21
$\begingroup$

I was in a question in Electronics SE, and a lot of people said something that I'm sure is wrong. They said, "The voltage between two points that are not part of the same circuit is undefined".

Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined".

Now, I'm 100% sure that the voltage between any two points in the universe is well defined. It doesn't matter if they are in different circuits, same circuits, or even the same galaxies.

It may be hard or impossible to calculate, but it is not undefined!

This was driving me crazy so I ask, am I wrong? Are they wrong?

A follow up question. If two pieces of the same metal, are under the same external electric fields (which may be zero), and carry the same excess charge (which may be zero), then the voltage between them must be zero, even if one piece is in Jupiter and the other one in an electric circuit in my back yard. Is this wrong?

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ I take two pieces of metal, no charge, zero applied electric field. One I leave on my work bench. One I place in a conducting sphere at the top of a Van de Graaff accelerator. Each piece continues to have no net charge and zero applied electric field. I now turn on the VdG machine. Without me telling you the voltage on the VdG machine, what is the potential difference between the two pieces of metal? $\endgroup$
    – Jon Custer
    Commented Nov 4, 2021 at 17:59
  • 4
    $\begingroup$ Also remember that the whole concept of the potential difference between two points becomes undefined as soon as you introduce unconfined time-varying magnetic fields to your system. $\endgroup$
    – The Photon
    Commented Nov 4, 2021 at 18:22
  • 16
    $\begingroup$ Would you be happier with the statement "Lacking any additional information, we cannot say what the voltage is between two points that are not part of the same circuit—the potential difference could be any number at all without introducing any inconsistencies."? $\endgroup$
    – Chemomechanics
    Commented Nov 4, 2021 at 19:21
  • 21
    $\begingroup$ As far as us electrical engineers are concerned, the voltage between two unconnected points is undefined, because there's no way you could know it, and it could change at any moment for any number of reasons. The real world is messy, not like the physicist's world of spherical cows. $\endgroup$
    – Hearth
    Commented Nov 5, 2021 at 2:43
  • 4
    $\begingroup$ What I find confusing about the question, most comments, and most answers is: What is the definition of "undefined"? These comments and answers seem to be referring to things that are "unknowable" or "indeterminate." But that doesn't mean the same thing as "undefined." We may define the meaning of terms (e.g., "potential difference") or the meaning of units (e.g., "volts") but we don't "define" physical quantities. They are what they are, even if we have no way to measure or calculate them. Do engineers define "undefined" in a way that would make these answers/comments make sense? $\endgroup$
    – Syntax Junkie
    Commented Nov 6, 2021 at 2:09

7 Answers 7

Reset to default
60
$\begingroup$

The voltage between to points in the universe is always defined. But the voltage between two circuits is undefined because (at least from electronics point of view) circuits are abstract models of closed systems.

Suppose we have two devices that consist of a battery connected a resistor - one resting on the ground - the other suspended from a balloon.

Schematic showing two independent battery-resistor circuits side by side

It would be possible to measure a real potential difference between any two points in those devices. And yet if we model these devices as different circuits any voltage between them cannot be calculated and is undefined because we haven't included any information about how those circuits are related.

What we can do is refine these circuits by including parameters such as resistance, capacitance and inductance of air between the devices. But when we include such information they cease to be two distinct circuits and become a single more complex circuit where voltage between any two points can be calcuated.

Schematic showing the same circuits as before connected by parallel R L C elements

Note: Here I used simple lumped circuit approximation to illustrate the point, but the principle remains the same even for more complex models.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Can you help me understand this part of your answer: "circuits are abstract models of closed systems." It would make sense to me if it read, "circuit diagrams are abstract models. Is that what you meant? But isn't the original question about actual circuits? $\endgroup$
    – Syntax Junkie
    Commented Nov 8, 2021 at 17:58
  • $\begingroup$ Circuits themselves are abstractions because everything can and does conduct electricity in one way or another. Air gaps are just weak capacitors or spark with high enough voltage, vacuum is part of a circuit in vacuum tubes and when your cell phone is handshaking it induces current in your speakers. The only way to tell if something is or is not part of particular circuit is whether it is convenient for us to do so, usually because the effects are expected to lower than some arbitrary threshold that we care about. But in nature $\mu A$ and $nA$ are conducted whether we like it or not. $\endgroup$
    – SilentAxe
    Commented Nov 8, 2021 at 21:14
  • $\begingroup$ @SyntaxJunkie Also related: Is every open circuit a capacitor? and Does alternating current (AC) require a complete circuit? $\endgroup$
    – SilentAxe
    Commented Nov 8, 2021 at 21:26
17
$\begingroup$

Is the voltage ever undefined?

