Charge moving with a constant linear velocity....

[Maciej Orman]

Charge moving in constant linear velocity does not produce magnetic field...
If not, please provide an explanation...

 

[Orodruin]

Charge moving in constant linear velocity does not produce magnetic field...

Yes it does.

If not, please provide an explanation...

Your question is unclear. You labeled this thread "A". Which part of Maxwell's equations do you not understand?

 

[vanhees71]

The most simple way to get the electromagnetic field (which of course has non-zero electric and magnetic components) is to use Lorentz invariance. Just solve Maxwell's equations for the charge at rest (leading of course to a Coulomb field) and then do a Lorentz boost. You'll get both electric and magnetic field components! Of course, it's easier to first use the four-potential and then take the derivatives to get the field-strength tensor.

 

[Maciej Orman]

OK, so if I place a magnetometer probe next to the running belt fo Van De Graff generator I should detect magnetic field, or should I? Also if constant linear velocity of a charge generate magnetic field then it means that absolute motion can be measured and the concept of aether is reality... Still one needs experimental confirmation but I cannot find any...

 

[vanhees71]

Of course not! There's no aether. Contrary to the Newtonian spacetime model special relativity (i.e., the Minkowskian spacetime model) precisely makes the description of electromagnetism possible without assuming a preferred reference frame (aka. "restframe of the aether" from the ancient days before Einstein).

Of course if there's a moving charge in a reference frame, the observer at rest wrt. this frame will observe both electric and magnetic field components. In a frame, where the charge stays at rest, he/she will observe only electric field components. The electromagnetic field is of course invariant, i.e., frame independent (it's an antisymmetric tensor field, the Faraday tensor).

 

[Maciej Orman]

Of course not! There's no aether. Contrary to the Newtonian spacetime model special relativity (i.e., the Minkowskian spacetime model) precisely makes the description of electromagnetism possible without assuming a preferred reference frame (aka. "restframe of the aether" from the ancient days before Einstein).

Of course if there's a moving charge in a reference frame, the observer at rest wrt. this frame will observe both electric and magnetic field components. In a frame, where the charge stays at rest, he/she will observe only electric field components. The electromagnetic field is of course invariant, i.e., frame independent (it's an antisymmetric tensor field, the Faraday tensor).

Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?

 

[Maciej Orman]

Just like to point that an abstract quantities like kinetic energy can change wrt to an observer but magnetic field is a physical quantity and does not change due to relative motion of an observer thus correct logic deduction suggests that constant linear velocity of charge particle does not produce magnetic field...
Also absence of magnetic field around charged particle would indicate an absolute stop motion... Feather more all objects with excess charge located on the Earth
would generated their own magnetic field due to the orbital motion of around 500 m/s...

 

[Ibix]

Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?

No. Whether a charge has a magnetic field or not is a frame-dependent question. If it has a magnetic field then it is in motion in that frame. If it does not have a magnetic field it is not in motion in that frame.

It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.

 

[Ibix]

magnetic field is a physical quantity and does not change

This is incorrect, as I wrote above.

 

[Maciej Orman]

No. Whether a charge has a magnetic field or not is a frame-dependent question. If it has a magnetic field then it is in motion in that frame. If it does not have a magnetic field it is not in motion in that frame.

It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.

Laws of physics must hold across all frames of motion and cannot be frame dependent...

 

[Maciej Orman]

This is incorrect, as I wrote above.

It would mean that charged particle in motion produces magnetic field which intensity is observer dependent thus completely hiding magnetic field for moving observer and showing magnetic field for non moving observer... That would be totally illogical...

 

[Ibix]

Laws of physics must hold across all frames of motion and cannot be frame dependent...

True. Whether or not a charge possesses a magnetic component to its electromagnetic field is not a law of physics. The relevant laws of physics are Maxwell's equations, which have the same form in all frames.

It would mean that charged particle in motion produces magnetic field which intensity is observer dependent thus completely hiding magnetic field for moving observer and showing magnetic field for non moving observer... That would be totally illogical...

If you think that then you need to revise your notion of what constitutes "logical". It's trivial to show that it happens by carrying out a Lorentz transform on the electromagnetic field strength tensor.

 

[Maciej Orman]

True. Whether or not a charge possesses a magnetic component to its electromagnetic field is not a law of physics. The relevant laws of physics are Maxwell's equations, which have the same form in all frames.
If you think that then you need to revise your notion of what constitutes "logical". It's trivial to show that it happens by carrying out a Lorentz transform on the electromagnetic field strength tensor.

