I am trying to understand forces in relativity:

Suppose there are two electrons A , B travelling side by side at 1 cm distance at .99 c (in space or at lHC, ignoring all other factors).

I know that $\gamma=7.089$, therefore energy-mass is 7 times greater in the lab frame whereas time and length of the electrons 7 times shorter in A-B frame, and Fe (A-B) at 1 cm is 2.3*10^-24 N. After 1/1000 second (in the lab frame, and 1 /7000 s in A-B frame) the two electrons will have drifted apart a few cm

enter image description here

In this article it is said that the ratio $F_B/F_E$ is $v^2/c^2$ for $v \ll c.$ If we need no relativistic corretion, according to that formula the magnetic force is .9801 times Fe in the lab frame and the net repulsive force will be 10^-26 N.

enter image description here

Now, as far as I can reckon, the difference in time is compensated by by the difference in mass, therefore I register a shorter distance in the lab frame than in the A-B frame.

Bounty question:

Can anyone fill in the second picture the correct values of all the forces acting between A, B and C in the lab frame and the resulting position of A and B after 1 /1000 second has elapsed in the lab frame ? If there were a difference in the outcome, could one deduce absolute motion?

please do not give formulae that are accessible on wiki or elsewhere

  • 3
    $\begingroup$ How is $0.99\ll1$? $\endgroup$ May 2, 2016 at 11:50
  • 5
    $\begingroup$ The downvote is independent of that shortcoming and represents an assessment of the question as a whole. Nevertheless, it is meaningless to quote the value 0.9801 as it simply does not apply. In the meantime, it seems you could use some time with a good relativity textbook and e.g. Purcell's chapter 5. For one, "if you spot a body approaching at near $2c$", then you can be sure that you're in a universe with different physics than the one we live in. $\endgroup$ May 2, 2016 at 12:21
  • 4
    $\begingroup$ If two bodies approach each other at near $c$ in the lab frame then, in our physics, each body will see the other approaching at near $c$, with a precise speed given by the relativistic velocity addition formula. This is explained in detail in any textbook on special relativity. $\endgroup$ May 2, 2016 at 13:30
  • 4
    $\begingroup$ I won't provide a tutorial on basic relativity - and indeed I won't respond again to basic queries. Your usual intuitions about time and space do not hold when objects move $c$ and the speed of light is a physical invariant; in particular, results in the lab frame do not translate directly into the perceptions of moving observers. For the details, again, see any introductory textbook on special relativity. $\endgroup$ May 2, 2016 at 13:54
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    $\begingroup$ Six different questions is too much for one post. $\endgroup$
    – David Z
    May 6, 2016 at 16:35

5 Answers 5


This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations.

As you mentioned, you can put your reference frame origin at one of the electron's position (though for actual calculations it might be easier to go into a center of mass system to use the symmetry of the situation). W.r.t. that reference frame, both electrons are at rest as the initial condition and hence there is only the electrostatic potential present and therefore no magnetic field. Therefore, in the rest frame chosen, the electrons will repulse each other, resulting in a divergence of their positions.

Now from the point of view of another reference system, for example an observer in a laboratory from whose point of view the electrons have a velocity of $.99c$ in say the $x$-direction, as you mention in your description, there indeed is a magnetic field present. Following special relativity, you write down the present field in the electrons rest frame (using the four-vector description of the electromagnetic potential) and use the appropriate Lorentz transformation to obtain the electromagnetic potential in the reference frame of the observer. The Lorentz transformation will mix up electric and magnetic potential, resulting in the interpretation that the magnetic field is the electric field in a different reference frame, though it would be more precise to formulate that using the vector potential $A$ as the physical entity.

I hope that clears up all the other questions you asked, as they really depend on the chosen reference frame and can be easily figured out having understood the situation in the rest frame of the electron.

EDIT: Note, that I'm only discussion the initial situation, i.e. I'm neglecting that the electrons will diverge resulting in a relative velocity and hence in a magnetic field present in the reference frame of one of the electrons. But I think your question regarded that initial situation where the electrons are at rest relative to each other and the fact that they are "moving from an observers point of view".

