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Sorcerer
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Obviously in the title I mention the user that recently got banned, but the reason I do is because s/he was having some trouble accepting that a B field transforms into a mixture of E and B fields per the Lorentz transformation (and other assorted quackery), so it got me thinking about why this must be true, hence this thread. Please tell me if this reasoning is justified.
So, my question is, is this example sufficient proof that E and B fields must transform into one another? (At first a paradox shows what happens if they don't transform into one another, then this is corrected to solve the paradox)
Consider a wire at rest with electron density -λ coulombs per meter has electrons moving to the right at speed v. The positive charges (of density λ coulombs per meter) in the wire remain at rest. The resulting current magnitude is therefore I = λv (the current is to the left, in this case, as the electrons moving to the right carry negative charge).
In addition to this, let there be a positive charge q outside the wire at a distance r, which is initially moving to the right at the same speed as the electrons, v.
What force does q feel? (at this stage we are neglecting special relativity for the sake of argument)
Because the negative and positive charge densities are equal, the wire is electrically neutral, and therefore there is no electric force on charge q. So we can disregard ##\vec F_E = q \vec E##. So the force on the charge will feel will be magnetic, given by ##\vec F_B = q \vec v × \vec B##.
The magnitude of the magnetic field (circling the wire) would thus be:
$$B = \frac{μ_0λv}{2πr}$$
which means the force magnitude the charge feels is:
$$F_B= \frac{qμ_0λv^2}{2πr}$$
Due to the direction of the magnetic field, the force causes the charge to be accelerated away from the wire. So the charge is initially moving parallel to the wire, and is then pushed away by the magnetic force.
Now consider the frame in which the charge is initially at rest.
In this frame, speed v = 0. Therefore there cannot be a magnetic force (because ##F_B = qvB##). Because the wire has equal positive and negative charge densities and so will experience no electric force either. Therefore the force on the charge must be zero.
But this leads to an impossible situation: in one frame of reference the positive charge feels a force equal to ##\frac{qμ_0λv^2}{2πr}## in a way that causes the charge to move in a circular motion (as magnetic forces do no work) and in another frame it feels a force equal to 0 and remains in its inertial motion.
So ignoring special relativity there appears to be a paradox.Solving the paradox using special relativity:In the frame in which the electrons are at rest (and the positive charge is initially at rest), the positive charges are therefore moving at speed v, which means distance measured by the frame of the electrons for the positive charges will be contracted, meaning the charge density λ will increase to:
$$\frac {λ}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Since we're dealing with a speed much lower than the speed of light (current is on the order of a few milometers per second, f I recall), this is approximately equal to:
$$λ + \frac{λv^2}{2c^2}$$
Likewise, in the frame where the electrons are at rest, their charge density must actually be less than it was in the frame where the electrons are moving. Specifically, it must decrease by:
$$\frac{λv^2}{2c^2}$$
giving a new negative charge density of
$$λ - \frac{λv^2}{2c^2}$$
Subtract this from the change in positive charge density to get what the charge density must be in the frame of the moving electrons:
$$λ + \frac{λv^2}{2c^2} - \left( λ - \frac{λv^2}{2c^2} \right) $$
$$λ + \frac{λv^2}{2c^2} - λ + \frac{λv^2}{2c^2} $$
$$\frac{λv^2}{2c^2} + \frac{λv^2}{2c^2} $$
$$\frac{λv^2}{c^2}$$
So now to get E:
Electric flux is related to charge by:
$$\int E. \, dA = \frac{q_{enclosed}}{ε_0}$$
That integral (using a cylindrical Gaussian surface) should give us the ##\frac{1}{2πr}## coefficient needed to make things correct. Just to cross the t's and dot the i's:
$$\int E. \, dA = \int_{ends} E. \, dA + \int_{sides} E. \, dA = \frac{q_{enclosed}}{ε_0}$$
$$\int E. \, dA = 0 + 2πrlE = \frac{q_{enclosed}}{ε_0}$$
qenclosed equals charge density times length ##l##, so, using the new charge density, canceling ##l## from both sides and solving for E leaves:
$$E = \frac{λv^2}{c^2} \frac{1}{2πrε_0}$$
So to get ##F_E##, multiply ##q## by ##E##:
$$F_E = qE = qλ \frac{v^2}{c^2} \frac{1}{2πrε_0} = \frac{qλv^2}{2πr} \frac{1}{c^2ε_0} = \frac{qλv^2}{2πr} μ_0$$
=>
$$F_E = \frac{qλv^2}{2πr} μ_0$$And of course, this is exactly the same value for the force of the magnetic field in the frame of reference where the charge is moving with speed v. Therefore there is no paradox, and the force the charge feels in the other frame is electric instead of magnetic.In summary:
(1) If magnetic fields and electric fields do not transform into one another, at least in this example we end up with a paradox.
