How Do Moving Charges Behave Differently in Different Reference Frames?

[lovetruth]

Consider two point charges in space: one positive(+Q) and other negative(-Q), lying on the y-axis and separated by distance 'r'. In frame A, both charges are at rest so, only attractive electrostatic force (F_elec) acts on both the charges which is defined by coulombs formula. In another frame B, both charges are moving with velocity 'v' in positive x-direction so forces acting on the charges are attractive electrostatic force(F_elec) and repulsive magnetic force(F_mag).
It can be seen that the net attractive forces acting on the charges is greater in frame A than in frame B by an amount F_mag since, F_elec is same in both frame A & B.

Also,
F_mag = - F_elec ,when v=c [this result can be obtained by using coulomb law, ampere law, and the fact that epsilon*mew=1/(c^2)]

Conclusion : So an observer in frame B will see that net attractive force acting on charges is reduced than in frame A. Also, the net attractive force in frame B tends to zero as 'v' approaches 'c'.

Q: How is the above situation compatible with SR. SR says nothin about relativity of forces only about relativity of length and time.

 

[Dale]

SR says nothin about relativity of forces only about relativity of length and time.

And what are the base units of force?

 

[lovetruth]

And what are the base units of force?

mass*length*time^-2

But that doesn't answer the question.

 

[Bill_K]

SR says nothin about relativity of forces only about relativity of length and time.

Of course it does, Lovetruth, special relativity says how force transforms from one reference frame to another, and also how electric and magnetic fields transform, which is what we need here. These may be found in any introductory relativity book:

E_∥' = E_∥
E_⊥' = γ(E_⊥ + v/c x B)
B_∥' = B_∥
B_⊥' = γ(B_⊥ - v/c x E)

where ∥ and ⊥ represent the components parallel and perpendicular to the relative motion. In your example, E' = γE, B' = γ(- v/c x E). In the moving frame the Lorentz force between the charges is

F' = q(E' + v/c x B') = q(γE + v/c x (γ(- v/c x E))) = qE γ (1-v²/c²) = qE

See? It's the same in the moving frame as it was in the original frame.

 

[lovetruth]

Of course it does, Lovetruth, special relativity says how force transforms from one reference frame to another, and also how electric and magnetic fields transform, which is what we need here. These may be found in any introductory relativity book:

E_∥' = E_∥
E_⊥' = γ(E_⊥ + v/c x B)
B_∥' = B_∥
B_⊥' = γ(B_⊥ - v/c x E)

where ∥ and ⊥ represent the components parallel and perpendicular to the relative motion. In your example, E' = γE, B' = γ(- v/c x E). In the moving frame the Lorentz force between the charges is

F' = q(E' + v/c x B') = q(γE + v/c x (γ(- v/c x E))) = qE γ (1-v²/c²) = qE

See? It's the same in the moving frame as it was in the original frame.

Can u provide any internet links which can show these E & B field transformation because I have not read this anywhere. As far as i kno, only length, time, mass, and energy transforms. Please give me the list of all the entity that can transform.

I have spotted an error in ur calculation, in the last step, u have assumed :
gamma*(1-v^2/c^2)= 1 which is incorrect.

 

[Bill_K]

I have spotted an error in ur calculation, in the last step, u have assumed :
gamma*(1-v^2/c^2)= 1 which is incorrect.

Sorry you're right, the formula I quoted for the Lorentz force is actually the 3-force F = dp/dt. The force which is the same in both frames is the 4-force, dp/dτ where τ is the proper time. They differ by a factor of γ.

Can u provide any internet links which can show these E & B field transformation because I have not read this anywhere.

Where have you read? A much better idea is to get a book and read that. The book I quoted from was Jackson's Classical Electrodynamics chap 12. It's also in Misner, Thorne and Wheeler chap 1.

 

[lovetruth]

Sorry you're right, the formula I quoted for the Lorentz force is actually the 3-force F = dp/dt. The force which is the same in both frames is the 4-force, dp/dτ where τ is the proper time. They differ by a factor of γ.

Where have you read? A much better idea is to get a book and read that. The book I quoted from was Jackson's Classical Electrodynamics chap 12. It's also in Misner, Thorne and Wheeler chap 1.

I have googled electromagnetic lorentz transformation. U have made a slight mistake in the formula, the electric and magnetic field terms do not have same units so, a factor of velocity should be multiplied to the B in the equation(v/c is dimensionless).
But u have made a remarkable attempt to solve the problem. Could u please solve the problem with corrected equation and show that force is same in both frame A & B.

