The question clearer: Is not the infinitesimal cube the absolute smallest infinitesimal volume?
(Sorry if people thought if it meant: "Is it possible and is it done in daily life to use anything other than the Cartesian volume element?" : I know the answer to this is of course yes and I know it's usefulness. But please note that the question title has not been changed at all! It stands.)

After the many discussions, now the questions stands at comparing infinitesimal volumes.

A holistic answer that addresses this will be appreciated. This involves a phrasing of what infinitesimals are, how an infinitesimal volume arises and what happens when such volumes are compared from two different coordinate systems. Is it OK to address the infinitesimal volumes as smaller versions of finite shapes? If it is fine, what is wrong in this gedanken?:

  • An infinitesimal is by definition a length that is really, really small. If then I multiply this length with the same but in two perpendicular directions, I get a cube. This is the infinitesimal volume in Cartesian coordinates.I.e., an infinitesimal volume should have all it's edges as infinitesimal lengths, right? Is any other infinitesimal volume theoretically correct? (I have trouble accepting cuboid shaped "infinitesimals" as well.)

I would highly appreciate people from physics background to answer this question in an intuitive, "Feynman lectures" way, for lack of better words. Everyone's time is appreciated!

My argument for the comparison of volume elements across different coordinates systems:

In any coordinate system, I can define an interval whose unit length I can define, right eg. $|ds|=\sqrt{dx^2+dy^2+dz^2} $. So, the infinitesimal volumes from any coordinate system can be compared. Given this and that infinitesimal volumes occur as a result, I would start of with 3 infinitesimal lengths with no possibility of curved surface. I should end up with a cube only.

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    $\begingroup$ Related: How to treat differentials and infinitesimals? , Rigorous underpinnings of infinitesimals in physics and links therein. $\endgroup$ – Qmechanic Sep 13 at 6:29
  • $\begingroup$ I think your question would be much easier to answer if you could boil it down to a few points which you didn't understand. As of right now, it is a large mess after the amount of edits you have made $\endgroup$ – Buraian Sep 13 at 17:58
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    $\begingroup$ Please make your question one cohesive post instead of tacking on edits at the end. An edit history is available for those who are interested. $\endgroup$ – BioPhysicist Sep 13 at 19:32
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    $\begingroup$ "An infinitesimal is by definition a length that is really, really small." You can't really start off epistemological discussions with assertions like that. On top of that, it is a vague/incohesive assertion. $\endgroup$ – Fattie Sep 14 at 17:34
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    $\begingroup$ Is there a physical definition of "infinitesimal" that I'm not aware of? Who said it has to be a length, rather than a pure, dimensionless, numerical value? $\endgroup$ – chepner Sep 14 at 20:15

13 Answers 13


Infinitesimal volume elements do not have to be cubes.

Some familiar examples come from typical solids of revolution problems from calculus 1/2. Typically one discusses using either the "disk/washer" or "cylindrical shells" methods for finding the volume of the solid. As you can guess, the former method uses infinitesimally thin disks/washers as volume elements, and the latter uses cylindrical shells with infinitesimal thickness.

Volumes that are finite in one or two dimension(s) and that are infinitesimal in a third dimension are still infinitesimal because an infinitesimal value multiplied by a finite value is still infinitesimal. You can also build up "non-cube" volume elements by integrating over certain variables from your "cube" volume elements. For example, you can get spherical shell volume elements by integrating over the azimuthal and polar coordinates: $$\text dV=\text dr\cdot\int_0^\pi\int_0^{2\pi} r^2\sin\phi\,\text d\phi\,\text d\theta=4\pi r^2\,\text dr$$

which as you can tell is the volume of a spherical shell of radius $r$ and thickness $\text dr$.

