propto \left(\rho_1*\rho_2\right)\left(1-v_1*v_2\right)[/tex]

So what is the inertial frame with respect to which [tex]v_1[/tex] and [tex]v_2[/tex] are determined? If they are both positive, then that the inertial frame cannot be the center of momentum frame. If that previous question is misguided, then are [tex]v_1[/tex] and [tex]v_2[/tex] relative to the observer?

For [tex]F[/tex] to be constant, I found 6 options:

* [tex]\rho_1[/tex] and [tex]\rho_2[/tex] vary to absorb the changes of [tex]I_1*I_2[/tex] that vary with the observer.
* [tex]\rho_1[/tex] and [tex]\rho_2[/tex] do not vary. [tex]I_1[/tex] and [tex]I_2[/tex] vary with each other inversely.
* There is a preferred inertial frame to determine [tex]v_1[/tex] and [tex]v_2[/tex].
* The formulas, [tex]I_1 = \rho_1 * v_1[/tex] and [tex]I_2 = \rho_2 * v_2[/tex], are wrong.
* The substitution of [tex]-\left(I_1*I_2\right)[/tex] for [tex]I^2[/tex] is wrong.
* The formula [tex]F\propto \rho^2-\left(I_1*I_2\right)[/tex], is wrong.
 
  • #7
I was probably a little terse with the early part of the argument. Here it is fleshed out a little:

We assume the beams are identical apart from speed. Then there exists a frame (COM) in which they appear symmetrically, with equal charge density and opposite currents. Now define [itex]\rho[/itex] as the apparent charge density of either beam in this frame, and define I as the apparent current in this frame. Then [itex]\pmI/\rho[/itex] will be the beam speeds, again in this frame: this is what you call [tex]v_1[/itex] and [tex]v_2[/itex].

These quantities [itex]\rho[/itex] and [itex]I[/itex] are now fixed for the problem in hand. They are just convenient constant quantities to fix the parameters of the problem. To find the charge densities and currents in an arbitrary frame, look to the boosted current vectors in the next paragraph. To repeat, these are
[tex]
j_1=(\rho_1,I_1)=\gamma(\rho-vI,-I-v\rho)//
j_2=(\rho_2,I_2)=\gamma(\rho+vI,I-v\rho)
[/tex]
Notice in particular that for appropriate choice of v, the currents, which are the second components of each vector, can be in opposite directions. The speeds of the beams in this arbitrary frame are [itex]\rho_i/I_i[/itex] for i=1,2.

The electric force is then proportional to [itex]\rho_1\rho_2[/itex], and the magnetic force proportional to [itex]-I_1 I_2[/itex]. The expressions above then give this in terms of the invariant parameters [itex]\rho,I[/itex]. This leads to the expressions in my previous post.

Summary: [itex]\rho_1,\rho_2,I_1,I_2[/itex] are the frame dependent quantities that afre the observed charge densities and currents of each beam. [itex]\rho,I[/itex] are the values for a specific special frame.

Hope that's a bit less confusing, sorry the earlier post wasn't very clear on the definitions.
 
  • #8
henry_m said:
I was probably a little terse with the early part of the argument. Here it is fleshed out a little:

We assume the beams are identical apart from speed. Then there exists a frame (COM) in which they appear symmetrically, with equal charge density and opposite currents. Now define [itex]\rho[/itex] as the apparent charge density of either beam in this frame, and define I as the apparent current in this frame. Then [itex]\pmI/\rho[/itex] will be the beam speeds, again in this frame: this is what you call [tex]v_1[/itex] and [tex]v_2[/itex].

These quantities [itex]\rho[/itex] and [itex]I[/itex] are now fixed for the problem in hand. They are just convenient constant quantities to fix the parameters of the problem. To find the charge densities and currents in an arbitrary frame, look to the boosted current vectors in the next paragraph. To repeat, these are
[tex]
j_1=(\rho_1,I_1)=\gamma(\rho-vI,-I-v\rho)//
j_2=(\rho_2,I_2)=\gamma(\rho+vI,I-v\rho)
[/tex]
Notice in particular that for appropriate choice of v, the currents, which are the second components of each vector, can be in opposite directions. The speeds of the beams in this arbitrary frame are [itex]\rho_i/I_i[/itex] for i=1,2.

The electric force is then proportional to [itex]\rho_1\rho_2[/itex], and the magnetic force proportional to [itex]-I_1 I_2[/itex]. The expressions above then give this in terms of the invariant parameters [itex]\rho,I[/itex]. This leads to the expressions in my previous post.