I hate the fact SR contains so many transformations!??!
I wonder whether the laws of physics and speed of light are also transformed(seriously).
I think the results of SR are quasi-true but there must be another simple theory with few equations which can predict the effects of SR without giving a obnoxious headache.

 

[lovetruth]

Hey bill_k, can u explain why electromagnetic transformation is not used in the feynmann derivation of relation between Electric, magnetic field and relativity.

Here is the link of the problem i am referring: http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html

In this case the EM transformation is ignored but the result of same force is obtained.

 

[Bill_K]

U have made a slight mistake in the formula, the electric and magnetic field terms do not have same units so, a factor of velocity should be multiplied to the B in the equation(v/c is dimensionless). But u have made a remarkable attempt to solve the problem. Could u please solve the problem with corrected equation and show that force is same in both frame A & B.

Nonsense, the formulas are correct. They're in Gaussian units.

 

[lovetruth]

Nonsense, the formulas are correct. They're in Gaussian units.

I have attached image for the correct equation of transformation. Ur formula is wrong even in gaussian unit.

 

[Dale]

As far as i kno, only length, time, mass, and energy transforms. Please give me the list of all the entity that can transform.

Everything transforms! The question is only how it transforms.

I hate the fact SR contains so many transformations!??!
I wonder whether the laws of physics and speed of light are also transformed(seriously).
I think the results of SR are quasi-true but there must be another simple theory with few equations which can predict the effects of SR without giving a obnoxious headache.

There is nothing particularly unusual about SR with respect to transformations. Coordinate transformations are common to all branches of physics. The only difference is that in SR the coordinate transforms are 4D instead of 3D and that the metric is Minkowski instead of Euclidean. The Lorentz transform is certainly easier than doing a transformation from Euclidean to spherical coordinates.

 

[lovetruth]

Everything transforms! The question is only how it transforms.

There is nothing particularly unusual about SR with respect to transformations. Coordinate transformations are common to all branches of physics. The only difference is that in SR the coordinate transforms are 4D instead of 3D and that the metric is Minkowski instead of Euclidean. The Lorentz transform is certainly easier than doing a transformation from Euclidean to spherical coordinates.

What about mass, energy, electric and magnetic field transformation?

 

[Dale]

What about mass, energy, electric and magnetic field transformation?

Yes, they all transform. Mass transforms as a scalar, energy transforms as a component of a vector, and the electric and magnetic fields transform as the components of a tensor. Everything transforms one way or another.

 

[bcrowell]

What about mass, energy, electric and magnetic field transformation?

It's all a lot simpler than you're imagining. Pretty much everything of physical interest in relativity transforms as some kind of tensor.

Rest mass is a rank-0 tensor.

Energy is part of the energy-momentum vector, which is a rank-1 tensor.

The electric and magnetic fields are the ingredients of the Maxwell tensor, which is a rank-2 tensor.

-Ben

 

[lovetruth]

It's all a lot simpler than you're imagining. Pretty much everything of physical interest in relativity transforms as some kind of tensor.

Rest mass is a rank-0 tensor.

Energy is part of the energy-momentum vector, which is a rank-1 tensor.

The electric and magnetic fields are the ingredients of the Maxwell tensor, which is a rank-2 tensor.

-Ben

Does the sum of electric and magnetic force changes on charges when reference frame are changed (like in the question i have posted)?

 

[pervect]

The 3-force, dp/dt, depends on your reference frame. So the short but possibly misleading answer is yes.

However, the 4-force, dp/dtau, does NOT depend on the reference frame. To be more precise, the 4-force transforms via the Lorentz transform.

There are two differences - the second formula uses proper time tau,rather than coordinate time t, and it uses the energy momentum 4-vector, rather than the momentum 3-vector.

The 4-force is actually d(energy-momentum-4-vector)/dtau, where the energy-momentum 4-vector is

E, px, py, pz

E being the energy, px, py, and pz, being the x, y, and z components of momentum.

Like the 4-force, or any 4-vector in SR, the energy momentum 4-vector transforms via the Lorentz transform.

I'd suggest picking up some textbook like Griffiths.

https://www.amazon.com/dp/013805326X/?tag=pfamazon01-20

You should be able to get a copy from your local library, possibly via inter-library loan. If you like it, you might even want to buy it, if you have the book budget money.

 

[bcrowell]

However, the 4-force, dp/dtau, does NOT depend on the reference frame. To be more precise, the 4-force transforms via the Lorentz transform.

I know some people like to think about it this way, and I'm sure it's a self-consistent logical system, but it's not the way I think about it myself. I would just point out to lovetruth that if you write a 4-force in terms of its (t,x,y,z) components in a particular frame, then act on it with a Lorentz transformation, its components in the new frame are different numbers. So in that sense, it *does* change, and the Lorentz transformation says how it changes.

