5 ms·

Are you being as pedantic as to ask for "through an area"? Because that's literally the defined relationship between current and charge.

contravariant 8y ago
This isn't mere pedantry, there is a difference between current and the time derivative of charge, I don't mind that they prefer to use one or the other, but if you insist on making a periodic table then you've got to take both into account. It's also an especially bad idea to claim the time derivative of current is the the double derivative of charge, given that these are two very different notions of derivative. To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level.
- lnx01 8y ago
  Current is unequivocally the time derivative of charge. There're no two ways about it. i(t) = dQ(t) / dt "To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level." This statement is wrong, while water in some cases provides a decent analogy for electricity, this is not one of them.
  - contravariant 8y ago
    If I have a piece of wire with a constant current through it the charge everywhere on the wire remains constant. In fact the charge will be more or less neutral in most cases. Note that the alternative would be that the wire somehow accumulates charge as it passes current, which clearly isn't happening.
    - tremon 8y ago
      No, you are wrong. Electrical current is the mathematical representation of a charged particle moving between two points. The value of the current is directly related to the amount of charge being moved. As for your "wire" example, you're conveniently ignoring that a single (open-ended) wire isn't a closed circuit and as such there can be no current (and hence no accumulating charge). And if it's a closed circuit, then there are other elements in the circuit that consume or generate charge.
      - drostie 8y ago
        They are not wrong. In the Maxwell equations both come in as fundamentally different terms. The claim is that current density J is different from the time rate of change of charge density dρ/dt. That is not to say they are unrelated; they are related by the continuity equation, dρ/dt = -∇·J. The distinction is real, because what you are calling current in the one case is actually a spatial derivative of current, as indicated by the ∇. I would actually go a step further than this and say that current is actually properly defined as the source of magnetic field. On the conventional definition of current, it is physically impossible for current to flow through a capacitor, but we speak of that all the time. So the True Current Density is just J + ε dE/dt in SI units. Actually taking that seriously, however, does require to committing to language which sometimes seems a little awkward, like saying electromagnetic radiation involves an AC current oscillation that propagates through empty space transverse to its oscillation.
        alxlaz 8y ago
        > I would actually go a step further than this and say that current is actually properly defined as the source of magnetic field. That's actually (still, and somewhat) how the Ampere is defined. There are ongoing efforts to change this though. I... prefer to avoid discussions like this one, but I thought you might appreciate this part :-).
      - contravariant 8y ago
        Fine, charge a circular piece of wire, place it in space and spin it around. Voila, current without any accumulation of charge. Honestly I'm confused why it's turning out to be such a controversial point that there's a difference between a change in charge and the movement of charges. Heck in their diagram they claim that the voltage across an inductor is the double derivative of its charge, but inductors can't even hold charge so it's unclear what they're trying to say. They also claim that the charge across a resistor is somehow related to the integral of the voltage across it, but again (ideal) resistors can't really hold charge. The only way to interpret this supposedly 'universal' periodic system is by interpreting charge and its derivatives in different ways depending on context, which isn't convincing in a supposedly universal system.
    - lnx01 8y ago
      "charge everywhere on the wire remains constant" - perhaps. But the charge is still flowing, it's still moving at a rate of dQ/t. What is dQ/t you may ask? It's the time derivative of charge. It's the change in charge over time, the rate of flow of charge. I=Q/t Current=Charge/Time Amps=Coulombs/Second I don't know how to make this clearer.
  - tntn 8y ago
