Talk:Kilogram/Archive6

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Archive6

Contents 1 Volt is unaffected by the kilogram? 1.1 Ready 1.2 section break 1.3 The problem

Volt is unaffected by the kilogram?

From an editor comment in the article: "The VOLT is indeed defined relative to the newton (which is derived from the kilogram) but the volt’s magnitude is unaffected by variations in the magnitude of the kilogram because the newton appears once in both its numerator and denominator (watt/amp). In turn, the TESLA too is not affected."

Is this true? It's not clear to me that the effect on the magnitude of the ampere with variations in the kilogram is linear. --Random832 (contribs) 16:51, 2 April 2008 (UTC)

Worked the math, it's _not_ linear, it's a square root. If the kilogram were to become 99% of its current value, the ampere would become 99.4987437 of it's current value. --Random832 (contribs) 16:54, 2 April 2008 (UTC)

• I haven’t seen your math Random832 so I can’t point out the flaw but I believe you are mistaken. If you follow the chain of dependency of the SI definitions and look at how a change in the kilogram affects things, I believe you will find that no matter what, the volt can be reduced to meters per second. This makes sense, since electricty can be viewed in terms of simple mechanical kinetics with electrons (magnetism is the magic paddle that links the mechanical world to that of the electron). The electron-volt (the very small unit of energy) is equal to 1.602176487 × 10^–19 J and the mass of the electron is 9.10938215 × 10^–31 kg. So the velocity of a single electron when it is accelerated to a kinetic energy of one electron-volt is 593,096 meter/second. That much is simple physics. It follows then that the velocity that an entire coulomb of electrons achieve when they pas through an electrostatic potential difference of one volt, in vacuo is 593,096 meter/second, during which time they gain precisely one joule of energy. I believe you’ll find that no matter how one changes the kilogram, the volt always reduces to 593,096 meter/second. Try it. But remember, as the kilogram changes, so too must the numeric value for the mass of the electron, which is a constant in absolute Planck terms (as are the meter and the second in this hypothetical). In the end, after all the terms are crossed out, the volt always comes out in the mathematical wash as 593,096 meter/second; which is to say: if the kilogram were to change, the absolute electrostatic potential of one volt would be unchanged at 9.5882 × 10^–28 Planck units of potential. Greg L (my talk) 23:05, 9 April 2008 (UTC)

  P.S. Good catch about the circular definition of “equals IPK + 42 µg.” Indeed, the correction would have to be expressed in terms of one of the parts-per notations such as percentage or ppb. Thanks. Greg L (my talk) 23:38, 9 April 2008 (UTC)

• (in specific response to your electron-volt example) But since the ampere would change, the numeric value for the charge of an electron would also change - and I believe the math below establishes that the change would not be in linear proportion to the other change. --Random832 (contribs) 15:26, 10 April 2008 (UTC)

so,

so, if

then

Is there a problem with this calculation? The 0.990 and 0.995's are each the "new" value of the unit they are attached to - obviously the kilogram won't ever lose a full percent of its mass, but it's easier to calculate. --Random832 (contribs) 15:07, 10 April 2008 (UTC)

• Hmmmm. Must study this… Greg L (my talk) 18:15, 10 April 2008 (UTC)

• Been busy in real life. Just enough time to argue on Talk:MOSNUM but not enough for this. I’ll have time this weekend. Greg L (my talk) 20:10, 11 April 2008 (UTC)

Ready

Sorry, I really couldn’t follow what you were doing with your mathematics above. Instead of trying to crunch actual values using a 1% change as you did, I elected to counter with yet another analysis. This time, I took care to ensure I used only BIPM definitions from the start, rigorously adhered to them, and used almost no numeric values in the analysis (just nearly pure, symbolic algebra).

This BIPM web site “defines” the watt, volt, and newton. This BIPM web site “defines” the ampere. This one defines the second. And this one defines the meter.

The ampere is defined as such:

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10^–7 newton per meter of length.

