# Ohm's Law Mellow - Comments

• Insights
anorlunda submitted a new PF Insights post

Ohm's Law Mellow Continue reading the Original PF Insights Post.
I disagree to the general Idea that Ohm's Law is not a law, yet I strongly support the importance of this issue which I think boils down to the very basics philosophy of what is a physical law, what are the therms under which it consider violated.
Following the article spirit, I would recommend title like: Ohm's hypothesis or even better - Ohm's convention.
I understood the solar device is given in order to provide real example for violation of ohm's law. But it is not. the rectification profiles comes from a diode like element in the equivalent circulate. Up to date there is no system that violates "Ohm's law" . Note for example that Memristor, Thermistor, Warburg element or any constant phase (CPE) or non-CPE do not violates Ohm's law.

Staff Emeritus
There are no rigid rules in science about the use of words like law, rule, theory, etc, so we are free to disagree

I view the definition of R as the ratio of V and I. As a definition, it can't be violated by definition (pun intended )

Newton's Laws and the conservation laws are all derivable from the fundamental symmetries of nature. I resist the idea of making Ohms Law comparable with those. But if you want to call it a law, go ahead and knock yourself out. But don't complain if I prefer different words.

vanhees71
Gold Member
2019 Award
Ohm's Law is derived from many-body theory. It's defining a typical transport coefficient in the sense of linear-response theory. It's the "answer" of the medium to applying an electromagnetic field, and defines the electric conductivity in terms of the induced current, ##\vec{j}=\sigma \vec{E}##, where in general ##\sigma## is a tensor and depends on the frequency of the applied field. So Ohm's Law is a derived law and has its limit of validity (particularly the strength of the electromagnetic field must not be too large in order to stay in the regime of linear-response theory).

sophiecentaur
Gold Member
I view the definition of R as the ratio of V and I. As a definition, it can't be violated by definition (pun intended )
I wholeheartedly agree. It's a formula / definition and says nothing about whether or not Ohm's law happens to apply to what's connected to the terminals on the 'black box' we're examining. R could change, or not as V,I or T changes. If it doesn't happen to change then the component is not following Ohm's Law. But one calculation wouldn't tell you one way or the other.
This puts me in mind of the SUVAT equations with which we learned to calculate motion under constant acceleration. We don't refer to them them as 'Laws' of motion and we wouldn't dream of suggesting that a measurement of the change in velocity of an object in a given time would be the same under all conditions. But somehow, R=V/I is referred to as Ohm's Law. Teachers and lecturers can be very sloppy about these things. Aamof, I don't remember the constant acceleration thing being emphasised to me in SUVAT learning days, either. They just drew a V/t triangle and did some calculations. It left me uneasy for quite a long while. But teenagers feel 'uneasy' about a lot of things so it was actually the last of my worries.

Staff Emeritus
Ohm's Law is derived from many-body theory. It's defining a typical transport coefficient in the sense of linear-response theory. It's the "answer" of the medium to applying an electromagnetic field, and defines the electric conductivity in terms of the induced current, ##\vec{j}=\sigma \vec{E}##, where in general ##\sigma## is a tensor and depends on the frequency of the applied field. So Ohm's Law is a derived law and has its limit of validity (particularly the strength of the electromagnetic field must not be too large in order to stay in the regime of linear-response theory).
That's true for the specialized case of a linear and uniform mediums. The article addresses the general case of circuits containing any components, linear/nonlinear, active/passive. As the article says, you can always linearize about a point, define R=V/I, then use linear circuit methods to solve it.

sophiecentaur
Gold Member
Ohm's Law is derived from many-body theory. It's defining a typical transport coefficient in the sense of linear-response theory. It's the "answer" of the medium to applying an electromagnetic field, and defines the electric conductivity in terms of the induced current, ##\vec{j}=\sigma \vec{E}##, where in general ##\sigma## is a tensor and depends on the frequency of the applied field. So Ohm's Law is a derived law and has its limit of validity (particularly the strength of the electromagnetic field must not be too large in order to stay in the regime of linear-response theory).
That's ramping it up a bit for a number of the audience, I think. But also, if σ changes with some other variable, the relationship breaks down so any 'Law' has hit the rails. A Law that's worth its salt will involve all the relevant variables - Ohm's law, when stated fully, fits that requirement.

