Simulating electron beam - what equations to use?

[Vdtta]

Hi,

I'm doing a simulation of an electron beam. I've done integration of classical electrodynamics equations where I calculate electric fields with Coulomb force and magnetic fields with Lorentz force. I would like to know how to make this simulation more correct by using SR equations.

What equation do I use to replace Coulomb force, and what is the SR equation for Lorentz force?


Thank you.

 

[clamtrox]

Lorentz force is completely compatible with special relativity; that's the whole point of SR. However for Coulomb force you need to take into account the finite signal speed by using retarded potentials. This is not going to look pretty.

 

[Vdtta]

Lorentz force is completely compatible with special relativity; that's the whole point of SR. However for Coulomb force you need to take into account the finite signal speed by using retarded potentials. This is not going to look pretty.

You are mistaken about Lorentz force, it's probably the most non-relativistic equation there is.

F= q(v x B)

It has no speed of light and no propagation delays. Even worse, the velocity there is completely non-relativistic, it's not even: F= q(v1-v2 x B) ...which might explain why Lorentz came up with his theory of aether, as this velocity has nothing to be relative to, except some absolute reference frame that turns out to be Earth inertial reference frame.

Anyway, are there some examples somewhere on the internet as how to implement retarded potentials? Is there any open source simulation of elementary charges that deals with both forces, so I can look it up there? Thanks.

 

[jtbell]

You are mistaken about Lorentz force, it's probably the most non-relativistic equation there is.

F= q(v x B)

On the contrary, it is relativistically valid and can be used to derive a manifestly Lorentz-covariant formula for the electromagnetic force on a moving charge. See for example

http://farside.ph.utexas.edu/teaching/em/lectures/node127.html

this velocity has nothing to be relative to, except some absolute reference frame that turns out to be Earth inertial reference frame.

This equation is valid in any inertial reference frame, using the velocity, electric field and magnetic field evaluated/measured in that reference frame.

 

[clamtrox]

It has no speed of light and no propagation delays. Even worse, the velocity there is completely non-relativistic, it's not even: F= q(v1-v2 x B) ...which might explain why Lorentz came up with his theory of aether, as this velocity has nothing to be relative to, except some absolute reference frame that turns out to be Earth inertial reference frame.

The formula you quote for the Lorentz force is incorrect, you are missing the part with the electric field. The thing that makes it come out invariant is that the electromagnetic field is not invariant under Lorentz transformations (remember that a stationary charge generates an electric field, moving charge generates magnetic field). I'm sure you can find the details in most electrodynamics textbooks.

Secondly, Lorentz force is mediated by the electromagnetic field, so it does not "propagate" anywhere; instead the EM field exerts Lorentz force LOCALLY onto a test particle. The fields propagation is of course limited by the speed of light, that was the second point I made in my first post.

 

[Vdtta]

The formula you quote for the Lorentz force is incorrect, you are missing the part with the electric field.

My formula is very correct, see Wikipedia or any decent textbook. You are suggesting "F= q(E+ v x B)" which is terrible notation as it hides the other well known force.

Lorentz force: F(B)= q(v x B)
Coulomb force: F(E)= k* q1*q2/r^2

F(combined) = F(B) + F(E)

Same thing, only this is more fair towards Coulomb and is especially useful if you are integrating it in an algorithm, since you have to split all your vectors to x,y,z components anyway, and so before we finally add those two vectors you would really need them to be separated one from another.

The thing that makes it come out invariant is that the electromagnetic field is not invariant under Lorentz transformations (remember that a stationary charge generates an electric field, moving charge generates magnetic field). I'm sure you can find the details in most electrodynamics textbooks.
.

So, as I gather, you want me to include "E" in my equation so that E can transform to B, right? I've heard that before and that is just nonsense. E can not possibly transform into B and vice versa.

B varies with velocity, E does not.
B force is PERPENDICULAR to velocity, E is not.

And most importantly there are two kinds of magnetic fields:

1.) Magnetic field due to spatial velocity of moving charge.
2.) Magnetic north and south pole fields due to spin of a charge.

