- #31
mattt
- 299
- 125
humbleteleskop said:
With constant velocity magnetic field is constant.
Are you serious or just trolling? :-)
Traveling electric and magnetic fields
- Thread starter letsenibeh
- Start date
Click For Summary
The discussion centers on the propagation of electric and magnetic fields in the context of a charged sheet moving at a constant velocity, as described in Feynman's Lectures on Physics. Participants debate whether the electric field can be eliminated when the magnetic field reaches a steady state, with some arguing that a charged body in motion generates time-dependent magnetic and electric fields, contrary to the assumption of a static magnetic field. The Biot-Savart Law is noted to apply only in magnetostatics, leading to the conclusion that a moving charge radiates electromagnetic waves. The conversation emphasizes the need for a full treatment of Maxwell's equations to understand the dynamic behavior of these fields. Ultimately, the consensus is that the electric and magnetic fields are interdependent and cannot be treated as static in this scenario.
Are you serious or just trolling? :-)
- #32
mattt
- 299
- 125
For example, imagine a xyz-inertial frame. With respect to this frame, we have a point charge whose equation of motion is:
(-t, 1, 0)
That is, the point charge is moving along the line of (implicit) equations y=1, z=0 (which is parallel to the x-axis, and at a distance=1 ) and the vector velocity of the point charge is
\vec{v}=(-1,0,0)
Can you compute \vec{B}(x,y,z,t)?
If you compute it (using that Biot-Savart formula), you'll get:
\vec{B}(x,y,z,t) = \frac{\mu_0}{4\pi}\frac{q}{((t+x)^2 + (y-1)^2 + z^2)^{\frac{3}{2}})}(0, z, 1-y)
That is
\vec{B}(x,y,z,t) = \frac{\mu_0}{4\pi}\frac{q}{((t+x)^2 + (y-1)^2 + z^2)^{\frac{3}{2}})}(z\vec{j} + (1-y)\vec{k})
At any given point P(x_0,y_0,z_0) (not in the line y=1, z=0), the \vec{B} field at that point mantain the same direction and sense for all time, but its magnitude obviously depends on time.
That is why, at any given point P(x_0,y_0,z_0), the magnitude of the vector \vec{B}(x_0,y_0,z_0,t) goes to zero when the point charge is too far away from this point (in the distant past and in the distant future), and this magnitude reaches its maximum value when the point charge is in the same yz-plane that the point P (i.e. when the point charge is closest to the point P).
(-t, 1, 0)
That is, the point charge is moving along the line of (implicit) equations y=1, z=0 (which is parallel to the x-axis, and at a distance=1 ) and the vector velocity of the point charge is
\vec{v}=(-1,0,0)
Can you compute \vec{B}(x,y,z,t)?
If you compute it (using that Biot-Savart formula), you'll get:
\vec{B}(x,y,z,t) = \frac{\mu_0}{4\pi}\frac{q}{((t+x)^2 + (y-1)^2 + z^2)^{\frac{3}{2}})}(0, z, 1-y)
That is
\vec{B}(x,y,z,t) = \frac{\mu_0}{4\pi}\frac{q}{((t+x)^2 + (y-1)^2 + z^2)^{\frac{3}{2}})}(z\vec{j} + (1-y)\vec{k})
At any given point P(x_0,y_0,z_0) (not in the line y=1, z=0), the \vec{B} field at that point mantain the same direction and sense for all time, but its magnitude obviously depends on time.
That is why, at any given point P(x_0,y_0,z_0), the magnitude of the vector \vec{B}(x_0,y_0,z_0,t) goes to zero when the point charge is too far away from this point (in the distant past and in the distant future), and this magnitude reaches its maximum value when the point charge is in the same yz-plane that the point P (i.e. when the point charge is closest to the point P).
- #33
- 24,488
- 15,057
This is obviously wrong, as you can see by comparing with the correct solution given in #26!
The reason is very simple! The charge and current density are given by
\rho(t,\vec{x})=q \delta^{(3)}(\vec{x}-\vec{v} t), \quad \vec{j}(t,\vec{x})=q \vec{v} \delta^{(3)}(\vec{x}-\vec{v} t), \quad \vec{v}=\text{const}.
Thus you have to use the retarded potentials (or Jefimenko's equations) to evaluate the fields. The result is, of course, the same as the one obtained in #26 via a Lorentz boost from the rest frame of the particle to the frame, where it moves with the constant velocity \vec{v}. The way over the retarded potentials or Jefmenko's equations are tedious compared to the elegant way via the Lorentz transformation!
Again, this is an example for the harm done by textbooks not emphasizing the relativistic covariance of electrodynamics and overemphasizing statics!