The ampere can thus be thought of as the product of two factors: [fixed experimental setup per meter of length] · 2 × 10^–7 N. For simplicity, let’s algebraically simplify and express the ampere as follows:

A = K · 2 × 10^–7 N

where…

K = [fixed experimental setup per meter of length]

N = newton of force

The most important thing to note in this simplification is that the fixed experimental setup term, K, contains only two conductors of fixed length at a fixed separation, there is no buried “kilogram” term within; that’s the important part.

Note too that it makes perfect sense to think of the ampere as simply a force. Regardless of the experimental setup (the arbitrary length of the conductors or the separation between them), flowing currents creates a force between conductors. We were all amazed when we got our first big fuel cells running in R&D because fuel cells are amp-rich and voltage-shy. We could entertain ourselves for quite a long time by simply watching 4-gauge cables flex back and forth as we turned the load on and off. In the case of the SI, the BIPM reversed the ampere=force relationship and elected to define the magnitude of the ampere in terms of the force it generates.

Let’s go through the chain of definitions as defined by the BIPM in the SI:

• V = ^W⁄_A
  
• W = ^J⁄_S = J·s^–1
  
• J = N·m
  
• N = m·kg·s^–2
  
• The second and the meter are the other SI base units (the ampere being the first) but I won’t bother with them since we all know what they are and they are immaterial to this discussion.

Thus, we can define the volt in all the following equivalencies:

V = ^W⁄_A = ^J·s–1⁄_A = ^{(N·m)·s–1}⁄_A = ^{(m·kg·s–2)·m·s–1}⁄_A = ^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 N} = ^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 (m·kg·s^–2)}

Now, note this fact about the last equivalency above: it features the kilogram only once in the numerator and denominator. This is the case because it is a term within the newton, which also appears precisely once in the numerator and denominator. That means the volt is invariant with respect to the kilogram. Now let’s look at something else that sheds light about how we think of what electric potential (the volt) is:

^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 (m·kg·s^–2)}

…the volt is a velocity: m·s^–1 = ^m⁄_s

…except in this case, I believe the actual velocity (here, it crunched to ¹⁄_{2 × 10^–7 m/s} = 5,000,000 m/s), is really a dimensionless quantity because the experimental setup, K, that defines the ampere is an arbitrary one: even though it’s “one meter of length for conductors one meter apart,” the force between the conductors would change non-linearly as the meter varied. I’m not sure, but I suspect that if you use all-Planck units, the “velocity” of a volt is probably equal to 1 (in the Planck unit of velocity, which is the speed of light). If you solve this velocity kinetically using the mass of the electron (9.10938215 × 10^–31 kg), you end up with 593,096 meter/second. To recap how one arrives at a velocity using mechanical kinetics, here is what I did:

The electron-volt (the very small unit of energy) is equal to 1.602176487 × 10^–19 J and the mass of the electron is 9.10938215 × 10^–31 kg. So the velocity of a single electron when it is accelerated to a kinetic energy of one electron-volt is 593,096 meter/second. It follows then that the velocity that an entire coulomb of electrons achieve when they pas through an electrostatic potential difference of one volt, in vacuo is 593,096 meter/second, during which time they gain precisely one joule of energy. It’s just billiards.

Anyway, the bottom line answer is that the volt, and therefore the tesla, are unaffected by variations in the magnitude of the kilogram. Further, the value of the volt can be expressed as 593,096 m/s for electrons. Further still, this value will not change unless the meter changes independently of the second; a change in the second can not affect the volt because the meter, which is keyed to the velocity of light, changes proportionally with the second.

