# V = IR . proof?

1. Dec 9, 2004

### Oliminator

can anyone proove that V = IR ?

2. Dec 9, 2004

### ZapperZ

Staff Emeritus
3. Dec 9, 2004

### rayjohn01

This equation supposes that you have defined 'V' , the formal definition being
'work done per unit charge' or 'energy per unit charge'. It also assumes you can measure this -- although we use a 'voltmeter' it could be done with a more basic instrument such as an 'electrometer' which only deals with charges and force due to charging.
Current 'I' is defined as 'charge per unit time ' passing some defined point.
So the equation relates a mechanism of 'loss of energy ' R to the other two.
It supposes that this mechanism is constant and in the case of a 'resitor' is due to the interaction of the 'charge carriers ' with the lattice which induce molecular motion which is lossed or equalised by heat loss.
Having defined the units it is a matter of measurement , most materials do NOT obey 'Ohms' law exactly for a variety of reasons , but if the eperimental errors are accounted for -- then for good linear conductors such a copper , constantin , etc the rule appears to apply pretty well .
In this sense the law is an idealised relationship in exchange of work. The first to charge something with excess carriers ( a battery ) , ther second to allow the discharge via a conductor -- and then to notice that in this exchange a loss occurs -- due to R. This represents the loss in heat in the exchange.
R also represent loss of energy from a system independant of how -- so for intance an antenna has a radiation resistance -- meaning loss of energy to space.
Ray.

Last edited: Dec 9, 2004
4. Dec 9, 2004

### ZapperZ

Staff Emeritus
I think the way you described it made it sounds as if Ohm's Law is purely and currently (no pun intended) just phenomenology. While it may be true in the beginning of its history, it is no longer that anymore. Again, if you follow the link that I gave, it can be derived from the Drude model of conduction electron being a "free electron gas" You derive, purely on statistical grounds, of the relationship between the current density, and the conductivity (or resistivity) as a function of the applied electric field. This is nothing more than Ohm's Law in a different outfit.

It is only when the assumption of a pure free electron gas breaks down do we begin to notice deviation from Ohm's Law (such as periodic potential requiring Bloch wavefunction). So the emergence of Ohm's Law is essentially the consequence of the free electron description for ideal conductors.

Zz.

5. Dec 9, 2004

### krab

IIRC, Ohm's law was given a firm footing by Onsager. If you can get Phys Rev, see here:

http://prola.aps.org/abstract/PR/v38/i12/p2265_1

Otherwise, just google Onsager and "Ohm's law" and eventually you'll find something.

6. Dec 10, 2004

### rayjohn01

To Zz
I have no problem with most of what you said -- but you used fields I wanted to use voltage ( part of the equation ) - also current can be due to things other than conduction electrons , and system loss be represented by R ( see antennas and radiation resistance ).
The Drude model describes an idealised situation in a simple conductor
but the law is employed over a wider range of cases especially semiconductors with various impurities . -- Studies of noise shows that most materials other that pure metallic samples show excess noise indicating other mechanisms at work.
Ray.

Last edited: Dec 10, 2004
7. Dec 11, 2004

### Andrew Mason

There is an observed linear relationship between V and I in (non-reactive) electrical circuits. The reason it is a linear relationship cannot be the subject of mathematical proof. It has to do with the physics of current flow in a conductor. So I don't see how it can be 'proven' mathematically other than to observe that R is defined as the proportionality constant between V and I.

AM

8. Dec 12, 2004

### ZapperZ

Staff Emeritus
You seemed to have missed the part where I said that the expression given for the current density is essentially Ohm's Law in a different outfit. That IS Ohm's law. It is just expressed in a form that is geometry independent. The electric field IS relatied to the potential difference since it has units of V/m. The resistivity becomes the "resistance" after you consider the geometry of your conductor. And the same can be said about the current density. So you DO already have Ohm' law!

The Drude model IS an idealized situation. I've said this already that the free electron gas is only valid for simple, conventional conductors. But in many instances, it is also a good approximation for most other conductors and semiconductors. We know this because Landau has done a tremendous service by his formulation of the Fermi Liquid theory. He has shown that even when there's a weak interaction among the charge carriers, these can still be "renormalized" into free particles but with the interaction lumped into a "quasiparticle mass". This is the origin on why we can still treat a lot of materials via the Drude model even when the "free electron gas" picture isn't quite correct.

However, in many instances also, it can deviate especially when you do not have enough charge carriers to uphold the "free electron gas" statistics, or when the mechanism of transport varies significantly from simple band theory. The Mott-Hubbard compound is an example. One can also look at tunneling conductance and see that Ohm's Law breaks down especially at low potential difference.