Logic also suggests that mathematical models have no effect on physics of reality thus incorrect models will not be supported by experimental evidence...
In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field then that will prove that your Lorentz transform correctly represents mathematical model of the phenomena in question...

 

[Ibix]

In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field

A straight wire carrying a constant current.

 

[vanhees71]

Indeed the electromagnetic field,
$$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
is a tensor field and thus invariant under Lorentz transformations. There's no way to infer an absolute coordinate frame, i.e., there's no aether in special relativity. That's how special relativity has been discovered by Lorentz, Fitzgerald, Poincare, and finally Einstein.

 

[Maciej Orman]

Indeed the electromagnetic field,
$$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
is a tensor field and thus invariant under Lorentz transformations. There's no way to infer an absolute coordinate frame, i.e., there's no aether in special relativity. That's how special relativity has been discovered by Lorentz, Fitzgerald, Poincare, and finally Einstein.

OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all? Also the above model should be solved for magnetic field not a force...

 

[Ibix]

OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all?

A straight wire carrying a constant current.

 

[Maciej Orman]

A straight wire carrying a constant current.

There is not a single electron moving in constant linear velocity in a wire carrying constant current...

 

[vanhees71]

Now let's finally do the calculation, which hasn't occurred in this thread. The most simple way is to just do a Lorentz boost of the four-potential.

So let ##x^{\mu}## inertial coordinates, where the charge is at rest sitting in the origin of the spatial frame. Then, in (1+3) notation we have simply a Coulomb field, which is described in Lorenz gauge by the potentials
$$A^0(x)=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}(x)=0. \qquad (1)$$
The four-potential in Lorenz gauge transforms as a four-vector, i.e.,
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\rho} A^{\rho}(x).$$
To make our life easier, instead of doing the combersome transformation we rather write everything in a covariant way. The situation is described in a covariant way by the charge ##q## of the particle (which is a scalar) and the particles' constant velocity, which is described covariantly by the four-velocity ##(u^{\mu})=(\gamma,\gamma \vec{v})##, where ##\gamma=1/\sqrt{1-\vec{v}^2}## (I set ##c=1## for convenience).

Thus, the building blocks for our field, given in the particles's restframe, where ##(u^{\mu})=(1,0,0,0)##, are only ##x^{\mu}## and ##u^{\mu}##. It's clear that ##A^{\mu} \propto u^{\mu}## and finally we need ##\vec{x}^2##. It's easy to project out the time component from the four vector ##x^{\mu}##:
$$(x^{\mu}-u^{\mu} u \cdot x)=(0,\vec{x})$$
and thus
$$-\vec{x}^2=[x-u(u \cdot x)]^2=x^2-(u \cdot x)^2.$$
So finally we have
$$A^{\mu}=\frac{q}{\sqrt{(u \cdot x)^2-x^2}} u^{\mu}.$$
In the reference frame where the particle moves with three-velocity ##v## we have
$$u^{\prime \mu}=\gamma (1,\vec{v})$$
and
$$(u \cdot x)^2-x^2=\gamma^2 (t'-\vec{v} \cdot \vec{x})^2-t^{\prime 2}+\vec{x}^{\prime 2}.$$
With this you get the field components in the primed frame by taking the usual differential operators,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$

 

[vanhees71]

OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all? Also the above model should be solved for magnetic field not a force...

Hm, that's a good question. I'm not aware of an experiment which has ever measured the electromagnetic field of a single charged body moving with constant velocity. Of course the measurable macroscopic fields used in everyday electrical engineering all follow Maxwell's equations, and thus we can be pretty sure that everything works also for this special case.

 

[Maciej Orman]

Hm, that's a good question. I'm not aware of an experiment which has ever measured the electromagnetic field of a single charged body moving with constant velocity. Of course the measurable macroscopic fields used in everyday electrical engineering all follow Maxwell's equations, and thus we can be pretty sure that everything works also for this special case.

A belt of Van De Graaff generator carries excess charge with constant linear velocity but I have never seen anybody detecting magnetic field around it...

 

[Maciej Orman]

A belt of Van De Graaff generator carries excess charge with constant linear velocity but I have never seen anybody detecting magnetic field around it...

Rowland disk experiment proves that accelerated static charge on a disk creates magnetic field...