The voltage, or potential difference, between any two points is always defined. It is defined as the work per unit charge required to move the charge between the two points.

"The voltage between two points that are not part of the same circuit is undefined".

It is not defined in the sense that one cannot determine what that voltage is by applying Kirchhoff's voltage laws to two independent circuits. But any voltage that may exist between them is still defined as above.

Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined".

I look at the question of voltage between a point isolated from ground and ground as a separate one. There is a known electric field between an isolated point in the atmosphere and earth ground. In areas of fair weather, the atmospheric electric field near the surface of the earth's surface typically is about 100 V/m directed vertically in such a sense as to drive positive charges downward to the earth. See:

https://glossary.ametsoc.org/wiki/Atmospheric_electric_field#:~:text=In%20areas%20of%20fair%20weather,charges%20downward%20to%20the%20earth

So if a point in the circuit is 1 meter above the ground, the voltage between it and the ground with air in between would be 100 V. However, due to air being a good insulator, the current density is only about 10 $^{-12}$ amperes per square meter parallel to the earth, so its influence on the circuit would be appear to be negligible. Its effect on you would also be negligible since you are a relative good conductor (relative to the air) so that when you stand on the ground you and the ground become equipotential surfaces. See.

https://www.feynmanlectures.caltech.edu/II_09.html

Regarding your follow up question if the two metal pieces have the same positive or negative charge, the voltage would be zero. If you connected the pieces with a conductor there would be no current in the conductor.

Hope this helps.

Improve this answer
$\endgroup$
12
  • 1
    $\begingroup$ This sparked my curiosity. Can this electric field be used to generate usable electricity? I'm guessing that not, or else things would be a lot easier in our world. But if there is such a potential difference for just 1 meter of height, why can't it be used with large enough antennae or something? $\endgroup$
    – Vilx-
    Commented Nov 5, 2021 at 10:50
  • $\begingroup$ Read the Feynman link I provided $\endgroup$
    – Bob D
    Commented Nov 5, 2021 at 12:02
  • 2
    $\begingroup$ @BobD Nice answer, but I disagree with the statement "The voltage, or potential difference, between any two points is always defined. It is defined as the work per unit charge required to move the charge between the two points." This is only strictly true in the absence of time varying magnetic fields, because $\nabla \times \mathbf E = -\partial \mathbf B/\partial t$ and conservative fields are irrotational. $\endgroup$
    – ummg
    Commented Nov 5, 2021 at 14:18
  • 1
    $\begingroup$ @PeteKirkham What I quoted to you was under the heading "Voltage". So it was a description of voltage. Now it's true they didn't use the word "defined", while they used the word defined for the next item under the heading "Current" which read "Electric current $i(t)$ through a surface is defined as the rate of charge transport through that surface". Whether or not they deliberately did not use the term "defined" for voltage is a matter of conjecture. $\endgroup$
    – Bob D
    Commented Nov 5, 2021 at 18:06
  • 2
    $\begingroup$ @Pete Kirkham For what it’s worth, during the course of 35yrs before retiring I both authored and contributed to many standards (UL, IEC, IEEE) and one of the hardest damn things to do was getting folks to agree on definitions. Even on a definition of the device covered by the standard! $\endgroup$
    – Bob D
    Commented Nov 5, 2021 at 21:10
16
$\begingroup$

I interpret that statement as "the voltage difference between those two disconnected circuits cannot be solved for using Kirchhoff's laws".

Improve this answer
$\endgroup$
10
$\begingroup$

I want to elaborate a little bit on what several comments and answers have stated: Undefined from an engineering perspective is different from undefined from the physics perspective.

An electrical engineer wants to control voltages and currents in a circuit to do something useful. What matters is whether they have control over the voltage between two points in the circuit. Another way to phrase this is: can they predict what a voltmeter would read if they connect two points in a circuit? If they cannot, then the voltage is undefined from an engineering perspective because an engineer cannot rely on that voltage to be any particular value. It's irrelevant whether electric potential is actually defined from a physics perspective.