vanhees71
Gold Member
2019 Award
That's true for the specialized case of a linear and uniform mediums. The article addresses the general case of circuits containing any components, linear/nonlinear, active/passive. As the article says, you can always linearize about a point, define R=V/I, then use linear circuit methods to solve it.
It's true for any medium in the linear-response regime. More completely written out the relation reads
$$\tilde{\vec{j}}(\omega,\vec{k})=\hat{\sigma}(\omega,\vec{k}) \tilde{\vec{E}}(\omega,\vec{k}),$$
where we have Fourier-transformed fields in the frequency-wave-number domain, and ##\hat{\sigma}## is a complex-valued symmetric 2nd-rank tensor obeying the analytic structure in the complex ##\omega## plane such that it is a retarded propagator.

If you have "active" elements and non-linearities, you have to extend the approximation beyond the linear-response level, as far as I know.

vanhees71
Gold Member
2019 Award
That's ramping it up a bit for a number of the audience, I think. But also, if σ changes with some other variable, the relationship breaks down so any 'Law' has hit the rails. A Law that's worth its salt will involve all the relevant variables - Ohm's law, when stated fully, fits that requirement.
I don't know, what you mean. Electric conductivity is a typical transport coefficient, describing the response of the medium to a small perturbation around equilibrium (in this case by a weak electromagnetic field). It's restricted to weak fields in order to stay in the linear-response regime. Of course, it has a range of validity, as has any physical law (except the ones we call "fundamental", because we don't know the validity ranges yet ;-)).

• LvW
sophiecentaur
Gold Member
I don't know, what you mean. Electric conductivity is a typical transport coefficient, describing the response of the medium to a small perturbation around equilibrium (in this case by a weak electromagnetic field). It's restricted to weak fields in order to stay in the linear-response regime. Of course, it has a range of validity, as has any physical law (except the ones we call "fundamental", because we don't know the validity ranges yet ;-)).
I just meant that your wording and representation takes it to a higher level of understanding and familiarity. Of course the equation is correct - but it doesn't pretend to be a Law. By the time one gets to the level that you are using to describe what happens, I doubt that one would bring in the term Law.
But I guess this will never lie down as it falls within the overlap between higher level Physics and down to Earth practicalities; the two have different agendas.

sophiecentaur
Gold Member
Yes, if it's an applied voltage. A voltage drop (symbol V) occurs when current passes through the resistor.
Conversely, with a voltage source, EMF (symbol E) produces the current that passes through the resistor.
Isn't that just a chicken and egg argument for describing a 'relationship' between two variables?

• cabraham
...and we can think about the meaning of the form: V=I*R.
We are using this form to find the "voltage drop" caused by a current I that goes through the resistor R.
However, is it - physically spoken - correct to say that the current I is producing a voltage V across the resistor R ?
(Because an electrical field within the resistive body is a precondition for a current I, is it not?)
Yes it is physically correct to say that current I produces voltage V in a resistance. It is also correct that a voltage V places across resistance results in current I.
Either one can give rise to the other.
No, an electric field across the resistance is not necessary for current to commence. A switch is closed, a battery has an E field due to redox chemical reaction. Charges move through the cables towards the resistor. Current is already commenced by battery redox. When the charges reach the resistance, they continue into the body but incur collusions between electrons & lattice ions. This results in e lectrons droppii g from conduction band down to valence band. Polarization occurs with photon emission. When current is in a resistance it gets warm from this energy conversion. The E field across the resistor happens when charges emitted from the battery arrive. Positive battery terminal attracts electrons from cable. An electron vacating its parent atom leaves a positive ion behind or hole if you prefer. The atom next in line emits an electron towards this hole. Reverse happens at negative battery terminal. The charges & the associated E field arrive at the resistor. Current already is established, as the charges are in motion before the resistor receives them. Charges proceed through the resistor colliding with lattice ions resulting in polarization & photon emission. Polarized charges have an E field, & the line integral of said E field over the distance is the voltage drop.
At equilibrium the equation J = sigma*E, or E = rho*J, which is Ohm's law in 3 dimensions. I will elaborate if needed.

