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Maciej Orman
Charge moving in constant linear velocity does not produce magnetic field...
If not, please provide an explanation...
If not, please provide an explanation...
Charge moving with a constant linear velocity....
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A charge moving with constant linear velocity does produce a magnetic field, contrary to the initial claim. This is explained through Maxwell's equations and Lorentz transformations, which show that the electromagnetic field is frame-dependent, meaning observers in different frames will perceive the presence or absence of magnetic fields differently. The discussion emphasizes that while a charge at rest only generates an electric field, a moving charge generates both electric and magnetic fields in the observer's frame. The concept of absolute motion or aether is dismissed, as special relativity does not support a preferred reference frame. Experimental evidence for these principles is acknowledged as lacking for single charged particles, yet macroscopic observations in electrical engineering align with theoretical predictions.
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Yes it does.Maciej Orman said:Charge moving in constant linear velocity does not produce magnetic field...
Your question is unclear. You labeled this thread "A". Which part of Maxwell's equations do you not understand?Maciej Orman said:If not, please provide an explanation...
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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...
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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).
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
vanhees71 said: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...
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
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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.Maciej Orman said:Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?
It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.
Ibix
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This is incorrect, as I wrote above.Maciej Orman said:magnetic field is a physical quantity and does not change
Maciej Orman
Laws of physics must hold across all frames of motion and cannot be frame dependent...Ibix said: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.
Maciej Orman
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 said:This is incorrect, as I wrote above.
Ibix
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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.Maciej Orman said:Laws of physics must hold across all frames of motion and cannot be frame dependent...
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 said: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...
Maciej Orman
Logic also suggests that mathematical models have no effect on physics of reality thus incorrect models will not be supported by experimental evidence...Ibix said: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.
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
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A straight wire carrying a constant current.Maciej Orman said:In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field
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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.
$$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
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
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...vanhees71 said: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.
Ibix
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Maciej Orman said:OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all?
Ibix said:A straight wire carrying a constant current.
Maciej Orman
There is not a single electron moving in constant linear velocity in a wire carrying constant current...Ibix said:A straight wire carrying a constant current.
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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,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$
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,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$
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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 said: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...
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...vanhees71 said: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
Rowland disk experiment proves that accelerated static charge on a disk creates magnetic field...Maciej Orman said: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...
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The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...
Maciej Orman
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...vanhees71 said: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,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$
Maciej Orman
Yes, but this is not an excess electric charge moving in constant linear velocity...vanhees71 said:The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...
Maciej Orman
It was Rowland who first shown this effect to Hertz and all that is charge accelerated in circular motion produces magnetic field...vanhees71 said:The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...
Ibix
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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.)Maciej Orman said: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...
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
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 said: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.
Ibix
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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 said: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...
Maciej Orman
But electrons in wire never move with constant linear velocity... What's constant is the average velocity...Ibix said: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.
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