  • $\begingroup$ "...Following special relativity, you write down the present field in the electrons rest frame (..l) and use the appropriate Lorentz transformation.." that is just what my question was about and what I was expecting you to show. Can you at least address the first aspect and specify what does the value of $.98 Fe$ become with the SR correction $\endgroup$
    – user104372
    May 1, 2016 at 6:32
  • $\begingroup$ What you seem to miss is that the fact that the electrons are moving with some speed close to the speed of light relative to some observer's rest frame, does not have any "influence" on their attractive/repsulsive behaviour, i.e. you can calculate if they attract or repulse each other in the centor of mass system as mentioned. Here, you can easily calculate either in an electrostatic approximation, or include the fact that, one accelerated, the elctrons will generate a magnetic field. The factor .99c (or just .99) will not appear in any quotient of some electric/magnetic fields in CM-system $\endgroup$ May 2, 2016 at 8:11
  • $\begingroup$ Sorry, I will not make any diagrams or detailed calculations as I don't have the time, I'm simply trying to help you understand the situation. And I'm telling you again: do all the calculation in the center of mass system, forget about the velocity $.99c$ that the elcetrons move w.r.t. to some oberserv's reference frame. The will get a definite answer, wheter the electrons will acctract of repulse. Then, if you are interested, you can transform into the observer's frame using Lorentz transformation. Following the principals of special relativity, the physics in both frames is the same! $\endgroup$ May 2, 2016 at 9:26

Lorentz Transformations

Suppose we call the lab frame the K-frame and a frame moving at velocity, $\mathbf{v}$, relative to the K-frame called the K'-frame. Then we can express the electromagnetic fields in the K'-frame in terms of the K-frame fields as: $$ \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \boldsymbol{\beta} \times \mathbf{B} \right) - \frac{\gamma^{2}}{\gamma + 1} \boldsymbol{\beta} \left( \boldsymbol{\beta} \cdot \mathbf{E} \right) \tag{1a} \\ \mathbf{B}' & = \gamma \left( \mathbf{B} - \boldsymbol{\beta} \times \mathbf{E} \right) - \frac{\gamma^{2}}{\gamma + 1} \boldsymbol{\beta} \left( \boldsymbol{\beta} \cdot \mathbf{B} \right) \tag{1b} \end{align} $$ where $\mathbf{E}(\mathbf{B})$ is the electric(magnetic) field in the K-frame, $\boldsymbol{\beta} = \mathbf{v}/c$, $c$ is the speed of light, and $\gamma$ is the relativistic Lorentz factor.

1) what is the actual value...?

The $k$ factor in that article is $\left( 4 \pi \varepsilon_{o} \right)^{-1} = 8.987552 \times 10^{9} \ m/F$, where $\varepsilon_{o}$ is the permittivity of free space. The $q^{2}$ value is just the elementary charge squared = $2.56697 \times 10^{-38} \ C^{2}$. Therefore, $k \ q^{2} = 2.307078 \times 10^{-28} \ V \ m \ C$ and now if you divide by $1 \ cm = 10^{-2} \ m^{-1}$ squared, then $F_{E} = 2.307078 \times 10^{-24} \ N$.

The factors in $F_{B}$, $\mu_{o} \ q/4 \pi$, has a magnitude of $1.602218 \times 10^{-26} \ N \ s^{2} \ C^{-1}$, where $\mu_{o}$ is the permeability of free space.

Intuitively it shouldn't be smaller, yet it can't possibly get greater than 1, can it?

The magnitude of $F_{B} = 4.7552 \times 10^{-14} \ N$, which is roughly $10^{10}$ times larger than the magnitude of $F_{E}$.

I am not sure why you think that the magnetic force could not exceed the electric force, but it depends upon the situation.

2) what happens to the trajectories of the particles due to the influence of the mutual magnetic field (Lorenz force), do they converge?

That depends upon the situation. In your specific example, you should realize that the solar wind is always permeated by a background magnetic field and that the particles generally follow the magnetic field.

3) their paths should also diverge since the repulsion (Coulomb force) is slightly stronger than the attraction (Biot-Savart)? What is the result of the interaction of all these forces, what happens to the particles? I tried to make a graph. Is it correct? what are the final values?

So charged particles are almost never isolated. In the solar wind (and nearly all plasmas in the universe), the particles' electrostatic fields are screened by oppositely charged particles over an average distance called the Debye length. Therefore, unfortunately, these two particles may not really act as two independent particles. In fact, they will most likely act as part of velocity distributions and exhibit a collective behavior more like a fluid than just two individual particles.

4) why does the magnetic field produce an attraction?

It's similar to the problem of two parallel current-carrying wires. The $\mathbf{j} \times \mathbf{B}$-force (related to the Hall effect) of the current from wire 1 under the influence of the magnetic field from wire 2 will produce a force on wire 1 directed toward wire 2. The same is true for wire 2 due to the influence of wire. Thus, the two wires could be said to attract each other.