The reason is very simple! The charge and current density are given by
\rho(t,\vec{x})=q \delta^{(3)}(\vec{x}-\vec{v} t), \quad \vec{j}(t,\vec{x})=q \vec{v} \delta^{(3)}(\vec{x}-\vec{v} t), \quad \vec{v}=\text{const}.
Thus you have to use the retarded potentials (or Jefimenko's equations) to evaluate the fields. The result is, of course, the same as the one obtained in #26 via a Lorentz boost from the rest frame of the particle to the frame, where it moves with the constant velocity \vec{v}. The way over the retarded potentials or Jefmenko's equations are tedious compared to the elegant way via the Lorentz transformation!
Again, this is an example for the harm done by textbooks not emphasizing the relativistic covariance of electrodynamics and overemphasizing statics!
- #34
mattt
- 299
- 125
I meant that even using "his" Biot-Savart formula for that situation, \vec{B}(x,y,z,t) depends explicitly on time.
- #35
humbleteleskop
- 114
- 0
mattt said:
I didn't invent that equation and I have nothing to say about it except that it works as a part of Lorenz force equation. By itself, it's incomplete. What that means and how it relates to whatever you are talking about, I don't know.
I notice now there is no actually any second object in the opening post, no interaction, no forces acting except within the sheet itself. Those are kinematic equations, and if the answer is not about calculating forces, then I was wrong to say those equations apply. I'm sorry for the confusion.
- #36
humbleteleskop
- 114
- 0
mattt said:
What you have there is an integral, integral over time. What I have in the algorithm is not integral, it is evaluation of instantaneous points in time, one at the time. When you run that on a computer in real-time, as recursive algorithm that it is, then it gets integrated over time, literally.
- #37
Born2bwire
Science Advisor
Gold Member
- 1,780
- 24
Assuming that at t=0, the electron is at x=-100 and that it moves at a velocity of 90 m/s in the +x direction, then the position as a function of time is
\mathbf{r'} = (90t-100)\hat{x}
Thus, the observational vector that we will use is
\mathbf{r}-\mathbf{r'} = (x+100-90t)\hat{x}+y\hat{y}+z\hat{z}
Now, given that we are ignoring retardation and relativistic effects, the electric field is the simple Coulombic field given as
\mathbf{E}(\mathbf{r},t) = \mathbf{E_{NR}}(\mathbf{r},t) = \frac{q}{4\pi\epsilon_0} \frac{1}{(x+100-90t)^2+y^2+z^2} \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \cdot \begin{bmatrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{bmatrix}
where
\theta = \cos^{-1} \left( \frac{z}{\sqrt{(x+100-90t)^2+y^2}} \right)
\phi = \tan^{-1} \left( \frac{y}{x+100-90t} \right)
As for the case where we have a speed of 0.99c, as already stated by others, then we will need to take into relativistic transformations and retardation. The exact electric field in this case will be
\mathbf{E}(\mathbf{r},t) = \frac{0.0199}{(1-0.0199\sin^2(\theta_v))^{1.5}} \mathbf{E_{NR}}\left(\mathbf{r}, t-\frac{\sqrt{ (x+100-90t)^2+y^2+z^2 }}{c}\right)
where
\theta_v = \cos^{-1} \left( \frac{x+100-90t}{\sqrt{(x+100-90t)^2+y^2}} \right)The Lorentzian boost now reduces the amplitude of the electric field, depending upon the angle with which we observe the electron with respect to its velocity. In addition, the retardation means that we have a time delay as we wait for the electromagnetic wave to reach us as it is traveling at the speed of light.
Now, I'm sure from here you can work out your resulting error in the magnetic field.
Of course the above is a reiteration of what matt has given and is another restatement of what vanhees posted in #26.
\mathbf{r'} = (90t-100)\hat{x}
Thus, the observational vector that we will use is
\mathbf{r}-\mathbf{r'} = (x+100-90t)\hat{x}+y\hat{y}+z\hat{z}
Now, given that we are ignoring retardation and relativistic effects, the electric field is the simple Coulombic field given as
\mathbf{E}(\mathbf{r},t) = \mathbf{E_{NR}}(\mathbf{r},t) = \frac{q}{4\pi\epsilon_0} \frac{1}{(x+100-90t)^2+y^2+z^2} \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0 \end{bmatrix} \cdot \begin{bmatrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{bmatrix}
where
\theta = \cos^{-1} \left( \frac{z}{\sqrt{(x+100-90t)^2+y^2}} \right)
\phi = \tan^{-1} \left( \frac{y}{x+100-90t} \right)
As for the case where we have a speed of 0.99c, as already stated by others, then we will need to take into relativistic transformations and retardation. The exact electric field in this case will be
\mathbf{E}(\mathbf{r},t) = \frac{0.0199}{(1-0.0199\sin^2(\theta_v))^{1.5}} \mathbf{E_{NR}}\left(\mathbf{r}, t-\frac{\sqrt{ (x+100-90t)^2+y^2+z^2 }}{c}\right)
where
\theta_v = \cos^{-1} \left( \frac{x+100-90t}{\sqrt{(x+100-90t)^2+y^2}} \right)The Lorentzian boost now reduces the amplitude of the electric field, depending upon the angle with which we observe the electron with respect to its velocity. In addition, the retardation means that we have a time delay as we wait for the electromagnetic wave to reach us as it is traveling at the speed of light.
