Derive Ohm's law?

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I'm still a intermediate science student. i'm not understanding how to derive the ohm's law. can anyone help me?
 

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  • #2
G01
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Laws are not derived. They are relationships describing something observed in nature. Newton's Laws are not derived, they come from observations. Ohm's Law comes from our observations of resistors and other materials that display constant resistance(i.e. a linear relationship between current and voltage). It is just a mathematical statement describing those observations. Ohm's Law also isn't a "universal law," since not all materials follow it. Diodes are good examples of materials that don't follow Ohm's law.

For your reading pleasure:biggrin::

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html
 
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^ thanks for the info.
 
  • #4
ZapperZ
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Actually, you CAN derive Ohm's law. It isn't derived in intro physics, but I think most people who take Solid State Physics see this in almost the first week of class when dealing with the Drude Model.

When you arrive at [itex]J = \sigma E[/itex], that is essentially Ohm's Law if you remember that J is current crossing an area A, [itex]\sigma[/itex] is [itex]1/\rho[/itex] where [itex]\rho[/itex] is the resistivity, and E is change in potential over a unit length. So you do end up with Ohm's Law.

Zz.
 
  • #5
jtbell
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It isn't derived in intro physics,
It's often done in intro physics at the college level, at least in the calculus-level courses (Halliday/Resnick et al.).
 
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Actually, you CAN derive Ohm's law. It isn't derived in intro physics, but I think most people who take Solid State Physics see this in almost the first week of class when dealing with the Drude Model.

When you arrive at [itex]J = \sigma E[/itex], that is essentially Ohm's Law if you remember that J is current crossing an area A, [itex]\sigma[/itex] is [itex]1/\rho[/itex] where [itex]\rho[/itex] is the resistivity, and E is change in potential over a unit length. So you do end up with Ohm's Law.

Zz.
It depends on what you call "Ohm's Law". I believe it is common to call [tex]{\bf J}=\sigma {\bf E}[/tex] Ohm's law, and V=IR is a special case of it. In this view, there is indeed no derivation as it is essentially observational (although I believe you can use statistical mechanics to justify it in some way).
 
  • #7
Gokul43201
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It depends on what you call "Ohm's Law". I believe it is common to call [tex]{\bf J}=\sigma {\bf E}[/tex] Ohm's law, and V=IR is a special case of it. In this view, there is indeed no derivation
Then what do you call the thing in Zz's link?
 
  • #8
ZapperZ
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It depends on what you call "Ohm's Law". I believe it is common to call [tex]{\bf J}=\sigma {\bf E}[/tex] Ohm's law, and V=IR is a special case of it. In this view, there is indeed no derivation as it is essentially observational (although I believe you can use statistical mechanics to justify it in some way).
Er.. that's puzzling. The whole thing was derived using the statistical distribution of free electrons, which is the starting point of the Drude model. So how is it not a derivation?

And I've only shown the simplest case. I could have easily pointed to the Boltzmann transport equation where a more rigorous derivation can be shown in which the Drude model is a special case.

Zz.
 
  • #9
olgranpappy
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I'm still a intermediate science student. i'm not understanding how to derive the ohm's law. can anyone help me?
I guess there has been some confusion in this thread about what "derived" means. Of course, it is a correct statement that the laws of physics are ultimately based on experimental observations... fine.

But, usually, when we say derived, we mean, "derived from well-known, more fundamental laws." Or something to that effect. Indeed, what else could we possibly mean?

Thus, I think that "derived" in this context (classical picture of currents and whatnot) should mean "derived from Newton's laws as applied to a classical picture of a solid as an assortment of mobile negative charges and fixed positive charges"

Indeed, all we need is [tex]F=\frac{dp}{dt}[/tex] and the definition
[tex]\vec j=-ne\vec v[/tex] where j is current density, n is number density, e is magnitude of electric charge, and v is the average velocity of the charges:

From Newton's 2nd Law the sum of all forces acting on the i-th electron (and here we are also in the presence of an external electric field, E) gives the time rate of change of that electron's momentum:
[tex]
\frac{d \vec p_i}{dt}=-e\vec E(\vec x_i)+F_i^{\textrm{other}}
[/tex]
where the electric field is evaluated at the location of the electron and the term F^other refers to all of the other forces on the electron other than the external field (such as the very complicated electric fields produced by the other electrons).

Now, we assume that the electric field is constant (at least as far as many of the electrons in the sample are concerned). I.e., it varies only on distances much much greater than the inter-electron spacing. Then we calculate the average time rate of change of the momenta, called p, of all the electrons:
[tex]
\frac{d \vec p}{dt}\equiv
\frac{1}{N}\sum_{i=1}^N\frac{d \vec p_i}{dt}
=-e\vec E + <F^{\textrm{other}}>
[/tex]
where the <F> term is very complicated because the N-body problem we are considering is in general very complicated. So, we have to make a simplifying assumption to continue otherwise we are doomed.

