Example A and B, where the only difference is the movement speed of the observer

The question I have about it is how we can get the same result of net force acting on the individual protons if we judge the system from 2 different reference frames. One using more of the magnetic part of the electromagnetism, other more of electrical part of it to calculate the forces.

So I'll try to explain my understanding of it trough these two examples in which I get different results of particle acceleration depending on the different reference frames named in the Picture as "observer reference frames".

For simplicity I've decided to ignore the magnetic field created by the particles spins. I think they do not matter for this mental excercise.

In example A, both protons are traveling in the same direction and with the SAME speed. If the observer is stationary, he sees both of them moving away from him at the same speed. Because the protons have the same charge, they are repelled away from eachother by electrostatic force. But because they are moving, the upper proton creates a magnetic field around itself and that field acts on the other proton in such a way to push him up (towards the other proton). The same is true for the other proton which magneticaly attracts the other proton. The direction net force is dependant on the speeds of protons and how far away from eachother they are. So in one case they might attract eachother (high speeds) and in other case repel (close together).

Now let's look at example B. The observer is traveling at the same speed as the protons and in the same direction. The protons according to the observers aren't moving. So magnetic force doesn't appear between the protons. The only thing the protons feel according to the observer is the repulsive electrostatic force between them, so the protons, no matter how far away from eachother they are or at what velocity they are traveling (according to the stationary observer from case A) Will always repel eachother.

Now this entirely contradicts itself. Does this have something to do with time dilation and space contraction effects? And if so, how?

I've read the examples of a current conducting wire and it's effect on a moving independant charge. That picture makes sense because of space contraction, so there are more of opposite (stationary) charges in the same volume of space (in a wire) than if the charges weren't moving relative to eachother. That creates an electrostatic force which can in different reference frame be considered a magnetic force. But here we do not have the "help" from those opposite charges. And I believe the same effect still occures. How?

  • $\begingroup$ see here physics.stackexchange.com/q/304901 $\endgroup$ – Mihai B. Jan 26 '17 at 11:03
  • $\begingroup$ I am sorry, but my knowledge of mathematics and electromagnetism is not so extensive to understand that. It would be fine if anyone could just explain in layman's terms because of which aspect of special relativity, both observers (moving and stationary) see the "same" events unfold. Because as far as I know, if you ignore particle's spin, if two particles aren't moving according to eachothers reference frames they shouldn't feel magnetic force and so no matter what velocity they travel at compared to let's say earth, they should never feel attractive force. $\endgroup$ – MaDrung Jan 26 '17 at 14:09
  • $\begingroup$ Closely linked physics.stackexchange.com/questions/306371/… $\endgroup$ – Rob Jeffries Jan 29 '17 at 9:06

Short answer:

Force is not a Lorentz invariant and neither is acceleration. The protons always repel each other, with a force that combines the electric and magnetic components of the Lorentz force and depends on the frame of reference of the observer, but which is maximised in their rest frame and which approaches zero as the protons become ultra-relativistic.


Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ is the E-field of a stationary proton.

In the lab frame, the electric field in the direction between the two protons is increased by the Lorentz factor to $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$ and where $\gamma \geq 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ This is exactly as required by the rules for transforming forces under special relativity. The force acting between the two protons is smaller in the lab frame and approaches zero as the protons become more and more relativistic.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

  • $\begingroup$ "The problem you may be having is in thinking that force is a relativistic invariant - it is not." is misleading - the observer relative velocity doesn't influence the Net force - just how much is attributed to E or B, which your next line states correctly: "Electric and magnetic fields are transformed when looking at them from a different frame of reference." $\endgroup$ – f5r5e5d Jan 29 '17 at 1:38
  • $\begingroup$ Read the whole answer @f5r5e5d , the net force is indeed changed by a factor of $\gamma$. Or are you claiming that forces are invariant under Lorentz transformations? sciencebits.com/Transformation-Forces-Relativity. It is important we get this straight, since there are several confused answers here. Did you read Purcell & Morin, or do you have another source for your claim? $\endgroup$ – Rob Jeffries Jan 29 '17 at 9:01
  • $\begingroup$ 1st principle of Special Relativity - The laws of physics are invariant (i.e. identical) in all inertial systems (non-accelerating frames of reference). $\endgroup$ – f5r5e5d Jan 29 '17 at 19:23
  • $\begingroup$ @f5r5e5d And so they are. The law $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ is true in both frames of reference. That does not mean that $\vec{F}$ is invariant (and it isn't) . Indeed normal vectors (like force, acceleration, velocity, electric and magnetic fields) CANNOT be invariant under Lorentz transformations. I suggest you consult any elementary textbook on special relativity before posting further comments that will confuse the OP. $\endgroup$ – Rob Jeffries Jan 29 '17 at 19:39
  • $\begingroup$ "box car gedankenexperiment" - spring scale measuring the force between the protons - the needle doesn't point to different forces for different observers $\endgroup$ – f5r5e5d Jan 29 '17 at 20:07

I feel you are missing the relativistic formulas for the electromagnetic fields spread by uniformly moving charges. These can be found as formulas (1538), (1539) in this link.

In this case the velocity of the moving charges matters in both moving frames so you should get the same physical phenomena.