Summary: [itex]\rho_1,\rho_2,I_1,I_2[/itex] are the frame dependent quantities that afre the observed charge densities and currents of each beam. [itex]\rho,I[/itex] are the values for a specific special frame.

Hope that's a bit less confusing, sorry the earlier post wasn't very clear on the definitions.

So if I reduce this system to simply two elementary charges going in the same or opposite directions, would it be possible to use [itex]q_1,q_2[/itex] instead of [itex]\rho_1,\rho_2[/itex], where [itex]q_1,q_2[/itex] deviate from [itex]q[/itex]?
 
  • #9
kmarinas86 said:
So if I reduce this system to simply two elementary charges going in the same or opposite directions, would it be possible to use [itex]q_1,q_2[/itex] instead of [itex]\rho_1,\rho_2[/itex], where [itex]q_1,q_2[/itex] deviate from [itex]q[/itex]?

Not quite, though the effect is basically the same. The system of two point particles a slightly different for a couple of reasons so we're not really comparing like with like.

You can think of the frame dependence of charge density as being due to a Lorentz contraction, so at greater speeds you appear to have more charge in a smaller place. The total charge of a lump of charged stuff stays the same, but the volume in which it's contained appears smaller, hence the apparent increase in density. So point particles should also retain the same charge in any frame.* But the electric field is still frame dependent. We need a slightly different argument though.

Complication number 1: some of the symmetry is lost. With two long beams, you can regard the charge density as being constant over the length of the beam. With a pair of particles, you don't have this symmetry so the situation changes as they fly past one another. For definiteness, we can think about only the moment of closest approach, when the velocities are perpendicular to the separation. (As long as we only consider frames related by boosts parallel to the motion of the particles, this is a frame-independent moment).

Complication number 2: point particles are points! Formally, charge densities and current densities are zero everywhere apart from on top of the particle, where they are infinite. This is expressed by an object called the Dirac delta-function, if you haven't met it. But we can avoid this kind of thing by considering the electric and magnetic potentials and fields directly.

It's worth doing the calculation. The easiest way to do it is to work out the field due to a single charged particle in uniform motion. Start off with the well-known expression for the scalar potential of a stationary charged particle [itex]\phi=q/|\vec{x}|[/itex], boost to another frame and calculate E and B from there. The result is that the electric field gets increased by a Lorentz factor, and receives an extra term which decreases the field ahead of and behind the particle, and a magnetic field appears.

Going back to the original question, looking at the electric and magnetic force on another charged particle moving past, we get the same conclusion as for the beams we were considering previously: The electric force is augmented by a gamma factor and a magnetic force compensates to make the result frame-independent.

Ultimately, the 'reason' for this, like charges appearing smaller in different frames or similar is a matter of interpretation, not physics. Physics is forces and accelerations and other measurable things. My personal interpretation would be that the charge of the particle is not frame-dependent, but the field it creates is.

* An aside: This is not necessarily a very meaningful statement: after all, the charge of a particle is probably defined by the field it creates when it's at rest, and the electric field when it's in motion is different from this. Saying the charge is frame-independent is really only a sensible extension of the definition of the charge to arbitrary frames. This is a good way of defining things: say what it is in the rest frame, and define it to be scalar. E.g. mass:=rest frame energy.
 
  • #10
henry_m said:
Not quite, though the effect is basically the same. The system of two point particles a slightly different for a couple of reasons so we're not really comparing like with like.

You can think of the frame dependence of charge density as being due to a Lorentz contraction, so at greater speeds you appear to have more charge in a smaller place. The total charge of a lump of charged stuff stays the same, but the volume in which it's contained appears smaller, hence the apparent increase in density. So point particles should also retain the same charge in any frame.*

[...]

* An aside: This is not necessarily a very meaningful statement: after all, the charge of a particle is probably defined by the field it creates when it's at rest, and the electric field when it's in motion is different from this. Saying the charge is frame-independent is really only a sensible extension of the definition of the charge to arbitrary frames. This is a good way of defining things: say what it is in the rest frame, and define it to be scalar. E.g. mass:=rest frame energy.

Let's say I have this situation:

*Observer 1 sees two elementary charges (q_1 and q_2) moving in opposite directions at 0.5 c.
*Observer 2 sees Observer 1 moving at 0.8 c [ = (0.5+0.5)/(1+(0.5*0.5)) c ] (q_1) and two elementary charges moving in the same directions at 0.5 c [ = (0.8-0.5)/(1-(0.8*0.5)) c ] (q_2) and 0.9286 c [ = (0.8+0.5)/(1+(0.8*0.5)) c ] (q_2).

