Question about electromagnetic force and mass

[Mårten]

According to Coulomb's law we have that the force between two charged particles could be described by

$$(1)~~~F = k \frac{Q_1 Q_2}{r^2}.$$

Now, a force is according to most sources something that causes a mass to accelerate (not a charge to accelerate), so the force in (1) above will solely act on charged bodys and charged particles with mass. So if we have a charged particle P_1 with charge Q_1 and without mass in (1) (and of course another particle P_2 with charge Q_2), P_1 will not be affected by the force resulting from (1), and therefore it will not start to move.

Am I right in this? That is, electromagnetic forces stemming from Coulomb's law, is like all other forces, they're acting on masses and not on charges?

I was thinking of this, when wondering about how electrostatic induction works. When a positive charged object is close to a neutral object, electrons in the neutral object will move to the side facing the positive charged object. And then I was thinking that if the electrons had no mass, they wouldn't move here, because a force is something which acts on masses not on charges. Or could it be that the EM force actually acts on the particular charge, and not on the mass of the particle holding the charge?

 

[tiny-tim]

According to Coulomb's law we have that the force between two charged particles could be described by

$$(1)~~~F = k \frac{Q_1 Q_2}{r^2}.$$

Now, a force is according to most sources something that causes a mass to accelerate (not a charge to accelerate), so the force in (1) above will solely act on charged bodys and charged particles with mass. So if we have a charged particle P_1 with charge Q_1 and without mass in (1) (and of course another particle P_2 with charge Q_2), P_1 will not be affected by the force resulting from (1), and therefore it will not start to move.

Am I right in this? That is, electromagnetic forces stemming from Coulomb's law, is like all other forces, they're acting on masses and not on charges?

Hi Mårten!

Electric forces only act on charges, and they act per mass.

The smaller the mass, the larger the acceleration (because a = F/m).

For example, an electron has the same charge (well, opposite) as a proton, but a much smaller mass. So the same electric field will accelerate the electron much more than the proton (and in the opposite direction).

 

[h0dgey84bc]

Are there any massless particles with charge? If you had a massless particle with the same charge as an electron, the electrostatic force would be the same on this particle, but the acceleration a=F/m (with m=0) would seem to be infinite.

With a photon F=ma doesn't apply, and you have to invoke gen rel to find out how the gravitational field acts on it remembering (remembering the equivalence of energy and mass, and that E=pc).

What would be the analogue of this for the E-field acting on a massless charged particle, to avoid the apparent problem of infinite acceleration?

 

[tiny-tim]

Are there any massless particles with charge?

Hi h0dgey84bc!

The m in F = ma is the relativistic mass, in other words the energy.

All particles (including photons) have energy, so there are no massless particles, and the problem of infinite acceleration doesn't apply.

 

[h0dgey84bc]

Hi Tiny-tim,

I was referring to zero "rest mass" particles that have charge.

F=ma doesn't apply when considering the *gravitational* force on a massless particle like a photon. The momentum-energy tensor is responsible for bending space-time, and the photon just follows the appropriate geodesic.

I really don't see why F=ma should automatically apply in the case of an ultrarelatvisitc charged particle in the EM case (since if it has zero rest mass, this hypothesised charged particle would always be moving at c, meaning Newtonian laws are in accurate)

 

[Redbelly98]

Are there any massless particles with charge?

I was referring to zero "rest mass" particles that have charge.

The only entities with zero rest mass (that I'm aware of) are photons and gravitons. And maybe one of the neutrinos. All of these have zero charge, however.

 

[h0dgey84bc]

Yep, sounds about right, I can't think of any others either that have zero rest mass.

Let's say there existed such a particle however, it has a zero rest mass and the charge of an electron. Obviously F=kQe/r^2, but how would one find the acceleration on this particle? surely the Newtonian F=ma, will not apply since the particle is moving at c.