 

[vanhees71]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

 

[Maciej Orman]

Now let's finally do the calculation, which hasn't occurred in this thread. The most simple way is to just do a Lorentz boost of the four-potential.

So let ##x^{\mu}## inertial coordinates, where the charge is at rest sitting in the origin of the spatial frame. Then, in (1+3) notation we have simply a Coulomb field, which is described in Lorenz gauge by the potentials
$$A^0(x)=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}(x)=0. \qquad (1)$$
The four-potential in Lorenz gauge transforms as a four-vector, i.e.,
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\rho} A^{\rho}(x).$$
To make our life easier, instead of doing the combersome transformation we rather write everything in a covariant way. The situation is described in a covariant way by the charge ##q## of the particle (which is a scalar) and the particles' constant velocity, which is described covariantly by the four-velocity ##(u^{\mu})=(\gamma,\gamma \vec{v})##, where ##\gamma=1/\sqrt{1-\vec{v}^2}## (I set ##c=1## for convenience).

Thus, the building blocks for our field, given in the particles's restframe, where ##(u^{\mu})=(1,0,0,0)##, are only ##x^{\mu}## and ##u^{\mu}##. It's clear that ##A^{\mu} \propto u^{\mu}## and finally we need ##\vec{x}^2##. It's easy to project out the time component from the four vector ##x^{\mu}##:
$$(x^{\mu}-u^{\mu} u \cdot x)=(0,\vec{x})$$
and thus
$$-\vec{x}^2=[x-u(u \cdot x)]^2=x^2-(u \cdot x)^2.$$
So finally we have
$$A^{\mu}=\frac{q}{\sqrt{(u \cdot x)^2-x^2}} u^{\mu}.$$
In the reference frame where the particle moves with three-velocity ##v## we have
$$u^{\prime \mu}=\gamma (1,\vec{v})$$
and
$$(u \cdot x)^2-x^2=\gamma^2 (t'-\vec{v} \cdot \vec{x})^2-t^{\prime 2}+\vec{x}^{\prime 2}.$$
With this you get the field components in the primed frame by taking the usual differential operators,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$

Yes, but I need simple equation shown magnetic field intensity at some point near the charge and on the right factor of velocity times factor of charge magnitude thus showing that magnetic field is proportional to charge and velocity...

 

[Maciej Orman]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

Yes, but this is not an excess electric charge moving in constant linear velocity...

 

[Maciej Orman]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

It was Rowland who first shown this effect to Hertz and all that is charge accelerated in circular motion produces magnetic field...

 

[Ibix]

A belt of Van De Graaff generator carries excess charge with constant linear velocity but I have never seen anybody detecting magnetic field around it...

Have an infinite flat sheet of charge lying in the x-z plane. The electric field is purely in the y-direction and is given by ##E_y=\sigma/2\epsilon_0## where ##\sigma## is the charge density. That means ##F^{ij}=0## for all i and j except ##F^{02}=-F^{20}=\sigma/2c\epsilon_0##. Lorentz transforming to a frame where the sheet is moving at speed ##v## gives a magnetic field in the z direction of ##B_z=\sigma v/2c^2\epsilon_0##. (Beware sign errors in that.)

I can't find an authoritative source for the charge density on the belt of a van de Graaff generator. http://www.chegg.com/homework-help/questions-and-answers/belt-van-de-graaff-generator-carries-surface-charge-density-7-00-mc-m-2-belt-0500-m-wide-m-q7847099 citing 7mC/m², and this one (also a homework question) gives 65μC/m². Taking the higher figure and its cited belt speed of 21m/s gives a field of around 9x10^-8T - about 0.2% the strength of the Earth's magnetic field. I'm sure it's possible to detect that, but it's not going to be done casually.

 

[Maciej Orman]

Have an infinite flat sheet of charge lying in the x-z plane. The electric field is purely in the y-direction and is given by ##E_y=\sigma/2\epsilon_0## where ##\sigma## is the charge density. That means ##F^{ij}=0## for all i and j except ##F^{02}=-F^{20}=\sigma/2c\epsilon_0##. Lorentz transforming to a frame where the sheet is moving at speed ##v## gives a magnetic field in the z direction of ##B_z=\sigma v/2c^2\epsilon_0##. (Beware sign errors in that.)