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  • $\begingroup$ I'm not sure this qualifies. The basic volume element is still a prism of dimensions $r d\phi$, $r \sin \phi d\theta$, and $dr$. $\endgroup$ – Ryan Thorngren Sep 15 at 20:04
  • $\begingroup$ @BioPhysicist I do know that cubes are not the only infinitesimal volumes. "... because an infinitesimal value multiplied by a finite value is still infinitesimal..." then the discussion moves to comparing infinitesimal volumes. $\endgroup$ – Sidarth Sep 16 at 2:09
  • $\begingroup$ @Sidarth Then ask a new question. Just make sure to be specific about the comparison you want. $\endgroup$ – BioPhysicist Sep 16 at 3:16
  • $\begingroup$ I have added the final edit to the question which points this out. Should I ask a new question? What about the activity on this one? Frankly, comparing infinitesimal volumes has been the question all along. $\endgroup$ – Sidarth Sep 16 at 4:52
  • $\begingroup$ @Sidarth If your own answer is what you are looking for then you can accept it. I feel like the goal posts have been changing for this question for a while. Technically if you realized you were wanting to ask a different thing you should have asked a new question from the beginning, not update your question so that current answers became somewhat invalid. $\endgroup$ – BioPhysicist 2 days ago

Your comments (and to a lesser extent, your Question) indicate a severe confusion about ever having an infinitesimal volume. You never construct an infinitesimal volume. Infinitesimal volumes appear at the end of a limiting process.

Where do the infinitesmial rectangular parallelepipeds you are discussing appear? They appear in the limit of an iterated triple integral. An iterated triple integral involves nested orthogonal partitions to construct Riemann sums. In the limit as the diameters of all the partitions decrease to zero, the resulting volume elements are the infinitesimal rectangular parallelepipeds you first describe.

Can there be other infinitesimal volumes? Of course; use a different coordinate system. If you have arranged your triple integral to be in spherical coordinates, then you (may, if your region of integration includes it,) have an infinitesimal sphere at the center and the rest will be volumes bounded by two radii, two longitudes (which bound a spherical wedge) and to latitudes (which bound a spherical segment). In the limit as all the partition diameters go to zero, you obtain infinitesimal versions of these volumes.

Notice that at no point during the taking of the limit do you ever have an infinitesimal volume. These infinitesimals only appear once the partition diameters finish going to zero. I'm not going to get in the philosophical difficulties of completed infinities and whether the results of infinite processes exist. The point is that we use non-infinitesimals to infer what would happen if we really could use infinitesimals.

As another example of a different infinitesimal volumes, consider cylindrical coordinates. Here, we have cylinders on the longitudinal axis, and, everywhere else, volumes bounded by two cylinders of constant radii, two planes of constant angle, and two planes of constant longitude. Let's give those last volumes a name: "fred"s. The cylinders and freds are not rectangular parallelepipeds. In the limit as the partition diameters go to zero, we end up with infinitesimal cylinders and infinitesimal freds.

There is a different idea -- using non-rectangular regions in the usual 1-dimensional Riemann sum. For instance, graph the function over the interval of interest, then pack the area between the curve and the $x$-axis with disks. Sum the areas of the disks. Then repeat the process in the limit as the radius of the disks goes to zero. What you find is that you do not get the same value as the usual integral. If you are careful in specifying your packing method, you will actually have a limit as the radii go to zero and the resulting total disk area will underestimate the actual integral due to the "gaps" between the disks.

In short, the method described in the first few paragraphs where we partition all of the space of integration into pieces is necessary -- don't leave gaps.

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    $\begingroup$ I had to read your answer multiple times to understand! It might be the way that I have been schooled that I have never thought of infinitesimal volume elements as the "end result" of anything. So, I take a coordinate system first, assign a shape in the central region (shape is obvious?) and then any volume is like a "skin" above it. An infinitesimal volume occurs when the grid lines cut into this thick skin, then make the grid lines take the limiting value. Is this correct? $\endgroup$ – Sidarth Sep 13 at 17:19
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    $\begingroup$ @Sidarth : Yes -- you don't have infinitesimals until you reach the limit of infinitesimally separated grid lines. $\endgroup$ – Eric Towers Sep 13 at 17:20
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    $\begingroup$ @Ruslan : "Iterated" does not by itself indicate the depth of the nesting. In the OP"s case the depth is always $3$ as indicated by "triple". When Fubini applies (which will be the case for anything OP is attacking), the non-different result of the non-distinct notation makes the distinction you are drawing a distinction without a difference. $\endgroup$ – Eric Towers Sep 13 at 20:07
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    $\begingroup$ When Fubini theorem applies, "iterated triple" is simply redundant. $\endgroup$ – Ruslan Sep 13 at 20:15
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    $\begingroup$ @Ruslan : Repeating what I said doesn't change anything. $\endgroup$ – Eric Towers Sep 13 at 20:15