In lovetruth's OP, I believe the conclusion is correct. Let three-force F act between the two charges in the original frame. The four-force in that frame is (0,0,F,0) (based on the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, and the charges have gamma=1 in the original frame). Lorentz-transforming to the new frame gives (0,0,F,0) again (because a boost in the x direction doesn't change a y-component). Again applying the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, the corresponding three-force is F/\gamma.

My calculation agrees with the statement in Lovetruth's #1 that the three-force is smaller in the new frame.

My calculation agrees with the calculation in Bill K's #4 that F' = qE γ (1-v²/c²), since this quantity equals qE/γ=F/γ. (As lovetruth and Bill_K agreed in #5,6, there was a trivial algebra mistake when Bill equated this to qE.)

-Ben

 

[pervect]

I might not have been too clear. I do think of 4-vectors as geometric objects, and when I'm thinking of them as geometric objects I think of them as "unchanging" in some larger philosophical sense.

But this is potentially confusing because, as Ben points out, the components of these geometric objects do change as you change the reference frames. It's just that they all change in a standardized manner.

I tried to address that in my post,but I might not have spelt it out too clearly.

What really doesn't change is the norm of the four-vector.

 

[bcrowell]

Hi, pervect,

I think we're basically in agreement. We just have different words that we use to express the same math. Just wanted to head off any possible confusion.

-Ben

 

[lovetruth]

I know some people like to think about it this way, and I'm sure it's a self-consistent logical system, but it's not the way I think about it myself. I would just point out to lovetruth that if you write a 4-force in terms of its (t,x,y,z) components in a particular frame, then act on it with a Lorentz transformation, its components in the new frame are different numbers. So in that sense, it *does* change, and the Lorentz transformation says how it changes.

In lovetruth's OP, I believe the conclusion is correct. Let three-force F act between the two charges in the original frame. The four-force in that frame is (0,0,F,0) (based on the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, and the charges have gamma=1 in the original frame). Lorentz-transforming to the new frame gives (0,0,F,0) again (because a boost in the x direction doesn't change a y-component). Again applying the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, the corresponding three-force is F/\gamma.

My calculation agrees with the statement in Lovetruth's #1 that the three-force is smaller in the new frame.

My calculation agrees with the calculation in Bill K's #4 that F' = qE γ (1-v²/c²), since this quantity equals qE/γ=F/γ. (As lovetruth and Bill_K agreed in #5,6, there was a trivial algebra mistake when Bill equated this to qE.)

-Ben

It will be great if someone can name an experiment in which reduced EM forces on charged particles are observed, like: Interaction of alpha and beta particles as a function of their velocities in a bubble chamber.

If the EM force is reduced by changing frame then, all 4 fundamental forces(EM,gravity,strong,and weak forces) should have 'magnetic counterparts force' which can reduce the net force by 1/gamma [like Gravito-magnetic(made-up term)]. These 'magnetic counterparts' must exist otherwise reality will not be same in every frame.

I also have a great doubt about the problem in my OP: In the new frame B, should I
1. Apply Newtons law of motion with reduced EM force and neglect time dilation
2. Use electrostatic force as in frame A and apply time dilation
3. Use reduced EM force and also apply time dilation
Some how i feel reduced EM force is tantamount to time dilation and therefore, applied individually(options 1 & 2).

 

[lovetruth]

I have read on the internet that net forces on charged bodies does not change when frames are changed. This result is also shown by Feynman in his books.
Links: physics.weber.edu/schroeder/mrr/MRRtalk.html
galileo.phys.virginia.edu/classes/252/rel_el_mag.html

So the question still remains: Whether net force on charged particles changes when frames are changed?

 

[bcrowell]

It will be great if someone can name an experiment in which reduced EM forces on charged particles are observed, like: Interaction of alpha and beta particles as a function of their velocities in a bubble chamber.

Cathode ray tubes are a good example.

If the EM force is reduced by changing frame then, all 4 fundamental forces(EM,gravity,strong,and weak forces) should have 'magnetic counterparts force' which can reduce the net force by 1/gamma [like Gravito-magnetic(made-up term)]. These 'magnetic counterparts' must exist otherwise reality will not be same in every frame.

You need to distinguish between forces and fields here. All four-forces transform the same way. That doesn't mean that all fields have the same behavior.

I also have a great doubt about the problem in my OP: In the new frame B, should I
1. Apply Newtons law of motion with reduced EM force and neglect time dilation
2. Use electrostatic force as in frame A and apply time dilation
3. Use reduced EM force and also apply time dilation
Some how i feel reduced EM force is tantamount to time dilation and therefore, applied individually(options 1 & 2).