Infact F=kQe/r^2 probably doesn't apply either, since this is an electrostatic equation (unless the test charge Q was also one of the hypothesised rest mass zero charged particles, also moving in the same frame at c)

 

[tiny-tim]

F=ma doesn't apply when considering the *gravitational* force on a massless particle like a photon. The momentum-energy tensor is responsible for bending space-time, and the photon just follows the appropriate geodesic.

Hi h0dgey84bc!

Gravity isn't the only force.

And, as you say, the momentum-energy tensor is responsible for bending space-time, and it is that, rather than F = ma, that governs the "gravitational" effect on all particles, not just photons.

Let's say there existed such a particle however, it has a zero rest mass and the charge of an electron. Obviously F=kQe/r^2, but how would one find the acceleration on this particle? surely the Newtonian F=ma, will not apply since the particle is moving at c.

Infact F=kQe/r^2 probably doesn't apply either, since this is an electrostatic equation (unless the test charge Q was also one of the hypothesised rest mass zero charged particles, also moving in the same frame at c)

EDIT: F=kQe/r^2 doesn't apply to a moving charge anyway:
we need the Lorentz force, F = q(E + v x B), or its relativity equivalent.

And the Newtonian formula, technically, isn't F=ma, it's F = d(mv)/dt, which works fine when translated into relativity.

From example, an "acceleration" in the same direction as a photon simply changes its frequency.

 

[Redbelly98]

Let's say there existed such a particle however, it has a zero rest mass and the charge of an electron. Obviously F=kQe/r^2, but how would one find the acceleration on this particle? surely the Newtonian F=ma, will not apply since the particle is moving at c.

Infact F=kQe/r^2 probably doesn't apply either, since this is an electrostatic equation (unless the test charge Q was also one of the hypothesised rest mass zero charged particles, also moving in the same frame at c)

For this hypothetical particle,

F = q ( E + VxB ) = dp / dt

where p is momentum.

Such a particle could be made to alter its direction, but not its speed. One could also change its de Broglie wavelength by changing the momentum's magnitude.

 

[h0dgey84bc]

Hi h0dgey84bc!

Gravity isn't the only force.

Of course, but what I was trying to illustrate is that F=ma doesn't apply to a photon (or any general rest-mass zero particle) when we are considering gravities action upon it. We can't just say a=F/m, and that m is the relativistic mass in the case when F is a gravitational force, thusly I don't see why we would if instead F was an electric force.

And, as you say, the momentum-energy tensor is responsible for bending space-time, and it is that, rather than F = ma, that governs the "gravitational" effect on all particles, not just photons.

I was not implying that the momentum-energy tensor was the source of gravitational force for just photons (or rest mass zero particles in general), merely that it was the momentum-energy tensor NOT F=ma we must invoke when considering the gravititational force on a particle (in this case a rest mass zero particle).

My implication was that by analogy, why would F=ma work on something intrinsically relativistic (a zero rest mass particle moving a speed c) in the EM case.

Even in special rel the expression is:
$\vec{f}= \gamma m \vec{a} + \gamma^3 m \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v}$

But here m is the rest mass something a photon doesn't have.
In the past I have only ever dealt with a gravitational force on a zero rest mass particle (since this is the only force in reality that acts on them) and thus have just used Gen rel and the momentum energy tensor, geodesics etc to work out its effect on them.

But how would a "charged photon" accelerate in an E-field?

Seems that F=dP/dt always holds, but p=E/c, which just leads to F=1/c dE/dt

hmm

 

[h0dgey84bc]

And the Newtonian formula, technically, isn't F=ma, it's F = d(mv)/dt, which works fine when translated into relativity.

very true, F=dP/dt is the general formula that works even with special relativity, but all the derivations from this I have seen have usually involved rest mass. If you can show me it for a photon I will be happy

 

[tiny-tim]

Hi h0dgey84bc!

For something at the speed of light (and therefore with zero rest-mass), (m,mv)/√(1 - v²/c²) is replaced by (|p|,p).