You can not equate an apple with watermelon and a pair of cherries.

This equation is valid in any inertial reference frame, using the velocity, electric field and magnetic field evaluated/measured in that reference frame.

Ok, here is my electron beam and there is you overrunning those electrons.

......e1 -> 

......e2 -> 

. . . . . . . . . . . . . . . . . . . .jtbell->

From the SR point of view this should be the same:

1.) Electrons travel at 10m/s and you are overtaking them with 20m/s

2.) Electrons are stationary and you are running past them at 10m/s.

3.) You are stationary and electrons go backwards at 10m/s.

4.) Electrons travel at 999910m/s and you are overtaking them with 999920m/s1.) F= q(10 x B)
2.) F= q(0 x B)
3.) F= q(10 x B)
4.) F= q(999910 x B)But, obviously it is not the same, so what do we do now? Relativistic formulas are very recognizable, they always have "c" in them, plus some retarded-time stuff or something as clamtrox was saying. This equation is very far from special relativity, I need something like Lienard-Wiechert potentials, but that does not seem to include Lorentz force, only Coulomb force.

 

[jtbell]

My formula is very correct, see Wikipedia or any decent textbook. You are suggesting "F= q(E+ v x B)" which is terrible notation as it hides the other well known force.

Lorentz force: F(B)= q(v x B)
Coulomb force: F(E)= k* q1*q2/r^2

F(combined) = F(B) + F(E)

The Coulomb formula for electric field E = k*q1/r^2 is valid only when the source charge q1 is stationary.

From the SR point of view this should be the same:

1.) Electrons travel at 10m/s and you are overtaking them with 20m/s

2.) Electrons are stationary and you are running past them at 10m/s.

3.) You are stationary and electrons go backwards at 10m/s.

4.) Electrons travel at 999910m/s and you are overtaking them with 999920m/s1.) F= q(10 x B)
2.) F= q(0 x B)
3.) F= q(10 x B)
4.) F= q(999910 x B)

In order to get equivalent results in all reference frames, you must include both E and B. They are not the same in different inertial reference frames. You have to transform them using the Lorentz transformation, when you switch from one reference frame to another. See for example

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html

 

[mikeph]

Clamtrox has the exact idea- retarded potentials. But this is messy.

B varies with velocity, E does not.
B force is PERPENDICULAR to velocity, E is not.

You can not equate an apple with watermelon and a pair of cherries.

This is what I'd expect someone who hasn't studied electrodynamics to say, because in relativity they are each manifestations of the exact same electromagnetic field.

 

[Vdtta]

The Coulomb formula for electric field E = k*q1/r^2 is valid only when the source charge q1 is stationary.

That seems wrong, we need a FORCE between two charges, not the FIELD. You can not have equations like that. You can not have distance without q2, you need two of something for distance. Field is potential, you can have it with only one charge, but the force is INTERACTION - attraction or repulsion, and so you need two, and then you automatically have relative position e.g. distance vector "r". -- Anyway, how does electric field vary with velocity? According to what formula and just how much error is in Coulomb law compared to this 'valid equation'?

In order to get equivalent results in all reference frames, you must include both E and B. They are not the same in different inertial reference frames. You have to transform them using the Lorentz transformation, when you switch from one reference frame to another. See for example

http://farside.ph.utexas.edu/teachin...s/node123.html

I'm including them both. Those are vectors, they can be separated, that does not make any difference, except a little bit of error you say I get with Coulomb law. But, my electrons are more or less parallel and have the same velocity, so their relative velocities are pretty much zero, and so Coulomb law should come up as quite correct for this situation as they can be considered stationary, right? Please look at the example below.

This is what I'd expect someone who hasn't studied electrodynamics to say, because in relativity they are each manifestations of the exact same electromagnetic field.

What kind of comment is that? I did study physics and I know enough math to know we can not add apples and oranges. So, if you think to know better, just show me. Can you show me please?