At the same time:

*Observer 2 sees two elementary charges (q_3 and q_4) moving in opposite directions at 0.5 c.
*Observer 1 sees these two elementary charges moving in the same directions at -0.9286 c [= x c, where -0.5=(0.8+x)/(1+0.8x)] (q_3) and -0.5 c [= x c, where 0.5=(0.8+x)/(1+0.8x)] (q_4) and also sees Observer 2 moving in the same direction at -0.8 c [= x c, where 0=(0.8+x)/(1+0.8x)].

Therefore, the force between q_1 and q_2 should equal the force between q_3 and q_4. However, to bring q_3 and q_4 to the state of q_1 and q_2, isn't work required to make that happen?

You had said that for a given [tex]F = \rho^2 + I^2[/tex], it is invariant with respect to the observer. But it would seem that by accelerating two charges, say with a voltage [tex]V[/tex], you must change [tex]I[/tex], and thus you must have a new [tex]F[/tex]. It should be possible to change [tex]I[/tex] for any two parallel, linear charge distributions kept at a fixed distance from each other, simply by applying a voltage [tex]V[/tex]. This of course requires acceleration of these charges.

It would seem that while (as you said) the force [tex]F[/tex] between two charges is invariant with respect to the inertial observer chosen, [tex]F[/tex] would have to change when the charges themselves change inertial frames by being accelerated. A similar question: Isn't the invariant mass only invariant with respect to the observer, rather than with respect to the body before vs. after acceleration? (Note that the latter, but not the former, involves proper acceleration of the charges.) Since net-work is required to change the inertial frame of a charge, isn't there a unique inertial frame in which a given charge possesses the minimum amount of observer-invariant mass in that the net-work done on the body is 0?
 
  • #11
I'm not really sure I follow your question.

When I talk about 'boosting to a new frame' I'm not doing anything to change the physical system itself. I'm only changing coordinates, and then the physics, like the force here, that come out of the calculation must be independent of the way chosen to describe it, or who is looking at it.

You seem to be suggesting that if you change the setup by pushing the particles around and accelerating them, you change the force. But this is hardly surprising; the force between two wires carrying 1 amp is not the same as the force between two wires carrying 2 amps. When we talk about invariant quantities (scalars), the invariance refers to frame/observer/coordinate independence.
 
  • #12
henry_m said:
I'm not really sure I follow your question.

When I talk about 'boosting to a new frame' I'm not doing anything to change the physical system itself. I'm only changing coordinates, and then the physics, like the force here, that come out of the calculation must be independent of the way chosen to describe it, or who is looking at it.

You seem to be suggesting that if you change the setup by pushing the particles around and accelerating them, you change the force. But this is hardly surprising; the force between two wires carrying 1 amp is not the same as the force between two wires carrying 2 amps. When we talk about invariant quantities (scalars), the invariance refers to frame/observer/coordinate independence.

Correct.

What I was wondering about were situations such as the following:

Say I accelerate a lab. Then I turn on two beams in opposite directions from opposite ends of the lab. Then I accelerate the lab over time such that the laboratory changes by a speed greater than that of either of the beams as measured from the initial frame of the laboratory. Therefore, both beams are going in the same direction relative to the initial frame of the lab, but they go in opposite directions according to the lab's final frame. I would like to know what would effect this would have on the Lorentz force. Would the Lorentz force be decreased as a result of the beams going in the same direction relative to the initial frame, or would the Lorentz force be increased as a result of going in opposite directions relative to the final frame? Or... is there perhaps a totally different frame not given by the problem, which when accounted for, which will result in the correct answer? If the latter is correct, what would that frame be?
 
  • #13
kmarinas86 said:
Correct.

What I was wondering about were situations such as the following:

Say I accelerate a lab. Then I turn on two beams in opposite directions from opposite ends of the lab. Then I accelerate the lab over time such that the laboratory changes by a speed greater than that of either of the beams as measured from the initial frame of the laboratory. Therefore, both beams are going in the same direction relative to the initial frame of the lab, but they go in opposite directions according to the lab's final frame. I would like to know what would effect this would have on the Lorentz force. Would the Lorentz force be decreased as a result of the beams going in the same direction relative to the initial frame, or would the Lorentz force be increased as a result of going in opposite directions relative to the final frame? Or... is there perhaps a totally different frame not given by the problem, which when accounted for, which will result in the correct answer? If the latter is correct, what would that frame be?

So??
 
  • #14
kmarinas86 said:
is there perhaps a totally different frame not given by the problem, which when accounted for, which will result in the correct answer? If the latter is correct, what would that frame be?
All frames will get the correct answer. The Lorentz force and Maxwell's equations are all Lorentz-symmetric.
 