Now, I'm sure from here you can work out your resulting error in the magnetic field.
Of course the above is a reiteration of what matt has given and is another restatement of what vanhees posted in #26.
Last edited:
- #38
- 24,488
- 15,057
Again: This is approximate at best (of course it's a quite good approximation for such small velocities). The field, however should read in this approximation
\vec{E}(t,\vec{x}) \simeq \frac{q}{4 \pi \epsilon_0} \; \frac{\vec{x}-\vec{y}(t)}{|\vec{x}-\vec{y}(t)|^3}, \quad \vec{y}(t)=\vec{x}_0+\vec{v}_0 t.
In this way, of course, you'll never get any radiation part for an accelerated charge. It's the crucial point of Maxwell theory: By adding the "displacement current" (to express it in the oldfashioned pre-relativistic language) the theory becomes Poincare invariant and adds electromagnetic waves to the game. Of course this was the discovery in the 19th century, unifying electricity, magnetism, and optics into one consistent theory, the Maxwell equations of the electromagnetic field!
\vec{E}(t,\vec{x}) \simeq \frac{q}{4 \pi \epsilon_0} \; \frac{\vec{x}-\vec{y}(t)}{|\vec{x}-\vec{y}(t)|^3}, \quad \vec{y}(t)=\vec{x}_0+\vec{v}_0 t.
In this way, of course, you'll never get any radiation part for an accelerated charge. It's the crucial point of Maxwell theory: By adding the "displacement current" (to express it in the oldfashioned pre-relativistic language) the theory becomes Poincare invariant and adds electromagnetic waves to the game. Of course this was the discovery in the 19th century, unifying electricity, magnetism, and optics into one consistent theory, the Maxwell equations of the electromagnetic field!
Last edited:
- #39
humbleteleskop
- 114
- 0
@Born2bwire
I don't see you calculated B field, what am I supposed to compare mine to? Are we even talking about the same thing? I'm not really sure if what you call "field" is the same thing I call "field potential". Don't they have different units in your and my equations?
I don't see you calculated B field, what am I supposed to compare mine to? Are we even talking about the same thing? I'm not really sure if what you call "field" is the same thing I call "field potential". Don't they have different units in your and my equations?
- #40
HomogenousCow
- 736
- 213
humbleteleskop said:
Take a coulomb field (stationary) and then boost it, magical things will begin to happen.
- #41
Born2bwire
Science Advisor
Gold Member
- 1,780
- 24
humbleteleskop said:
I don't know what a field potential is. Fields and potentials are two different things. A potential is the kernel for the resulting field. We all have only been discussing the electromagnetic field. Nobody has used the phrase "field potential" except yourself and it is obvious from the equations that you post that you are talking about the field.
Our units are the same. Going into the magnetic field is superfluous at this point as we first need to discuss whether or not you understand what is wrong with your expression for the electric field (though matt has already given you the non-relativistic magnetic field for a similar case). Besides, the electric field defines the electromagnetic field in this case via Maxwell's Equations. So once we correct the electric field, we will have the correct magnetic field as a result.
You keep treating the electron like it is stationary in space. Your electromagnetic field states that no matter where the electron lies, the field is infinite at the origin. It takes into no account where the electron is at any point in time nor where the observer lies. Surely you understand that the Coulombic field is centered at the location of the electron, not at the origin of the coordinate system?
- #42
humbleteleskop
- 114
- 0
Born2bwire said:
I'm talking about field potentials. Electric potential is to electric force what gravitational potential is to gravity force. And that is what is in my equations. Same for magnetic field potential and magnetic force, only magnetic field potential has different topology, that is different orientation of gradient vectors. A point charge has electric and gravity field potential that look like a ball, magnetic field potential of a point charge looks like a doughnut.
Can you express that magnetic field with either some equation or numerically, or not?
I'm not treating anything, I'm just using equations from my textbook some other people invented hundreds of years ago. If they are wrong it's not my fault, it's just surprising how they still manage to work.