The way to approach this without getting confused is to stop using three-vectors and just use four-vectors, which transform in a consistent way. For example, you can express the equivalent of Newton's second law in terms of four-vectors.

-Ben

 

[bcrowell]

I have read on the internet that net forces on charged bodies does not change when frames are changed. This result is also shown by Feynman in his books.
Links: physics.weber.edu/schroeder/mrr/MRRtalk.html
galileo.phys.virginia.edu/classes/252/rel_el_mag.html

So the question still remains: Whether net force on charged particles changes when frames are changed?

I've gone through both pages carefully, and I don't see any statement that "net forces on charged bodies does not change when frames are changed."

Fowler's page at virginia.edu presents a simplified version of a derivation by Purcell, referring the reader to Purcell at the end for more details. One of the simplifications is that Fowler makes an approximation at "[...]so relativistic contraction will increase their density to[...]" Purcell does not make the approximation. This means that all of Fowler's results are only valid to leading order in v. At the end, he says, "This purely electrical force is identical in magnitude to the purely magnetic force in the other frame! So observers in the two frames will agree on the rate at which the particle accelerates away from the wire, but one will call the accelerating force magnetic, the other electric." He does not state this as a general law that forces and accelerations are frame-invariant (which would be false). He is also only saying this in the context of an approximation to leading order in v. His statement can't be true to all orders in v, because the 3-force f and the 3-acceleration a do not satisfy F=ma in all frames, if m is kept constant. What is the same in all frames is F=mA, where F is the four-force and A is the four-acceleration:
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
In the present case, where the velocity is perpendicular to the force, F=mA reduces to f=m\gamma a. This is the well known result for what some older books call the "transverse mass:"
http://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass
Fowler is only doing a calculation of f and a to leading order in v. To leading order in v, \gamma=1, and this is the justification for his claim that, in this context, f=ma holds in both frames.

If you want to see an exact treatment, check Purcell out of the library. It's a wonderful book.

-Ben

 

[lovetruth]

Cathode ray tubes are a good example.

Cathode ray are acted by electric and magnetic source which are stationary. I wanted an experiment in which electric and magnetic field is produced by a moving charged particle.
The best experiment would be observing the trajectories of alpha and beta particles which have same velocity and are at some distance to each other. The trajectories could be seen in cloud or bubble chamber. If anyone here is nuclear physicist by any chance then, this will be a great experiment.

 

[lovetruth]

Is there a way of using only 3-force with some changes and avoiding 4-force concept.

 

[bcrowell]

Cathode ray are acted by electric and magnetic source which are stationary. I wanted an experiment in which electric and magnetic field is produced by a moving charged particle.

Huh? All magnetic forces are produced by moving charges acting on moving charges.

Is there a way of using only 3-force with some changes and avoiding 4-force concept.

Sure, some books use them and some don't. What I would really suggest is that you stop trying to learn this stuff just by googling and asking questions on PF. There is no substitute for a book that presents the material in an organized way. The questions you're asking make me think that the following two books would probably be good ones for you to look at:

Purcell, Electricity and Magnetism
Taylor and Wheeler, Spacetime Physics

I think you'll have a lot more luck if you dig into one of those, approach the subject systematically, and ask questions here if there's something you don't understand.

-Ben

 

[lovetruth]

I know some people like to think about it this way, and I'm sure it's a self-consistent logical system, but it's not the way I think about it myself. I would just point out to lovetruth that if you write a 4-force in terms of its (t,x,y,z) components in a particular frame, then act on it with a Lorentz transformation, its components in the new frame are different numbers. So in that sense, it *does* change, and the Lorentz transformation says how it changes.

In lovetruth's OP, I believe the conclusion is correct. Let three-force F act between the two charges in the original frame. The four-force in that frame is (0,0,F,0) (based on the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, and the charges have gamma=1 in the original frame). Lorentz-transforming to the new frame gives (0,0,F,0) again (because a boost in the x direction doesn't change a y-component). Again applying the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, the corresponding three-force is F/\gamma.

My calculation agrees with the statement in Lovetruth's #1 that the three-force is smaller in the new frame.

My calculation agrees with the calculation in Bill K's #4 that F' = qE γ (1-v²/c²), since this quantity equals qE/γ=F/γ. (As lovetruth and Bill_K agreed in #5,6, there was a trivial algebra mistake when Bill equated this to qE.)