Different coordinate systems have different kinds of volume elements; The volume elements are a consequence of how the grid lines of the coordinate system are set. Volume element can be generated by nudging the parameters which describe points in the space by infinitesimal amounts and figuring out the volume of the region generated as a consequence. This is especially useful in multi-variable volume integrals and in the application of some vector-calculus results such as the divergence theorem.

On some more thought, I'd like to add one more point. Yes, you are correct that infinitesimals are small quantities, however you're missing a crucial point. Depending under what constraint you put while your quantity small, the actual structure of this 'small quantity' would be different. This would be understandable using the references I have given in the bottom.

As a more direct example, suppose you have a large cube and you keep scaling down the dimension till you get some sort of infinitesimal volume cube, and now for contrast, consider a large sphere and imagine scaling it down till you get a tiny infinitesimal sphere. These two things are infinitesimal volume elements but the volume each contains is different due to the actual object which you are shrinking being different.

Deriving Volume Element for spherical co-ordinates

Lecture series which shows the concept described above using 3-d animations

For Understanding the ideas of linear transformation noted in the previous lecture better

Answer V2.0 based on op's new details of the question:

  1. and 2.) You can relate the volume elements between different co-ordinate systems using the determinant of jacobian. In a way, the Jacobean is the ratio of n-volume in one system to the n-volume in another. Also do not forget that some transformations do not behave the same globally, for example it is easy to understand that the 'natural unit' vector of polar co-ordinates scales up as you move further away from the origin (*)

  2. I'm not sure on this on what exactly you mean by 'smallest'. You need an absolute measuring scale to measure the concept of smallest. If I were to guess, the smallest volume element would be a singular linear transformation which squishes space into a point and hence literally have zero volume.

  3. and 4.) Not gonna comment on hyperreal numbers as I have not done much of it and this concept was dealt with in Dave's answer in much detail already.

  4. Yes, the properties of a shape other than n-volume measures should be invariant under uniform scaling. For example, consider similar triangles.

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  • $\begingroup$ 1) Could you explain how this is going to affect the infinitesimal volume:"...Also do not forget that some transformations do not behave the same globally..." 2) In any coordinate system, I can define an interval whose unit length I can define, right eg. |ds|=\sqrt{dx^2+dy^2+dz^2} . So, the infinitesimal volumes from any coordinate system can be compared. 4) Give this and that infinitesimal volumes occur as a result, I would start of with 3 infinitesimal lengths with no possibility of curved surface. I should end up with a cube only. $\endgroup$ – Sidarth Sep 16 at 2:23

An infinitesimal is by definition a length that is really, really small.

I think that your question arises due to a misunderstanding of what infinitesimals are. Infinitesimals are not easy to understand, they can be understood either as a limit as a quantity goes to zero or in terms of the hyperreal numbers. As the hyperreal concept is relatively new compared to the limit, it is not often taught, but it does have some clarity which I find helpful.

The hyperreal line is the real line augmented with infinities whose absolute values are larger than any real numbers and their reciprocals, the infinitesimals, whose absolute values are smaller than any positive real number.

The thing about the infinitesimals is that as individual numbers (not as sets) they can be manipulated with all of the same operations as reals. You can multiply an infinitesimal by a real number and get another infinitesimal. The infinitesimals can themselves be ordered, meaning that if $dx$ is an infinitesimal then $2 dx$ is larger than $dx$, but still smaller than any positive real and therefore still a perfectly valid infinitesimal.

This is important because infinitesimals can preserve their relationships to each other. They are all smaller than any positive real, even if some infinitesimals are larger than other infinitesimals. So $dx \ dy \ dz$ is half the volume of $dx \ dy \ (2 dz)$, but they are both infinitesimal.

