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I can easily understand what divisive units imply, but not what multiplicative units imply.

What I mean is, when I read "$12 \:\mathrm{eggs/carton}$", I mentally convert it to, "There are 12 eggs for each carton". I get that.

But, when I see units like joules * seconds, $\:\mathrm{J \cdot s}$, it really bothers me for some reason. I'm not sure what it's saying. It's not saying the amount of joules expended over a certain time-frame, otherwise it would be joules/second, wouldn't it? So, when I read units like joules * second, I try to relate it to my egg-carton analogy, and it comes out like this:

"There are 12 eggs for each reciprocal of a carton."

It just doesn't make any sense for me. Is this a problem for anybody else? Should this be on math SE? Is this even a valid question or am I just asking why $2 + 2 = 4$?

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    $\begingroup$ If it helps, translate $J s$ into $J/Hz$. $\endgroup$
    – Johannes
    Dec 22, 2014 at 4:42
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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/32096 $\endgroup$
    – Johannes
    Dec 22, 2014 at 5:14
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    $\begingroup$ You might like also to contemplate how you make sense of m * m (unit of area) and the kg * m part of kg * m / (s * s) (unit of force). $\endgroup$
    – Steve Jessop
    Dec 22, 2014 at 15:25
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    $\begingroup$ You could consider $J*s$ as a unit with still no name. Work is $N*m$, which also makes no sense. But this has been called Joules. And defined as energy. Now it makes sense again: Work is not how many "newtons times meters" you add, but rather "how many joules" you add. If you are adding an amount of energy and multiplying with the time it took you, then you can define this new proporty as something. And call the odd unit $J*s$ something more sensible. $\endgroup$
    – Steeven
    Dec 22, 2014 at 17:55
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    $\begingroup$ @Steeven, although the unit has no name (except "joule seconds"), the quantity is called action (or angular momentum). $\endgroup$
    – The Photon
    Dec 22, 2014 at 19:04

5 Answers 5

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Seems valid to me...

I think the interpretation of your example could be that you are doing something with some joules, for some period of time, and that the product of the number of joules and the number of seconds has a certain value, and the units would be... joule-seconds!

I can picture the operator of an energy storage facility quoting a price for storing energy. Storing $10,000\text{ joules}$ for $400\text{ seconds}$ would cost a certain amount. Storing double the energy for half the time would use the same resources, and cost the same.

The most common example, of course, is the use of $\text { kilowatt-hour}$ as a unit of energy. In this case, you are using the rate of energy output multiplied by the time over which the output exists, to represent the total energy.

It has been common for some time to represent the amount of work in a job in $\text{ man-hours}$. The more people you can throw at the project, the sooner it will be done...

A less obvious unit is the measurement of the length of, say, a sailboat, in $\text{ knot-seconds}$...

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    $\begingroup$ Don't forget the light-year... $\endgroup$
    – octonion
    Dec 22, 2014 at 5:39
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    $\begingroup$ @PatrickM maybe in software engineering, but not in e.g. gardening. (If it takes one person 10 hours to mow a lawn, then five people can do it in 2 hours, give or take a few minutes, and assuming they have access to five lawnmowers) $\endgroup$
    – user253751
    Dec 22, 2014 at 6:44
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    $\begingroup$ @immibis - Actually, Brooks was talking about some tasks being indivisible or otherwise poor candidates for breaking down. He coined the phrase "9 women can't make a baby in a month" in that book. Manual labor tasks can make good candidates for using more resources, but you do have to balance against the overhead of coordinating all those resources. (i.e. You need managers) You may find that adding the management structure reduces the output below what could be accomplished with fewer resources. (Strangely, this exactly the type of problem CompSci exists to solve!) $\endgroup$
    – 64BitBob
    Dec 22, 2014 at 13:10
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    $\begingroup$ -1. I think you are missing the point of the question. All your examples are logically understandable and not of the questioned kind. Storing 10,000 joules for 400 seconds would cost a certain amount; there is no joules times seconds here. $\mathrm{kWh}$ is not joules times time, it is joules per time times time, which is still joules. Makes sense. Man-hours is not $man \cdot hour$ it is only a name for $hours/man$. Your final example hits the point (but is not explained): $knot*seconds$ makes no sense. Just like $Newton*Meters$ makes no sense. And this is what should be explained to the OP. $\endgroup$
    – Steeven
    Dec 22, 2014 at 17:47
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    $\begingroup$ What is nonsensical about the physics equation "Distance = velocity x time"? And if velocity is in knots and time is in seconds, what are the obvious units for distance? $\endgroup$
    – DJohnM
    Dec 22, 2014 at 17:57
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Reciprocal cartons for storing eggs might make little sense, reciprocal seconds (Hz) do.