    This is what happens when someone with a clear idea of a lumped element model slams into someone with a clear idea of Maxwell's equations.
  - OJFord 8y ago
    That equation is unequivocally true...under certain assumptions. Contravariant is simply saying that it's being stated and used as if it holds generally; but that's not the case. This thread quickly dismissed 'through an area' as pedantic, but as far as I can tell that really is the crux of the debate: this simplified form of the equation only holds for scalar flux - viz. constant over and perpendicularly through a fixed flat surface. In the generalised form, `I` is a function of the area at a point.
- DougBTX 8y ago
  > To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level. My electro mag skills are a little rusty, but I don’t think that’s the right analogy. You wouldn’t say that the speed of a runner is the derivative of their height, similarly when someone is running you wouldn’t try to say that their speed isn’t a derivative of their height, as their height is more or less constant “for the length of the run”. Similarly, it isn’t the height of the runner, but the distance they’ve covered, it isn’t the depth of the river, but how much water has passed through, it isn’t the static charge of the wire, but how much charge has passed through.
  - contravariant 8y ago
    I think you're just rephrasing what I'm trying to say in several different ways. We can probably agree that the amount of charge passing through is different from the change to the total charge. And in my opinion something like a 'periodic system' can't just treat both the same. It's especially bad to derive identities by taking the time derivative (or anti-derivative) on one side and switching between current and charge on the other.
  - drostie 8y ago
    It's only a little bit off. To get the true hydrodynamic analog to a capacitor, connect the bars of two big pistons together so that volume accumulated in one comes at the expense of the other, then hit the bar with a spring so that its motion comes with some energy cost that can oppose a constant pressure. The point is that this component alone ties flow to accumulation, whereas generally your other components (resistors=thin pipes, wires=thick pipes, batteries=Archimedes screws, inductors=turbines connected to flywheels) do not accumulate volumes of water inside of them. Flow needs to make sense even without accumulation due to flow-balance, just like force needs to make sense even in situations where velocity stays constant due to force-balance.
jkabrg 8y ago
The amount of charge in an area stays constant because of Kirckhoff's laws, so dQ/dt=0 always. Wouldn't a better definition be: I = q v . A where q is charge density. A is an area (actually, a normal vector to some flat cross-section, with magnitude equal to size of area), and v is velocity of the charged particles. I'm using the dot product. If we check the units: (coulombs * metres^{-3}) * (metres * seconds ^{-1}) * metres^{2} = coulombs / seconds By the way, I fudged the above to make "charge density" have units $coulombs * metres^{-3}$. I'm not the best person at physics.
- majewsky 8y ago
  > The amount of charge in an area stays constant because of Kirckhoff's laws, so dQ/dt=0 always. Eh what? No. There seems to be some huge misunderstanding here. First of all, charge is not "in an area". It's in volumes, and flows through areas. And the amount of charge in a volume is absolutely not constant (in general). If you think it is, then please explain what happens when you rub a balloon on your hair, or why/when lightning strikes.
  - jkabrg 8y ago
    I was referring to Kirchhoff's first law which only applies to circuits -- that example is enough to show that I=dQ/dt is wrong. I wasn't considering the cases you're talking about (lightning etc.) because only one example is needed to prove an equation wrong. I could easily produce a single specific example where dQ/dt=0 but there is clearly a non-zero current present. And you can talk about charge in an area (sort of) if you multiply charge density by area; the result has units coulombs * metre^{-1}. I thought counting charge in a cross-section of wire was something people did, one way or another, even if the units aren't Coulombs but Coulombs*metre^{-1}. I could very easily have misunderstood something. With the clarifications above, I'm not clear on what though.
    - majewsky 8y ago
      Okay, that's much more nuanced than "dQ/dt = 0 always".