  • #15
kmarinas86 said:
Correct.

What I was wondering about were situations such as the following:

Say I accelerate a lab. Then I turn on two beams in opposite directions from opposite ends of the lab. Then I accelerate the lab over time such that the laboratory changes by a speed greater than that of either of the beams as measured from the initial frame of the laboratory. Therefore, both beams are going in the same direction relative to the initial frame of the lab, but they go in opposite directions according to the lab's final frame. I would like to know what would effect this would have on the Lorentz force. Would the Lorentz force be decreased as a result of the beams going in the same direction relative to the initial frame, or would the Lorentz force be increased as a result of going in opposite directions relative to the final frame? Or... is there perhaps a totally different frame not given by the problem, which when accounted for, which will result in the correct answer? If the latter is correct, what would that frame be?

I'm not really sure what you're getting at here. The point I've been trying to make is that the Lorentz force does not depend at all on the reference frame. Your original post had a very sensible question: you indicated that the magnetic force can't possibly be the same in all frames of reference, and asked how that was compatible with the principles of relativity. I've tried to explicitly demonstrate how we get this consistency, which is down to the electric/magnetic splitting of the Lorentz force being observer dependent, but the total force still being independent.

I don't know how this last question adds to it: the fact that you're physically accelerating the lab to a speed v rather than just comparing the point of view of two observers at relative speed v has no material effect on the argument or conclusions.
 
  • #16
DaleSpam said:
All frames will get the correct answer. The Lorentz force and Maxwell's equations are all Lorentz-symmetric."

Let's recall the proportionality you gave:

[tex]F^2 \propto \rho^2 + I^2[/tex]

Where [itex]F[/itex], [itex]\rho[/itex], and [itex]I[/itex] are frame-invariant parameters.

While time-variant changes of [itex]F(t)[/itex] can be explained by changes of the relative velocity between the two currents over time, this does not account for the total value of [itex]F(t=given)[/itex]:

  • One may have two (-) beam currents which carry [itex]I_1=1A[/itex] and [itex]I_2=2A[/itex], respectively.
  • One may also have two (-) beam currents which carry [itex]I_3=2A[/itex] and [itex]I_4=3A[/itex], respectively.
  • The relative velocity between the first two currents can be the same as the relative velocity between the latter two currents, as measured from the center of momentum frame of each case, for any necessary compensation to allow this to be the case can be achieved by having different charge densities [itex]\rho_i[/itex] for each current.
  • However, because the linear force density integrated over the length observed in any inertial frame offered as given (i.e. force) is frame-invariant, it is also deduced that the total electromagnetic fields of 1A and 2A as measured from their combined COM frame are not the same intensity as the electromagnetic fields of 2A and 3A as measured from their combined COM frame. The deduction remains valid only if the electromagnetic field intensity vectors can by themselves account for the force vectors.
  • A current element observed to have a length [itex]l[/itex] in the 2A current (i.e. as viewed from the former case's COM frame) cannot look the same as a current element observed to have a length [itex]l[/itex] in the 3A current (i.e. as viewed from the latter case's COM frame). Furthermore, there must be differences in the observed charge densities not solely attributed to any the aforementioned relative velocities. Deductively, such differences cannot be totally accounted solely by the Lorentz contractions implied by these relative velocities.

Conclusion and afterthought: The time variation of the relative velocities of charges, currents, observers, etc. are enough to account for the changes in the frame-invariant value of the time-variant [itex]F(t)[/itex], but they are not enough to account for the total of the frame-invariant value of [itex]F(t=given)[/itex]. Whatever may account for the total value of [itex]F(t=given)[/itex] must go beyond the discussion of relative velocity. A physical, frame-invariant difference in the field intensity, whether weighted by distance or not, must exist between the two aforementioned example cases. The latter case would require more energy in its field than in the former. Whatever is attaining energy and carrying it along must have more than one degree of freedom. The collection of charges (even it is a pair of (-) charges) as a whole should be what carries this energy, not just the individual charges. Still, it is not right to say that you can explain the total energy with just velocities alone. With that, you can only explain a difference in the force. Thus, to think that SR can account for the whole force is a misunderstanding.
 
  • #17
I don't know what you are talking about.

http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

Electromagnetism can be written in a manifestly covariant form, including the Lorentz force.

Btw, the correct term is "covariant", not "invariant". The four-force is a vector, so it transforms as a rank-1 tensor, i.e. covariantly. Only rank 0 tensors (scalars) transform trivially, or in other words are invariant.
 
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  • #18
DaleSpam said:
I don't know what you are talking about.

http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

Electromagnetism can be written in a manifestly covariant form, including the Lorentz force.

