How Do Moving Charges Behave Differently in Different Reference Frames?

In summary, the two point charges in space are attracted to each other more in the moving frame than in the original frame.
  • #36
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.
 
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  • #37
bcrowell said:
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).
 
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  • #38
lovetruth said:
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.
 
  • #39
lovetruth said:
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.
 
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  • #40
atyy said:
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).
 
  • #41
lovetruth said:
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" ).
 
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  • #42
atyy said:
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
 
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  • #43
lovetruth said:
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.
 
  • #44
lovetruth said:
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.
 
  • #45
PAllen said:
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.
 
  • #46
atyy said:
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.
 
  • #47
The LW formulas are more general. They include accelerating charges whereas the transformation approach would only give you constant velocity charge.
 
  • #48
lovetruth said:
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.
 
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  • #49
atyy said:
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.
 
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  • #50
lovetruth said:
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 rp(tr):
R=r-rp(tr), where tr=t-R/c.

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

I also define the displacement vector from the current position:
R'=r-rp(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.
 
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  • #51
atyy said:
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 rp(tr):
R=r-rp(tr), where tr=t-R/c.

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

I also define the displacement vector from the current position:
R'=r-rp(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.

Thank you for the derivation of E field using Xiao's eqn. which perfectly agrees with the fitzpatrick eqn. 1538.
I am wondering whether the lienard-Weichert potential has been proven experimentally like coulomb law.
 
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  • #52
Also, are xiao eqn and jeffimenko equation are same?
 
  • #53
lovetruth said:
I am wondering whether the lienard-Weichert potential has been proven experimentally like coulomb law.
The Lienard-Weichert potential follows directly from Maxwell's equations, so yes. In fact, it is much more general than Coulomb's law, which is only an approximation for a stationary charge. The LW reduces to Coulomb's law in that situation.
 
  • #54
lovetruth said:
Thank you for the derivation of E field using Xiao's eqn. which perfectly agrees with the fitzpatrick eqn. 1538.
I am wondering whether the lienard-Weichert potential has been proven experimentally like coulomb law.

lovetruth said:
Also, are xiao eqn and jeffimenko equation are same?

I don't know about the Xiao and Jeffimenko equations, and I don't have detailed knowledge of an experimental proof of the Lienard-Wiechert potentials, but I would suggest looking at applications of the (generalized) http://en.wikipedia.org/wiki/Larmor_formula" .

More generally, the Lorentz transformations are a http://www.physics.ox.ac.uk/users/iontrap/ams/teaching/rel_B.pdf" of Maxwell's equations.

Einstein's contribution was not to propose the Lorentz transformations, but to modify Newton's second law to be consistent with the Lorentz symmetry of Maxwell's equations. Einstein was applying the same sort of reasoning you used when you said:

lovetruth said:
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 don't know if gravity has a "magnetic" component (I don't think http://en.wikipedia.org/wiki/Gravitomagnetism" , Eq 22).
 
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  • #55
It's probably a bit off topic, but the Bel decomposition of the Riemann does break it down into four parts. It's common to call one the electric part, (or the electrogravitic tensor), another the magnetic part,and the third the topological part. I'm not sure about the fourth part - perhaps it's zero.

See https://www.physicsforums.com/showpost.php?p=1347429&postcount=10, the electric part of the Riemann, sometimes called the tidal tensor, is just R_{abcd} u^b u^d, where u is a timelike vector of some observer.

Chris mentions taking the "right hodge dual" to get the magnetic part, I'm not quite sure how that's done..
 
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  • #56
pervect said:
It's probably a bit off topic, but the Bel decomposition of the Riemann does break it down into four parts. It's common to call one the electric part, (or the electrogravitic tensor), another the magnetic part,and the third the topological part. I'm not sure about the fourth part - perhaps it's zero.

See https://www.physicsforums.com/showpost.php?p=1347429&postcount=10, the electric part of the Riemann, sometimes called the tidal tensor, is just R_{abcd} u^b u^d, where u is a timelike vector of some observer.

Chris mentions taking the "right hodge dual" to get the magnetic part, I'm not quite sure how that's done..

That sounds really interesting! The only introductory material on that I've found so far is Senovilla's http://arxiv.org/abs/gr-qc/0010095.
 
  • #59
atyy said:
Looks like a great set of lectures - thanks for the link!

You are wellcome. I am not sure if you have already discovered his QM courses, but here is the link anyway,



http://www.cosmolearning.com/courses/quantum-physics/

also the profs wikipedia page



http://en.wikipedia.org/wiki/V._Balakrishnan_(physicist )
 
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  • #60
atyy said:
I don't know about the Xiao and Jeffimenko equations, and I don't have detailed knowledge of an experimental proof of the Lienard-Wiechert potentials, but I would suggest looking at applications of the (generalized) http://en.wikipedia.org/wiki/Larmor_formula" .

More generally, the Lorentz transformations are a http://www.physics.ox.ac.uk/users/iontrap/ams/teaching/rel_B.pdf" of Maxwell's equations.

Einstein's contribution was not to propose the Lorentz transformations, but to modify Newton's second law to be consistent with the Lorentz symmetry of Maxwell's equations. Einstein was applying the same sort of reasoning you used when you said:



I don't know if gravity has a "magnetic" component (I don't think http://en.wikipedia.org/wiki/Gravitomagnetism" , Eq 22).

Thanks for the insight.
 
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