As a concrete example, consider a digital NOT gate. Let's say we work with 5V logic, so passing a 5V signal into the input of a NOT gate gives 0V and passing 0V to the input of the gate gives 5V. Now, what happens to the output if the input of the digital NOT gate was not connected to any part of the circuit? Physically, the input will be at some voltage relative to the circuit ground, but you cannot predict what this voltage is as it will vary depending on the environment. Thus, for an engineer, the input voltage is indeterminate and of no engineering use, hence "undefined". The consequence is that sometimes the output will be 5V, sometimes it will be 0V, and the circuit designer has no control over when this happens, which is highly problematic if you're trying to make the circuit do something useful. Circuit designers take great care to avoid these indeterminate voltages. For example, this is why you have pull-up and pull-down resistors to make sure that the inputs to digital gates are never "floating". Hence, there is need to have vocabulary to talk about whether you have practical control over the potential difference between two parts of a circuit.

Improve this answer
$\endgroup$
15
  • $\begingroup$ Very good explanation, except for the last part. Electric potential and voltage = difference of potentials is well defined even for inductor with time-varying current, via the Coulomb potential formula. It's just that this potential by itself isn't sufficient information to determine total electric field (because of the induced electric field component). Voltage on perfect inductor is $LdI/dt$, both from physics and engineering perspective. There is no difference between the two perspectives. However maybe electrical engineers are more likely to well understand this than physicists. $\endgroup$
    – Ján Lalinský
    Commented Nov 5, 2021 at 10:55
  • $\begingroup$ What we have is that $E=−\nabla V−\frac{\partial A}{\partial t}$ and now we have gauge freedom to choose what $V$ means and, importantly, affect what $\Delta V$ can be. We can choose the Coulomb gauge, but we don't have to. Consequently, $\Delta V$ is no longer uniquely defined. But agreed that the explanation is a bit confusing. $\endgroup$
    – user318039
    Commented Nov 5, 2021 at 14:24
  • $\begingroup$ In practice we do, it's always the Coulomb potential in AC circuits, even when they include inductors. $\endgroup$
    – Ján Lalinský
    Commented Nov 5, 2021 at 17:37
  • $\begingroup$ A digital NOT gate implemented in TTL is not really floating in that sense (the effective pull-up resistance is ~5 kΩ). I suppose you mean a MOSFET NOT gate or similar. The time constants may be long (due to capacitances and high resistances), but if it was isolated from external influences/fields, wouldn't it eventually settle (the insulation resistance is not infinite)? Perhaps add an approximate settling time (backed up by a simple calculation) to demonstrate that it is impractical? $\endgroup$
    – Peter Mortensen
    Commented Nov 6, 2021 at 15:14
  • $\begingroup$ @JánLalinský no. When a variable magnetic field is present voltage is NOT equal to potential difference. In the same way we can decompose the total electric field in the contributions of the coloumbian electric field and the induced electric field, voltage becomes the contribution of the path integral of the coloumbian part (which can be expressed as potential difference) and the path integral of the induced part (which is the contribute of the emf). See for example Popovic "Introductory Electromagnetics", sec 14.4 Potential difference and voltage in a time-varying electric and magnetic field. $\endgroup$
    – Peltio
    Commented Nov 6, 2021 at 16:27
6
$\begingroup$

The concept of potential difference, as it is used in circuit theory, is only defined in what is called the lumped circuit approximation. This approximation requires that the extent of the circuit is physically small enough that light-speed delays between the elements of the circuit are insignificant. It also requires that there are no time-varying magnetic fields passing through the wires interconnecting the circuit.

With some effort and hand-waving circuit theory can be extended to accommodate transmission lines (where the light speed delay is not insignificant), but we generally still must pretend (even if it isn't true), that the ground potential is constant between the two ends of the transmission line.

Both of the components of the lumped circuit approximation are invalidated in the case where one part of your circuit is on Earth and one on Jupiter, so you can not expect the potential difference between two points on the two parts of the circuit to be well defined.

