# Will a beam of protons and a beam of electrons attract or repel each other?

Consider a beam of protons and a beam of electrons moving in parallel in the same direction - will they attract or repel each other?

The answer according to my teacher has been to assume the beam of electrons and protons as current in a conductor, and thus they repel. But the explanation has two problems:

1. Isn't a proton beam very different from current in a conductor, as the conductor as a whole is electrically neutral and has no electric field outside but a proton beam does have a strong electric field?

2. Doesn't a moving charge experience both electrostatic and magnetic forces according to the Lorentz force?

You are correct; your teacher is wrong.

Consider protons and electrons moving parallel in the same direction and with the same speed. In the inertial frame of the charges, we clearly have an attractive electrostatic force that will make the beams bend towards each other, and no magnetic force.

The attraction will be there also in our frame of reference, in which we will measure both a slightly higher electrostatic attraction and a small magnetic repulsion (which in the end will give the same behavior).

• What is electrostatic attraction higher than and why? Commented Aug 30, 2020 at 11:19
• @Protein. It's higher in our frame of reference than in the inertial frame of the charges because the charges are denser in our frame due to length contraction. Commented Aug 30, 2020 at 11:34
• Length contraction occurs at higher velocities. Our frame has a lower velocity than that of proton's. Shouldn't length contraction occur in proton's frame of reference? Commented Aug 30, 2020 at 13:49
• @Protein. Length contraction occurs at all velocities, but it's hardly noticeable at low velocities. Also, length contraction is a relative phenomenon. If the distance between protons (in the direction of movement) is $L$ in the frame of the protons, in our frame the distance will be $L\sqrt{1-v^2/c^2}.$ Commented Aug 30, 2020 at 13:57
• This formula tells that length measured from our frame will contract in proton frame of reference and lengths measured from proton's frame of reference will contract in our frame of reference? I am lost . Commented Aug 30, 2020 at 14:05

In questions like these its always helpful to imagine the scenario from their (protons' and electrons') point of view. In their reference frame theyre not moving at all and thus have a speed of 0, which means that for them there is NO magnetic field and they attract each other.

However, since time is running slower in a reference frame which is in motion relative to the observer, their attractive forces will also be weaker. This difference equals the magnetic repulsion you would notice from the observers point of view

• How does time affect force?I am not familiar with relativity !pls explain. Commented Aug 30, 2020 at 11:24
• @Protein I posted another answer to this thread answering your question in detail (since my comment was a bit too long) Commented Aug 30, 2020 at 13:44
• Thanks for making so much effort. 🙏 Commented Aug 30, 2020 at 13:54
• @Protein You're welcome buddy! ;) Commented Aug 30, 2020 at 13:54

Since my answer was too long for a comment, I decided to post another answer on this thread answering the request "How does time affect force? I am not familiar with relativity! pls explain." asked by @Protein.

The Coloumb Force of a charge $$q$$ in the electric field $$E$$ is defined as $$\frac{dp'}{dt'}=qE'$$. Since time runs slower for the charge $$q$$ relative to the observer, the time increment $$dt$$ is longer by a factor of $$\gamma=(1-\beta^2)^{-\frac{1}{2}}$$ (Where $$\beta=\frac{v}{c}$$). In addition to that, the impulse $$p'$$ of the charge is different to the impulse $$p$$ measured by the observer, since mass increases with velocity. Both effects lead to the equation $$\frac{d(m_{0}\gamma v)}{\gamma dt}=qE$$. Deriving the impulse with the product rule ($$\gamma$$ is a function of $$v$$ and therefore also a function of time!) and then simplifying the equation results in $$\gamma^2 m_{0} a=q E$$. The factor $$F=ma$$ is the force measured by the observer and $$F'=qE'$$ the force measured by the moving reference frame. Substituting results in the equation $$F=\frac{F'}{\gamma^2}$$ This means that all forces $$F'$$ measured by the moving reference frame are weaker by a factor of $$\gamma^2$$ for the observer.

We can further derive an equation for the difference of the force $$F$$ measured by the observer and of the force $$F'$$ measured by the moving reference frame:

$$\Delta F=F'-F=F'-\frac{F'}{\gamma^2}=F'(1-\frac{1}{\gamma^2})=F'(1-(1-\beta^2))=\beta^2F'$$ This apparent additional force $$\Delta F$$ that we measure IS in fact the repulsive magnetic force caused by the motion of the charges. This means that

$$\vec{F_{B}}=-\frac{v^2}{c^2} \vec{F_{E}}$$ where $$\vec{F_{B}}$$ is the force caused by the magnetic field and $$\vec{F_{E}}$$ is the force caused by the electric field.

This is an effect we observe in particle accelerators. When accelerating a beam of electrons to near the speed of light the attracitve magnetic force $$\vec{F_{B}}$$ between the electrons approaches $$\vec{F_{B}}=\lim_{v\to c}\Bigl(-\frac{v^2}{c^2} \vec{F_{E}} \Bigr)=-\vec{F_{E}}$$ which compensates the repuslive electric force. This is pretty neat since because of that we can have pretty narrow and therefore precise electron beams or beams of different charged particles. Also notice how this effect does not only apply to electric fields, but to force fields in general!

• fixed some major spelling mistakes. Should be correct now Commented Aug 30, 2020 at 14:00
• It will take some time for me to digest all this. Commented Aug 30, 2020 at 14:07
• @Protein How advanced are you in physics? Commented Aug 30, 2020 at 14:12
• Let's talk in chatroom Commented Aug 30, 2020 at 14:18