  • $\begingroup$ But how can you determine that they are moving? Relatively to eachother they aren't. So no magnetic force should be felt by them and hence no attraction. But if you look at them from some diferent reference frame they actualy "crash" together because they are moving a lot faster than the observer and so they feel eachothers magnetic fields. Something is not right with this. You can't really say at what speed something is traveling unless you're calculating the speed relative to something else. But in one reference frame they feel attractive magnetic force, in other there is no attraction. $\endgroup$ – MaDrung Jan 26 '17 at 14:14
  • $\begingroup$ I still think you haven't done the maths yet with the given formulas above. The thing is that you assume in the "stationary observer frame" the electric/magnetic field is the same as an static charge/uniform and infinite current can create and that is not true. When you input the velocity, relative to the observer in that frame, the balance among electric and magnetic forces should be the same as in the frame with the observer moving along with the two particles. $\endgroup$ – Juan Luis Gómez González Jan 26 '17 at 16:02
  • $\begingroup$ Force is not a relativistic invariant. $\endgroup$ – Rob Jeffries Jan 28 '17 at 17:51
  • $\begingroup$ @MaDrung: Let's assume the protons move close to speed of light seen by a 1. observer. According to the formulas above, the Colomb repulsion almost cancelled out by the magnetic attraction, so both protons will slowly move towards each other, because the cancellation is almost complete. Now you go in the rest frame of the protons. You see both particles fast moving towards each other due to only Colomb, why? In the rest system of the protons the time is dilated, it flows very slowly, and in this slowly flowing time units indeed the slow motion seen by the first observer appears finally fast. $\endgroup$ – Frederic Thomas Jan 28 '17 at 22:01
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    $\begingroup$ @MaDrung: I am sorry for the confusion I created. I recomend to you at the formulas given in the answers, you have a repelling Colomb force between the protons in the reference frame where they are at rest, (simply Colomb law), and in a reference frame where the protons are moving perpendicular to the direction of the force, the force will be weaker until it get's cancelled out if the protons moved at speed of light .But the magnetic force will be never stronger than the Colomb force in any reference system. $\endgroup$ – Frederic Thomas Jan 31 '17 at 19:26

Now this entirely contradicts itself. Does this have something to do with time dilation and space contraction effects? And if so, how?

Forces are different in different frames, and that is not a problem. If force was zero in some frame and non-zero in another frame, that would be a problem, but luckily there is a repulsive force between those two protons in all frames.

In the proton pair's frame it takes some time for the protons to move some distance apart. In another frame that time is longer, this is the time dilation effect. If the protons are moving very fast, then the force between them is very small, and it takes very long time for the proton's to move some distance apart.

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    $\begingroup$ So electrostatic force is so big that no matter the speed of the protons, it would never be overcome by magnetic force (which attracts in this case)? And all those effects can be explained with time dilation? $\endgroup$ – MaDrung Jan 30 '17 at 7:00
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    $\begingroup$ "So electrostatic force is so big that no matter the speed of the protons, it would never be overcome by magnetic force (which attracts in this case)?" YES. "And all those effects can be explained with time dilation?" NO. ... OR MAYBE YES, AS WE CAN VERY EASILY CALCULATE THE DISTANCE BETWEEN THE PROTONS AT SOME TIME USING THE IDEA OF TIME DILATION. ... BUT THEN THE ONLY THING THAT GETS EXPLAINED IS THE MOTION OF THE PROTONS, IT IS EXPLAINED BY A COULOMB FORCE AND A TIME DILATION EFFECT. $\endgroup$ – stuffu Jan 30 '17 at 14:23

In case of 2 equal charged particles of charge $e$ (as the proton) at rest in a reference frame their mutual force is according to Colomb's law the well-known formula (in SI-units): $F=\frac{e^2}{4\pi \epsilon r^2}$

$r$ is the distance between both, and $\epsilon$ the dielectricity constant of the medium. If both particles move with velocity $v$ if observed from another reference frame its force (both electric and magnetic force are considered in the following) is

$F=\frac{e^2}{4\pi \epsilon r^2}\left(1-\frac{v^2}{c^2}\right)$ (to be multiplied by $\times\gamma$, see below)

where $c$ is the speed of light. If $v=0$ the old Colomb's law is recovered. On the other hand in case both protons move at speed of light $c$ the force between is zero. In other words: at speed of light electric colomb force and magnetic Lorentz force cancel out each other (in this particular case). This is how electrodynamics works. Length contraction only happens in the direction of motion which is perpendicular of the direction of force, it does not play any role here. You could imagine that due to the time dilation which infinite at speed of light that the repulsion is in fact frozen, it fits with the intuitive view one might have on this effect. A derivation of this formulas can be found in the standard textbooks, Jackson for instance.

Edit: Apparently the second formula has to be corrected by a factor $\gamma$. Then it turns out to be $F=\frac{e^2}{4\pi \epsilon r^2}\sqrt{1-\frac{v^2}{c^2}}$ The qualitative argumentation is not changed, however.

  • $\begingroup$ Could you say where in Jackson this is derived. $\endgroup$ – Rob Jeffries Jan 28 '17 at 18:06
  • $\begingroup$ plug equation (11.152) in (12.1). Then you get it. $\endgroup$ – Frederic Thomas Jan 28 '17 at 18:33
  • $\begingroup$ I'm afraid I don't. $\endgroup$ – Rob Jeffries Jan 28 '17 at 18:43
  • $\begingroup$ From Jackson it is not so evident. I recommend you to google "space charge of proton beams", you will find lecture notes which will explain it. $\endgroup$ – Frederic Thomas Jan 28 '17 at 21:39
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    $\begingroup$ In the meantime I looked up some other source and it seems to be that your formula is correct. The argument with the length contraction of the line density apparently explains the $1/\gamma$ discrepancy. $\endgroup$ – Frederic Thomas Jan 29 '17 at 0:41

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