E-field of a constant velocity charged particle

[aliinuur]

TL;DR Summary

With assumption that information about position of the particle travels at speed of light i've got following e-fields. Where am i wrong? In the figure a particle is traveling at half the speed of light.

 

[Dale]

When drawing figures it is important to be clear what the various lines mean. I have no idea what those lines are supposed to represent because nothing is labeled.

The field of a generally moving point particle is given here:

https://en.m.wikipedia.org/wiki/Liénard–Wiechert_potential

 

[aliinuur]

Lines are E-fields, black dots are trace of the particle from bottom to top.

 

[Vanadium 50]

And what does the frisbee represent?

 

[Ibix]

Lines are E-fields, black dots are trace of the particle from bottom to top.

The easy way to do this is to write down the E-field in the particle rest frame, plug it in to the Faraday tensor and Lorentz transform (remembering to transform the coordinates).

 

[Dale]

Lines are E-fields, black dots are trace of the particle from bottom to top.

Then it is clearly wrong. You have field lines intersecting where there is no charged particle

 

[renormalize]

What does your graph tell us that's not conveyed by the standard depiction of the electric field of a moving charge?

(from J.D. Jackson, Classical Electrodynamics, 2nd ed., pg. 555)

 

[Ibix]

It would help, @aliinuur, if you told us exactly what you've plotted. Is it a 3d plot? What quantity (field strength, one field component,...?) did you plot. How did you calculate the value?

Note that you can use $\LaTeX$ here, delimited by double # marks for inline and double $ signs for paragraph maths. See the LaTeX Guide linked below the reply box for more details.

 

[aliinuur]

It would help, @aliinuur, if you told us exactly what you've plotted. Is it a 3d plot? What quantity (field strength, one field component,...?) did you plot. How did you calculate the value?

Note that you can use $\LaTeX$ here, delimited by double # marks for inline and double $ signs for paragraph maths. See the LaTeX Guide linked below the reply box for more details.

it is 2D plot, black dots are the particle trace from bottom to top. Lines are the electric field lines

 

[Ibix]

it is 2D plot, black dots are the particle trace from bottom to top. Lines are the electric field lines

Electric field lines should start on the charge; some of yours are closed loops. What are they?

 

[Dale]

it is 2D plot, black dots are the particle trace from bottom to top. Lines are the electric field lines

So it is clearly wrong for the reason I said above.

 

[aliinuur]

Electric field lines should start on the charge; some of yours are closed loops. What are they?

i used circles to draw electric fields, no need to mind them

 

[Ibix]

i used circles to draw electric fields, no need to mind them

I see.

Those field lines are wrong - compare the diagram @renormalize provided in post #7. How did you calculate them? Your OP is very vague.

 

[aliinuur]

I see.

Those field lines are wrong - compare the diagram @renormalize provided in post #7. How did you calculate them? Your OP is very vague.

that figure is based on that every time the particle changes its positon information aboout it travels with the speed of light like waves on a puddle, thats how i got curved lines. It feels for me that straight lines mean infinite information travel velocity

 

[Vanadium 50]

So your figure superposes various time slices on the same graph?

@Dale and @Ibex have made a very reasonable request - that you label your graph. I don't understand why you don't want to do this, and more importantly, you don't want to make it easy for the people who are trying to help you to figure out what you are saying.

 

[Dale]

It feels for me that straight lines mean infinite information travel velocity

How would you propose to send any information via the straight lines from the E field of an inertially moving charge? They are straight. They do not send information faster than c

 

[PeterDonis]

information aboout it

So your graph is not a graph of "electric field". It's a graph of "information about the position of the particle". Which just raises the question: what do you mean by "information about the position of the particle"? Math would help here.

 

[Dale]

Labels and a good description would be good. So far the labels are nonexistent and the descriptions are vague

 

[aliinuur]

How would you propose to send any information via the straight lines from the E field of an inertially moving charge? They are straight. They do not send information faster than c

If E-field lines are straight it means even at infinited distance space derivative of E-field points in direction of the particle. Curved lines however point where the particle was in distance/c ago.
I guess the solution is that you can know where the particle in real time unless its not acceleration.