That’s my analysis. I’m glad you challenged me to this exercise because I didn’t put nearly this much diligence into my previous analysis. In both cases though, I started with the actual BIPM definitions (I didn’t go get “kibibytes” from Wikipedia, nor other secondary opinion or sources) and stuck with straightforward algebra the entire way. Thus, this isn’t original research; it’s math, which is fair game as anyone can analyze the same facts and crunch them (and goof at it too). To me, the algebraic result passes my “grin” tests when I think about electricity in terms of what is going on with the ampere (a quantum phenomenon enlarged to gigantic, observable proportions due to the vast quantity of electrons moving at the same time), and with the volt. Further, it matches perfectly with my view of electricity from a mechanical (kinetics) viewpoint. Let me know if you find a flaw in any of this.

My old boss at the fuel cell lab used to be the head engineer at an electrical utility. He once read a book on quantum mechanics in a lawn chair on the beach in Hawaii for recreation. He and I got all sideways on the above as he simply could not get his mind off the engineering equivalencies and think about the units strictly in terms of the definitions within the SI. This was not easy (for me anyway). I’ve spent hours here on this post, digging into the facts, analyzing consequences, committing the logic to written form, and checking what I wrote to satisfy myself that it seems correct. I’m glad I did though, because I wasn’t entirely sure about leaving the volt and tesla off the list of units that are affected by the kilogram. I trust you’ll let me know if I did a brain fart here.

Greg L (my talk) 20:13, 12 April 2008 (UTC)

---

What I was trying to get across with my first response above to your electron-volt example is the fact that the number of electrons in a coulomb - and therefore, the mass of a coulomb of electrons, and therefore the amount of energy that they would gain from being accelerated to a particular velocity - would also change. --Random832 (contribs) 22:22, 12 April 2008 (UTC)

If you have a problem with the use of concrete numeric values, try this, where "x" is the ratio between the new mass of the kilogram and the old one. The use of radical signs should make things more clear, even though I thought I'd gotten across that 0.995 here is the square root of 0.990:

this is, I believe, how the Ampere is actually defined in relation to the newton. This is not only what Ampère's force law on here says, but also what I remember from my physics text in high school. I didn't go to primary sources, but can you point out if this is incorrect - or is not how the ampere is defined? k_m is an invariant physical constant, whose numeric value in Nm/A² is fixed to define the Ampere. The fact that the units here include only one kilogram term, but _two_ ampere terms, is what clued me in to the fact that it's a square root relationship rather than linear. - I checked the website you linked defining the Ampere, it confirms what I say here. k_m is μ₀/2π. --Random832 (contribs) 22:45, 12 April 2008 (UTC)

so,

therefore,

and, if

then

even though it’s “one meter of length for conductors one meter apart,” the force between the conductors would change non-linearly as the meter varied. - um, who said anything about the METER varying? --Random832 (contribs) 22:22, 12 April 2008 (UTC)

section break

A simplification, if you don't want to read all of the above

"A = K · 2 × 10^–7 N" - this is where you have a "brain fart" as you put it. There are two wires, both carrying a current, and the force is proportional to the product of the two currents. The correct relationship would be A² = K * 2 × 10^–7 N. That's why I did the numerical analysis with 0.990 - when everything's 1, it's easy to forget where you've taken its square root. --Random832 (contribs) 22:22, 12 April 2008 (UTC)

Incidentally, this also means the farad, siemens, and henry - and any unit whose fundamental terms include A²/kg or kg/A², would not be affected. --Random832 (contribs) 22:40, 12 April 2008 (UTC)

• Not at all. The test as to whether or not a derived unit is affected by a change in magnitude of the kilogram is whether or not the kilogram appears an equal number of times (and at the same power) in the numerator and denominator. You cited the siemens above. That is a reciprocal ohm. Look at the ohm. It doesn’t balance so both it and the seimens are affected. Pretty much all the electrical units are affected because the kilogram underpins the SI base unit of electricy: the ampere, and everything else flows from that. The notable exception is the volt. A 10% bigger kilogram doesn’t affect the volt by virtue of the fact that the newton (and therefore the kilogram) appears once each in both the watt and the ampere. Therefore, the tesla too is unaffected. Greg L (my talk) 23:48, 12 April 2008 (UTC)