Btw, the correct term is "covariant", not "invariant". The four-force is a vector, so it transforms as a rank-1 tensor, i.e. covariantly. Only rank 0 tensors (scalars) transform trivially, or in other words are invariant.

Remember what henry_m said?

henry_m said:
When we talk about invariant quantities (scalars), the invariance refers to frame/observer/coordinate independence.

When I mentioned the word "invariant" here, I was speaking of this "frame/observer/coordinate independence". Hence, I often used it in a hyphenated way, such as when I write "frame-invariant" or "time-invariant". "frame invariant" OR "invariant frame" OR "frame invariance" returns greater than 10x as many results than "frame covariant" OR "covariant frame" OR "frame covariance" and a greater part of them are not taken out of context from sentences like, "In the Cartesian frame, covariant and contravariant vectors are indistinguishable, so that it is natural to define covariant components for the vector by the equation B_sub_i = B_sup_i (37.2)" (See http://books.google.com/books?id=MSBB_ENAhT4C).

In support of your thread derail, I quote from http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors:

Wikipedia: Covariance and contravariance of vectors said:
In physics, vectors often have units of distance or distance times some other unit (such as the velocity), whereas covectors have units the inverse of distance or the inverse of distance times some other unit. The distinction between covariant and contravariant vectors is particularly important for computations with tensors, which often have mixed variance. This means that they have both covariant and contravariant components, or both vectors and dual vectors.

In looking in the article you linked to, I find this:

Wikipedia: Covariant formulation of classical electromagnetism said:
The four-current is the contravariant four-vector which combines electric current and electric charge density
[...]
The electromagnetic stress-energy tensor is a contravariant symmetric tensor which is the contribution of the electromagnetic fields to the overall stress-energy tensor
[...]
The bound current is derived from the magnetization and electric polarization which form an antisymmetric contravariant magnetization-polarization tensor

Going back to the article I brought up, I just found this:

Wikipedia: Covariance and contravariance of vectors said:
For a vector (such as a direction vector or velocity vector) to be coordinate system invariant, the components of the vector must contra-vary with a change of basis to compensate. That is, the components must vary in the opposite way (the inverse transformation) as the change of basis. Vectors (as opposed to dual vectors) are said to be contravariant.
[...]
For a dual vector, (such as a gradient) to be coordinate system invariant, the components of the vector must co-vary with a change of basis to maintain the same meaning. That is, the components must vary by the same transformation as the change of basis. Dual vectors (as opposed to vectors) are said to be covariant.

Per the quotes, the general property of invariance seems to be more fundamental than this covariance you mention.
 
  • #19
Covariant vectors and contravariant vectors are duals of each other and just differ in whether the indices are up or down. They can be switched from one to the other simply by multiplying by the metric. But vectors are not invariant. It doesn't matter how many times more often the word shows up in a search. Henry was careful to apply the word invariant to a scalar, not a vector.

I am not trying to derail the thread, just trying to help you communicate better.

And the main point remains: simply by writing the laws of EM in a manifestly covariant form you guarantee that if they are satisfied in one frame they are satisfied in all frames.
 
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  • #20
DaleSpam said:
Covariant vectors and contravariant vectors are duals of each other and just differ in whether the indices are up or down. They can be switched from one to the other simply by multiplying by the metric. But vectors are not invariant. It doesn't matter how many times more often the word shows up in a search. Henry was careful to apply the word invariant to a scalar, not a vector.

I am not trying to derail the thread, just trying to help you communicate better.

They're not "invariant"? Let's see what another article in Wikipedia has to say about that:

Wikipedia: Covariance and contravariance of vectors said:
For a vector (such as a direction vector or velocity vector) to be coordinate system invariant, the components of the vector must contra-vary with a change of basis to compensate.

So vectors can be "coordinate system invariant".

Covariant formulation of classical electromagnetism said:
The four-current is the contravariant four-vector which combines electric current and electric charge density.
[...]
The electromagnetic stress-energy tensor is a contravariant symmetric tensor containing the electric potential and magnetic vector potential
[...]
The bound current is derived from the magnetization and electric polarization which form an antisymmetric contravariant magnetization-polarization tensor
[...]
which determines the bound current
[...]
If this is combined with [itex]F^{\mu \nu}[/itex] we get the antisymmetric contravariant electromagnetic displacement tensor which combines the electric displacement and the H-field

If Joseph Henry were alive, he might tell you that the correct term for causes of the Lorentz force (i.e. the above properties) are "contravariant".