In fact even if $x$ and $y$ are finite real numbers $ x \ y \ dz$ can be an infinitesimal volume. An infinitesimal volume merely needs to be smaller than any positive real volume, not smaller than other infinitesimal volumes. For that a single infinitesimal in the product suffices. A spherical shell from radius $r$ to $r+dr$ is a completely legitimate and valid infinitesimal volume $4 \pi r^2 dr$ even though its surface area is finite $8 \pi r^2$. This all follows from the properties of hyperreal numbers.

Infinitesimals can be formed into a hyperreal plane and into vectors, and those vectors can have norms and dot products, so you can have arbitrary infinitesimal shapes. You can have right angles, but you can also have arbitrary other angles. There is nothing magical about right angles that allows them and forbids other angles. You can have straight lines, but you can also have arbitrary curved lines. There is no restriction to right angles and straight lines.

Since you realize that infinitesimals can be orthogonal to each other, it should not be surprising that there is no limitation to other angles and thence to arbitrary shapes. The same rules that allow you to construct orthogonal infinitesimals allow you to construct other shapes. Again, all of this follows from the hyperreals.

Since many people are not familiar with hyperreals, here are some introductory sites (by no means complete or optimal):






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    $\begingroup$ "...An infinitesimal volume merely needs to be smaller than any positive real volume, not smaller than other infinitesimal volumes...." I did not know this! Very non -intuitive since I was initially thinking that the infinitesimal volume has to be vanishingly small, meaning there is only one possible infinitesimal volume, one made of infinitesimal lengths. $\endgroup$ – Sidarth Sep 14 at 0:28
  • $\begingroup$ @Sidarth I am glad it helped. I have only recently learned of hyperreal numbers and now wish that calculus were taught using them $\endgroup$ – Dale Sep 14 at 0:48
  • $\begingroup$ Yes. Definitely it has helped. But I also feel that an extra structure has been imposed in order to make something that was initially non-intuitive seem more plausible. I feel that it did not address the problem straightforwardly but rather justified somethings by using extra definitions. (I am not a mathematician. So this approach is a little different to me!) $\endgroup$ – Sidarth Sep 14 at 0:52
  • $\begingroup$ @Sidarth yes, I agree. But structures that make initially non-intuitive things seem plausible are my kind of structures! Particularly if the new structures are consistent with the established but non-intuitive approach, and are easier to use. The extra definitions are then worth learning. Of course, that is pure personal opinion and you are under no obligation to agree! $\endgroup$ – Dale Sep 14 at 1:56
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    $\begingroup$ Right. I think it's safe to say that I am completely in rigorous mathematics now? $\endgroup$ – Sidarth Sep 14 at 2:16

It’s not so much a question of what is theoretically correct, more a question of which shape of region allows us most easily to pass to the limit and derive a differential equation or an integral (which is usually the goal of this step).

The choice of region often depends on the symmetry of the problem. In problems with cylindrical symmetry it is common to use a cylindrical shell. In problems with spherical symmetry a spherical shell is often used.

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To take an entirely different approach to the various integration-related approaches of other answers ....

You appear to be perfectly comfortable defining "an infinitesimal cube", as a cube with sides of infinitesimal length.

Let's go one step further ... lets say:

"the infinitesimal cube with a vertex at the origin, and lying within the positive octant of 3 dimensional space w.r.t. the origin, comprises all points whose x-, y- and z- co-ordinates lie within an infinitesimal distance from 0."

That seems a perfectly reasonable definition of an infinitesimal shape, and I can't really imagine any other definition of a shape that you can "have", "get" or "take", that doesn't amount to essentially the same thing.

Ok, so we're happy with the infinitessimal cube.

Consider this definition:

"all points whose Cartesian distance from the origin, lies within an infinitesimal distance from 0."

Voila ... an infinitessimal sphere :) i.e. a sphere with infinitessimal radius.

Obviously all of this gets lost when you start to think of things in terms of mathematical limits. But this presents a answer w.r.t. the initial question, as asked.