The ohm, then, would _also_ be unaffected. My point this entire time is that the kilogram does NOT appear once in the ampere - it only appears one HALF time (that is, the 0.5 power - the square root - of the kilogram appears). It appears once in the ampere-squared. The ampere must appear twice as many times (or, rather, at twice the power) in the numerator as the kilogram does in the denominator, or vice versa, for it to "balance". --Random832 (contribs) 00:05, 13 April 2008 (UTC)

• A volt is a W/A. How do you keep coming up up squares and square roots? One deals with squares and square roots primarily if you are trying to figure out a change in resistance from volts and watts, or a change in watts from amps and resistance, or a change in amps from watts and resistance. But that’s not what needs to be done here and that’s not what I’m doing but somehow seems to be what you’re doing and I don’t know why. And you’ve twice now started with Ampère's force law to somehow discern a relationship between the kilogram and the volt. One can’t possibly figure out the answer to the question using the formula you chose. The newton appears twice in the BIPM definition of the volt. The following parenthetical is the newton, expressed entirly in SI base units: (m·kg·s^–2). Now let’s look at the volt, expressed entirely in SI base units. How many “kg” do you see in it below?

V = ^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 (m·kg·s^–2)}

I count two. Greg L (my talk) 00:35, 13 April 2008 (UTC)

But the formula "K · 2 × 10^–7 N" is incorrect, I pointed that out above - for your proposed "K", an ampere would be "√(K · 2 × 10^–7 N)" - are you even paying attention to my argument? --Random832 (contribs) 00:45, 13 April 2008 (UTC)

V = ^{(m·kg·s–2)·m·s–1}⁄_{√(K · 2 × 10^–7)(m^0.5·kg^0.5·s^–1)}. --Random832 (contribs) 00:49, 13 April 2008 (UTC)

you’ve twice now started with Ampère's force law to somehow discern a relationship between the kilogram and the volt. because that law is how the relationship between the kilogram and the AMPERE is defined! --Random832 (contribs) 00:52, 13 April 2008 (UTC)

• Don’t shout and settle down. I showed you every step of the way above and you aren’t seeing it. Show your math. Just how in the world do to you start with V = ^W⁄_A and end up with V = ^W⁄_√A ? (as shown in your 00:49, 13 April 2008 (UTC) post) Greg L (my talk) 01:07, 13 April 2008 (UTC)

The problem

You've somehow convinced yourself that the definition of an ampere contains an entire newton term (instead of a N^0.5), you keep using that as a premise, and you are not listening when I keep pointing out that it's wrong. --Random832 (contribs) 00:57, 13 April 2008 (UTC)

You seem to think that because this isn't one of the situations where squares come up in everyday electrical problems, that it's somehow inappropriate to be talking about square roots. The problem is, you're ignoring that the force between two current-carrying conductors (in terms of which is how the ampere is defined) is proportional to the _product_ of the two currents (when the currents are the same, the product of them is the _square_ of that value) divided by the distance between them. --Random832 (contribs) 01:03, 13 April 2008 (UTC)

• This is not complex. The BIPM defines the volt as V = ^W⁄_A. I choose that definition over yours (as shown in your 00:49, 13 April 2008 (UTC) post), where V = ^W⁄_√A Greg L (my talk) 01:10, 13 April 2008 (UTC)
  • !!! I never said "V = ^W⁄_√A" Please go back and re-read my argument if you don't understand this. --Random832 (contribs) 01:22, 13 April 2008 (UTC)
  • A = √(K · 2 × 10^–7 N), period. --Random832 (contribs) 01:19, 13 April 2008 (UTC)

• I am too defending it; above (twice) and here again. Here is what the BIPM says about what the ampere equals.

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10^–7 newton per meter of length.