Joules times seconds (J.s) is equivalent to Joules per Hertz (J/Hz). It denotes angular momentum, the amount of energy per unit of rotational frequency.

Similarly, reciprocal meters (wave numbers) also make perfect sense. So units like Newton times meter (N.m) are also encountered in physics.

Alternatively, one can view units like N.m (= J) as an accumulation: when I push an object with a given force, the amount of work performed increases proportional to the path length.

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    $\begingroup$ I like your example about being a conversion factor for angular momentum. Similarly, Plank's constant is ~6.6e-34 J*s, and as you suggest, we can think of this as J/Hz. That is, Joules of energy for a photon with a certain frequency (in Hz). So, in this way, Plank's constant can be viewed as a simple conversion factor. $\endgroup$
    – DanHickstein
    Dec 22, 2014 at 18:13
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    $\begingroup$ Reciprocal cartons tend to break a lot of eggs. $\endgroup$
    – Tom Robinson
    Dec 23, 2014 at 3:59
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This answer will make more sense if you've had some basic Calculus (derivatives and integrals).

Draw a graph of some random function $E = f(t)$, where the horizontal axis $t$ is time (in seconds), and the vertical axis $E$ is energy (in joules). It doesn't matter what shape the graph has (but let's stay in the upper-right quadrant, so $t$ and $E$ are positive quantities).

  • At any given time, $t$, you can find the energy, $E$, at that time by tracing a vertical line from the origin to the function. The distance of that line represents the number of joules at a given instant.

  • Now look at the slope of the graph at that point. The angle of that slope represents the instantaneous change in the energy over time. It is the rate of change, in joules/second. In calculus terms, this is a derivative.

  • Now pick two times along the $t$ axis ($t_1$ and $t_2$). Find the area enclosed between $t_1$ and $t_2$, and above the $t$-axis, and under the function $f(t)$. This enclosed area represent the sum of all the energy at each instant, over a period of time (from $t_1$ to $t_2$). This area is measured in joule*seconds. In calculus terms, this is an integral.

Thus, while divided units represents a rate of change in one variable relative to another, multiplied units represent an accumulation of one variable as another is changing.

Seen this way, you can view distance as merely the accumulated effect of a velocity over time (e.g. if velocity is held constant, you accumulate distance linearly). Similarly, the area of a shape can be seen as an accumulation of distance measured in one direction as distance in another direction changes.

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  • $\begingroup$ Thanks, it totally clicked for me with the calculus example :) $\endgroup$
    – Trevor Krumrine
    Sep 22, 2022 at 8:30
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To understand what Joules $\times$ seconds means, one must understand the terms by themselves. The Joule is a unit of energy, and the second is a unit of time. So, what 500 J $\times$ s means, is that there is a source of energy that I can use in the following manner: 500 Joules for 1 second; or 250 Joules for 2 seconds; or 100 Joules for 5 seconds; or 1 Joule for 500 seconds; or in any combination of numbers who's product is 500!

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Building on the answer that Johannes gave:

Imagine you have a broad band source of energy, transmitting on multiple frequencies. It might be the sun, or it might be a RF transmitter, or ...

Now we are going to measure the energy transmitted in each wave band for a certain time, and plot this as a function of the frequency observed. This leads to a plot with Hz along the X axis and J/Hz (the energy density per unit frequency, that is the energy measured divided by the bandwidth of my receiver) along the Y axis.

And boom - out of nowhere I have a unit J/Hz (or J sec) which makes physical sense...

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