(2) Length contraction changes the charge density, allowing an electric field to arise in one of the frames which yields a force with the same magnitude as the other frame.I understand this is probably the simplest set up possible, so does this make sense? Did I miss something important? Thanks for taking the time to look at this.
So, my question is, is this example sufficient proof that E and B fields must transform into one another? (At first a paradox shows what happens if they don't transform into one another, then this is corrected to solve the paradox)
Consider a wire at rest with electron density -λ coulombs per meter has electrons moving to the right at speed v. The positive charges (of density λ coulombs per meter) in the wire remain at rest. The resulting current magnitude is therefore I = λv (the current is to the left, in this case, as the electrons moving to the right carry negative charge).
In addition to this, let there be a positive charge q outside the wire at a distance r, which is initially moving to the right at the same speed as the electrons, v.
What force does q feel? (at this stage we are neglecting special relativity for the sake of argument)
Because the negative and positive charge densities are equal, the wire is electrically neutral, and therefore there is no electric force on charge q. So we can disregard ##\vec F_E = q \vec E##. So the force on the charge will feel will be magnetic, given by ##\vec F_B = q \vec v × \vec B##.
The magnitude of the magnetic field (circling the wire) would thus be:
$$B = \frac{μ_0λv}{2πr}$$
which means the force magnitude the charge feels is:
$$F_B= \frac{qμ_0λv^2}{2πr}$$
Due to the direction of the magnetic field, the force causes the charge to be accelerated away from the wire. So the charge is initially moving parallel to the wire, and is then pushed away by the magnetic force.
Now consider the frame in which the charge is initially at rest.
In this frame, speed v = 0. Therefore there cannot be a magnetic force (because ##F_B = qvB##). Because the wire has equal positive and negative charge densities and so will experience no electric force either. Therefore the force on the charge must be zero.
But this leads to an impossible situation: in one frame of reference the positive charge feels a force equal to ##\frac{qμ_0λv^2}{2πr}## in a way that causes the charge to move in a circular motion (as magnetic forces do no work) and in another frame it feels a force equal to 0 and remains in its inertial motion.
So ignoring special relativity there appears to be a paradox.Solving the paradox using special relativity:In the frame in which the electrons are at rest (and the positive charge is initially at rest), the positive charges are therefore moving at speed v, which means distance measured by the frame of the electrons for the positive charges will be contracted, meaning the charge density λ will increase to:
$$\frac {λ}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Since we're dealing with a speed much lower than the speed of light (current is on the order of a few milometers per second, f I recall), this is approximately equal to:
$$λ + \frac{λv^2}{2c^2}$$
Likewise, in the frame where the electrons are at rest, their charge density must actually be less than it was in the frame where the electrons are moving. Specifically, it must decrease by:
$$\frac{λv^2}{2c^2}$$
giving a new negative charge density of
$$λ - \frac{λv^2}{2c^2}$$
Subtract this from the change in positive charge density to get what the charge density must be in the frame of the moving electrons:
$$λ + \frac{λv^2}{2c^2} - \left( λ - \frac{λv^2}{2c^2} \right) $$
$$λ + \frac{λv^2}{2c^2} - λ + \frac{λv^2}{2c^2} $$
$$\frac{λv^2}{2c^2} + \frac{λv^2}{2c^2} $$
$$\frac{λv^2}{c^2}$$
So now to get E:
Electric flux is related to charge by:
$$\int E. \, dA = \frac{q_{enclosed}}{ε_0}$$
That integral (using a cylindrical Gaussian surface) should give us the ##\frac{1}{2πr}## coefficient needed to make things correct. Just to cross the t's and dot the i's:
$$\int E. \, dA = \int_{ends} E. \, dA + \int_{sides} E. \, dA = \frac{q_{enclosed}}{ε_0}$$
$$\int E. \, dA = 0 + 2πrlE = \frac{q_{enclosed}}{ε_0}$$
qenclosed equals charge density times length ##l##, so, using the new charge density, canceling ##l## from both sides and solving for E leaves:
$$E = \frac{λv^2}{c^2} \frac{1}{2πrε_0}$$
So to get ##F_E##, multiply ##q## by ##E##:
$$F_E = qE = qλ \frac{v^2}{c^2} \frac{1}{2πrε_0} = \frac{qλv^2}{2πr} \frac{1}{c^2ε_0} = \frac{qλv^2}{2πr} μ_0$$
=>
$$F_E = \frac{qλv^2}{2πr} μ_0$$And of course, this is exactly the same value for the force of the magnetic field in the frame of reference where the charge is moving with speed v. Therefore there is no paradox, and the force the charge feels in the other frame is electric instead of magnetic.In summary:
(1) If magnetic fields and electric fields do not transform into one another, at least in this example we end up with a paradox.
(2) Length contraction changes the charge density, allowing an electric field to arise in one of the frames which yields a force with the same magnitude as the other frame.I understand this is probably the simplest set up possible, so does this make sense? Did I miss something important? Thanks for taking the time to look at this.