===== EXAMPLE - BEAM A

........e1-> v=100m/s
                 |
                 | r=0.01m                    
                 |
........e2-> v=100m/sClassical: 
-----------
F(B)= q(100 x B) 
F(E)= k* q1*q2/0.01^2 

F(total)= F(B) + F(E)Special Relativity:
--------------------
F(B)= ??
F(E)= ??

F(total)= ??

===== EXAMPLE - BEAM B

........e1-> v=999m/s
                 |
                 | r=0.01m                    
                 |
........e2-> v=999m/sClassical: 
-----------
F(B)= q(999 x B)
F(E)= k* q1*q2/0.01^2

F(total)= F(B) + F(E)Special Relativity:
--------------------
F(B)= ??
F(E)= ??

F(total)= ??

MY "UNEDUCATED" ANALYSIS IS THIS:

1.) We have two electrons with constant separation between them, therefore they have no relative velocity and so can be considered stationary.

2.) From this we conclude electrostatic repulsion will be constant and Coulomb law should work pretty well in this case.

3.) But magnetic attraction will increase with their "individual" velocity, relative only to one reference frame, which is static observer or Earth inertial reference frame.

4.) Even if there was some relative velocity between those electrons, we have no way of including it in Lorentz force equation, it is simply not relativistic at all.

http://en.wikipedia.org/wiki/Relative_velocity
- "In kinematics, relative velocity is the vector difference between the velocities of two objects, as evaluated in terms of a single coordinate system, usually an inertial frame of reference unless specifically stated otherwise."

Lorentz equation: F= q(v x B) <--- absolute velocity

Imaginary equation: F= q(v1-v2 x B) <--- relative velocity

Please provide Special Relativity solution,
and point if you believe there is any flaws in my analysis. Thank you.

 

[pervect]

Maxwell's equations _are_ totally relativistic. Example: the speed of light according to Maxwell's equations is always "c", never any other number. If you solve Maxwell's equations correctly, you'll have a relativistic solution.

The conflict between Maxwell's equations and the rest of physics is what inspired relativity - Maxwell's equations turned out to be correct, and the rest of physics had to be modified, slightly.

I'm not convinced you're necessarily solving Maxwell's equations from what little I can make out of your approach though.

The Lorentz force law, F = q(E + v x B) is totally relativistic, as many have pointed out - I have no idea why you think it isn't.

However, as others have pointed out (with references), Coulomb's law is NOT relativistic, it's only gives the electric field of a stationary charge. It does not give the electric field of a moving charge. You have to transform the field properly.

See for instance
http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

also

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_13.pdf
explains how the electric field transforms without using tensors.

The tensor approach has a lot to recommend it - I'm guessing from your posts that you haven't been introduced to the topic yet - that's the only reason I point out that the webpage doesn't use them.

You should probably find a textbook like Griffiths ((or, if you like math, Jackson :-))) that covers relativistc electrodynamics.

You can also work the problem out using Lineard Wiechart potentials, as others have remarked.

 

[Phrak]

Lorentz force: F(B)= q(v x B)
Coulomb force: F(E)= k* q1*q2/r^2

F(combined) = F(B) + F(E)

Look carefully at what you have here.

Your Lorentz force is local. It describes the force on a charge due to the local value of the magnetic field.

The Coulomb force is non-local requiring adjustments for propagation delay. Nevermind that it needs to be replaced by a vector equation.

The combined force is disjoint in two ways. One is the local action of a field, the other is the non-local action of a charge.

 

[Bob S]

In the lab, the transverse E and B fields are given by the kinematic transformations
http://pdg.lbl.gov/2009/reviews/rpp2009-rev-electromag-relations.pdf
and the length of the pulse in time is contracted by gamma, so the integrated effect is unchanged. Beam detectors can be made so that they have equal sensitivity to the transverse E field and the transverse v x B field, so that if the beam reverses direction, no (or very little) signal is observed. These are called directional couplers.
Bob S

 

[Vdtta]

The Lorentz force law, F = q(E + v x B) is totally relativistic, as many have pointed out - I have no idea why you think it isn't.