I also see the following:

[PLAIN]http://en.wikipedia.org/wiki/Lorentz_covariant said:
In[/PLAIN] [Broken] standard physics, Lorentz symmetry is "the feature of nature that says experimental results are independent of the orientation or the boost velocity of the laboratory through space".[1] Lorentz covariance is a related concept, covariance being a measure of how much two variables change together.

Lorentz covariance (from Hendrik Lorentz) is a key property of spacetime that follows from the special theory of relativity. Lorentz covariance has two distinct, but closely related meanings:

  • A physical quantity is said to be Lorentz covariant if it transforms under a given representation of the Lorentz group. According to the representation theory of the Lorentz group, these quantities are built out of scalars, four-vectors, four-tensors, and spinors. In particular, a scalar (e.g. the space-time interval) remains the same under Lorentz transformations and is said to be a Lorentz invariant (i.e. they transform under the trivial representation).
  • An equation is said to be Lorentz covariant if it can be written in terms of Lorentz covariant quantities (confusingly, some use the term invariant here). The key property of such equations is that if they hold in one inertial frame, then they hold in any inertial frame (this follows from the result that if all the components of a tensor vanish in one frame, they vanish in every frame). This condition is a requirement according to the principle of relativity, i.e. all non-gravitational laws must make the same predictions for identical experiments taking place at the same spacetime event in two different inertial frames of reference.

Note: this usage of the term covariant should not be confused with the related concept of a covariant vector. On manifolds, the words covariant and contravariant refer to how objects transform under general coordinate transformations. Confusingly, both covariant and contravariant four-vectors can be Lorentz covariant quantities.

The last bit is priceless.

And people wonder why science education is going downhill. It's much less costly to teach students up to B.S. level but not through it, especially when such obviously terrible nomenclature is being maintained at graduate levels as though it were scientifically better to leave it that way. With a little bit more effort, flawed nomenclature can be abolished just like how Pluto's "status as a planet" was "abolished".
 
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  • #21
So do you now understand that the four force is a covariant vector not an invariant scalar?
 
  • #22
DaleSpam said:
So do you now understand that the four force is a covariant vector not an invariant scalar?

We might not have been speaking of the four force. henry_m stated:

henry_m said:
Now here's the critical paragraph, and this is the kind of argument that will make relativity easy. Notice that the force is in fact the Lorentz inner product of the current four-vectors, which must be scalar. This generalises to any parallel charge and current distributions (current-carrying wires and beams of particles). If you can notice this from the very beginning then you don't have to do any work at all! Though it is nice to see how the force splits into electric and magnetic components in different frames.
 
  • #23
You will have to take that up with Henry. I was talking about the covariant formulation of the Lorentz force law that I linked to earlier.
 
  • #24
Can we at least agree that some charge does not have any [itex]I[/itex] at all, while others will? In other words, I am requesting verification that not all charge of quantity [itex]q[/itex] will produce the same invariant parameter [itex]I[/itex].

Isn't the state of being "current" due to applying a proper acceleration on a charge? A charge that has received a proper acceleration should be different in some way, or otherwise it may in no way explain the invariant parameter [itex]I[/itex]. What is the invariant parameter [itex]I[/itex] of a charge if charges do not share the same one?

You would think that something else besides the Lorentz force would increase together with it, wouldn't you?

As far I know from what is being taught, the electron won't change its rest mass because it lacks degrees of freedom. So while its Lorentz force for a charge must be changed after an event by the same amount in every frame of reference, there is no accepting that electron may gain rest mass.

So what about the electron changes together with the Lorentz force it generates? As henry_m stated, the Lorentz force is

henry_m said:
thankfully, independent of [itex]v[/itex].

So whatever the value of the Lorentz force is, it must be dependent on some variable that is independent of the [itex]v[/itex] of the Lorentz boost frame. However, then this rules out relativistic kinetic energy, because that is relative to the observer, according to what is taught anyway, which is akin to saying that there are some frames in which an electron is seen to have less energy after being driven in a particle accelerator to 0.99 c relative to the operator.

While individual electrons might not have internal degrees of freedom, electron beams certainly do. This can only mean that when electrons are accelerated in groups, they absorb energy. Then proper acceleration would have the effect of adding energy on a per electron basis, and at the same time changing the sum [itex]\rho^2+I^2[/itex] rather than simply converting [itex]\rho[/itex] into [itex]I^2[/itex]. So for electron beams it could be said that this ability to stored energy in some way accounts for the change of its [itex]\rho^2+I^2[/itex] (i.e. the energy change accounts for the change of the Lorentz force). This leaves an irriating inconsistency remaining for electrons having no internal degrees of freedom, for there may exist no energy change of the electron, as it is unable to absorb energy.
 