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    $\begingroup$ I wish this answer got more attention, it displays a nice idea! $\endgroup$ – Buraian Sep 15 at 10:19
  • $\begingroup$ The first definition is very useful and is what I picture of as well. The second part of the answer is correct of its own accord. But it does not help me. Sure, I have defined an infinitesimal volume but I have lost the ability to compare it with the cube.(right?) The highly voted answer does not really completely answer my question that is why I have not ticked it. The question is still there. Please contribute in your own different way. I would love if someone could explain to me without things like for e.g. the mathematical "tricks" eg, with new definitions to justify what is being done. $\endgroup$ – Sidarth Sep 16 at 2:35

You asked: "Does it make sense to take an infinitesimal volume of shape other than a cube?"

Yes, and no. If you have an integral $$\int\int\int \text{something} \mathrm d\alpha \mathrm d\beta\mathrm d\gamma$$ then you do have cubes in $\alpha\times\beta\times\gamma$.

But those cubes in $(\alpha, \beta, \gamma)$ will be distorted into other shapes depending on your coordinate system.

The main point is that all the shapes together cover the whole region and do not overlap (except for sets with measure to zero volume, typically, boundaries.

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  • $\begingroup$ This is very helpful. So what you're saying is that I take a cube, say it's infinitesimal and then apply some transformations due to which the cube gets distorted in shape but still preserving the volume. This is fine, if I can actually fix the size of an infinitesimal length and volume - which could be problem specific. But, given that an infinitesimal length is boundlessly small,(hence fixed in magnitude but unknown) can I even bend it and do things - and will it give the same volume? (there is a chance it might give a smaller volume!, right?) $\endgroup$ – Sidarth Sep 13 at 15:43
  • $\begingroup$ @Sidarth have you ever wondered why, when changing coordinates of an integral, the determinant occurs? The determinant is a volume function, and corrects for the distorted volume. I guess your question is rooted in math, not so much in physics. $\endgroup$ – Gyro Gearloose Sep 13 at 15:48
  • $\begingroup$ @Sidarth Take a piece of chalk and put it on the blackboard, flat, all the length of it. Then fix one end and rotate the chalk stick around it in a circle. You will notice, that the chalk marks decrease with distance to the center. This compensated by the determinate of the switching to polar coordinates. $\endgroup$ – Gyro Gearloose Sep 13 at 15:54
  • $\begingroup$ I guess so. When I started this question, I did was not aware. But coming back to the point, how am I allowed to say (theoretically), the new volume element is actually the infinitesimal? (Answer: If I fix the length of an infinitesimal thats the same in every coordinate system, then during the transformation, the new volume element can be different but is the tiniest volume element possible in that coordinate system -hence, it is the infinitesimal volume element of that coordinate system, am I correct?) $\endgroup$ – Sidarth Sep 13 at 15:56
  • $\begingroup$ @Sidarth I guess what you are missing is the mathematical definition of a limit. Such as, for example, the definition of a derivative. $\forall_{x<\epsilon}\dots$ $\endgroup$ – Gyro Gearloose Sep 13 at 16:09

Yes, it absolutely does sometimes make sense to have infinitesimals be shapes other than cubes. Particularly when the metric space being used is not necessarily Euclidean.

In Walter Rudin's Principles of Mathematical Analysis, in the early part of his formulation of the general form of Stoke's Theorem (i.e., for arbitrary finite dimensional metric spaces), he builds a general integral calculus for an arbitrary metric spaces using parallelepipeds. I'd say, in mathematical circles at least, that's a fairly famous case of non-cubic infinitesimals.

You might get much better answers by posting this question on Math.SE. For anyone wondering what areas of physics might use arbitrary non-Euclidean metric spaces, I'm personally not sure, but I know manifolds are used in Physics, and I believe General Relativity is at least sometimes studied or modeled with non-Euclidean coordinates.

Related to this part of your question:

Some one can come and ask then why are you not making the length in the third dimension also the same

A parallelepiped can have all the lengths of its edges be equal. What makes it not a cube is that its angles are not necessarily equal.