In any case, to answer the original question, there IS a statistical derivation of Ohm's Law, and it is derived from the statistical treatment of a free electron gas (or free charge gas, if you prefer). Fermi Liquid theory has generalized that into "free quasiparticle gas" (a bit redundant).

Zz.

9. Dec 12, 2004

### ZapperZ

Staff Emeritus
One CAN define R as the ratio of V and I. However, one cannot define a LINEAR relationship between V and I if this ratio isn't a constant. Ohm's Law defines such relationship.

However, there is also a separate issue here. Given that V = IR and that we have the observed linear relationship, can one DERIVE, based on the microscopic picture, of this expression? Most macroscopic theory is seldom "complete" without a derivation of it's description from the microscopic picture. Without this, the macroscopic description is purely phenomenological. A clear parallel example of this would be the London equation describing the macroscopic phenomenon of superconductivity, which isn't complete till the BCS theory came along that started off right at the microscopic level.

Again, Ohm's Law may have started off being purely phenomenological (as most theories do), but it CAN now be derived from the free electron gas model. Every student taking a class in Solid State Physics will have to do this straight away in Chapter 1. And then, if one is using Ashcroft and Mermin as the text, in Chapter 3, you encounter a topic titled "Failure of the Free Electron model".

Zz.

10. Dec 12, 2004

### rayjohn01

To Zz

Your pointing out of the various models does not constitute a 'proof' of Ohms law -- if the models prove reasonably accurate then they have value -- a model can only be internally consistent IT PROVES NOTHING.
Any proof is a physical test of the proposed law, and we know that such tests reveal subtle discrepancies , many times of no importance. However noise tests show complex phenomena as 'excess noise' ( not simple Gaussian )
even a straight material such as carbon shows excess noise -- so do thin film metallic resistor which exhibit random migration.
Theory is fine -- but not a proof , also to formulate such a proof is not simple the equation relates VOLTAGE to CURRENT ( not fields ) to do this requires to create a 'voltage ' source some sort of standard, you must be able to change this in some accurate way , and you must be able to measure 'current ' also as a standard.
People have worked on this for eons devising ways to create the standards required -- it has little to do with pointing out some 'ideal theory' even if it is a good one.

The relation V=I.R has no definition of size it could equally be 1 microvolt or 1 Mega volt so non linearity is included it's just that 'r' becomes a dependant variable , with scale.

Notice however that this expression says something a little odd -- IF 'R' is zero it states that 'I' --- > infinity for a finite voltage , but voltage is defined as the energy per unit charge -- i.e. 1 volt says that 1 coulomb has 1 joule of energy -- but massive charge of infinite quantity has infinite energy ?????

Ray.

Last edited: Dec 12, 2004
11. Dec 13, 2004

### ZapperZ

Staff Emeritus
Then we have a different interpretation of the original question of this thread. To me, when someone ask for "proof", it asks for "can one derive this from a more fundamental principle?"

Since this is physics and not mathematics, we cannot have, as a starting point, a "self-evident" principle to start with. Thus, there has to be a microscopic explanation for an observation that started of being purely phenomenological. It is clearly wrong to leave Ohm's Law AS IS, because like classical thermodynamics, ALL of its observations can be explained and derived from classical statistics. So this is clearly a search for the "cause-and-effect". The classical free electron gas is the "cause; Ohm's Law is ONE of it's many "effects". This reduces Ohm's Law as one of many consequences of the free electron gas model. Knowing what causes the emergence of Ohm's Law clearly is an extremely useful knowledge, because we can then know when and WHY certain material do not follow those.

I don't understand the rest of your "argument", though. You cannot get to "R=0" in a "real" conductor, even in an idealized case. This is because, for example, if you tried lowering the temperature to reduce R, at some point, the assumption of the classical gas breaks down (there aren't enough collisions to maintain the Maxwell-Boltzmann distribution). And if you're refering to superconductors as an example of "R=0", then we certainly can throw out Ohm's Law because it doesn't work at all there! So I find this example completely off-topic. People dealing with transport in materials do not worry about the application of Ohm's Law in such situation because this relationship breaks down well before we get even close to that stage!

BTW, if I give you the expression

$$j = \sigma E$$

where $$j$$ is the current density, $$\sigma$$ is the conductivity (which is the reciprocal of resistivity), and $$E$$ is the applied electric field, are you telling me that you do not see how this is really Ohm's Law?

Zz.