-Ben

In my first post(#1), I have derived the the net attractive force on each charged particle in frame B is:
F=qE(1-v^2/c^2)=qE/(gamma)^2 where, E=Electric field which is same in frame A &B

The net force in frame B as calculated by Bill K & bcrowell using 4-force is :
F=qE/(gamma)

There is a difference between the force i have calculated using simple EM and the force calculated by 4-force concept by a factor of gamma. The 2 forces derived using 2 different approaches should yield same results otherwise, relativity is not self-consistent.

 

[lovetruth]

Huh? All magnetic forces are produced by moving charges acting on moving charges.

All I wanted was an experiment that is equivalent to the situation in frame B: two charged particle with same velocity at some perpendicular distance 'r'. The trajectory of these charged particle would then give a precise amount of force acting on them at each instant. I wanted this experiment and not any other kind. In CRT u will have plethora of charged particles moving in all directions and are at time-varying distance to the electrons. This will be not be good for understanding dynamical forces between just 2 charged particles.

 

[bcrowell]

The 2 forces derived using 2 different approaches should yield same results otherwise, relativity is not self-consistent.

Just because different people get different answers to a problem, that does not mean that relativity is not self-consistent. Most likely it means that one of those people messed up. We have three results showing the same thing: mine, Bill_K's, and Wikipedia's equation for the transverse mass. Your result is the one that is different. Since I have seen the transverse mass result in multiple places, I'm convinced that the mistake is yours. If you want help finding your own mistake, you could ask nicely. Beating your chest and proclaiming relativity to be inconsistent does not constitute asking nicely.

In my first post(#1), I have derived the the net attractive force on each charged particle in frame B is:
F=qE(1-v^2/c^2)=qE/(gamma)^2 where, E=Electric field which is same in frame A &B

And now looking back at your #1 I see that there is no such material in it. The variable gamma never appears in that post.

 

[lovetruth]

Just because different people get different answers to a problem, that does not mean that relativity is not self-consistent. Most likely it means that one of those people messed up. We have three results showing the same thing: mine, Bill_K's, and Wikipedia's equation for the transverse mass. Your result is the one that is different. Since I have seen the transverse mass result in multiple places, I'm convinced that the mistake is yours. If you want help finding your own mistake, you could ask nicely. Beating your chest and proclaiming relativity to be inconsistent does not constitute asking nicely.


And now looking back at your #1 I see that there is no such material in it. The variable gamma never appears in that post.

OK I will derive the net force on charged particle as seen by an observer in frame B,
F_net = q(E+v*B)=[(q/r)^2]*[{1/4(pi)(epsilon)}-{(mew)*(v^2)/4(pi)}]
=qE[1-(epsilon)*(mew)*(v^2)]=qE[1-v^2/c^2]=qE/(gamma)^2

I thought you could see that simple result by yourself. This expression is legitimate because I have used physical laws that are valid in all inertial frame(one of the hypothesis of relativity).
how could you say that you are convinced that my results have errors and not yours without giving a proper justification; it can be the other way around.

 

[Dale]

I thought u could see that simple result by urself. This expression is legitimate because I have used physical laws that r valid in all inertial frame(one of the hypothesis of relativity).
how could u say that u r convinced that my results have errors and not urs without giving a proper justification; it can be the other way around.

SMS messaging shorthand is not acceptable here. Please review the rules that you agreed to when you signed up.

The burden of proof is always on the person challenging mainstream science. Einstein didn't get special treatment in that regard nor should you.

 

[lovetruth]

SMS messaging shorthand is not acceptable here. Please review the rules that you agreed to when you signed up.

The burden of proof is always on the person challenging mainstream science. Einstein didn't get special treatment in that regard nor should you.

Ok I have edited my post to remove shorthand. Dale, what is your views about the problem that I have posted.

 

[Dale]

Ok I have edited my post to remove shorthand.

Thanks, it is much more readable now.

Dale, what is your views about the problem that I have posted.

I agree completely with Ben's statement above:

The way to approach this without getting confused is to stop using three-vectors and just use four-vectors, which transform in a consistent way. For example, you can express the equivalent of Newton's second law in terms of four-vectors.

The effort that you invest to learn about four-vectors will be well worth your while. I didn't understand relativity for about 7 years until I stumbled on four-vectors. Then everything suddenly "clicked". I highly recommend it.

 

[bcrowell]

OK I will derive the net force on charged particle as seen by an observer in frame B,
F_net = q(E+v*B)=[(q/r)^2]*[{1/4(pi)(epsilon)}-{(mew)*(v^2)/4(pi)}]
=qE[1-(epsilon)*(mew)*(v^2)]=qE[1-v^2/c^2]=qE/(gamma)^2

I thought you could see that simple result by yourself. This expression is legitimate because I have used physical laws that are valid in all inertial frame(one of the hypothesis of relativity).
how could you say that you are convinced that my results have errors and not yours without giving a proper justification; it can be the other way around.