5) lastly, but most important, I read that the magnetic force is interpreted through relativity as the electric field in a different frame, is that right? But here the two electrons are in the same frame, and in that frame they are not moving (apart from a slight divergence) and they should not feel any magnetic force.

See the Equations 1a and 1b above...

6) how can a repulsive force transform into an attractive one in a different frame?

A force is not a Lorentz invariant. Meaning, the direction and magnitude of a force can change in a different reference frame. Forces conform to what are called Lorentz transformations. So forces can qppear different in different reference frames.

can an observer C moving with the CM deduce that they all (A,B,C) are moving even if the are not aware, and even find out the actual speed they are moving at, just observing that the value of the repulsion is different from the one predicted by Coulomb?

No, motion is not something you can deduce. Think about riding in a car, train, or air plane at constant speed. Without looking out a window you could not determine that you were moving. You can measure electric and magnetic fields in the frame of reference of the detector. Without a measurement in a different reference frame, you can only discuss the measurement in the measurement frame. You can apply a Lorentz transformation to those measurements to infer what you should observe in any given frame, but you cannot determine that you are moving based upon electric and magnetic field measurements alone.

Example Reversal

For simplicity, let us use the non-relativistic limit (i.e., $\gamma \rightarrow 1$) and assume: $$ \begin{align} \mathbf{v} & = \left( 0, v_{y}, 0 \right) \\ \mathbf{u} & = \left( u_{x}, 0, 0 \right) \\ \mathbf{E} & = \left( E_{x}, 0, 0 \right) \\ \mathbf{B} & = \left( 0, 0, B_{z} \right) \end{align} $$ where $\boldsymbol{\beta} = \mathbf{v}/c$, $\mathbf{u}$ is the instantaneous particle velocity in the K-frame, and we assume $u_{x} > 0$ and $B_{z} > 0$. Then we can show that the Lorentz force in the K-frame is given by: $$ \begin{align} \mathbf{F} & = q \ \left( \mathbf{E} + \frac{\mathbf{u}}{c} \times \mathbf{B} \right) \tag{2a} \\ & = q \ \left( E_{x}, - \frac{u_{x} \ B_{z}}{c}, 0 \right) \tag{2b} \end{align} $$

Performing Lorentz transformations and velocity additions, we can show that the relevant 3-vectors in the K'-frame are: $$ \begin{align} \mathbf{u}' & = \left( u_{x}, -v_{y}, 0 \right) \\ \mathbf{E}' & = \left( E_{x} + \frac{v_{y} \ B_{z}}{c}, 0, 0 \right) \\ \mathbf{B}' & = \left( 0, 0, B_{z} + \frac{v_{y} \ E_{x}}{c} \right) \\ \mathbf{F}' & = q \ \left( \mathbf{E}' + \frac{\mathbf{u}'}{c} \times \mathbf{B}' \right) \tag{3a} \\ & = q \ \left( E_{x} \ \left[ 1 - \frac{v_{y}^{2}}{c^{2}} \right], - \frac{u_{x}}{c} \ \left[ B_{z} + \frac{v_{y} \ E_{x}}{c} \right], 0 \right) \tag{3b} \end{align} $$

Then to reverse the sign of the $\hat{\mathbf{y}}$-component of Equation 2b, the $\hat{\mathbf{y}}$-component of Equation 3b must satisfy: $$ \begin{align} E_{x} & < 0 \ \text{ & } \ v_{y} > - \frac{c \ B_{z}}{E_{x}} \\ & \text{OR} \\ E_{x} & > 0 \ \text{ & } \ v_{y} < - \frac{c \ B_{z}}{E_{x}} \end{align} $$

So in principle, it is possible for the sign of a component of the 3-vector force to reverse.

