Say you had a new physical quantity you wanted to determine the units for. How do you go about this?

For the strength of an electromagnet for example, you could carry out a simple experiment like the one here and find that the strength of an electromagnet is proportional to both the current through the circuit and the number of turns in the coil around the iron core. So at best this experiment gives us that $$\text{units electromagnet strength}=\text{amps}\times \text{units of X}$$ for some unknown constant $X$. (Where we ignore the units of "turns of coil".) This constant $X$ could have any units at all! So we are at a loss for the units of electromagnet strength. (Note: I am aware that $X$ is given by $N\times \mu$ where $\mu$ is the permeability of the medium: Ampere's law ) Is there another experiment that specifically seeks to determine the units of electromagnet strength? Or can it be found from theoretical considerations, or maybe some other experiment that is different from the electromagnet one above?

Basically I am interested in any experiments (no need to be about electromagnetism) that specifically seek to find the units of some physical constant. Are there experiments like this?

  • 1
    $\begingroup$ In your specific case it's the exact other way around. The unit for current, the Ampere is defined by the mechanical force per unit length that two parallel conductors exert on each other. That way the permeability for free space becomes a trivial mathematical constant. See SI units for details. $\endgroup$
    – CuriousOne
    Commented May 8, 2015 at 13:25
  • $\begingroup$ @CuriousOne: Sorry I'm not sure I understand that. How would we know that "units of electromagnetic strength" only depend on A? I'm not seeing how the fact how amperes are defined immediately tell us the units for $\mu$. (I'm working on the assumption that we don't know the units of electromagnet strength yet.) $\endgroup$
    – user45220
    Commented May 8, 2015 at 14:40
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    $\begingroup$ The definition for Ampere basically says that a current of one Ampere flowing trough a pair of conductors of negligible diameter in vacuum will exert a force of 2e-7N per meter. That the force is proportional to each current and depends inversely of the distance between the conductors is, of course, an experimentally learned relationship. The constant for the permeability of the vacuum is fixed arbitrarily, which basically defines the unit for the current, since the units for force and distance are already defined. An Ampere is no more special than a Newton... they are both definitions. $\endgroup$
    – CuriousOne
    Commented May 8, 2015 at 14:50
  • $\begingroup$ @CuriousOne: Thanks, but I'm still failing to see how this tells us the unit of electromagnet strength! Maybe I'm just dumb $\endgroup$
    – user45220
    Commented May 8, 2015 at 15:03
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    $\begingroup$ It's not a dumb question, the answer is just not as satisfying as you may think: units are conventions. What a meter is was defined by a metal rod in Paris. A kg is still defined by a chunk of metal sitting in a safe. The speed of light c was a measured quantity until we found that defining c and using it to measure the meter instead was the better idea. Why did we do that? Because the second can be defined extremely well by measuring the frequency of atomic transitions... in the end all of this is driven by the quantities that are most easily measured with the highest precision. $\endgroup$
    – CuriousOne
    Commented May 8, 2015 at 15:08

1 Answer 1


It is a fairly good question.

Physical quantities don't have dimensions. They have one or more scales, and it is the scales that have dimensions. It is perfectly legitimate to set $\epsilon=\mu=1/c$, which would mean that the unit of charge $Q^2 = \text{Joule-seconds}$. This would derive for the fps system, $1\ \text{verber} = \frac{1}{94.55}\ \text{coulombs}$, and $1\ \text{galvin} = 3.98\ \text{volts}$. This system has been explored in the past by Fitzgerald, by Kennelly, and by myself.

The number of dimensions or free units necessary to resolve all current systems is to have three electrical units, two of which are set to unity in SI, and a different two set to unity in the CGS. These extra dimensions have meaning, as well.

You might note there is the speed of light, and a separate constant, represented by equating the size of the forces in Ampere's law, to the Coulombic law of two uniformly charged wires. When they are equal, then the current in Ampere's law is $1\ \text{verber}$ in $T$ seconds, and in Coulomb's law, $1\ \text{verber}$ in $L$ feet. The measure $L/T$ feet per second is the "electromagnetic velocity constant" needed to convert $\mathrm{esu}$ to $\mathrm{emu}$.

It was wild speculation that the EMV and speed of light were the same number, and this was proven in 1863 by Maxwell, when he used fluid-dynamic equations to derive a number of point-wise "boundary" conditions, and showed the result of this, electromagnetic waves travel at the EMV. It was Hertz who proved that waves produced by a rotating magnet acted like light.

Writing an equation like Ampere's law as $\frac{F}{l} = \frac{I^2}{R}$ (weber), or $\frac{F}{l} = \frac{2I^2}{R}$ (Maxwell = emu) or $\frac{F}{l} = \frac{I}{2\pi R}$ (Lorentz) are all perfectly valid, and all give the dimensions of current as $\sqrt{\text{force}}$.

It's only when you write $\frac{F}{l} = \frac{k I^2}{R}$ that the $F$, $I$ and $R $ can be all defined separately, and the constant can be a non-unit value equal to $\frac{F}{I^2}$, and the existence of the multiple equations in front all suggest a new dimension is called for.

By lining up the CGS systems and MKS systems, one can show that over LMTQ, two extra quantities are needed. One could have, e.g. LMTQI rad, for example, where $Q$ and $I$ could be separately defined, and $\frac{Q}{I} = \kappa T$, for example.

The other thing that affects units is to eliminate uncertainties in the constants and calculation errors. One fires up cyclotrons in volts, and at various voltages, one gets things like spontaneous electron-pair creation, etc. These are masses measured in 'equivalent volts' or in modern parlance, electron-volts. The FPSC unit would be electron-galvins.

Likewise, one might not know the exact value of the constant, and express the equation in the form $E = j M \Theta$ or $F = g M$, by representing $E\ \text{(energy)} = j H\ \text{(heat)}$, and heat is proportional to $\text{mass}\times\text{temperature}$, e.g. $\mathrm{Btu} = \mathrm{lb}\,\mathrm{°F}$, $\mathrm{cal} = \mathrm{g}\,\mathrm{°C}$, $\mathrm{Cal} = \mathrm{kg}\,\mathrm{°C}$.

NASA does not use the weight of the earth in $\mathrm{lbs}$ or $\mathrm{kg}$, but a unit $GM = g R^2$, for which the value of $g R^2$ is reliable to eight digits, and $G$ or $M$ individually to four. $GM$ for the earth is $3.986004\times 10^{14}\ \mathrm{m^3/s^2}$, and for curvature issues, $\frac{GM}{c^2} = 4.435028\ \mathrm{mm}$, while $G$ can only be relied to three or four digits.

Generally, you pick the form of the equation which gives the most exact reproducible values, and not rely on the constants in the theory.

So in the case, e.g. of "magnetic charge". SI has no unit for this, but there are formulae which lead to what was magnetic charge = pole strength ($P$), and an SI unit can be found (eg $\text{Weber}$)

The equation $F = P H$ gives $P = \frac{\text{newton metres}}{\text{ampere}} = \text{weber}$. This is the symmetric Kennelly form.

The equation $F = P B$ gives $P = \frac{\text{newton metre}^2}{\text{weber}} = \text{ampere metres}$, this is the form suggested by Somerville (of "fine structure constant" fame).

So you need to write your equation, define something like $F = \text{"}P\text{"} H$, and the solution of the algebraic blobs called "units" will give the unit and/or dimensions.


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