Please tell me just where it the BIPM says anything about the square root of the force? The BIMP says V = ^W⁄_A, not V = ^W⁄_√A Greg L (my talk) 01:25, 13 April 2008 (UTC)

• If there is a flaw in my math or logic, it must be here. It would be terribly helpful if you didn’t redefine the first equivalency; that is after all, what the BIPM says the volt is equivalent to.

V = ^W⁄_A = ^J·s–1⁄_A = ^{(N·m)·s–1}⁄_A = ^{(m·kg·s–2)·m·s–1}⁄_A = ^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 N} = ^{(m·kg·s–2)·m·s–1}⁄_{K · 2 × 10^–7 (m·kg·s^–2)}

Greg L (my talk) 01:18, 13 April 2008 (UTC)

Your flaw is BEFORE that, "A = K · 2 × 10^–7 N" is wrong and I explained that it was wrong and why it was wrong as the very first statement in my argument above. --Random832 (contribs) 01:20, 13 April 2008 (UTC)

I used proper algebraic simplification from a real starting point to arrive at "A = √(^m·N⁄_km)" where k_m is a real physical constant. You waved your hands to arrive at "A = K · 2 × 10^–7 N". --Random832 (contribs) 01:28, 13 April 2008 (UTC)

• OK, I’ll come down here and post what the BIPM says. Please tell me where the ampere is equal to the square root of the newtons of force.

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10^–7 newton per meter of length.

Greg L (my talk) 01:29, 13 April 2008 (UTC)

Exactly. It's the current maintained in TWO straight parallel conductors. If the kilogram (and therefore the newton) changes, the current in BOTH conductors therefore has to change. And the Ampere force law says that the force is proportional to the product of BOTH currents (since the currents are the same, that product is the square one of the currents). --Random832 (contribs) 01:32, 13 April 2008 (UTC)

• I’m thinking about this. Greg L (my talk) 01:37, 13 April 2008 (UTC)

P.S. An ampere is equal to a certain, small amount of force, in newtons; not the square root of newtons. Isn’t it? Greg L (my talk) 01:31, 13 April 2008 (UTC)

no, it's an ampere _squared_ (i.e. one ampere in one wire multiplied by one ampere in the other wire) that is equal to that certain small amount of force in newtons [per metre of the length of the wires, but you knew that] --Random832 (contribs) 01:37, 13 April 2008 (UTC)

For something that could be experimentally verified if you had the stuff to set up such an experiment - imagine, against all plausibility, that the kilogram (and the newton) have become four as large, and that the ampere is now defined in terms of this new super-newton. The amount of current to go through each wire to generate a force of four times as much is only twice as much, not four times. I.e. the current that needs to go through two parallel conductors to generate four (present-day) newtons of force is only two amperes, rather than four amperes. --Random832 (contribs) 01:55, 13 April 2008 (UTC)

I believe you are correct Random832. The volt and the tesla should be affected as the square root of the change in the kilogram. I wish we had found the key to communicating a bit faster. Are the volt and tesla the total extent of the effect of this? Greg L (my talk) 02:21, 13 April 2008 (UTC)

• Random832, I corrected the article per your teachings (∆ here). Believe me, while the debate process leading up to making the correction was painful, I am very appreciative that you saw my editors note and my error of omission has been corrected (although my hidden editors note explaining why the volt and tesla weren’t included in the list was proven to be a bunch of horse crap). My primary objective is that this kilogram article be the best, most authoritative source on the kilogram there is for a general-interest readership. I’ll take intervention from someone who is right (v.s. a bone cone who messes things up) any day. Wikipedia and its readers are the beneficiaries. Thanks very much. Greg L (my talk) 02:40, 13 April 2008 (UTC)

I believe that also some other units (ohm, henry, siemens, farad and any i've missed) listed as affected would not be affected, since their terms include kg/A^2 or A^2/kg. The weber would be affected whereas the logic that excluded the volt would also exclude it, but it was already listed as affected. I've edited the article accordingly. --Random832 (contribs) 03:05, 13 April 2008 (UTC)