I can't find an authoritative source for the charge density on the belt of a van de Graaff generator. http://www.chegg.com/homework-help/questions-and-answers/belt-van-de-graaff-generator-carries-surface-charge-density-7-00-mc-m-2-belt-0500-m-wide-m-q7847099 citing 7mC/m², and this one (also a homework question) gives 65μC/m². Taking the higher figure and its cited belt speed of 21m/s gives a field of around 9x10^-8T - about 0.2% the strength of the Earth's magnetic field. I'm sure it's possible to detect that, but it's not going to be done casually.

Using the above equation for charge of 1C and speed of 1m/s I calculated field of 5.55556 E-18 so for non relativistic speeds large charge o 1C has no detectable magnetic field... Thank you it is the answer I have been looking for...

 

[Ibix]

Using the above equation for charge of 1C and speed of 1m/s I calculated field of 5.55556 E-18 so for non relativistic speeds large charge o 1C has no detectable magnetic field... Thank you it is the answer I have been looking for...

That formula isn't applicable to point charges. Both E and B fields vary with position in that case. But for a point charge the B field at any point will be weaker by a factor of at least ##\gamma v/c^2## than the E field measured at the same point, so the same comments apply. It's only when you get a large number of electrons in motion (like in a wire) that the magnetic field is detectable.

 

[Maciej Orman]

That formula isn't applicable to point charges. Both E and B fields vary with position in that case. But for a point charge the B field at any point will be weaker by a factor of at least ##\gamma v/c^2## than the E field measured at the same point, so the same comments apply. It's only when you get a large number of electrons in motion (like in a wire) that the magnetic field is detectable.

But electrons in wire never move with constant linear velocity... What's constant is the average velocity...

 

[Ibix]

But electrons in wire never move with constant linear velocity... What's constant is the average velocity...

So what? If each electron has instantaneous velocity ##v_i<<c## then its contribution to the magnetic field is proportional to ##v_i##. Add up the total fields and the result is proportional to ##\sum v_i## and hence proportional to the average velocity.

I'm modelling the wire as much thinner than the distance at which you measure the field. Feel free to integrate over a cylindrical wire if you like.

 

[Maciej Orman]

So what?

Magnetic field generated by charge moving with accelerated motion is well known... One can generate large magnetic field just by waving an electret which has low charge in uC... The Rowland disk experiment shows that... Still, electret produces no magnetic field when with constant linear velocity motion due to 500 m/s orbital motion of the Earth...

 

[Ibix]

Still, electret produces no magnetic field when with constant linear velocity motion due to 500 m/s orbital motion of the Earth...

Motion of the Earth with respect to what? The charge's motion with respect to your sensors is what matters for whether there is a magnetic field detected or not.

 

[Dadface]

OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all? Also the above model should be solved for magnetic field not a force...

I think that's a good point and I think a little bit of research is needed to answer your question about experimental evidence. But i would like to add a couple of points that may be relevant.

Firstly, your Van de Graaf belt is analogous to a straight current carrying wire. Using ordinary electric currents eg from a power pack you can easily detect the field around wire geometries such as solenoids. But it's most difficult to detect it around a straight wires, simply because that geometry gives the weakest field. To give a convincing demonstration you have to use quite high currents and depending on what you use that can lead to blown fuses or molten wires.

Secondly, although the Van de Graaf can produce high voltages the currents can be low, so any magnetic field will be difficult to detect.

I suggest that you do some research, find out what a typical current carried by a V de G belt is and then Biot and Savart to estimate the B field due to that current. I suspect that your magnetometer might not be sensitive enough to detect That field.

 

[Maciej Orman]

I think that's a good point and I think a little bit of research is needed to answer your question about experimental evidence. But i would like to add a couple of points that may be relevant.

Firstly, your Van de Graaf belt is analogous to a straight current carrying wire. Using ordinary electric currents eg from a power pack you can easily detect the field around wire geometries such as solenoids. But it's most difficult to detect it around a straight wires, simply because that geometry gives the weakest field. To give a convincing demonstration you have to use quite high currents and depending on what you use that can lead to blown fuses or molten wires.

Secondly, although the Van de Graaf can produce high voltages the currents can be low, so any magnetic field will be difficult to detect.

I suggest that you do some research, find out what a typical current carried by a V de G belt is and then Biot and Savart to estimate the B field due to that current. I suspect that your magnetometer might not be sensitive enough to detect That field.

But one can arrange the belt to move around the circle thus making the motion of the charge accelerate and magnetic field can be easily detected even with a compass... See the Rowland disk experiment...