# What does all these strange collection of units means? [duplicate]

For example let's take a simple example of velocity that is the object displaced per unit time or rate of change of position. It has some meaningful units, i.e. $$\rm m/s$$, which means that the position is changed by $$1$$ meter per unit second.

However, there are some units which puts many of the novices like me in the confusion, such as the unit of impulse i.e $$\rm N\:s$$ (Newton-second) the another example we can go for is the unit of planks constant '$$\rm J\:s$$' (joule-second). (We also get those in the case of some proportionality constants). So what does it signify? And how to interpret all these types of units so that they can make some sense while dealing with them?

• have you tried looking at the derivations of those units on wikipedia? May 24 at 20:37
• @nielsnielsen yes, I have but for me I think there is some problem in understanding the language they've used (yes obviously it's in English but it's getting very hard to understand for me, what are they actually trying to say) May 24 at 20:46
• Does this answer your question? Meaning of a product of two physical units (and links therein) May 25 at 10:23
• Obligratory xkcd references: Unit cancelation is weird, Dimensional Analysis May 25 at 11:33

Not every unit is supposed to have a direct interpretation. During a computation, you can find yourself combining physical quantities in varied ways, and when you analyse the result to determine its unit, you usually find something that isn't particularly illuminating.

For example, let's take electrical resistance. We usually call its unit "ohm", but it's actually kg.m$$^2$$.s$$^{−3}$$.A$$^{−2}$$... The latter has very little use in terms of physical interpretation. But as you develop an intuition of what electrical resistance is, you get used to what 1 $$\Omega$$ is, without any need to return to basic units.

Another example, you mention Planck constant's unit, J.s, which is also kg.m$$^2$$.s$$^{-1}$$... That won't help much to "get" what that means. Two very different physical quantities share that unit:

• angular momentum
• action

There's very little in common between those two, so there's nothing deep to get from the unit alone.

So I wouldn't expect the unit to tell me much about the physical quantities, most of the time.

• My favourite example: light intensity is Watts per square metre, which is kilograms per cubic second.
– J.G.
May 24 at 22:08
• One that I love to show to my students to blow their minds: when studying plasma frequency, it's sometimes useful to separate the plasma density from the rest of the formula: $f_p=\Gamma\sqrt{n}$. But then, $\Gamma$'s unit is m$^\frac{3}{2}$.s$^{-1}$. The non-integer exponent can be disturbing! May 24 at 22:16
• And I like the units of $\mu_0$ being newton per square ampère. May 25 at 9:57
• @Miyase And my favourite is the unit of the rate constant which is variable with the order of equation i.e $(\frac{mol}{dm^3} )^{1-n} s^{-1}$ where $n∈W$. May 26 at 17:53

A quantity's units arise directly from the quantity's definition. For example,

Impulse due to a constant force F acting for a time $$\Delta t$$ is $$\mathbf F\ \Delta t$$.

So if $$\mathbf F=20\ \text{N North}$$ and $$\Delta t = 3.0$$ s, then Impulse = 60 Ns North.

The units N and s are parts of the quantities force and time and must be carried along with them as algebraic symbols when the quantities are multiplied together. Far from units being a nuisance, if you forget what 'impulse' means, you can pretty much work this out if you do happen to know that the units are Ns.

A student usually first meets the Planck constant as the proportionality constant, $$h$$, in the relation $$E=hf$$ between a photon's frequency, $$f$$ and its energy, $$E$$. So if we know that light of frequency $$5.090 \times 10^{-34}$$ Hz has photons of energy $$3.186 \times 10^{-19}$$ J, this gives a value for the Planck constant of $$h=\frac Ef=\frac{3.186 \times 10^{-19}\ \text J}{5.090 \times 10^{-34}\ \text{Hz}}=6.63 \times 10^{-34}\ \text {J Hz}^{-1}= 6.63 \times 10^{-34}\text {J s}$$
The last step, substituting s for Hz$$^{-1}$$, is perhaps the tricky one. But we know that frequency is number of cycles per unit time, so Hz means s$$^{-1}$$. But since $$\text{Hz}=\text s^{-1}$$, then $$\text{Hz}^{-1}=\text s$$, showing that $$\text{Hz}^{-1}$$ can be written neatly as s.

Look, too, at how nicely the units work out when you do the calculation the other way round, using the accepted value for $$h$$ ... For example, the energy of a photon of light of frequency $$4.656 \times 10^{14}$$ Hz is $$E=hf=6.626 \times 10^{-34}\ \text{Js} \times 4.656 \times 10^{14}\ \text{Hz}=3.09 \times 10^{-19}\ \text J.$$ The last step relies on $$\text{Hz}=\text s^{-1}$$, as established earlier.

The lesson here is that units are our friends. If you are substituting the values of quantities into a formula, substitute the unit as well as the number. When you then evaluate the formula, don't just perform the arithmetical operations on the numbers, but simplify the units, treating them like algebraic quantities. Then the answer will automatically emerge with its correct units – a useful check that you're using the formula correctly.