Let's say that we have defined a certain physical quantity from a particular relationship and then we find another relationship and define the physical quantity again.

For example, $$v = u + at$$ $$\text{and }v = \sqrt{u^2+2as}$$ where $v$ denotes the final velocity, $u$ denotes the initial velocity, $a$ is the acceleration, $s$ denotes the displacement and $t$ denotes the time.

Why are the dimensions of the physical quantity, when evaluated using the first relation, the same as those when evaluated using the second relation?

I know that this might sound like a silly question and the answer to this is most likely trivial but it seems like I have some misconception that is preventing me from fully grasping it, which I hope to clarify.

  • 5
    $\begingroup$ Why do you think they are different? $\endgroup$ Jul 27, 2020 at 4:53
  • 2
    $\begingroup$ @SuperfastJellyfish I don't. I'm just wondering how the definition of any physical quantity in multiple ways somehow gives the very same dimension and never leads to any inconsistency. Is that understandable? $\endgroup$ Jul 27, 2020 at 4:55
  • 5
    $\begingroup$ The dimensions should be the same. In your example the dimensions of the two quantities are the same. Perhaps you can explain why the dimensions don't seem to be the same for you. $\endgroup$ Jul 27, 2020 at 4:55
  • 2
    $\begingroup$ @flippiefanus I am aware that the dimensions of final velocity in the particular example that I've mentioned are consistent. But, I'm asking more about the 'why'. We defined dimensions as something that tell us how a physical quantity can be represented in terms of the base quantities. Now,when we define that quantity in multiple ways, it still is the same when represented in the base quantities. I jut can't grasp it completely. :( $\endgroup$ Jul 27, 2020 at 5:01

4 Answers 4


Basically, it's because your two equations aren't definitions, they're results.

It's like saying, "What if we define $100 = 10 \times 10$ and then define $100 = 80 + 20$? How do we know if these definitions are consistent with each other?" The question doesn't actually make sense. Those equations aren't definitions, they're results.

In physics, you derive results by starting with definitions (for example, velocity is defined as $v = dx/dt$) and applying mathematically valid operations, such as multiplying both sides by something. All of these steps keep the dimensions of both sides the same, so the final results have consistent dimensions as well.

  • $\begingroup$ Thanks for the answer, first of all! What about defining force as $F = ma$ and $F = G\dfrac{m_1m_2}{r^2}$? Would you consider them to be separate definitions? $\endgroup$ Jul 27, 2020 at 5:30
  • 7
    $\begingroup$ @RajdeepSindhu That's different. The first one is a definition of force. The second one is an example of a specific force. Since $G$ is a new quantity, and we've already defined force, applying dimensional analysis to the second equation actually just tells us what the dimensions of $G$ are. $\endgroup$
    – knzhou
    Jul 27, 2020 at 5:49
  • $\begingroup$ @RajdeepSindhu in addition to what knzhou has pointed out in the comment above, $F=ma$ is a defining differential equation and the specific form of $F$ is different in different cases. For gravity you plug in the inverse square form in the LHS of the differential equation. $\endgroup$ Jul 27, 2020 at 8:09
  • $\begingroup$ Adding to the above pair of comments, it is often useful to write the first equation as $a=F/m$ (treating the applied force as known and solving for the resulting acceleration). The form $F=ma$ most directly gives you the force that must be acting on the system to generate an observed acceleration. $\endgroup$
    – RLH
    Jul 27, 2020 at 19:03
  • 1
    $\begingroup$ I generally write them as $\sum F = ma$ and $F_g = G\frac{m_1 m_2}{r^2}$ to emphasize that the quantities on the left-hand side are not the same. But that's not really something to get too far into in the comments. @RajdeepSindhu you could post a followup question based on your first comment, if you like, and then everyone can offer their answers. Or if you prefer I can migrate this whole conversation to chat and people can continue it there. $\endgroup$
    – David Z
    Jul 28, 2020 at 0:08

The answer is to think of it backwards. We don't start by saying $u+at$ and $\sqrt{u^2+2as}$ have equivalent units. We start by saying that, fundamentally, we think of "velocity" as a thing which is a physical quantity. If two expressions for the same physical quantity yield different units, we strongly question whether one of them is fundamentally wrong.

Over the years, we have developed an axiomatic model of how units work. The traditional calculus for quantities defines the concept of a unit Z, and a quantity, which is $\mathbb R \times [Z]$ (a real number "multiplied" by a unit). From there, they define how that multiplication should distribute over other arithmetic operations, such as $$x\times[Z_1] + y\times[Z_1] = (x + y) \times [Z_1]$$ $$x\times[Z_1] \cdot y\times[Z_2] = (xy) \times ([Z_1]\times[Z_2])$$ $$\sqrt{x\times[Z_1]^2} = \sqrt x \times [Z_1]$$

and so forth. And, of course, we defined the concept of unit multiplication and division that we are now used to. We defined "dimensionality" to capture whether it was meaningful to add treat units as different "spellings" of the same quantities, or if they were fundamentally different. Several common dimensionalities are length, time, area (length squared), and speed (length divided by time)

Over time, what we found was that equations which were consistent with this particular treatment of units could be "right," while those which were found inconsistent basically never were. So we declared these to be the "right" way to handle units, and added constants to handle any oddities that might occur.