There is a mistake after the second equals sign, where you assume that the electric field of a moving charge is given by Coulomb's law. For the correct expression, see equation (1538) on this web page: http://farside.ph.utexas.edu/teaching/em/lectures/node125.html

 

[lovetruth]

There is a mistake after the second equals sign, where you assume that the electric field of a moving charge is given by Coulomb's law. For the correct expression, see equation (1538) on this web page: http://farside.ph.utexas.edu/teaching/em/lectures/node125.html

Sorry for my ignorance of the subject matter; I must read more.
But I have found that the E & B field with relativistic effect accounted doesn't lead to the force that you have derived using 4-force.
I will be using subscript 'a' or 'b' to differentiate the field & forces either in Frame A or B respectively.
The force on charged particle in Frame B is:
F_b = q(E_b+v*B_b)
But we know from equation(1539),
B_b = (v*E_b)/c^2
Therefore,
F_b = [q*E_b][1-v^2/c^2] = [q*E_b]/(gamma)^2
But from equation(1538),
E_b = E_a[(gamma)/{1+(v*gamma/c)^2}^(3/2)]
therefore,
F_b = [q*E_a]/[(gamma){1+(v*gamma/c)^2}^(3/2)]
therefore, the final result we get is:
F_b = F_a/[(gamma){1+(v*gamma/c)^2}^(3/2)]

But you have derived using 4-force in post #17 that:
F_b = F_a/(gamma)

Clearly, the relationship between F_b & F_a are not same in the two equations using E & B fields and 4-force separately. Please tell me if I made some mathematical or conceptual error in my calculations. If possible, derive the realtionship between the net force in frame B and frame A using only E & B fields and lorentz force(I know I am asking for too much).

 

[bcrowell]

Bill_K's calculation in #4, as corrected in #6, is similar in its approach to your latest attempt. We can be pretty sure that it's right, because his result matches up with Wikipedia and it also matches up with what I got using four-vectors. Therefore I would suggest that you try to correlate yours with his and try to find where the remaining errors are in yours by looking for places where your intermediate results disagree with his intermediate results.

 

[lovetruth]

Bill_K's calculation in #4, as corrected in #6, is similar in its approach to your latest attempt. We can be pretty sure that it's right, because his result matches up with Wikipedia and it also matches up with what I got using four-vectors. Therefore I would suggest that you try to correlate yours with his and try to find where the remaining errors are in yours by looking for places where your intermediate results disagree with his intermediate results.

I have no doubt in your calculation of net force in frame B by using 4-force concept. In my approach I have used electromagnetic force(corrected for relativistic effect, thanks to your link). I believe that 4-force is a lorentzian transformation i.e. it relates force in frame A(F_a) and force in frame B(F_b). In order to use 4-force, an observer in frame B has to know force in frame A. What is so special about the force in frame A? Nothing as all inertial frames are equivalent. Therefore, the observer in frame B can find the force by using electromagnetic field without any knowledge of force in frame A. So the point I am stressing here is that observers in frame A or B should be able to find out the force independently using EM fields without ever knowing the forces in each others frame. The force calculated independently by observers in frame A & B should be in agreement with 4-force(which is apparently not the case in the problem being considered).

I am looking for error in my calculation but in my derivation, everything is very straightforward so i doubt it will be easy to spot. I have used Lorentz force, and substituted E & B fields directly from the link you have posted. If you can't understand the steps of my derivation, you can ask me(I know I can't write mathematical expression nicely using HTML codes). It would be best if u can post the derivation of force in Frame B as I have attempted using EM fields(corrected for relativistic effects) and without touching 4-force entirely. I will be fully satisfied once you can show that EM force and 4-force give the same result(so that i can have peace of mind).

 

[bcrowell]

It would be best if u can post the derivation of force in Frame B as I have attempted using EM fields(corrected for relativistic effects) and without touching 4-force entirely. I will be fully satisfied once you can show that EM force and 4-force give the same result(so that i can have peace of mind).

I'm satisfied with my own result. If you want to debug your own result, I've suggested how to do it. I'm not interested in doing it for you.