  • $\begingroup$ I don't think you have really answered the question 5 (the interesting one :)), have you? $\endgroup$
    – Ilja
    May 1, 2016 at 17:31
  • $\begingroup$ @user11374 - I answered 5 by showing those equations. Let $\mathbf{E} = 0$ and $\mathbf{B} \neq 0$ in the K-frame. You can see from those equations that $\mathbf{E}' \neq 0$ so long as there is an orthogonal component of $\mathbf{v}$ to $\mathbf{B}$. $\endgroup$ May 2, 2016 at 13:34
  • $\begingroup$ @user11374 - I also answered 2 and 3 but perhaps I did not explain it clearly enough. You cannot treat a plasma as single particles. The behavior of single particles is not the same as the ensemble velocity distribution. This point brings a great deal of confusion but it is true. Distributions evolve according to Boltzmann-like equations and single particles to the Lorentz force (Note: the Lorentz force is a component of Boltzmann-like equations for charged particles but not everything). $\endgroup$ May 2, 2016 at 13:37
  • $\begingroup$ @user11374 - As for 6, I could not think of a specific example last night but will play around with this a bit if I have a few moments of free time today... $\endgroup$ May 2, 2016 at 13:38
  • $\begingroup$ @user11374 - I think CuriousOne's comment answers the question in the link you placed in your last comment. Unfortunately, I do not really understand the figure in your question. I do know that the particles found in the interplanetary medium cannot be treated as single particles because they are part of distributions. It seems like a subtle point but it is actually quite critical in kinetic theory and statistical mechanics, which is why I answered the way that I did (assuming particles between Mars and Jupiter, as you suggested in your question). $\endgroup$ May 6, 2016 at 12:20

A few comments before doing the calculation:

  • In the CM frame, there is only an attractive force, while in the given frame, there is both an attractive and a repulsive force. This is no more mysterious than the fact that a vertical object in my frame can look tilted to somebody with rotated axes. Going from the CM frame to your frame mixes the electric field into the magnetic field in the exact same way.
  • Given the above, why can't you detect absolute motion by just measuring if there's a magnetic field or not? You can't. You can only ever measure the Lorentz force, which has both electric and magnetic contributions; for more detail see here.
  • Even though the presence of an electric or magnetic field is frame-dependent, you're right that the particle trajectories are not: if they eventually hit each other in one frame, they better do the same in another frame.

First let's do the calculation in the CM frame. Let the $x$ and $y$ axes be horizontal and vertical as in your pictures, and let the objects have mass $m$ and charge $q$. Then $$F = q E_y$$ which implies $$a_y = qE_y / m.$$ The particles accelerate away from each other.

Now let's do the same calculation in your given frame. By length contraction, the electric field is increased by a factor of $\gamma$, so $$E_y' = \gamma E_y.$$ Using the field transformations as shown here the magnetic field satisfies $$\mathbf{B} = \frac{\mathbf{v} \times \mathbf{E}}{c^2}$$ which implies $$B_z' = -v E_y' / c^2 = -\gamma v E_y / c^2.$$ Next, we need to compute the acceleration. We can't use the Lorentz force law, because it doesn't have relativistic corrections. The right force law is $$ma^\mu = q F^{\mu\nu} u^\nu$$ where $a$ is the four-acceleration, $u^\nu = (\gamma, \gamma v, 0, 0)$ is the four-velocity, and $F$ is the electromagnetic field tensor. This is just a matrix multiplication; the part we care about is $$a'_y = (q/m)(\gamma E_y' + \gamma v B_z').$$ Now substitute all our previous expression in, for $$a'_y = (q/m)(\gamma^2 E_y - \gamma^2 (v^2/c^2) E_y) = (q/m)\gamma^2 (1-v^2/c^2) E_y = q E_y / m.$$ This exactly agrees with our previous result: the accelerations are the same in both frames, so nobody can detect which frame is "really" moving. If we didn't add the magnetic field, the result wouldn't have come out right.

By the way, if you look at our computation of $a'_y$, you can see that the ratio $F_B/F_E$ is exactly $v^2/c^2$, for any $v$, which answers your first question. But both of these quantities increase with $v$, so the total stays the same.

  • $\begingroup$ @user11374 This is addressed by my second bullet point: you can't tell if there are "two forces", because any measurement you do involves both of them. In fact, since the electromagnetic field tensor is one object which includes the electric and magnetic fields as components, it's actually exactly like tilting a vector: you are resolving the same object into components in two different ways. $\endgroup$
    – knzhou
    May 2, 2016 at 6:58
  • $\begingroup$ @user11374 Well, I showed explicitly that the total force stays exactly the same, and having two forces vs. one is not a physically meaningful statement. $\endgroup$
    – knzhou
    May 2, 2016 at 7:10
  • $\begingroup$ does your rep allow you to vote for reopening ? I added a picture to show what I expect in an answer $\endgroup$
    – user104372
    May 7, 2016 at 8:58

The observer moving with the CM will measure that the force of repulsion the electrons is given by $F=\frac{e^2}{4 \pi \epsilon_0 d^2}$ ($d$ is their separation), he can only make measurements in his reference frame (that is moving with speed $v$), and will not be able to be determine this speed.

