Nature of the electric field

diagopod

I've been learning the basics of Maxwell's laws, and was especially interested to see that there was a Gaussian formulation of Newtonian gravity in terms of the negative divergence of the gravitational field, which in turn helped me understand the divergence of the electric field a bit more clearly. But the gravitational field g is quite straightforward for me to understand, as its simply acceleration, or at least reduces to units of acceleration. E, on the other hand, doesn't seem to have such simple units. Even using Gaussian units, it it's not much clearer to me than SI's Force/charge or Volts/meter, although it's nice not to have Epsilon0 to contend with. In any event, are there any units that treat the E-field in a more conceptually accessible way, like g, as something like acceleration, or velocity? Thanks for any help on this, appreciate it.

LostConjugate

The electric field is a force which is an acceleration. It is in SI units of newtons. newtons per coulomb to be exact. It is conservative just like gravity.

elect_eng

The two quantities are completely analogous, but you are looking for something that just doesn't work out. The gravitational field is force per unit mass, while the electric field is force per unit charge. Again this is perfectly analogous.

However, gravitation is fundamentally different than electric-static force in that inertial mass is exactly equal to gravitational mass, and cancels out in the equations. This is why the static Newtonian gravitational field is an acceleration field. The best you can do is say that electric field is like ma/q, where m is mass, a is acceleration and q is charge. Gravity is like ma/m, if you want to make the comparison.

To highlight the difference, consider that both an electron and a proton have the same charge magnitude, but something like a factor of 1830 difference in mass. You can't even choose a dimensional unit convention that will work in all cases.

Personally, I think that electric field is very intuitive when thought of as force per unit charge. It represents the force experienced by a test charge. It is also the negative gradient of the voltage, and voltage is potential energy per unit charge, which is also very intuitive, when compared to mechanical equations. (i.e. work is force integrated through a distance)

diagopod

Thanks, I see your point. I suppose if force were defined in terms of charge instead of mass, I'd be asking the same question about the gravitational field.

LostConjugate

Huh? Force is the change of the change of position, mass is only a coefficient.

Born2bwire

Well, technically it's the change in momentum which is related to the change in mass and the change in velocity. However, outside of examples like rocketry, there is no change in mass. But yes, diagopod's comment does not make sense as the the charges that the electrodynamic forces act on have mass to and thus they respond in the same way as a particle does to a gravitational force.

diagopod

True, true, but I should probably clarify what I was trying to say and save a little bit of face :) I meant that if, for example, charge and the laws of electrostatics had been discovered/defined first, and "force" had been defined accordingly not as F = ma, but rather F = qa, and thus not meaning force as we now know it, then the e-field would simply be acceleration and g, once discovered, would adopt the units qa/m, which would be as awkward to me conceptually as I find the e-field to be now (as ma/q). Of course, I'm probably still flubbing it, but just trying to explain what I was trying to say :)

espen180

I wouldn't expect that definition to hold water. What about chargeless particles like neutrons? Also acceleration does not retain it's original mathematical formulation here. Electrons and protons have the same ammount of charge but are accelerated at different rates in an electric field. How do you account for that?

elect_eng

There is no need to save face. We all ask this question when we first contemplate this. This issue kept Galileo, Newton and Einstein busy enough. So you are in good company.

However, you can now crystalize the notion that gravity is very unique and quite amazing in its behavior. Electrostatic force and Newtonian static gravity are very very similar, but there is the fundamental difference that gravitational mass and inertial mass are absolutely identical as far an all experimental evidence indicates. This has the result that the gravitational field can be interpreted as an acceleration field, and nothing can make electrostatic charge exhibit this behavior.

This is not a trivial thing and it is a key concept that led Einstein to general relativity. GR is to Newtonian static gravity, what Maxwell's equations are to electrostatics. GR and Maxwell's Equations are full dynamic classical theories backed by experimental evidence. It is interesting that this so called "equivalence principle" links acceleration and gravity and is a key component to GR.

kcdodd

It can be fun to play around with different dimensional interpretations to give insight to questions like this, even if they have little practical value.