Last edited: Dec 13, 2004
12. Dec 13, 2004

### rayjohn01

To Zz

I have liitle problem with what you say except the definition of proof.
If you explain at a deeper level some law by another more complex theory -- that to me is great -- but now how do you prove THAT theory
Not by simple agreeing with the first.
To me there are two types of theory
a) purely abstact ( like pure maths)
b) a model of some physical phenomenon
in the latter case to me 'proof' is an agreement with measurements and as such is not a finished proof , since discrepancies will arise .
All we say is that the theory is good as far as , measurements go.
In this sense one can never prove Ohms Law in a total sense , but one can say that 'IT' and deeper explanations agree with measurements in a variety of circumstances to within some degree of accuracy.
Ray

13. Dec 16, 2004

### Pieter Kuiper

The validity of Ohm's law is not limited to free-electron-gases or Fermi liquids. It also holds for hopping conductors and for ionic conductors.

Ohm's law just means that there is dissipation which results in a constant velocity for charged particles in an electric field. It is analogous to the constant speed of raindrops in air. Relaxation times, viscosity, dissipation, friction - really all the same thing.

14. Dec 16, 2004

### ZapperZ

Staff Emeritus
But the amount of raindrops can change. In small band gap semiconductors, as the potential difference increases, at some point, you have an addition of charge carriers into the conduction band. So with increasing "V", the charge carrier density changes, resulting in a non-linear I-V curve.

Also note that Ohm's Law also assumes that one has an elastic scattering of either scattering centers and/or each other. Magnetic impurities are notorious for inducing inelastic scattering (example: Kondo effect). Again, you will not get a linear I-V curve if you happen to be in such a regime.

In such cases, the strict Fermi Liquid description breaks down.

Zz.

15. Dec 16, 2004

### Pieter Kuiper

Of course, voltage-dependent resistors do exist.
Of course, for Ohm's law to work, energy has to be transferred to the lattice as heat. Elastic-inelastic, does it matter what you call it?

Kondo just means that there is a minimum in the resistivity as a function of temperature (at low temperatures of gold with a little bit of iron in it, for exemple). But current is still proportional to voltage, and resistivity is well defined.

Fermi liquid, Kondo - theoretical overkill !

(The limitations on Ohm's law that I encounter occur when sample dimensions are smaller than the mean free path.)

16. Dec 16, 2004

### rayjohn01

Also--

Ohm's Law also works for radiation into space where no charges are involved.

17. Dec 16, 2004

### ZapperZ

Staff Emeritus
Not to prolong this any longer than it should, but I don't think this is a "theoretical overkill". And I'm saying that as an experimentalist too.

The whole issue of when Fermi Liquid/quasiparticle description is valid is a major issue in condensed matter. The normal state of optimally doped high-Tc superconductor, for instance, shows no sign of having the expected Fermi liquid quasiparticles - at least from both optical conductivity and ARPES spectra measurements. Furthermore, when the scattering rate is that high, the lifetime of such excitation becomes a factor, because you really do not have a "well-defined quasiparticle" to start with, and all the assumptions made in the FL theory simply breaks down. Since practically everything we know of in conventional transport problem depends on such a description, it isn't a matter of esoteric "overkill" to know of such boundaries, or what kind of a description we are using that results in such-and-such observation.

Zz.

18. Dec 16, 2004

### Pieter Kuiper

I am not saying that Fermi liquid theories or Kondo theories are not important or interesting. I am just saying that high theory has very little to do with the observation that current is generally proportional to voltage. Of course, in order to explain the magnitude of the proportionality factor (ie the resistivity) you need theory, which also should account for its frequency and temperature dependence.

19. Dec 16, 2004

### Gokul43201

Staff Emeritus
Quick question : Does all FL behavior go away ? I recall reading that you could explain most of the transport behavior by assuming a marginally FL behavior, but making the linewidth of the quasiparticle excitation different.

20. Dec 16, 2004

### ZapperZ

Staff Emeritus
That's a difficult question to answer because it is not clear where the boundary is between FL and MFL. Besides, MFL doesn't tell you any more than what we already know, because it is essentially phenomenology - a stop-gap measure to describe the property in the superconducting state (the normal state is still non-Fermi Liquid).

My old work on the highly overdoped high-Tc superconductors tends to show that they are more "Fermi-liquid-like" than any other doping level. At the very least, one has well-defined quasiparticle even in the normal state. But one still does not have the typical T^2 dependence of the scattering rate that Fermi Liquid requires, so one can essentially argue that that is still an MFL.

Zz.