See post #3, #6 and #9, see examples BEAM A & BEAM B. Lorentz force was discovered before SR and is a part of classical electromagnetism. Why do you think it's relativistic? What is that velocity "v" relative to, aether?

Relative velocity is the vector difference between the velocities of TWO objects. Where is the second object? Where is the vector difference?Do magnetic fields propagate faster than the speed of light?
If not, then where is retarded-time, where is the "c" in that equation?

Look carefully at what you have here.

Your Lorentz force is local. It describes the force on a charge due to the local value of the magnetic field.

The Coulomb force is non-local requiring adjustments for propagation delay. Nevermind that it needs to be replaced by a vector equation.

The combined force is disjoint in two ways. One is the local action of a field, the other is the non-local action of a charge.

Lorentz force or magnetic field propagate without the delay?

Can you solve examples BEAM A & BEAM B (post #9)?

In the lab, the transverse E and B fields are given by the kinematic transformations
http://pdg.lbl.gov/2009/reviews/rpp2...-relations.pdf
and the length of the pulse in time is contracted by gamma, so the integrated effect is unchanged. Beam detectors can be made so that they have equal sensitivity to the transverse E field and the transverse v x B field, so that if the beam reverses direction, no (or very little) signal is observed. These are called directional couplers.
Bob S

What particular equations are you suggesting, how to use them?

Can you solve examples BEAM A & BEAM B (post #9)?

 

[Ben Niehoff]

Read a book such as Griffiths or Jackson and stop yelling at people who are trying to help you.

 

[Vanadium 50]

stop yelling at people who are trying to help you.

Thank you.

OP, your question was answered in post 2. If you don't understand the answer, please ask questions about it. If you don't believe the answer, there is little to be gained by arguing about it, and the thread should be closed.

 

[Vdtta]

Read a book such as Griffiths or Jackson and stop yelling at people who are trying to help you.

Those books, if anyone here actually read them, do not seem to be helpful as no one is able to apply whatever it is you are referring to is written there and is supposedly not known about 100 year old theory, as if it was not written anywhere else. Why is it hard to actually put equations down, integrate variables from the given example and print the results? As long as people do not understand what is the difference between relative and non-relative velocity, and as long as they think classical electromagnetism equations are compatible with special relativity, it is me who is helping here.
Now, let me help you too.

F = q(E + v x B)

You see that velocity "v" there? It is absolute velocity, it is not relative to anything but one frame of reference, which is static observer aka Earth inertial frame of reference. You do not need to read any books to figure that out, you only need to know what relative velocity actually is. So, stop yelling at me, go to school or read Wikipedia.

http://en.wikipedia.org/wiki/Relative_velocity
I gave the most simple example there is - TWO ELECTRONS - and if you can not apply your supposed knowledge and show by example what you preach, I'm afraid you just wasted your time with whatever books you have read and your advice can not be taken seriously, but at least you have a chance now to learn what relative velocity actually is, and you're welcome, my dear.

OP, your question was answered in post 2. If you don't understand the answer, please ask questions about it. If you don't believe the answer, there is little to be gained by arguing about it, and the thread should be closed.

My question was not answered. Lorentz force equation is a part of classical electromagnetism. How in the world can anyone think it is relativistic? There was no explanation given, no reason or argument... only to go away and read some book?!

Where is retarded-time, where is the "c" in Lorentz equation? Why no one can not apply what they preach and actually solve the simplest case scenario - TWO ELECTRONS?

 

[Vanadium 50]

As long as people do not understand what is the difference between relative and non-relative velocity, and as long as they think classical electromagnetism equations are compatible with special relativity, it is me who is helping here.



Now, let me help you too.

F = q(E + v x B)

You see that velocity "v" there? It is absolute velocity, it is not relative to anything but one frame of reference, which is static observer aka Earth inertial frame of reference.

I'm afraid that's simply not true.

This thread is done.