Relativistic Transformation of Lorentz Force (E + v x B)

Click For Summary
The discussion revolves around the relativistic transformation of the Lorentz force experienced by a test charge between two charged capacitor plates. Initially, it is noted that in the rest frame of the plates, the Lorentz force simplifies to q[E], as the magnetic field B is zero. However, when analyzing the frame of the moving charge, the electric field transforms to E', but the force does not equate as q[E] does not equal q[E']. The discrepancy arises because the Lorentz force is a three-vector and is not invariant across different inertial frames, influenced by time dilation and length contraction. Ultimately, the conversation highlights the complexities of relativistic physics and the necessity for further study to fully grasp these concepts.
mairzydoats
Messages
40
Reaction score
3
Here is a problem:
Imagine two equally charged capacitor plates parallel to the x-y axis, whose area is large enough compared to the distance between them that fringe effects can be ignored. The bottom plate (at z=0) is + charged, and the top is - charged. The vector field E is therefore directed upward ('upward' in this case being the positive 'z' direction) from the + plate to the - plate, and its magnitude should be equal to [surface charge density/epsilon].

If we examine the Lorentz force q[E + vxB] experienced by a test charge situated between the two plates, and moving with constant velocity parallel to the plates as seen from the frame of reference in which the plates are at rest, relativistic theory tells us we should get the same total Lorentz force (albiet with different E and B values) when we examine it in the frame of the moving charge, in which the test charge is still, and the plates are in motion.

Well then, first let's look at the frame where the plates are at rest and the test charge moves. Because the plates are at rest, B must be zero, and therefore the vxB term must also be zero. The Lorentz force reduces to q[E].

But now let's look at the frame of the moving test charge. In this frame, there is a B-field due to the moving plates, but because we are moving with the test charge, it is at rest in this frame ... and therefore magnitude of B doesn't effect the force it experiences. The vxB term once again vanishes, leaving the Lorentz force to be q[E'], where E' means the electric field of the first reference [E] frame as transformed by the lorentz factor of [1/(1-v squared/c squared)].

But q[E] does NOT equal q[E']!

What is wrong with this analysis?
 
Physics news on Phys.org
mairzydoats said:
we should get the same total Lorentz force
The Lorentz force is a three vector, so it is not invariant.
 
  • Like
Likes bcrowell
DaleSpam said:
The Lorentz force is a three vector, so it is not invariant.
Thank you for your reply. I apologize if my understanding of the subject matter is not advanced enough for me to know what you mean. How can the force on something, which is a 'real event', change simply because change our inertial frame of obervation?
 
mairzydoats said:
Thank you for your reply. I apologize if my understanding of the subject matter is not advanced enough for me to know what you mean. How can the force on something, which is a 'real event', change simply because change our inertial frame of obervation?
Force is mass times acceleration, and acceleration is the second derivative of distance with respect to time. Due to time dilation and length contraction acceleration in one frame is different from acceleration in another frame, therefore so is force.
 
DaleSpam said:
Force is mass times acceleration, and acceleration is the second derivative of distance with respect to time. Due to time dilation and length contraction acceleration in one frame is different from acceleration in another frame, therefore so is force.

but the force in my question is created by an E field whose only component is in the positive z-direction, and this, along with any acceleration it may create, is also in the positive z-direction. The transformation of these frames parallel to the x-y plane the test charge moves in doesn't contract length parallel to z, correct?
 
mairzydoats said:
but the force in my question is created by an E field whose only component is in the positive z-direction, and this, along with any acceleration it may create, is also in the positive z-direction. The transformation of these frames parallel to the x-y plane the test charge moves in doesn't contract length parallel to z, correct?

oh wait a minute ... [duhh] ... time still dilates. ha-ha. I guess a clock moving with my test charge would tick less than a clock situated on one of the plates as the charge is accelerated in the positive z direction by either E or E'. I still really have to mull this over & study before I'm quite comfortable. But thanks for pointing me in the right direction!
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 101 ·
4
Replies
101
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K