 

[atyy]

Sorry for my ignorance of the subject matter; I must read more.
But I have found that the E & B field with relativistic effect accounted doesn't lead to the force that you have derived using 4-force.
I will be using subscript 'a' or 'b' to differentiate the field & forces either in Frame A or B respectively.
The force on charged particle in Frame B is:
F_b = q(E_b+v*B_b)
But we know from equation(1539),
B_b = (v*E_b)/c^2
Therefore,
F_b = [q*E_b][1-v^2/c^2] = [q*E_b]/(gamma)^2
But from equation(1538),
E_b = E_a[(gamma)/{1+(v*gamma/c)^2}^(3/2)]
therefore,
F_b = [q*E_a]/[(gamma){1+(v*gamma/c)^2}^(3/2)]
therefore, the final result we get is:
F_b = F_a/[(gamma){1+(v*gamma/c)^2}^(3/2)]

But you have derived using 4-force in post #17 that:
F_b = F_a/(gamma)

Clearly, the relationship between F_b & F_a are not same in the two equations using E & B fields and 4-force separately. Please tell me if I made some mathematical or conceptual error in my calculations. If possible, derive the realtionship between the net force in frame B and frame A using only E & B fields and lorentz force(I know I am asking for too much).

http://farside.ph.utexas.edu/teaching/em/lectures/node125.html Eq 1538 defines E in terms of r, whereas Eq 1529 defined E' in terms of r', so try checking your expression for E in terms of E'. Eq 1537 gives r' in terms of r.

Edit: Actually, it seems easier to just use Eq 1534 & 1536.

 

[lovetruth]

http://farside.ph.utexas.edu/teaching/em/lectures/node125.html Eq 1538 defines E in terms of r, whereas Eq 1529 defined E' in terms of r', so try checking your expression for E in terms of E'. Eq 1537 gives r' in terms of r.

Edit: Actually, it seems easier to just use Eq 1534 & 1536.

Thanks atty, you made me realize my mistake. For my derivation i will be using the convention as followed on the link.

F = q[E+v*B]
but from eqn 1539, B = v*E/c^2
F = qE[1-v^2/c^2]
F = qE/(gamma)^2
but from eqn 1538 & 1529 and taking u_r = 0,
E = E'(gamma)
therefore,
F= qE'/(gamma)
therefore,
F = F'/(gamma)

This result matches with 4-force prediction. I am not surprised by the agreement of EM force and 4-force because EM field of the moving charges is itself derived from the Lorentzian transformation. So any result obtained by using EM field of moving charges are bound to be in agreement with that of Lorentzian 4-force.
The eqn 1538 & 1539 shows that E & B field of moving charge is anisotropic as well as dependent on velocity of the moving charge. Is there any experiment proving that eqn 1538 & 1539 are true. For example, Coulomb law and Ampere law are easily proved by experimentations in a Lab. Note that Ampere law and eqn 1539 are not same(Ampere law does not include the gamma term).

 

[atyy]

Thanks atty, you made me realize my mistake. For my derivation i will be using the convention as followed on the link.

F = q[E+v*B]
but from eqn 1539, B = v*E/c^2
F = qE[1-v^2/c^2]
F = qE/(gamma)^2
but from eqn 1538 & 1529 and taking u_r = 0,
E = E'(gamma)
therefore,
F= qE'/(gamma)
therefore,
F = F'/(gamma)

This result matches with 4-force prediction. I am not surprised by the agreement of EM force and 4-force because EM field of the moving charges is itself derived from the Lorentzian transformation. So any result obtained by using EM field of moving charges are bound to be in agreement with that of Lorentzian 4-force.
The eqn 1538 & 1539 shows that E & B field of moving charge is anisotropic as well as dependent on velocity of the moving charge. Is there any experiment proving that eqn 1538 & 1539 are true. For example, Coulomb law and Ampere law are easily proved by experimentations in a Lab. Note that Ampere law and eqn 1539 are not same(Ampere law does not include the gamma term).

Although http://farside.ph.utexas.edu/teaching/em/lectures/node125.html" ).

 

[lovetruth]

Although http://farside.ph.utexas.edu/teaching/em/lectures/node125.html" ).

I have seen that electric and magnetic field of moving particle is also given by lienard-Wiechert potential. But this formula is different from that given inutexas.edu link.
http://en.wikipedia.org/wiki/Liénard–Wiechert_potential

 

[PAllen]

I have seen that electric and magnetic field of moving particle is also given by lienard-Wiechert potential. But this formula is different from that given inutexas.edu link.
http://en.wikipedia.org/wiki/Liénard–Wiechert_potential

First, to compare you should start with the E field formula in the section:

"Corresponding values of electric and magnetic fields"

Second, note that the wikipedia formulas are written in terms of retarded position (as described therein), while the Texas.edu link formula is in terms of current position. Finally, note that the second big term of the wikipedia E formula vanishes for inertial motion of the charge, which is the only case covered by the Texas formula. Also, different units allow you to ignore various 'noise' constants. I haven't worked it all out, but I am sure if take all of this into account you would find they are equivalent.