 

[h0dgey84bc]

Hi h0dgey84bc!

For something at the speed of light (and therefore with zero rest-mass), (m,mv)/√(1 - v²/c²) is replaced by (|p|,p).

Are you referring to the four momentum $$<BR> p^{\mu} </BR>$$ ?

Also why is my tex not working

 

[tiny-tim]

Are you referring to the four momentum $$p^{\mu}$$ ?

Also why is my tex not working

testing: $$A = B$$

(it's not you … it isn't working on other threads, either )

yes, that's the 4-momentum for speed-of-light particles …

it doesn't need an m, only a p and an E = |p|.

 

[h0dgey84bc]

yes, that makes sense then, so you end up with F_mu=d/dt (P_mu)

the latin spatial comps of the four force: F_i=d/dt p_i...so I guess that the problem simply reduces to equating this with the i'th comp of the (4 vec version) Lorentz force, and solving this to find p as a function of time.

wonder why there are no massless particles with charge/colour etc

 

[Redbelly98]

Perhaps there is something pathological about the electric field of a charge moving at c, which prevents it from existing. If I weren't 20+ years removed from my last electromagnetism course, I might even attempt the calculation

 

[h0dgey84bc]

Perhaps there is something pathological about the electric field of a charge moving at c, which prevents it from existing. If I weren't 20+ years removed from my last electromagnetism course, I might even attempt the calculation

I think you may be on to something perhaps, def got me thinking anyway. It's kind of interesting that the E field only propagates its affects at c, so if the generating particle is also moving at c in the opposite direction: relativistic velocity addition formula v=w-u/(1-wu/c^2), if w and u are both c you get something undefined.
time to try and find my electrodyn notes maybe...

 

[h0dgey84bc]

hmm sorry to be a stickler on this problem, but I've been trying to derive the acceleration a massless particle would undergo in an E field (or for any force field except gravitational) and I can't derive a single relation without using rest mass somewhere, four vectors methods or otherwise. Even if you get the momentum as a function of time, I can't seem to relate this back to velocity/acceleration for a massless particle
...this problem is really bugging me now, haha

 

[Mårten]

Hi tiny-tim and others!

Thanks for thorough replies!

Electric forces only act on charges, and they act per mass.

Okey, that was also my theory, even though I didn't formulate it so well... Thanks for confirming this!

The smaller the mass, the larger the acceleration (because a = F/m).

But maybe, if the mass equals zero, then maybe NewtonII won't apply. Otherwise we would have packages of vacuum, with zero mass flying around us with infinite acceleration all the time. Maybe we have, we would never notice them...

 

[tiny-tim]

The smaller the mass, the larger the acceleration (because a = F/m).

But maybe, if the mass equals zero, then maybe NewtonII won't apply. Otherwise we would have packages of vacuum, with zero mass flying around us with infinite acceleration all the time. Maybe we have, we would never notice them...

Hi Mårten!

If … if … if …

NetwonII is basic … if you're going to throw it out, anything could happen …

massless pigs could fly! ​

 

[h0dgey84bc]

Newton II isn't that basic, especially when considering something inherently relatavistic (like a particle with zero rest mass traveling with c).

I'm seriously beginning to wonder if an E-field could effect this hypothesised "charged photon" at all, or perhaps just alter its frequency.

Where F,p,E represent the magnitudes of the force, momentum and Energy resepectivley:
p=E/c => F=dp/dt=1/c dE/dt=h/c df/dt

Another issue I've come across when trying to use the four vector approach, is that the four velocity is actually an undefined quantity for a photon (or other rest mass less quantity). Since the basic definition of the four velocity is that is the e_0 basis vector in an MCRF, but since there is no frame in which a photon is at rest there is no e_0 vector that is momentarily tangent to the photons world line.

