For a lack of a better title and phrasing of my question, I am asking: do units have to make sense in physics?

Let me give an example of what I mean. Let's say that I have some arbitrary units (for argument's sake):

$$\frac{\mathrm{meters}^{2} \cdot \mathrm{seconds} \cdot \mathrm{henry}}{\mathrm{coulomb}^{3}}$$

And I want this to be equal to charge let's say,

$$C = \frac{\mathrm{meters}^{2} \cdot \mathrm{seconds} \cdot \mathrm{henry}}{\mathrm{coulomb}^{3}}$$

Which has units of Coulombs. But the units don't add up to equal one coulomb as you can see.

So my real question is that do you have to do some manipulation with the units in order for them to equal one coulomb? What are the rules and conventions in physics for this?

I have seen in EE (electrical engineering) that if you can cancel out all units, then you can give it any units you want in the end. It is the same for the physics field? (EDIT: I may have misunderstood this part and it is probably not true).

Or can you set the units to whatever you want even if they are not equal to one coulomb?

I have heard that the constant in the EFE's is a conversion factor to get the right units:

$$\frac{8\pi G}{c^4} T_{\mu \nu }$$

Where $\frac{8\pi G}{c^4}$ is the conversion factor. This is a real example that I found that I want to understand as well.

In summary:

  • Can you set the units to whatever you want even if they are not equal to one coulomb?
  • If you can cancel out all units, then can you give it any units you want in the end?
  • Do you have to do some kind of manipulation of units to get the right units?
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    $\begingroup$ Can you give a link to the EE post that says "If you can cancel out all units, then can you give it any units you want in the end?". My guess is you are either misunderstanding what they said or misquoting (both reasonable things to do when you're confused and don't know what's going on). $\endgroup$
    – jacob1729
    Mar 29 at 10:46
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    $\begingroup$ You can make arbitrary expressions, but sometimes there's an implicit conversion that makes it work out. For example, if a speeding ticket costs $100 + $10 * mph over limit, the apparent units are $ + $ mph which doesn't make sense. But there's an implicit conversion, $100 + k * speed over limit‘ where k` is 10 USD / mph. $\endgroup$
    – MooseBoys
    Mar 29 at 23:23
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    $\begingroup$ There is a good example of two things that coincidentally have the same unit but aren't related: torque (Newtons times metres) and energy (also Newtons times metres). Even though they are both Newtons times metres, they are different metres, they are completely different measurements and you can't convert torque into energy. $\endgroup$
    – user253751
    Mar 30 at 9:24
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    $\begingroup$ @user253751 not quite true: torque is a vector (newtons crossed with meters), energy is a scalar (newtons dotted with meters). Those are not the same thing. $\endgroup$
    – PGmath
    Mar 30 at 13:08
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    $\begingroup$ @PGmath fine then, magnitude of torque $\endgroup$
    – user253751
    Mar 30 at 13:16

Units in physics are units of the real world. They must be consistent, because the real world is. After choosing a definition of necessary fundamental units (as defined by the SI system e.g. - I am not considering natural units here), all other units must be derived from those in a mathematically sensible manner.

Because, you move a distance in metres and carry an amount of Joules as kinetic energy and have inertia as an amount of kilograms. If you agree that you can't suddenly move a distance of "seconds" and have an inertia in "metres per second" and drink coffee that is "ten kilograms" hot, then you agree with this consistency of the world.

The fundamental properties like time, position, mass etc have fundamental units. All other properties are derived from those, and with units that are likewise derived from their units. Your speed is a change in position over time, so the distance-unit over the time-unit, metres-per-second. Speed can never have units that are not distance-units-over-time-units. This consistency must necessarily always be the case regardless of complexity. If not, then you have found a new fundamental property.

Note though: Now and then you might indeed meet equations with inconsistent units on either side of the equal sign. So that the unit of the left-hand-side isn't the unit of the right-hand-side. This is typically seen for emperical findings. It is mathematically sloppy but might work fine in applications. Nevertheless, it is mathematically sloppy.

As you also mention, a coefficient of sorts ought to be introduced to correct for unit mismatch. Typically, such unit correction is added to a constant or coefficient that is involved. For instance, in Hooke's Law for spring forces, we see proportionality between displacement from natural length and spring force - we thus invent a proportionality constant k:

$$F_{spr}=k\;\Delta x$$

What should the units of this constant $k$, which is called stiffness or simply the spring constant, be? If unitless, then this equation shows a number of Newtons equal to a number of metres. A mismatch. So, let's invent this $k$ to have units that make it all match: that would be Newton's-per-metre. And this definition of $k$ matches very well with a physical understanding of $k$ as stiffness - it is a material property that tells us something about how much the spring can be stretched before reaching a Newton of spring force; it does indeed tie distance with force, metres with Newton.

As another example, in Amontons' Law of kinetic friction at low normal pressures, we find proportionality between normal force and kinetic friction force - again we invent a coefficient to account for the proportionality constant:

$$f_k=\mu n$$

$\mu$ is called a friction coefficient. What unit should we give it? On the left-hand-side we have units of Newtons. On the right-hand-side we already have units of Newtons as well. By keeping $\mu$ unitless (or you can say "Newtonts-per-Newton" if you wish), this relationship fits. And so it went.