 

[atyy]

I have seen that electric and magnetic field of moving particle is also given by lienard-Wiechert potential. But this formula is different from that given inutexas.edu link.
http://en.wikipedia.org/wiki/Liénard–Wiechert_potential

Please compare carefully with Xiao's lecture notes that I linked to.

 

[lovetruth]

First, to compare you should start with the E field formula in the section:

"Corresponding values of electric and magnetic fields"

Second, note that the wikipedia formulas are written in terms of retarded position (as described therein), while the Texas.edu link formula is in terms of current position. Finally, note that the second big term of the wikipedia E formula vanishes for inertial motion of the charge, which is the only case covered by the Texas formula. Also, different units allow you to ignore various 'noise' constants. I haven't worked it all out, but I am sure if take all of this into account you would find they are equivalent.

Can you tell me about the 'noise constants' that you are referring. I really doubt that lienard weichart electric field is similar to that of the electric field derived by Lorentzian transformation(texas.edu). It will be nice if someone can show that the two equations involving retarded and current position are same.

 

[lovetruth]

Please compare carefully with Xiao's lecture notes that I linked to.

I can see that Xiao article include Lienard-Weichart formula. But I don't see how Lienard-Wiechart formula for Electric field & Electric field derived from Lorentzian transformation(texas.edu link) are equivalent.

 

[Dale]

The LW formulas are more general. They include accelerating charges whereas the transformation approach would only give you constant velocity charge.

 

[atyy]

I can see that Xiao article include Lienard-Weichart formula. But I don't see how Lienard-Wiechart formula for Electric field & Electric field derived from Lorentzian transformation(texas.edu link) are equivalent.

Try to compare http://physics.usask.ca/~xiaoc/phys463/notes/note19.pdf" 's Eq 1538.

 

[lovetruth]

Try to compare http://physics.usask.ca/~xiaoc/phys463/notes/note19.pdf" 's Eq 1538.

In my post #40, I have derived Electric field at a distance ‘r’ in y-direction of a charge moving with velocity ‘v’ in x-direction using Fitzpatrick eqn. 1538 that,
E = E’(gamma)
Where, E’ is the electric field when charge is at rest and , E is the electric field when charge is moving.
Now I will derive Electric field at a distance ‘r’ in y-direction of a charge moving with velocity ‘v’ in x-direction using Lienard-Weichert formula. I will assume velocity ‘v’ to be very small than c.
n = 0i + 1j
B = (v/c)i + 0j
Therefore, n – B = -(v/c)i +1j ~ 1j
Therefore, n.B = 0
Therefore, (1-n.B)^3 = 1
Using lienard-weichert formula for electric field,
E = E’/(gamma)^2

Clearly the electric field derived from Fitzpatrick (eqn 1538) and lienard-weichert formula(Xiao Eq 10.65) are different by a large amount.

 

[atyy]

In my post #40, I have derived Electric field at a distance ‘r’ in y-direction of a charge moving with velocity ‘v’ in x-direction using Fitzpatrick eqn. 1538 that,
E = E’(gamma)
Where, E’ is the electric field when charge is at rest and , E is the electric field when charge is moving.
Now I will derive Electric field at a distance ‘r’ in y-direction of a charge moving with velocity ‘v’ in x-direction using Lienard-Weichert formula. I will assume velocity ‘v’ to be very small than c.
n = 0i + 1j
B = (v/c)i + 0j
Therefore, n – B = -(v/c)i +1j ~ 1j
Therefore, n.B = 0
Therefore, (1-n.B)^3 = 1
Using lienard-weichert formula for electric field,
E = E’/(gamma)^2

Clearly the electric field derived from Fitzpatrick (eqn 1538) and lienard-weichert formula(Xiao Eq 10.65) are different by a large amount.

I'm going to try to start from http://physics.usask.ca/~xiaoc/phys463/notes/note19.pdf" 's formula, and see if it matches the Lorentz transformed version.

Xiao's formula contains the retarded position r_p(t_r):
R=r-r_p(t_r), where t_r=t-R/c.

For a charge moving at constant velocity, the current position r_p(t):
r_p(t)=r_p(t_r)+v.(t-t_r)

I also define the displacement vector from the current position:
R'=r-r_p(t)

Writing in terms of R':
R'=R-Rβ
R'=((R-Rβ).(R-Rβ))^1/2=R/γ (R'.β=0 for a perpendicular location)
n-β=R'/γR'
1-n.β=(1-γR'.β/R')/γ²=1/γ² (R'.β=0 for a perpendicular location)

Xiao's Eq 10.65 has n-β/((1-n.β)³.R².γ²), which for a perpendicular location works out to γR'/R'³.