I've never come across this problem, since obviously in the real universe massless particles are only deviated by gravity not the other 3 forces, but general relativity dictates this nicely. I've no idea how the inertia caused by the relativistic mass of a massless particle would manifest itself when the source of force is non gravitational.

Please let me know if you think I'm missing something, would appreciate seeing a few lines of derivation if anyone has the time...

P.S. is the latex working now?

 

[tiny-tim]

… the four velocity is actually an undefined quantity for a photon (or other rest mass less quantity). Since the basic definition of the four velocity is that is the e_0 basis vector in an MCRF, but since there is no frame in which a photon is at rest there is no e_0 vector that is momentarily tangent to the photons world line.

Hi h0dgey84bc!

For something at the speed of light (and therefore with zero rest-mass), the 4-velocity is (|p|,p).

 

[h0dgey84bc]

Hi h0dgey84bc!

For something at the speed of light (and therefore with zero rest-mass), the 4-velocity is (|p|,p).

That's not true tiny tim, that is the four momentum. The four velocity does not exist for a photon(as I've just varied with the help of Schutz, cf around p52, or if you don't have a copy i can paste the short derivation if the latex is up and running)

 

[tiny-tim]

That's not true tiny tim, that is the four momentum. The four velocity does not exist for a photon(as I've just varied with the help of Schutz, cf around p52, or if you don't have a copy i can paste the short derivation if the latex is up and running)

oops!

should be …

For something at the speed of light (and therefore with zero rest-mass), the 4-velocity is (c,v).

 

[h0dgey84bc]

For something at the speed of light (and therefore with zero rest-mass), the 4-velocity is (c,v).

OK, I will try and explain to you why this is not the case:

The fundamental definition of the 3-velocity is that is a vector tangent to a particles path through space, similarly we fundamentally define the 4-velocity as a vector tangent to the world line of a particle, and of such a length it stretches one unit of time in that particles frame.
For a uniformly moving (non accelerated) particle, look at the above definition in the inertial frame at which the particle is at rest. Then the four velocity points parallel to the time basis vector and is one unit long. It is easy to see that this implies the 4-velocity vector be equivalent to the $$\vec{e}_0$$ basis vector in this intertial frame.

An accelerated particle however of course has no inertial frame at which it is always at rest. However there is an inertial frame which momentarily, the particle is at rest; the Momentarily Co-moving Reference Frame (MCRF). The four velocity of an accelerated particle is defined to be the $$\vec{e}_0$$ of its MCRF at that event in spacetime.

Now we can also approach this by considering the 4-velocity as a derivative:
If a particle makes an infinitesimal displacement in spacetime $$\vec{dx}=(dt,dx,dy,dz)$$ then the magnitude of this displacement will be
$$\vec{ds^2}=(-dt^2,dx^2, dy^2, dz^2)$$ . Also the proper time is defined as $$d \tau^2= -\vec {dx} \cdot \vec {dx}$$ , this isn't hard to see when you remember dot products of four vectors give Lorentz scalars, and then just going into the rest frame.

Now consider the vector $$\frac {\vec{dx}}{ \vec{d \tau} }$$. It has magnitude
$$\frac {\vec{dx}}{ \vec{d \tau} } \cdot \frac {\vec{dx}}{ \vec{d \tau} }=-1$$
. Since it is also a multiple of $$\vec{dx}$$ it is tangent to the particles world line.
It is therefore a tangent to the worldline vector of unit magnitude -1. Hence we can see this is identical to our above definition of the 4-velocity.

So we have $$\vec{U} =\frac {\vec {dx} }{ \vec{ d \tau } }$$.

But now note that photons move on null world lines, hence $$\vec {dx} \cdot \vec {dx}=0$$ and thus the proper time is also zero. By our definition of the four velocity you can see that this would lead to an undefined quanitity, hence the four velocity of a photon is undefined.

Another way of saying this is that there is no MCRF, since there is no frame a photon can be a rest in, even momentarily.