contravariant 8y ago

This isn't mere pedantry, there is a difference between current and the time derivative of charge, I don't mind that they prefer to use one or the other, but if you insist on making a periodic table then you've got to take both into account. It's also an especially bad idea to claim the time derivative of current is the the double derivative of charge, given that these are two very different notions of derivative. To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level.

lnx01 8y ago
Current is unequivocally the time derivative of charge. There're no two ways about it. i(t) = dQ(t) / dt "To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level." This statement is wrong, while water in some cases provides a decent analogy for electricity, this is not one of them.
- contravariant 8y ago
  If I have a piece of wire with a constant current through it the charge everywhere on the wire remains constant. In fact the charge will be more or less neutral in most cases. Note that the alternative would be that the wire somehow accumulates charge as it passes current, which clearly isn't happening.
  - tremon 8y ago
    No, you are wrong. Electrical current is the mathematical representation of a charged particle moving between two points. The value of the current is directly related to the amount of charge being moved. As for your "wire" example, you're conveniently ignoring that a single (open-ended) wire isn't a closed circuit and as such there can be no current (and hence no accumulating charge). And if it's a closed circuit, then there are other elements in the circuit that consume or generate charge.
    - drostie 8y ago
      They are not wrong. In the Maxwell equations both come in as fundamentally different terms. The claim is that current density J is different from the time rate of change of charge density dρ/dt. That is not to say they are unrelated; they are related by the continuity equation, dρ/dt = -∇·J. The distinction is real, because what you are calling current in the one case is actually a spatial derivative of current, as indicated by the ∇. I would actually go a step further than this and say that current is actually properly defined as the source of magnetic field. On the conventional definition of current, it is physically impossible for current to flow through a capacitor, but we speak of that all the time. So the True Current Density is just J + ε dE/dt in SI units. Actually taking that seriously, however, does require to committing to language which sometimes seems a little awkward, like saying electromagnetic radiation involves an AC current oscillation that propagates through empty space transverse to its oscillation.
      - alxlaz 8y ago
        > I would actually go a step further than this and say that current is actually properly defined as the source of magnetic field. That's actually (still, and somewhat) how the Ampere is defined. There are ongoing efforts to change this though. I... prefer to avoid discussions like this one, but I thought you might appreciate this part :-).
    - contravariant 8y ago
      Fine, charge a circular piece of wire, place it in space and spin it around. Voila, current without any accumulation of charge. Honestly I'm confused why it's turning out to be such a controversial point that there's a difference between a change in charge and the movement of charges. Heck in their diagram they claim that the voltage across an inductor is the double derivative of its charge, but inductors can't even hold charge so it's unclear what they're trying to say. They also claim that the charge across a resistor is somehow related to the integral of the voltage across it, but again (ideal) resistors can't really hold charge. The only way to interpret this supposedly 'universal' periodic system is by interpreting charge and its derivatives in different ways depending on context, which isn't convincing in a supposedly universal system.
  - lnx01 8y ago
    "charge everywhere on the wire remains constant" - perhaps. But the charge is still flowing, it's still moving at a rate of dQ/t. What is dQ/t you may ask? It's the time derivative of charge. It's the change in charge over time, the rate of flow of charge. I=Q/t Current=Charge/Time Amps=Coulombs/Second I don't know how to make this clearer.
- tntn 8y ago
  This is what happens when someone with a clear idea of a lumped element model slams into someone with a clear idea of Maxwell's equations.
- OJFord 8y ago
  That equation is unequivocally true...under certain assumptions. Contravariant is simply saying that it's being stated and used as if it holds generally; but that's not the case. This thread quickly dismissed 'through an area' as pedantic, but as far as I can tell that really is the crux of the debate: this simplified form of the equation only holds for scalar flux - viz. constant over and perpendicularly through a fixed flat surface. In the generalised form, `I` is a function of the area at a point.
DougBTX 8y ago
> To clarify, claiming current is the derivative of charge is like claiming the current of a river is the derivative of it's water level. My electro mag skills are a little rusty, but I don’t think that’s the right analogy. You wouldn’t say that the speed of a runner is the derivative of their height, similarly when someone is running you wouldn’t try to say that their speed isn’t a derivative of their height, as their height is more or less constant “for the length of the run”. Similarly, it isn’t the height of the runner, but the distance they’ve covered, it isn’t the depth of the river, but how much water has passed through, it isn’t the static charge of the wire, but how much charge has passed through.
- contravariant 8y ago
  I think you're just rephrasing what I'm trying to say in several different ways. We can probably agree that the amount of charge passing through is different from the change to the total charge. And in my opinion something like a 'periodic system' can't just treat both the same. It's especially bad to derive identities by taking the time derivative (or anti-derivative) on one side and switching between current and charge on the other.
- drostie 8y ago
  It's only a little bit off. To get the true hydrodynamic analog to a capacitor, connect the bars of two big pistons together so that volume accumulated in one comes at the expense of the other, then hit the bar with a spring so that its motion comes with some energy cost that can oppose a constant pressure. The point is that this component alone ties flow to accumulation, whereas generally your other components (resistors=thin pipes, wires=thick pipes, batteries=Archimedes screws, inductors=turbines connected to flywheels) do not accumulate volumes of water inside of them. Flow needs to make sense even without accumulation due to flow-balance, just like force needs to make sense even in situations where velocity stays constant due to force-balance.

jkabrg 8y ago

The amount of charge in an area stays constant because of Kirckhoff's laws, so dQ/dt=0 always. Wouldn't a better definition be: I = q v . A where q is charge density. A is an area (actually, a normal vector to some flat cross-section, with magnitude equal to size of area), and v is velocity of the charged particles. I'm using the dot product. If we check the units: (coulombs * metres^{-3}) * (metres * seconds ^{-1}) * metres^{2} = coulombs / seconds By the way, I fudged the above to make "charge density" have units $coulombs * metres^{-3}$. I'm not the best person at physics.

majewsky 8y ago
> The amount of charge in an area stays constant because of Kirckhoff's laws, so dQ/dt=0 always. Eh what? No. There seems to be some huge misunderstanding here. First of all, charge is not "in an area". It's in volumes, and flows through areas. And the amount of charge in a volume is absolutely not constant (in general). If you think it is, then please explain what happens when you rub a balloon on your hair, or why/when lightning strikes.
- jkabrg 8y ago
  I was referring to Kirchhoff's first law which only applies to circuits -- that example is enough to show that I=dQ/dt is wrong. I wasn't considering the cases you're talking about (lightning etc.) because only one example is needed to prove an equation wrong. I could easily produce a single specific example where dQ/dt=0 but there is clearly a non-zero current present. And you can talk about charge in an area (sort of) if you multiply charge density by area; the result has units coulombs * metre^{-1}. I thought counting charge in a cross-section of wire was something people did, one way or another, even if the units aren't Coulombs but Coulombs*metre^{-1}. I could very easily have misunderstood something. With the clarifications above, I'm not clear on what though.
  - majewsky 8y ago
    Okay, that's much more nuanced than "dQ/dt = 0 always".