For instance, if, instead of defining a universal gravitational constant G, one were given a gravitational acceleration formula of the form:

[tex] a_g = \frac{1}{m_2} \frac{m_1 m_2}{r^2}[/tex]

and performed dimensional analysis one might conclude that mass must have dimensions of L^3/T^2 (L = length, T = time).

eg

[tex] L/T^2 = \frac{1}{L^3/T^2} \frac{L^3/T^2 L^3/T^2}{L^2}[/tex]

Now, L^3/T^2 looks like an accelerating quantity of 3D space itself. Is it a coincidence that GR could be interpreted as space flowing into mass?

Likewise, if one were handed a formula for electric acceleration of the form

[tex] a_e = \frac{1}{m_2} \frac{q_1 q_2}{r^2}[/tex]

and taking the previous dimensionality of mass, one might conclude charge has the same dimensions. However, now you have a modifying ratio of charge to mass: q_2/m_2, which would end up being a dimensionless attribute of the particle in question.

A step further would be to recognize that the dimensions of space and time are actually related by the speed of light: L = cT. So, charge and mass could be described by a single dimensional quantity such as L^3/T^2 = c^2L, or just L if you take c =1. Now, this is interesting, since now all fundamental dimensions (charge, mass, time, length) can be reduced to a single dimension, with all the dynamics described by dimensionless quantities (such as electron charge to mass: e/m_e), with all accelerations having dimensions of 1/L, which looks like the dimensions of some type of space/time derivative.

Even a step further would be to ask, given the above assumptions, what are the dimensions of the electric field? If one is given something like

[tex] E = \frac{q}{r^2}[/tex]

the dimensions of E would be

[tex] [E] = \frac{L^3/T^2}{L^2} = L/T^2 = 1/L[/tex]

given the above assumptions, E would have the same dimensions as acceleration, except one would have to keep in mind that it is modified by that dimensionless quantity when acting on a particle.

What about a magnetic field given by something like

[tex] B = \frac{q*v}{r^2}[/tex]

[tex] [B] = \frac{(L^3/T^2)*(L/T)}{L^2} = L^2/T^3 = 1/L [/tex]

Which has the same fundamental dimensionality, again keeping in mind the dynamics would be governed by the dimensionless quantities.

Anyway, just food for thought, take from it what you will.

espen180

is a_g the acceleration of body 2? If so , are you working with natural units? This is fine and all, but natural units are inconsistent with respect to conservation of units. Even in the famous space-time equivalence, units are preserved.

kcdodd

What I wrote here is not a new idea. If the only difference between two dimensions is a constant, then why even bother defining them to be different. One can see that every quantity can be reduced to a single dimensionality, and every dynamical field can be reduced to the a single dimensionality.

LostConjugate

Are you saying that F = (m)(d/dt)(v) = (d/dt)(mv) ?

diagopod

You're certainly right, it doesn't add up, I was just trying to say that if it had been true then my conceptually misunderstanding would have been reversed. Either way, point taken.

diagopod

Thanks, glad to know, appreciate the kind words.

Thanks for the further explanation. Btw, regarding the equivalence of inertial and gravitational mass, when we see precise values for the mass of the electron and proton, did the experiments that derived those values test for gravitational or inertial mass? If inertial, is there any way to test for the gravitational mass of the electron and proton, or can we safely deduce those by extrapolation from the macroscopically confirmed equivalence of i and g mass?

elect_eng

These are good questions, but they are outside my area of expertise. I hope someone else can help answer these questions because I'm curious too. My own guess is that gravitational force from atomic particles is too small, and hence difficult (or maybe impossible) to measure. If I were to try and measure proton and electron mass (either absolute in kg or relative to each other), I would measure inertial mass by using electric or magnetic fields to apply force, and then note the deflection from a straight path.

espen180

You are entering the domain of Quantum Gravity. There is to date no theory which accurately predicts gravitational effects when quantum effects are important. The reason is that we have not been able to measure anything at this level. Much of the problem lies in probing the gravitational effects, as the gravity of the lab equipment is many orders of magnitude higher than the effect to be measured.