Now I do note that these are incomplete. There's two corner cases where people disagree on the best way to handle units. One of them is angles. Technically radians are dimensionless -- they are a length divided by a length. However, many people have found it convenient to treat radians as having a dimensionality of "angle." This catches more mistakes, but runs into problems like the small angle approximation $sin(x\times[rad]) \approx x$ for small $x$. This obviously runs into trouble if radians have a dimensionality that we can't just handwave away.

The second area that causes problems are trancendentals. Decibels (dB) is a famously troublesome case because there is a logarithm in the equations for it. To date, we do not have an axiomization for such extended units, only the 7 major dimensions that we are used to from SI, so we have to admit that our quantity calculus is incomplete. For a handling on these issues, I recommend the article from Metrologia, On quantity calculus and units of measurement if you can access it.

So in the end, the math works because we spent a lot of time finding math that fit reality. And, when necessary, we fudge it and create incomplete rules to keep it in line with reality. I wish there was a more precise answer, but that's the best we've managed over hundreds of years of scientific inquiry!

  • 1
    $\begingroup$ Exactly what I was looking for, thanks! $\endgroup$ Jul 28, 2020 at 3:43
  • 2
    $\begingroup$ In a small angle approximation you can handwave the unit away because it is an approximation. But where angles really can't be dimensional is when you convert angular and orbital velocities: $v = \omega r$ and $\omega$ has to be just $[\mathrm{s}^{-1}]$ as this equation actually uses the definition of angle as ratio. $\endgroup$
    – Jan Hudec
    Jul 28, 2020 at 5:36
  • 4
    $\begingroup$ There are also more fundamental differences when dimension are or are not considered the same. The natural units used in particle physics have the same dimension for time and distance, with dimensionless velocity ($c = 1$), and consequently same unit for mass and energy (usually $\mathrm{eV}$), which is physically meaningful as mass and energy do change into each other in that domain. But in human-scale world they don't, so it is reasonable to treat them as different dimensions in most other contexts. $\endgroup$
    – Jan Hudec
    Jul 28, 2020 at 5:45
  • 2
    $\begingroup$ @JanHudec Good call on the angular velocity example. Small angle approximation just happened to be on my mind. Years ago I wrote a programming language that handled units. Unfortunately, many of the engineering equations which were programmed into it contained small angle approximations baked in, so that particular case showed up more than all other unit issues put together. I'm not bitter... years later... $\endgroup$
    – Cort Ammon
    Jul 28, 2020 at 6:16
  • 1
    $\begingroup$ Great answer, but aren't decibels technically just a representation of a ratio with respect to a reference level, and as such pretty similar to the angle problem? $\endgroup$
    – llama
    Jul 28, 2020 at 15:54

The reason is probably not anything deep. It is simply the fact that when you have a specific quantity, such as velocity in this case, the dimensions of quantity is determined by what it physically represents, and not by the equations that one can use to compute it in different scenarios. Still, the equations must produce dimensions that match those of the quantity. This is because these equations represent physical situations where these dimensions have physical meaning. I don't know how to explain this any better.


One may add to @knzhou's answer that dimensions can be modelled mathematically in a consistent way as explained in Terry Tao's post here: https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/

This is a refined version of the following intuition: We do not model the space of "length" as $\mathbb{R}$, but instead as $L = \mathbb{R} \mathrm{m}$. The element $1\, \mathrm{m}$ defines a basis for $L$, just as $1\, \mathrm{ft}$ does. The change of basis matrix is given by the equation $1\, \mathrm{ft} = 0.3048\, \mathrm{m}$.

If similarly $M = \mathbb{R} \mathrm{kg}$ is the space of masses, you can define the space "mass $\times$ length" as $M \otimes L$, and $\mathrm{kg} \otimes \mathrm{m}$ is the canonical basis induced by the given bases on $M$ and $L$.

In this view, the choice of units becomes as arbitrary as the choice of bases in linear algebra; and the dimension of a quantity is reflected in the vector space in which it lives (e.g. $M \otimes M \otimes L \otimes L \otimes L$). As a side remark, the canonical basis element of the dual $L^*$ evaluates to one on $1\, \mathrm{m}$ and hence can be thought of as $\frac{1}{\mathrm{m}}$, explaining how inverse dimensions can be modelled.

  • $\begingroup$ I like this idea, but I don't think it answers the question. $\endgroup$
    – mr_e_man
    Jul 27, 2020 at 17:00
  • $\begingroup$ Yes, this was just supposed to expand on the accepted answer's last sentence: "All of these steps keep the dimensions of both sides the same, so the final results have consistent dimensions as well". The consistency of dimension assignments is an (easy) mathemetical result which can be proven. $\endgroup$
    – wandersam
    Jul 27, 2020 at 20:20
  • 1
    $\begingroup$ Hi and welcome wandersam :-) Our rule here is that only things which answer the question should be posted as answers. That being said, this actually does seem like an answer to me - I mean, it's not the most clear, and I can understand how people would think otherwise, but I believe it is addressing the question in an appropriate way to be considered an answer. Building on someone else's answer is totally legitimate. $\endgroup$
    – David Z
    Jul 28, 2020 at 0:10
  • $\begingroup$ I know that this is minor, but for SI-units there should be a space between the value and the unit, so it is "0.3048 m" (without quotes). bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-EN.pdf section 5.4.3. $\endgroup$ Jul 28, 2020 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.