  • $\begingroup$ Could you describe all forces and how the position/trajectories vary with time? Coulomb Feis 2.3*10^-24 N,right? and Fm is .98 that value using the rough formula, what does it become with SR correction? and what is Lorentz force? gamma= 7, so mass is as many times and 1 second is .14 sin the lab frame, if the motion of the particles is circular., right? It would be cool if you could make a graph with composition of forces and show what happens after, say, 1/1000 second $\endgroup$
    – user104372
    May 1, 2016 at 15:19

The principle of relativity says that there is no experiment that can determine absolute motion. So all observers, regardless of relative motion, need to agree on the outcome of any experiment. Because to the relativity of observers' measuring devices, they may not numerically agree on the measurements. By applying the laws of relativity they will be able to deduce what measurements each other make, and they will agree on big picture results like "the particles mutually repel".

I would check out chapter 12 of Griffiths' undergraduate Electrodynamics book. It's a good introduction to using SR and EM together.

Co-moving observer

An observer who is at rest relative to the electrons is moving at the same speed and in the same direction as the electrons. In this co-moving reference frame the electrons produce only electric fields. There is no magnetic field. The force between the two electrons is given by their Coulomb interaction. The charges repel, and the net force on each is $$ F_\mathrm{net} = \frac{1}{4\pi\,\epsilon_0} \frac{e^2}{d^2}, $$ where $d$ is their separation and $e$ is the electron charge.

Stationary Observer

A stationary observer sees the two electrons move with velocity $v$, relative to itself. Lets call the direction of motion the $\hat{x}$ direction. This observer will say that the moving charges produce a both electric and magnetic fields. This observer is moving relative to the source charges, so the electric field is modified in this frame.

One way to think about this is that electric field lines that are perpendicular to the motion ($\hat{y}$ or $\hat{z}$ components) will be squeezed together by the Lorentz contraction. This makes the Coulomb force as measured by the stationary observer greater than that measured by the co-moving.

The stationary observer will see the electric field of one at the location of the other is $$ E^\prime = \frac{1}{4\pi\,\epsilon_0} \frac{\gamma\, e}{d^2}, $$ where $\gamma$ is the Lorentz factor for the electrons' motion.

The stationary observer will also see a magnetic field produced by each electron due to its motion. The $B$-field strength of one at the location of the other is $$ B^\prime = \frac{\mu_0}{4\pi} \frac{\gamma\, v\, e}{d^2}. $$

We can calculate the Lorentz force with the relativistic correction: $$ F^\prime_\mathrm{net} = \gamma\,e(\vec{E}\,^\prime + \vec{v}\times\vec{B}\,^\prime ) = \frac{1}{4\pi\,\epsilon_0} \frac{\gamma^2\, e^2}{d^2} - \frac{\mu_0}{4\pi} \frac{\gamma^2\, v^2\, e^2}{d^2} = \frac{1}{4\pi\,\epsilon_0} \frac{e^2}{d^2} \left(1 - \frac{v^2}{c^2} \right)\gamma^2. $$ For the last equality we use $\mu_0\epsilon_0 = 1/c^2$. You can work out the direction of the $E$ and $B$-fields and the cross product to convince yourself that the electric force is repulsive and the magnetic force is attractive.

The net force in the stationary frame is thus: $$ F^\prime_\mathrm{net} = \frac{1}{4\pi\,\epsilon_0} \frac{e^2}{d^2} = F_\mathrm{net} $$

The two particles will move away from each other at the same rate in both. So you cannot distinguish absolute motion by observing the rate of separation of two charges.

  • $\begingroup$ Length contraction occurs in the direction of motion. My comment about squeezing lines is related to the common interpretation of comparing distance between lines to field strength. Lines in the direction of motion get no closer together. Lines perpendicular to the direction of motion get closer together. The change in the Coulomb force comes from the change in the E-field which is in any standard SR treatment. $\endgroup$
    – Paul T.
    May 2, 2016 at 12:47
  • $\begingroup$ Measuring the changing distance between the two particles is equivalent to measuring the net force. Any real measurement will be non-instantaneous in either frame. I showed the instantaneous situation for simplicity. The forces are all geometrical vectors. Any force is the same geometrical object in both frames. The two observers simply measure it differently because they are using different measuring tools. I'm not sure I understand what you mean in the statement about "how forces interact in the different frames". $\endgroup$
    – Paul T.
    May 2, 2016 at 12:59