How can I just assume that this is going to give me the smallest possible volume?

Infinitesimals need not be "the smallest possible volume" at all. While I'm not an excellent mathematician, I'm pretty sure the requirements for an infinitesimal for use in integral calculus is that they be composed of a volume that is both fixed and arbitrary. The questions you need to be able to answer about an infinitesimal are, "can you calculate the volume of it?" and "can you choose the volume of it?"

The common rigorous way to answer the two questions I posed in the previous paragraph is to define a linear transformation that "chooses the volume" and ensure there exists another linear transformation that "calculates the volume", the latter transformation usually being a well-known formula for calculating volume.

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  • $\begingroup$ Non-euclidean will be tough for me. I am trying to visualize it Todd. Qmechanic pointed me to a source that talked about this with mathematical sophistication but I got lost. I said cuboidal. Not paralleopiped. Your explanation about the smallest volume is very useful for my question - the arbitrariness - but it sounds more to do with numerical methods and accuracy rather than analysis - which there is some absoluteness. Infinitesimal means smallest, right? $\endgroup$ – Sidarth Sep 13 at 15:04
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    $\begingroup$ @Sidarth Infinitesimal is a special term of art used in calculus to mean a mathematical object that has certain properties. In this context, it is not helpful, in my humble opinion, to reduce the definition of the word infinitesimal to "smallest". I know you said cuboidal. I was giving you an example of a non-cuboidal infinitesimal (a parallelepiped) that is used in a very famous book. In my answer, the word "analysis" means, essentially, "calculus". Calculus is basically numerical methods (the mathematical topic) with limits (the mathematical concept) applied. $\endgroup$ – Todd Wilcox Sep 13 at 15:12
  • $\begingroup$ So, calculus finally comes down to getting a numerical answer so the arbitrariness of the smallness of the elemental volume depends on the problem, is that what you are saying? $\endgroup$ – Sidarth Sep 13 at 15:20
  • $\begingroup$ @Sidarth I think the "shape" of the infinitesimal depends on the problem, and the "shape" is created by the linear transformation that defines the infinitesimal, which also depends on the problem. How small (or large) an infinitesimal can be made to be does not depend on its shape. We can make all kinds of shapes arbitrarily small. In some cases, it's harder to use a cube/cuboid than it is to use a different shape, which is why we don't always use cubes/cuboids. $\endgroup$ – Todd Wilcox Sep 13 at 15:26

The two most important properties of an infinitesimal volume used for integration are

a) its side is shorter than that of any other volume you care to specify

b) the value of the property that is a function of its dimensions, this is the thing you're integrating over a volume, is the same on any edge or vertex of the volume. That means it doesn't matter whether you sum f(x), f(x+dx) or f(x=dx/2), they are all the same as far as the integration, or the summation in the limit of dx->0, are concerned.

It doesn't matter what coordinate system you choose, cartesian, spherical or cyclindrical polar, a 'small' volume in any coordinate system will have those properties.

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  • $\begingroup$ I don't understand point-2 $\endgroup$ – Buraian Sep 15 at 10:19
  • $\begingroup$ @Buraian so if it mattered whether you took the lefthand side or the righthand side of the infinitessimal volume, then how would you get to the integral by taking the summation to a limit? $\endgroup$ – Neil_UK yesterday

Some important points I have gathered from discussions, that might help someone to completely dispel my doubts. Thanks to all!

  1. One does not set out to construct a volume element,rather, it occurs as a result of the grid system we are using , through a limiting process. Naturally, the shape of the volume element will depend on the system. A question on this: Is it ok to compare volume elements (magnitude) from different systems? (they are related by the determinant and will obviously be different, so, asking which is smaller is mathematically possible and my OP question still remains.)Related

Volumes that are finite in one or two dimension(s) and that are infinitesimal in a third dimension are still infinitesimal because an infinitesimal value multiplied by a finite value is still infinitesimal.

Fine, but what if I want to compare those different infinitesimal volume elements? This has been the question from the beginning.

  1. The infinitesimal volume element in a particular coordinate system will turn out to be the smallest volume possible in that grid system.