Hope that helps explain, it's my attempt to rip out the relevant proofs from Schutz's intro to Gen Rel, I also have an electronic copy of the book if you would like me to send it to you

 

[h0dgey84bc]

Arghh more formatting has messed up. Anyway the essential points are that four velocity can be defined as $$\vec{U}= \frac {\vec{dx} }{ \vec{d \tau} }$$,and that $$d \tau =\sqrt { -\vex{dx} \cdot \vec{dx} }$$. Since photons move on null world lines: $$\vex{dx} \cdot \vec{dx} =0$$. Hence the four velocity is undefined for a lightlike particle.

 

[tiny-tim]

By our definition of the four velocity you can see that this would lead to an undefined quanitity, hence the four velocity of a photon is undefined.

Another issue I've come across when trying to use the four vector approach, is that the four velocity is actually an undefined quantity for a photon (or other rest mass less quantity). Since the basic definition of the four velocity is that is the e_0 basis vector in an MCRF, but since there is no frame in which a photon is at rest there is no e_0 vector that is momentarily tangent to the photons world line.

ah … I see what you're getting at …

yes, the 4-velocity of a photon depends, up to a scalar factor (the "red-shift" factor), on the velocity of the observer.

So you can say that it's undefined, or defined except for a scalar factor, or defined for each observer.

But I don't really see why we would want to use it … don't we use the 4-momentum for all physical applications?

 

[h0dgey84bc]

ah … I see what you're getting at …

yes, the 4-velocity of a photon depends, up to a scalar factor (the "red-shift" factor), on the velocity of the observer.

So you can say that it's undefined, or defined except for a scalar factor, or defined for each observer.

hmm not quite, the 4-velocity of a photon is undefined for every observer. This is a direct consequence of the fact photons have no rest frame and therefore no definition of proper time can be made, and proper time is essential in defing the 4-velocity.

Another way of looking at it is that photons move on null lines in spacetime, there are obviously plenty of tangent vectors to the 45 degree path, but all of which have vanishing magnitude (because of the Minkowski metric)

But I don't really see why we would want to use it … don't we use the 4-momentum for all physical applications?

Yes, I want to use 4- momentum not 4-velocity, the above was just an aside. But the problem is in the problem we were initially discussing before digressing about 4 velocity, I can find the compents of the 4-momentum, but I can't relate this back to finding the actual trajectory the charged rest-massless particle would follow, how can one get the acceleration, since four velocity is undefined things like $$\vec{P}= m_0 \vec{U}$$ don't apply.

The real question I'm asking, is "how does the inertia of the relativistic mass(energy) of a charged massless particle manifest itself, when it is experiencing an E field?"

A mean take the simple example of a positivley charged photon moving at c in the x direction, then we switch on an E field in the y direction. How does the charged photon move? what are the equations?

$${d {p^y}}/{dt} =q E_y$$ then?

and the above analysis would also lead to $${d {p^x}}/{dt} =0$$ , implying x comp of p remains constant whilst the y compenent is increasing. But the charged photon must continue to obey $$E^2=c^2 (p^2{}_x+p^2{}_y$$ . Thus if x comp is constant and y is increasing this implies the energy is increasing, and therefore the frequency. There must be a limit to how much this can increase?

more to the point, how does one go onto find the actual trajectory once one has the comps of p as a function of time, in the case of this rest mass less photon?

 

[tiny-tim]

But I don't really see why we would want to use it … don't we use the 4-momentum for all physical applications?

I can find the compents of the 4-momentum, but I can't relate this back to finding the actual trajectory the charged rest-massless particle would follow, how can one get the acceleration, since four velocity is undefined things like don't apply.

Hi h0dgey84bc!

But why do you want the acceleration?

Again, we don't use it in physical applications …

we use good ol' Newton's second law, which isn't about acceleration, but about momentum:

Force = rate of change of momentum.