Improve this answer
$\endgroup$
11
  • $\begingroup$ I suppose we can always calculate the energy required to bring a charge along a path from point A to point B. Right?Then the question becomes, under what circumstances is the energy path-dependent? $\endgroup$
    – Gilbert
    Commented Nov 4, 2021 at 18:35
  • $\begingroup$ @Gilbert, in the presence of changing magnetic fields, the energy required will depend on the path chosen. Then you can't define a potential difference between the two points. $\endgroup$
    – The Photon
    Commented Nov 4, 2021 at 19:09
  • 1
    $\begingroup$ So, there is an approximation to the value of potential difference, used in circuitry, that is only valid under certain circumstances. But: the actual physical non-approximated value of potential difference, is that also undefined under those circumstances? $\endgroup$
    – Juan Perez
    Commented Nov 4, 2021 at 19:25
  • 1
    $\begingroup$ @JuanPerez, see my reply to Gilbert. The approximation is to assume the potential difference exists at all, even when there are actual differences in the energy required to get from A to B depending on the path. $\endgroup$
    – The Photon
    Commented Nov 4, 2021 at 20:38
  • 1
    $\begingroup$ Electric potential can be always defined via the Coulomb potential formula, whether magnetic fields are present or not. It just isn't always sufficient for determining other things, such as total electric field. $\endgroup$
    – Ján Lalinský
    Commented Nov 5, 2021 at 10:49
3
$\begingroup$

People seem to be overlooking something that I consider pretty important. The voltage is not always well-defined, even in principle. By definition, voltage is defined by $$ V(x) = - \int_P^x \vec{E} \cdot d\vec{x} $$ where $P$ is some chosen reference point. This definition only makes sense if the integral does not depend on the path you take to get between $P$ and $x$. For smooth enough $\vec{E}$, this is equivalent to demanding $\nabla \times \vec{E} = 0$. But Maxwell's equations tell us that $$ \nabla \times \vec{E} = -\frac{\partial B}{\partial t} $$ So whenever there are time varying magnetic fields, we cannot expect to have well-defined voltages, at least not according to the usual definition.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Actually, voltage is well defined - if you specify the path. You just can no longer define a potential function to associate to the total electric field. One can decompose the total field into its conservative ans solenoidal components - following Helmoltz - and then associate a potential function to the conservative part alone. Of course that alone is not sufficient to completely characterize the field, and you need to supply the vector potential A as well. $\endgroup$
    – Peltio
    Commented Nov 6, 2021 at 23:45
  • $\begingroup$ Incidentally, if you just change voltage into potential (and potential difference) into the above answer my objection ceases to exists. Also see electropedia.org/iev/iev.nsf/… and electropedia.org/iev/iev.nsf/… $\endgroup$
    – Peltio
    Commented Nov 7, 2021 at 0:01
  • $\begingroup$ I think its a safe assumption that there are non-zero time varying magnetic fields everywhere (even if extremely minuscule), and thus by your answer, voltage is undefined everywhere. Or is defined-ness a gradient, from well to ill, as the magnetic fields get more significant? $\endgroup$
    – Juan Perez
    Commented Nov 7, 2021 at 13:36
  • 1
    $\begingroup$ How much the voltage depends on the path of integration can vary. $\nabla \times \vec{E}$ in some sense measures this path-dependence of voltage. Though, as Peltio points out, there is nothing stopping us from defining voltage by $V(x) = - \int_P^x \vec{E} + \frac{\partial \vec{A}}{\partial t}$, which will always be well-defined (up to some choice of gauge), if hard to calculate. I'm not sure what the interpretation or significance of $V(x)$ defined this way would be though. $\endgroup$
    – Charles Hudgins
    Commented Nov 7, 2021 at 14:23
3
$\begingroup$

Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined

Undefined in the mathematical sense? No.
Undefined in the a real world physical measurement sense? Most likely Yes.

But it depends on who you are talking to. An engineer will probably use the term undefined meaning unmeasurable or unknowable.

Floating voltages are in most cases very hard to measure because as soon as you attach a meter to a floating voltage, the resistance of the thing that you are using to measure the voltage changes the 'circuit' (air with it's really high resistance can be part of a circuit) and so a floating voltage could be considered unknowable because there isn't a good way to know what the voltage is.

From a mathematical standpoint, 'no' all the equations and values are defined.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.