- #43
mattt
- 299
- 125
humbleteleskop said:
Dude, it is the formula you cited (the Biot-Savart approximate formula for the Magnetic FIELD created by a point charge moving with velocity \vec{v} ) :
\vec{B} = \frac{\mu_0}{4\pi}\frac{q}{r^3} \vec{v}\times\vec{r}
That is a VECTOR FIELD, not a potential anything...
- #44
Born2bwire
Science Advisor
Gold Member
- 1,780
- 24
- #45
humbleteleskop
- 114
- 0
mattt said:
So our equations are the same, what are we arguing about then?
How to call it? It is generally a type of vector field, yes, but in the case of EM fields they are more precisely referred to as "field potentials". You gave us the equation for magnetic field potential and I'll give us the equation for electric field potential.
http://en.wikipedia.org/wiki/Electric_potential
-"In classical electromagnetism, the electric potential (a scalar quantity denoted by Φ, ΦE or V and also called the electric field potential or the electrostatic potential) at a point of space is the amount of electric potential energy that a unitary point charge would have when located at that point. "
Whether that is the same thing you call "field", that I don't know.
- #46
mattt
- 299
- 125
Again, are you serious or just trolling?
If trolling, I would like to tell the moderator.
\vec{B} is NOT the magnetic vector potential, neither the scalar magnetic potential.
It is the Magnetic Field.
So you put a formula (Biot-Savart) and you don't even know what the terms in that formula are?
Now at least it is more clear why you don't see that magnetic field \vec{B} (created by a point charge that is moving with constant velocity) is time-dependant (varies with time). You don't even know the terms in the formula. :-(
If trolling, I would like to tell the moderator.
\vec{B} is NOT the magnetic vector potential, neither the scalar magnetic potential.
It is the Magnetic Field.
So you put a formula (Biot-Savart) and you don't even know what the terms in that formula are?
Now at least it is more clear why you don't see that magnetic field \vec{B} (created by a point charge that is moving with constant velocity) is time-dependant (varies with time). You don't even know the terms in the formula. :-(
- #47
Born2bwire
Science Advisor
Gold Member
- 1,780
- 24
humbleteleskop said:So our equations are the same, what are we arguing about then?
How to call it? It is generally a type of vector field, yes, but in the case of EM fields they are more precisely referred to as "field potentials". You gave us the equation for magnetic field potential and I'll give us the equation for electric field potential.
http://en.wikipedia.org/wiki/Electric_potential
-"In classical electromagnetism, the electric potential (a scalar quantity denoted by Φ, ΦE or V and also called the electric field potential or the electrostatic potential) at a point of space is the amount of electric potential energy that a unitary point charge would have when located at that point. "
Whether that is the same thing you call "field", that I don't know.
You have been giving us the fields, what you just quoted above is the electric potential and that is not what you have been giving us in your equations made obvious by the fact that the units are different and that the electric potential is a scalar, not a vector.
And no, your magnetic field equation is not what we have derived. Again, do you understand why your electric field equation is not correct compared to the electric field that I gave you?
- #48
humbleteleskop
- 114
- 0
mattt said:
I don't know what your "B" is, I'm telling you what E and B in my equations are, and they are called "electric potential" and "magnetic potential".
http://en.wikipedia.org/wiki/Electric_potential
http://en.wikipedia.org/wiki/Magnetic_potential
http://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field
- "The electric potential is a scalar field, while the magnetic potential is a vector field."
Last edited:
- #49
mattt
- 299
- 125
OK, you are obviously trolling. Bye, bye.
- #50
humbleteleskop
- 114
- 0
Born2bwire said:
1.
2.
The equation mattt posted is the same equation I posted the first time around. I was talking to him, I was not talking about any equation that you have posted.
I'm just quoting Wikipedia. I was not deriving any equation, I copy-pasted them all directly from Wikipedia. If you believe they are wrong it has nothing to do with me.
Last edited:
- #51
mfb
Mentor
- 37,395
- 14,232
This thread leads to nothing and letsenibeh did not come back so far. I closed the thread.
@humbleteleskop: I suggest to look for a good book as an introduction to electromagnetic fields.
@humbleteleskop: I suggest to look for a good book as an introduction to electromagnetic fields.
Similar threads
- · Replies 9 ·
- · Replies 5 ·
High School
A moving magnet in a linear electric field
- · Replies 8 ·
- · Replies 60 ·
Undergrad
Electric field of ring at any point
- · Replies 2 ·
- · Replies 4 ·
- · Replies 20 ·
- · Replies 4 ·
Graduate
Origin of Hertzian dipole radiation
- · Replies 21 ·
- · Replies 1 ·