 

[Ibix]

If E-field lines are straight it means even at infinited distance space derivative of E-field points in direction of the particle. Curved lines however point where the particle was in distance/c ago.

So, let's say the particle is moving along the y axis at velocity v, so $x=0$, $y=vt$. You want to draw the field at $t=0$, so you drew a circle of radius $1c$ centered on $(x,y)=(0,-v)$ with vectors all around it pointing at its center. Then you drew a circle of radius $2c$ centered on $(0,-2v)$, with vectors all around it pointing at its center. And so on. And then you tried to draw the integral curves of this field?

No, that won't work. There is also a time varying magnetic field in the frame where the charge is moving which is not present when the charge is stationary. That makes the E field different from some stitched-together slices of a Coulomb field.

I guess the solution is that you can know where the particle in real time unless its not acceleration.

Well, you don't know that it hasn't accelerated. I wouldn't say that the field points at where the particle is now. Rather, it points at the forecast of where the charge should be now if it maintained its inertial motion in the time since our delayed image of it. The forecast happens to be correct in this case.

 

[cianfa72]

No, that won't work. There is also a time varying magnetic field in the frame where the charge is moving which is not present when the charge is stationary. That makes the E field different from some stitched-together slices of a Coulomb field.

You mean that in the frame/coordinate chart where the source charge is moving there is an Electric field and a magnetic field (due to the motion of the charge in that frame) as well. One can calculate both Electric and magnetic fields in that frame using Liénard–Wiechert potential.

 

[aliinuur]

So, let's say the particle is moving along the y axis at velocity v, so $x=0$, $y=vt$. You want to draw the field at $t=0$, so you drew a circle of radius $1c$ centered on $(x,y)=(0,-v)$ with vectors all around it pointing at its center. Then you drew a circle of radius $2c$ centered on $(0,-2v)$, with vectors all around it pointing at its center. And so on. And then you tried to draw the integral curves of this field?

No, that won't work. There is also a time varying magnetic field in the frame where the charge is moving which is not present when the charge is stationary. That makes the E field different from some stitched-together slices of a Coulomb field.

Well, you don't know that it hasn't accelerated. I wouldn't say that the field points at where the particle is now. Rather, it points at the forecast of where the charge should be now if it maintained its inertial motion in the time since our delayed image of it. The forecast happens to be correct in this case.

First I draw a circle with radius c*5t and center at the origin, then a circle with radius c*4t pointing at (0,vt), then with radius c*3t pointing at (0, v*2t) and so on. The point is circle with radius c*t points where the particle was t ago.

 

[aliinuur]

So, let's say the particle is moving along the y axis at velocity v, so $x=0$, $y=vt$. You want to draw the field at $t=0$, so you drew a circle of radius $1c$ centered on $(x,y)=(0,-v)$ with vectors all around it pointing at its center. Then you drew a circle of radius $2c$ centered on $(0,-2v)$, with vectors all around it pointing at its center. And so on. And then you tried to draw the integral curves of this field?

No, that won't work. There is also a time varying magnetic field in the frame where the charge is moving which is not present when the charge is stationary. That makes the E field different from some stitched-together slices of a Coulomb field.

Well, you don't know that it hasn't accelerated. I wouldn't say that the field points at where the particle is now. Rather, it points at the forecast of where the charge should be now if it maintained its inertial motion in the time since our delayed image of it. The forecast happens to be correct in this case.

Magnetic field is actually electric field stressed along one of the 3 dimensions, there are no magnetic fields in special relativity.

 

[Ibix]

First I draw a circle with radius c*5t and center at the origin, then a circle with radius c*4t pointing at (0,vt), then with radius c*3t pointing at (0, v*2t) and so on. The point is circle with radius c*t points where the particle was t ago.

That's what I said, just with a different origin. Fine. And you have electric field vectors on those circles pointing at the center of each one?