  • #25
kmarinas86 said:
Can we at least agree that some charge does not have any [itex]I[/itex] at all, while others will?
Certainly. If a charge is stationary it will not have any current, if it is moving it will.

kmarinas86 said:
Isn't the state of being "current" due to applying a proper acceleration on a charge?
No, a charge moving inertially will have a current despite not having any proper acceleration.

kmarinas86 said:
A charge that has received a proper acceleration should be different in some way, or otherwise it may in no way explain the invariant parameter [itex]I[/itex].
[itex]I[/itex] is neither invariant nor covariant. You may want to read up on the four-current which is composed of the charge density and the current density: http://en.wikipedia.org/wiki/Four-current

You may also want to read about the covariant formulation of classical electromagnetism: http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

And the Lorentz force law: http://en.wikipedia.org/wiki/Lorentz_force

I don't know what henry_m was saying and he seems to be non-responsive.
 
  • #26
DaleSpam said:
Certainly. If a charge is stationary it will not have any current, if it is moving it will.

No, a charge moving inertially will have a current despite not having any proper acceleration.

[itex]I[/itex] is neither invariant nor covariant. You may want to read up on the four-current which is composed of the charge density and the current density: http://en.wikipedia.org/wiki/Four-current

You may also want to read about the covariant formulation of classical electromagnetism: http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

And the Lorentz force law: http://en.wikipedia.org/wiki/Lorentz_force

I don't know what henry_m was saying and he seems to be non-responsive.

At this point, it would seem that the only way I can make sense of the magnetic field is that it doesn't belong to anyone charge but instead is something that exists between charges as an independent entity. In other words, charge motion through "space" doesn't create the magnetic field, but rather, a magnetic field is something that exists as a result of the transformation of connections between charged particles—connections too small to be directly observed.
 
  • #27
kmarinas86 said:
At this point, it would seem that the only way I can make sense of the magnetic field is that it doesn't belong to anyone charge but instead is something that exists between charges as an independent entity.
Yes, I would not say that a field "belongs" to a charge. However, I am not sure what you are implying by the "between charges" comment. It doesn't take multiple charges to produce a magnetic field but if there are multiple charges in some set up then in the spaces between charges there would be fields.

Have you read the links? Can you ask some specific questions?

kmarinas86 said:
In other words, charge motion through "space" doesn't create the magnetic field, but rather, a magnetic field is something that exists as a result of the transformation of connections between charged particles—connections too small to be directly observed.
Magnetic fields can certainly be observed, and they certainly are created by a charge moving through space.
 
  • #28
DaleSpam said:
Yes, I would not say that a field "belongs" to a charge. However, I am not sure what you are implying by the "between charges" comment. It doesn't take multiple charges to produce a magnetic field but if there are multiple charges in some set up then in the spaces between charges there would be fields.

I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges. That might explain its "relative" nature.

DaleSpam said:
Have you read the links? Can you ask some specific questions?

I have read some of what they say, but I do not have any specific questions about them at this time.

DaleSpam said:
Magnetic fields can certainly be observed, and they certainly are created by a charge moving through space.

But is the moving "through space" itself responsible for the generation of this field? Saying that it is responsible for that appears to imply that there is some space to move through, by which a field may or may not produced, regardless of the observer. Movement "through space" sounds like a statement that belongs to discussions about an ether frame. My previous comments were not suggesting such an ether frame, but rather, a dynamical field the exists between charges without reference to any notion of a "rest" or "space" frame. So, the idea is that differences in the time-variations of coupling between a sender and multiple receivers explains why different observers may see a different magnetic field.

In a sense, this answers my question:

"Are the directions of charge flow (=current?) absolute?"

The answer being: No. They are relational, not absolute.
 
  • #29
kmarinas86 said:
I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.
The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.


kmarinas86 said:
But is the moving "through space" itself responsible for the generation of this field?
Yes, according to Maxwell's equations, see: http://en.wikipedia.org/wiki/Liénard–Wiechert_potential

kmarinas86 said:
Saying that it is responsible for that appears to imply that there is some space to move through, by which a field may or may not produced, regardless of the observer. Movement "through space" sounds like a statement that belongs to discussions about an ether frame. My previous comments were not suggesting such an ether frame
Nor were my comments suggesting an ether frame. "Moving through space" simply refers to velocity, which is obviously relative to the given reference frame. I don't know how you jumped from a simple statement about velocity to the aether.

kmarinas86 said:
, but rather, a dynamical field the exists between charges without reference to any notion of a "rest" or "space" frame.
Sure, but that field is the entire electromagnetic field tensor. It is NOT the magnetic field. The magnetic field is just a component of this tensor, so it only exists with reference to a specific frame.

kmarinas86 said:
In a sense, this answers my question:

"Are the directions of charge flow (=current?) absolute?"