  • 5
    $\begingroup$ Great answer! My preconceived notions about units were baseless, and I learned that there is as much rigor for units as there is for actual solutions of equations. Thanks. $\endgroup$
    – Tachyon
    Mar 29 at 14:57
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    $\begingroup$ measuring distance in seconds is usual practice as is measuring distance in kilograms (or mass in meters? I do not remember which one...). The only requirement for units is that in every equation units in each term are equivalent to each other so that we are working with quantities that are defined at the common scales. $\endgroup$
    – Umaxo
    Mar 29 at 20:37
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    $\begingroup$ @Umaxo Yes, I might add to the answer that I am assuming usual units and not natural units or other alternative unit systems. $\endgroup$
    – Steeven
    Mar 30 at 10:15
  • $\begingroup$ you can't suddenly move a distance of "seconds" - when doing relativity we routinely use units where $c=1$. So length and time are measured with the same units. $\endgroup$
    – Kostya
    Mar 31 at 10:45
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    $\begingroup$ @Kostya They are not the same units. That $1$ is a dimensional $1$: it provides the conversion factor from a time measured in seconds to a length measured in light-seconds. $\endgroup$
    – Ian
    Mar 31 at 13:58

I suppose the Wikipedia page on dimensional analysis might give you a better context. Anyway:

  1. Can you set the units to whatever you want even if they are not equal to one coulomb? - Absolutely no, unless see point #3.
  2. If you can cancel out all units, then can you give it any units you want in the end? - no you can't. I suppose you mean what happens if you construct an expression with no units, such as meter/meter. You just get a dimensionless quantity.
  3. Do you have to do some kind of manipulation of units to get the right units? - Not sure about the "manipulation" term, but you can use dimensional constants in some cases. For example, you're absolutely sure that in your problem where you are looking for a speed of a relativistic electron $v$, it has to depend on a dimensionless parameter $\alpha$. Then you can write $v=\alpha c$, where $c$ is the speed of light. It still does not guarantee that your expression is correct (it can be $v=\alpha^2 c$ or $v=c/\alpha^4$) but at least this is a reasonable assumption. So in this example, we used a dimensional constant $c$ to create something with a dimension of a velocity.

Of course this will not help you to identify numerical factors like $4\pi$ or so.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Apr 2 at 8:58

Let's distilling to the very basics of what we are doing when we are doing physics. (1) First, we measure some quantities, using some measuring devices and get a bunch of numbers. (2) Then, we substitute these numbers in some mathematical formula and retrieve the result. (3) The result predicts what we'll measure using the same measuring devices if we'd perform an experiment that accords with the formula.

Now, say we have a formula with the dimensional properties you are describing $$ Q = \frac{l^2tI}{q^3}$$ Following the distillation above, it tells you that you have to (1) use some measuring devices to get the values of some charge $q$, some length $l$, some time interval $t$ and some inductance $I$. Then you (2) substitute the values into your formula and get (3) a prediction of some other charge $Q$ that you should get using your measuring device.

So far so good. But now I'm coming with a different charge measuring device. It turns out that my measuring device gives exactly twice as large a number as the one that you are using. So, when I substitute the value for $q$ in the formula - it predicts the value $1/2^3 = 1/8th$ of the value that you predict. On the other hand, the experiment itself didn't change, so when I measure the resulting value of $Q$, I should get twice the value. Meaning that the formula didn't predict the value correctly for my measuring device.

  • 2
    $\begingroup$ This really is the best answer. Units (SI units at least) are all about referencing the discussion of expected measurements relative to other well-understood "standard" systems. You choose units according to preference (typically, to attempt to make your discussion clearest), and use them consistently following rules of logic and physical systems, if you want your discussion to make any sense at all. $\endgroup$ Mar 29 at 23:19
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    $\begingroup$ Also, in practice, if the units don't match, but the formula suggests the correct scaling relative to input values (which typically correspond to functions of the experimental setup or its results), all you do is append a constant of proportionality. (And then that constant is simply another thing to measure :) ). $\endgroup$ Mar 29 at 23:22

do units have to make sense in Physics?

Absolutely. Also one can use natural unit system, where some value is the ratio of a quantity with some known constant. I.e. imagine that you are expressing speed in terms of $c$ - spaceship is moving at $1/2~c$, or - sphere has $200~e$ charge. You can supply absolute value for $c,e$ and you will know exact measurement value in SI, CGS system or whatever. Or you can simply assign $c=1,e=1$ and just express every measurement as the ratio to some base unit.

if you can cancel out all units, then you can give it any units you want in the end

Really ? So you say, that if I measured that concrete lightning-bolt has struck with a charge of $1000\times$ (no units) of an average lightning-bolt power in terms of charge,- then you can say that it has struck with $1000\times$ of an average milk fat content ? What's the point of such nonsense ?

  • 3
    $\begingroup$ Natural units are still units they are just expressed as fractions of the speed of light (or equivalent). $\endgroup$
    – jacob1729
    Mar 29 at 12:47
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    $\begingroup$ You have a very good point about assigning units if there are no units. I guess I have misunderstood what I saw in EE. $\endgroup$
    – Tachyon
    Mar 29 at 14:50
  • $\begingroup$ @jacob1729 Yes, they are. I've fixed my wording a bit. $\endgroup$ Mar 29 at 17:41
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    $\begingroup$ @Tachyon you can make sense of that statement in terms of unit conversions. Say there are two dimensionful quantities A and B such that B doubles every time A is doubled. Then B=KA for some other dimensionful K. The dimensions of K are chosen as to perfectly cancel those of A on favour of those of B. A lot of constants of nature play this role. You can also define a new quantity A’=KA, and just write B=A’. The classic example is when we decide to measure time in units of distance in relativity. Instead of using t, you use t’=ct in all your formulas. (c=1 is a shorthand) $\endgroup$
    – Andrea
    Mar 30 at 7:41
  • 1
    $\begingroup$ @Andrea Unit conversion doesn't prove that any unit assignment without a physical reason of doing so is meaningful or valid. $\endgroup$ Mar 30 at 9:06

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