 

[h0dgey84bc]

I want to calculate the "charged photons" trajectory assuming it passes at the origin at time t=0 and then initially moves along the x-axis at speed c so has 3 velocity (c,0,0) and its position is given by r=(ct,0,0). Now after moving like this for some time we switch on an electric field of magnitude E in the y-direction.


What is the "charged photons" position and velocity as a function of time after the E field is switched on? What trajectory does it follow?

Perhaps you'll see the difficulties I keep encountering (unless I'm missing a little something, but if that's the case I'll also be happy )

 

[tiny-tim]

Perhaps you'll see the difficulties I keep encountering …

honestly, no …

you just measure everything in the same frame as you're measuring the E field in.

(you must always measure everything in the same frame, anyway)

 

[h0dgey84bc]

show me the money, this is a simple situation so shouldn't be a problem if you got the physics down...

and of course I'm measuring everything in the same frame

 

[h0dgey84bc]

$$F^{\mu} = \frac{dp^{\mu}}{d\tau}$$

firstly what is the form of $$F^{\mu}$$ in the scenario I've described above (i.e. external electric field in y direction)?

secondly, say we end up with an equation "like" (obviously it will depend on the form of F) :

$$qE_y = \frac{dp^{y}}{d\tau}$$ implying that $$p^y -> qEt$$
and another equation $$0=\frac{dp^{x}}{d\tau}$$ implying that $$p^x -> constant= p^x{}_0$$, and similarly $$0=\frac{dp^{z}}{d\tau}$$ implying that $$p^z -> constant= 0$$, now where do we go from here?

Obviously for a normal situation where the particle in question has mass, once you have the momentum you got the velocity too just by division by $$\gamma m_0$$, and then with knowledge of velocity getting position as a function of time (trajectory) is trivial. How do we proceed to get these things from knowledge of the momentum for a rest massless particle.

If you feel otherwise please post the few lines proceeding from the above, where you would take it next.

 

[Wiemster]

And the Newtonian formula, technically, isn't F=ma, it's F = d(mv)/dt, which works fine when translated into relativity.

Technically and nontechnically Newton's second law written as F=ma works as fine as F = d(mv)/dt. Actually F = d(mv)/dt = vdm/dt + mdv/dt =ma as it only applies to systems not changing their mass. Both equations need revising with a gamma factor in relativity, although your version a derivative less.

 

[h0dgey84bc]

Technically and nontechnically Newton's second law written as F=ma works as fine as F = d(mv)/dt. Actually F = d(mv)/dt = vdm/dt + mdv/dt =ma as it only applies to systems not changing their mass. Both equations need revising with a gamma factor in relativity, although your version a derivative less.

yeah adding in the gamma factor:

F = d(mv)/dt

$$F=\frac{ d\gamma m_0 v}{dt}$$
$$F= m_0 \frac{d\gamma v}{dt} +\gamma v \frac{ d m_0}{dt}$$
$$F=m_0 \frac{d\gamma v}{dt}$$
$$F=m_0 v \frac{d\gamma }{dt} +m_0 \gamma \frac{dv }{dt}$$
$$F=m_0 \gamma^3 v^2 \frac{dv}{dt} +m_0 \gamma \frac{dv }{dt}$$

but what if $$m_0 =0$$ ?

so seems we are left with the generic $$F^{\mu}=\frac{d p^{\mu} }{dt}$$ to use. But as my previous post adressed, seems this allows us to get the four momentum as a function of time quite easily. But how does one then infer the particles trajectory velocity etc from this quantity, since $$p^{\mu}=m_0 U^{\mu}$$ does not apply (for one we have the same problem of $$m_0 =0$$, and two the four velocity is undefined for a photon as I discussed above (if you don't believe me on this one please see Schutz or another intro Gen Rel book, he states it clear as day))

sigh, can someone just kill this thread and put me out of my misery

If someone could just show me the equations for my simple example of a "charged photon" moving in a uniform E-field in the y direction, I'd be supremeley grateful .