Claude

Isn't that just a chicken and egg argument for describing a 'relationship' between two variables?
Yesz it is. I & V generally h ave a circular relation. Either can come first & produce the other.

sophiecentaur
Gold Member
Yesz it is. I & V generally h ave a circular relation. Either can come first & produce the other.
We are so used to using Batteries, which are essentially Voltage Sources, that it is hard to avoid think of Voltage as the senior member of the VI pair.

• cabraham
We are so used to using Batteries, which are essentially Voltage Sources, that it is hard to avoid think of Voltage as the senior member of the VI pair.
I would agree. One thing worth noting is that a battery can be produced for constant current as well. A short across the terminals results in the off state, or no load. But losses would be greater than an open voltage source. So primary cells have been built for constant voltage operation for over a century.
Nuclear fission batteries have been produced, searching for them using "nucell" will give details. These nuclear cells are not only current sources, but a.c. instead dc. An a.c. current source battery, that is different. Apparently nuclear cells function better as a.c. current sources. They use fissionable material, so I won't hold my breath waiting for them to be available to the general public.

Claude

• sophiecentaur
Staff Emeritus
We are so used to using Batteries, which are essentially Voltage Sources, that it is hard to avoid think of Voltage as the senior member of the VI pair.
Try thinking of a superconducting loop with a magnetically induced current. No voltage in the loop before or after the current is induced.

We have disagreed before on the topic of teaching electricity. I think that we should stick to the 3 valid levels, QED, Maxwells, and Circuit Analysis (CA) [including Kirchoff's laws]. One of the key assumptions of CA is
• The time scales of interest in CA are much larger than the end-to-end propagation delay of electromagnetic waves in the conductors. [A simple rule of thumb, for 60 hertz AC circuits should have lengths of <500 km.]
For Voltage to start before Current, explicitly violates this assumption. That's OK in Maxwell's equations, but we should not mention it within the context of using CA. That is not helping students of CA, it is feeding them contradictory and confusing information.

Staff Emeritus
I don't know, what you mean. Electric conductivity is a typical transport coefficient, describing the response of the medium to a small perturbation around equilibrium (in this case by a weak electromagnetic field). It's restricted to weak fields in order to stay in the linear-response regime. Of course, it has a range of validity, as has any physical law (except the ones we call "fundamental", because we don't know the validity ranges yet ;-)).
I think you ignored the following paragraph from the article that inspired me to write the article in the first place.

Students often forget that limits exist. A frequent (and annoying) student question is, “So if I=V/R, what happens when R=0. Ha ha, LOL.” They think that disproves the “law” and thus diminishes the credibility of science in general. Their logic is false.
Students wouldn't ask that dumb question if they understood that Ohm's Law only applies to a limited region. I don't believe that their teachers understand that. I suspect that the teacher's teachers don't understand that. In basic electricity Ohm's Law is being taught as absolutely true as if it had a foundation like the principle of least action underlying Newton's Laws of Motion. Somehow, the message that limited ranges are obvious is not getting passed down the ladder. Perhaps is is related to the fact that conduction in bulk materials requires quantum effects to accurately describe and that is just too difficult for most students and most teachers. That is what this article tried to address.

Even grad students and profs could stand a reminder and a moment of reflection on the fact that there is not physical principle that says that there has to be any wide region where voltage and current are linearly proportional. It could have been nonlinear all the way. If that were true, then simple algebra could not have been used to analyze simple circuits, and the evolution of electricity, electronics and computers in the 20th century would have taken significantly longer. If computers had been delayed, so would all of science. Therefore, IMO we should all thank our lucky stars for the accident that Ohm's Law is useful at all.

• Asymptotic, vanhees71 and cnh1995
sophiecentaur