  2. "Infinitesimals" is a special construct in mathematics and should not be taken to mean "smallest" and need not correspond with my intuitive thoughts e.g. one the idea that one particular volume element being the absolutely smallest possible volume across all coordinate grid systems. (even though it seems to be a very plausible thing!)

  3. Infinitesimals behave like real numbers, with ordering as well but are actually hyper real. The definition of an infinitesimal volume is such that one infinitesimal in the product is sufficient to call it an infinitesimal - this completely makes my edit#2 and edit #3 insignificant(and even the first two points above) but does not answer straightforwardly: it's not really the minimum possible infinitesimal length. Still, however the product is going to be smaller than any real number,only that we are now comparing real and hyper real numbers (this is new to me since it feels like coming up with a definition to justify something.)

  4. The point is that we use non-infinitesimals to infer what would happen if we really could use infinitesimals.

This greatly relates to me. From the beginning, I have been thinking about properties of big shapes and have assumed it as the properties of their infinitesimal versions. ( Is this wrong and if so, why?). The quote below also tells that this supposition of mine is wrong:

How small (or large) an infinitesimal can be made to be does not depend on its shape. We can make all kinds of shapes arbitrarily small.

(I apologise if some of the points above are redundant/repetitive)

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    $\begingroup$ Re 1: Sure, why not? In 2D polar coordinates a volume (area in 2D) element is a sector of an annulus, and its area at $(r,\theta)$ is $rdrd\theta$. If $(r,\theta)$ has the rectangular coordinates $(x,y)$, then the area of the corresponding volume element is $dxdy$. Which is bigger? Well, it depends on what $r,dr,d\theta,dx,$ and $dy$ are! If you choose $dr,d\theta,dx,dy$ to all equal your favorite infinitesimal, then the polar area element is larger than the rectangular when $r>1$ and smaller when $r<1$. $\endgroup$ – Kevin Arlin Sep 15 at 20:30
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    $\begingroup$ It's basically unavoidable, by the way, that an infinitesimal not be "the smallest possible length", if you want to preserve any connection with differentiation. Here we say things like $(x+dx)^2\approx x^2+2xdx$" to calculate the derivative of $x^2$, but to do even this we must be able to multiply an infinitesimal by $2$! Thus we must have another infinitesimal which is twice as big...And what if we divide it by 2? The hyperreal perspective has the benefit that the hyperreals have all the same basic properties as the reals, just plus some infinitesimals. So your comment after (5) is right. $\endgroup$ – Kevin Arlin Sep 15 at 20:33
  • $\begingroup$ 1st comment) @KevinArlin This is the first time I'm pointed to the size of the infinitesimal volume itself changing as the place I take it changes! i.e., the r coordinate. '...favourite infinitesimal..." Do you mean assign a numerical value irrespective of the fact that angles are dimensionless? So the polar case is actually smaller than the Cartesian one in r<1 case. It might become still smaller in the 3d case because of another r dependence. $\endgroup$ – Sidarth Sep 16 at 2:45
  • $\begingroup$ w.r.t 2nd comment) Then, I want to compare the hyper real infinitesimals.Not infinitesimals with finite things. I don't have a problem with an infinitesimal being multiplied by 2 and being called an infinitesimal. That just refers to the fact that we are dealing with small things compared with our integration limits(?) .In the hyper real infinitesimals also, I can have the smallest value - the "infinitesimal even in the hyper real numbers". Then I can use this. (I hope no one says there is something now called hyperhyper real numbers!) $\endgroup$ – Sidarth Sep 16 at 2:50

Let's think about considering your question in the context of an abstract Riemannian manifold $(M,g)$. In local curvilinear coordinates $(x_1,x_2,\cdots,x_n)$ we have $dV=\sqrt{\vert g\vert}dx_1dx_2\cdots dx_n$.

Differential geometers like to say that the volume form on a manifold has no local structure. This is the gist of a famous theorem of J. Moser, published in 1965. In non-technical phrasing, we can find a change of coordinates (local diffeomorphism) so that $dV$ is identical to that of the volume form on $\mathbb{R}^n.$ So after a change in coordinates, the volume form looks like the standard one based on infinitesimal cubes.