We make the assumption ("relaxation time" assumption) that <F> can be roughly expressed by introducing one positive parameter, the relaxation time, tau. Then, by dimensional analysis and common sense (to get the right sign): [tex]<F>=-p / \tau[/tex] and we may simply solve our equation for p to find:
[tex]
\vec p = -e\vec E\tau(1-e^{-t/\tau})+\vec p_0 e^{-t/\tau}\;.
[/tex]

Now, we are interested in times much larger than tau so we simply have:
[tex]
\vec p = -e\vec E \tau
[/tex]
thus
[tex]
\vec j =\frac{Ne^2 \tau}{mV}\vec E
[/tex]
I.e., j is proportional to E, which is Ohm's Law with [tex]\sigma=\frac{ne^2\tau}{m}[/tex].

Then, to get the other form of ohm's law just integrate the above over the area of a wire of length L to find
[tex]
I=\frac{\sigma A}{L}V
[/tex]
or
[tex]
V=IR
[/tex]
with R = L/(\sigma A).
 
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  • #10
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Er.. that's puzzling. The whole thing was derived using the statistical distribution of free electrons, which is the starting point of the Drude model. So how is it not a derivation?

And I've only shown the simplest case. I could have easily pointed to the Boltzmann transport equation where a more rigorous derivation can be shown in which the Drude model is a special case.

Zz.
Sorry! I misunderstood your answer (and didn't notice your link). I assumed that by "derivation" Gigacore was asking how Ohm's Law follows from the laws of electrodynamics, i.e. Maxwell's equations, and nothing more complicated than that.

Anyway, as a kind of related comment on the statistical mechanical models, I always had the feeling that the agreement with the empirically-established Ohm's Law was taken as a justification of the original assumptions, and not that I was seeing a convincing derivation. Even with the Boltzmann equation I seem to recall that you need some simplifying assumptions to get anything useful, which are then justified by the result. It was awhile ago that I encountered these things, so please correct me if this feeling is way off.
 
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  • #11
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Well, the easiest way to "derive" it is to Taylor-expand the voltage across the resistor in terms of the current around a voltage of zero, drop the quadratic and higher order terms, and note that the current should be zero when the voltage is zero. Then, identify the resistance as dV/dI, and you're set. (Obviously, this doesn't give you much physics.)
 
  • #12
ZapperZ
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Sorry! I misunderstood your answer (and didn't notice your link). I assumed that by "derivation" Gigacore was asking how Ohm's Law follows from the laws of electrodynamics, i.e. Maxwell's equations, and nothing more complicated than that.
Er.. you can't derive Ohm's Law from maxwell equations because of one important thing - you need to know how to treat a bunch of charges moving at random, which isn't contained in Maxwell equation. That's why statistical physics comes in here. Without knowing the kind of distribution assumption, you don't have <v>.

Anyway, as a kind of related comment on the statistical mechanical models, I always had the feeling that the agreement with the empirically-established Ohm's Law was taken as a justification of the original assumptions, and not that I was seeing a convincing derivation. Even with the Boltzmann equation I seem to recall that you need some simplifying assumptions to get anything useful, which are then justified by the result. It was awhile ago that I encountered these things, so please correct me if this feeling is way off.
I'm not saying Ohm's Law didn't have any empirical history. I'm saying that we now know more about how it came about. Many things started off empirically or phenomenologically that we can now derive from First Principle. That's an example of the progress in physics.

And yes, you do have to make many simplification. Simply by stating that it can be derived via the Drude Model itself is a simplification, because we know that the Drude model has many limitations where it simply breaks down. If you open Ashcroft and Mermin, right in Chapter 3 is already a topic on the failure of the free-electron model. So one doesn't have to wait that long to know that this is a highly simplified model, which means Ohm's Law works very well in conventional conductors and may not in more exotic systems or conditions.

Zz.
 
  • #13
olgranpappy
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Well, the easiest way to "derive" it is to Taylor-expand the voltage across the resistor in terms of the current around a voltage of zero, drop the quadratic and higher order terms, and note that the current should be zero when the voltage is zero. Then, identify the resistance as dV/dI, and you're set. (Obviously, this doesn't give you much physics.)
uh. yeah.

and you can also get the non-linear corrections... they're just
[tex]
\frac{d^n V}{dI^n}
[/tex]
where n is greater than one.:wink:
 
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  • #14
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Er.. you can't derive Ohm's Law from maxwell equations because of one important thing - you need to know how to treat a bunch of charges moving at random, which isn't contained in Maxwell equation. That's why statistical physics comes in here.
Yes, I know that. That was my answer to what I thought his question was. My criticism of your response was only because I didn't look closely enough at your post.
 
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  • #15
G01
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Actually, you CAN derive Ohm's law. It isn't derived in intro physics, but I think most people who take Solid State Physics see this in almost the first week of class when dealing with the Drude Model.

When you arrive at [itex]J = \sigma E[/itex], that is essentially Ohm's Law if you remember that J is current crossing an area A, [itex]\sigma[/itex] is [itex]1/\rho[/itex] where [itex]\rho[/itex] is the resistivity, and E is change in potential over a unit length. So you do end up with Ohm's Law.

Zz.
Interesting, I guess I spoke to soon... I apparently didn't cover this in intro physics. I'll have to read up on this.
 
  • #16
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thanks for the info guys.

now i know how to do it.
 

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