The answer being: No. They are relational, not absolute.
Definitely.
 
  • #30
DaleSpam said:
kmarinas86 said:
I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.
The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.

I'm not sure if a charge can accelerate in isolation. If it is not even a physical possibility, this may as well be a non-issue.
 
  • #31
kmarinas86 said:
I'm not sure if a charge can accelerate in isolation. If it is not even a physical possibility, this may as well be a non-issue.
There are physically possible scenarios. E.g. A single charge accelerating due to a EM wave. In fact, in QM all EM interactions are built up from a series of this basic interaction.

In any case, did you understand my point above about the magnetic field being a component of the EM field tensor?
 
  • #32
DaleSpam said:
kmarinas86 said:
DaleSpam said:
kmarinas86 said:
I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.
The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.

I'm not sure if a charge can accelerate in isolation. If it is not even a physical possibility, this may as well be a non-issue.
There are physically possible scenarios. E.g. A single charge accelerating due to a EM wave. In fact, in QM all EM interactions are built up from a series of this basic interaction.

The keyword was isolation.

Also, I never saw a way to interact with EM waves that didn't involve a mass on the receiving end. This mass invariably has internal charges, which is also true even if the receiver is a neutron. Is there a way to take a measurement of a magnetic field [err... I mean an EM field] by a means that doesn't involve a mass? I don't think so.

DaleSpam said:
In any case, did you understand my point above about the magnetic field being a component of the EM field tensor?

Yes I did.
 
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  • #33
kmarinas86 said:
The keyword was isolation.
That was your word, not mine. I just said a single charge, I never said that it was isolated, just that there wasn't another charge.
 
  • #34
DaleSpam said:
That was your word, not mine. I just said a single charge, I never said that it was isolated, just that there wasn't another charge.

How is there not going to be another charge when we observe a charge?

I said:

kmarinas86 said:
I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.

You replied:

DaleSpam said:
The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.
 
  • #35
kmarinas86 said:
How is there not going to be another charge when we observe a charge?

I said:
You replied:
Yes, I agreed with that first sentence. Did you not see that it was only the second sentence I was objecting to?
 
<h2>1. What is charge flow and how is it related to current?</h2><p>Charge flow refers to the movement of electric charge through a conductor. This movement of charge is what creates an electric current, which is measured in units of amperes (A).</p><h2>2. Is the direction of charge flow always the same as the direction of current?</h2><p>No, the direction of charge flow and the direction of current can be different. This is because electric current is defined as the flow of positive charge, while in reality, negatively charged electrons are the ones that move through a conductor.</p><h2>3. Can the direction of current change?</h2><p>Yes, the direction of current can change depending on the direction of the electric field or the direction of the moving charges. In a direct current (DC) circuit, the direction of current remains constant, while in an alternating current (AC) circuit, the direction of current changes periodically.</p><h2>4. Is the direction of charge flow absolute?</h2><p>No, the direction of charge flow is not absolute. It can vary depending on the reference point or frame of reference. For example, the direction of current may appear to be different to an observer at rest compared to an observer in motion.</p><h2>5. How is the direction of current determined?</h2><p>The direction of current is determined by the direction of positive charge flow. In most cases, this is the direction of conventional current, which is the direction that positive charge would flow if it were able to. However, in some cases, such as in semiconductors, the direction of current may be opposite to the direction of positive charge flow.</p>

1. What is charge flow and how is it related to current?

Charge flow refers to the movement of electric charge through a conductor. This movement of charge is what creates an electric current, which is measured in units of amperes (A).

2. Is the direction of charge flow always the same as the direction of current?

No, the direction of charge flow and the direction of current can be different. This is because electric current is defined as the flow of positive charge, while in reality, negatively charged electrons are the ones that move through a conductor.

3. Can the direction of current change?

Yes, the direction of current can change depending on the direction of the electric field or the direction of the moving charges. In a direct current (DC) circuit, the direction of current remains constant, while in an alternating current (AC) circuit, the direction of current changes periodically.

4. Is the direction of charge flow absolute?

No, the direction of charge flow is not absolute. It can vary depending on the reference point or frame of reference. For example, the direction of current may appear to be different to an observer at rest compared to an observer in motion.

5. How is the direction of current determined?

The direction of current is determined by the direction of positive charge flow. In most cases, this is the direction of conventional current, which is the direction that positive charge would flow if it were able to. However, in some cases, such as in semiconductors, the direction of current may be opposite to the direction of positive charge flow.

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