Keeping this all in the arena of Euclidean space as you are doing, if you did use some other geometrical underpinning for your infinitesimal volume form, then after a change in coordinates, you can bend this infinitesimal geometry to look like infinitesimal cubes.

So long story short--by Moser, locally any two infinitesimal geometries that you might use to define volume are equivalent, from the viewpoint of a differential geometer.

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  • $\begingroup$ What do you mean when you say no local structure? I think you'd help op if you got into geometry of volume $\endgroup$ – Buraian Sep 15 at 10:04
  • $\begingroup$ @Buraian: A good example of local structure in Riemannian geometry is curvature, which is a local invariant. Perhaps a familiar example of a geometry that has no local structure is symplectic geometry. This follows from Darboux's theorem. Moser's 1965 paper gives an excellent proof. Suppose one person has conviction that infinitesimal cubes are "the volume element" while another person has conviction that another geometry is, e.g., the spherical volume element. Both need to accept that there exists a local volume preserving diffeomorphism converting one viewpoint to the other. $\endgroup$ – user52817 Sep 15 at 18:28

We can generalize the concept of integration thusly: given a space $S$ with a measure $m$ and given a function $S \rightarrow \mathbb R$, for each natural number $n$, separate $S$ into disjoint subsets, none of which has measure greater than $2^{-n}$. For each subset, take the supremum of $f$ over that subset, multiply it by the measure of the subset, and then sum the results over all subsets. Now take the limit as $N goes to infinity. Then do the same thing, except with the infinum rather than supremum. If those two limits are the same, then that is the integral.

If $S$ has a projection into n-dimensional Euclidean space (i.e. has n-dimensional coordinates), we can take these subset as being the Cartesian product of intervals of these coordinates (the concept of infinitesimals, to simplify it a bit, is taking the limit as the length of those intervals goes to zero). If $S$ actually is a Euclidean space, and the measure is the normal Euclidean one, then the measure of each subset is simply the product of the length of the intervals. Otherwise, we need to include a term representing this measure. In the limit, this reduces to the Jacobian.

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  • $\begingroup$ If you could write meaning of these mathematical terms you use as a footnote or smthn this would be an easier read $\endgroup$ – Buraian Sep 15 at 10:01

Since infinitesimal volumes are primarily used in the context of integration, I will answer from that perspective.

The infinitesimal volume elements used in Riemann integration are always generalised rectangles in their respective coordinate systems, since the Riemann integral is only defined over generalised rectangles.$^*$ For example, a volume element that looks like a wedge when represented in Cartesian coordinates could actually be a prism when represented in spherical coordinates (of course you would still need the determinant of the Jacobian to get the volume correct). This is because a generalised rectangle is just a Cartesian product of intervals, i.e. $\prod_i [a_i, b_i]$, in some coordinate system.

However, why should we restrict ourselves to just Riemann integration? The Lebesgue integral does not concern itself with unnecessary structures like coordinate systems or a notion of shapes, instead looking only at the value the integrand takes on measurable sets and the measures of those sets. In this case, it is the measure that gives us a notion of volume. Thus, we can see that mathematically it makes sense to use infinitesimal volume elements of arbitrary shape, but what about physically? Since matter is actually discrete, one might wonder if it even makes sense to discuss any infinitesimal volume of a physical object, but we know that physical objects can be modelled as being continuous in $\mathbb R^n$, so it makes sense that we can use that we can use the mathematics associated with $\mathbb R^n$ to describe said objects.

$^*$There is a way to extend the Riemann integral to more general Jordan domains, but that is still done by integrating over a generalised rectangle. A Jordan domain is a set whose boundary has Jordan content zero.

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  • $\begingroup$ If you define measure and measurable sets, this would be easier to understand for a non-math person and also what a Jordan domain is. If not, add some references to what you're saying. $\endgroup$ – Buraian Sep 15 at 10:03
  • $\begingroup$ @Buraian I added some links and explained some terms. $\endgroup$ – Sandejo Sep 15 at 14:49

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