Magnetic field is actually electric field stressed along one of the 3 dimensions, there are no magnetic fields in special relativity.

This is not correct. $B^2-E^2$ is an invariant, so if this is positive (i.e. $|\vec B|>|\vec E|$) you can never make $\vec B$ zero by changing frames. In fact, although there is a certain amount of observer dependence in electric and magnetic fields, all six components of the two vectors (or the six independent components of the Faraday tensor, equivalently) are needed to describe an EM field. If you don't believe me, start in a frame where there is a uniform magnetic field pointing in the z direction and zero E field and try to find a frame in which this is purely an electric field.

Independent of that, if you want to work in a frame where a magnetic field is present then you have to deal with the magnetic field. You may sometimes be able to find a frame in which the magnetic field is zero (e.g. the rest frame of the particle in this case), but you are explicitly not doing that.

 

[aliinuur]

That's what I said, just with a different origin. Fine. And you have electric field vectors on those circles pointing at the center of each one?

This is not correct. $B^2-E^2$ is an invariant, so if this is positive (i.e. $|\vec B|>|\vec E|$) you can never make $\vec B$ zero by changing frames. In fact, although there is a certain amount of observer dependence in electric and magnetic fields, all six components of the two vectors (or the six independent components of the Faraday tensor, equivalently) are needed to describe an EM field. If you don't believe me, start in a frame where there is a uniform magnetic field pointing in the z direction and zero E field and try to find a frame in which this is purely an electric field.

Independent of that, if you want to work in a frame where a magnetic field is present then you have to deal with the magnetic field. You may sometimes be able to find a frame in which the magnetic field is zero (e.g. the rest frame of the particle in this case), but you are explicitly not doing that.

I dont get how one possibly get magnetic field with zero E-field. For me, this is like gravitaional field without a mass, or electric field without a charged particle

 

[Ibix]

I dont get how one possibly get magnetic field with zero E-field.

The field around a straight current carrying conductor is a magnetic field with no electric field.

For me, this is like gravitaional field without a mass, or electric field without a charged particle

The standard response to that is to point out that physics doesn't really care how unintuitive something seems to you. It is the way it is.

 

[aliinuur]

The field around a straight current carrying conductor is a magnetic field with no electric field.

The standard response to that is to point out that physics doesn't really care how unintuitive something seems to you. It is the way it is.

there is electric field though.

 

[Ibix]

Inside the wire? Sure. Why is this relevant to either your original question or your mistaken belief that there is no such thing as a magnetic field?

 

[aliinuur]

Inside the wire? Sure. Why is this relevant to either your original question or your mistaken belief that there is no such thing as a magnetic field?

yeah, how is magnetic field relevant to my question? you have started talking about magnetic field

 

[aliinuur]

What does your graph tell us that's not conveyed by the standard depiction of the electric field of a moving charge?
View attachment 346647
(from J.D. Jackson, Classical Electrodynamics, 2nd ed., pg. 555)

yeah, you got it right

 

[cianfa72]

The field around a straight current carrying conductor is a magnetic field with no electric field.

I've a doubt here. In the model of straight conductor carrying current (a flow of electrons moving with constant velocity) is the net charge inside each volume/segment of the conductor supposed to be zero (due to the presence of positive metal ions) ?

 

[aliinuur]

I've a doubt here. In the model of straight conductor carrying current (a flow of electrons moving with constant velocity) is the net charge inside each volume/segment of the conductor supposed to be zero (due to the presence of positive metal ions) ?

you are right, even the hydrogen atom is a dipole, i.e. has a electric field

 

[renormalize]

you are right, even the hydrogen atom is a dipole, i.e. has a electric field

Since the 1s ground state of the hydrogen atom is spherically-symmetric, can you tell us in which direction this electric field points?

 

[aliinuur]

Since the 1s ground state of the hydrogen atom is spherically-symmetric, can you tell us in which direction this electric field points?

its dipole moment has equal chance of pointing in any direction :)

 

[PeterDonis]

Thread closed for moderation.