# A Notation Thing: d^3x

1. Mar 28, 2007

### Swapnil

I have seen some EM texts use this symbol: $$d^3 x$$. This just means the same thing as $$dV$$, right? Also, are there any advantages of using $$d^3 x$$ other than that it confuses people? :tongue:

2. Mar 28, 2007

### leright

eh? I've never seen that symbol before.

3. Mar 28, 2007

### cristo

Staff Emeritus
Yes, it does mean dV. As to whether there are any advantages? Erm.. pass.

4. Mar 28, 2007

### cesiumfrog

The real advantage is that it's shorter than writing "dxdxdx" everytime you do what is actually a sequence of three successively nested integrals. Technically you should use three integrand signs at the front and at least a couple sets of brackets between to make the old notation completely unambiguous, but we don't always bother with that either. It saves some writing whenever you have multidimensional integrals or higher order gradients/derivatives.

Compared to dV, the "advantage" is less abstraction. Conceptually, dV conveniently (and potentially confusingly) lets you imagine doing "just one" integral over volume, but when you actually have to get a pen and explicitly solve it, what you actually must do is integrate over each of the *three* axes *in turn*.

Last edited: Mar 28, 2007
5. Mar 28, 2007

### Swapnil

I see. But what if your volume is not in cartesian coordinate. Say you are working in spherical coordinates, would you still use the notation $$d^3x$$ for the infinitesimal volume or is there a seperate notation for that?

6. Mar 29, 2007

### leright

but if you're integrating in cartesian you're not integrating wrt to dxdxdx, but dxdydz, so that doesn't make much sense. I don't like that notation at all.

7. Mar 29, 2007

### nrqed

well, it's a matter of taste. It's like $\frac{dy}{dx}$. It does not make sense since a derivative is not a ratio of two quantities but a limit. Yet, it's accepted as a notation. I guess that one has to be a bit flexible with these things.

Regards

8. Mar 29, 2007

### dextercioby

Funny thing is that physics that d^3 x notation is very common.

9. Mar 29, 2007

### Swapnil

What about the above question?

10. Mar 29, 2007

### masudr

No. One reason we have $d^3 \vec{x}$ is to remind you about the units. So you have a volume, and therefore under the integrand you need a per volume thing (i.e. a density) for it to make sense.

EDIT: OK, so people have made this point above. I was trying to say that it just means a generic small change in whatever variable, not specifically along the x-axis.

Last edited: Mar 29, 2007
11. Mar 29, 2007

### Crosson

I see this notation in context where one must switch back and forth between volumes and hyper-volumes $$d^4 x$$.

12. Mar 30, 2007

### leright

I consider dy/dx very much a ratio. It is the limit of a ratio. I don't know why people cannot just think about derivatives as ratios. it's an infinitessimal increment of dy over an infinitesimal increment of dx. There's nothing inherently wrog with thinking of dy/dx as a ratio of infinitesimals. Take an infinitesimal change in x, and determine the corresponding infinitesimal change in y, take the ratios, and you have a derivative.

In physics this is the way to think of things intuitively.

Now, d^3x doesn't make things more intuitive at all...it just seems flat out wrong and confusing.

13. Mar 30, 2007

### arildno

$$\frac{\partial{y}}{\partial{x}}\frac{\partial{z}}{\partial{y}}\frac{\partial{x}}{\partial{z}}=-1$$
(Yes, the minus sign belongs there..)

14. Mar 30, 2007

### CPL.Luke

notice how he was speeking of ordinary derivatives in his post.

partials are significantly different.

15. Mar 30, 2007

### ObsessiveMathsFreak

This notation is just wrong, on so many levels.

It should really be $$dx^3$$ or if you really wanted to be penandtic $$(dx)^3$$, but that seems a little off to. Personally I like to use $$d\mathbf{x}$$, where $$\mathbf{x}$$ is understood to be "n" coordinates. But this too will give you problems if you scale coordinates, because if you make $$\mathbf{x}=2\mathbf{y}$$ you'll have $$d\mathbf{x}=2^n d\mathbf{y}$$, so be careful there.

OK, higher order derivatives are denoted as follows.

$$\frac{d^3 y}{dx^3}$$

Note that the placement of the 3 in superscript is in different locations. There's a very good reason for this. The notation essentially means something like
$$\frac{d^3 y}{dx^3}\equiv\frac{d^3 y}{(dx)^3}\equiv\frac{d^3 y}{dxdxdx}\equiv\equiv \frac{d d d y}{dx dx dx}$$

OK that last part there was especially horrific, but my point here,(with my highly nonstandard terms) is to get across that the superscript positions mean different things. The upper one means that the "d", differentiation, is being applied twice. The superscript on the bottom means that the differentiation is being taken with respect to dx in each case. The canonical expansion is of course:
$$\frac{d^3 y}{dx^3}\equiv\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(y\right)\right)\right)$$
Note on the "top", lines "d" alone is repeated twice, and on the bottom all of "dx" is repeated.

Analagously, you should really use $$dx^3$$ for integrals.

$$\iiint f(x) dx^3 \equiv \iiint f(x) (dx_i)^3 \equiv \iiint f(x) dx_1dx_2dx_3$$
OK a lot of that is nonsense, but if you wrote
$$\iiint f(x) d^3x$$
It seems to imply that you mean to integrate three times with respect to the same variable x, not with respect to three different variables. Anyway this is why I prefer $$d\mathbf{x}$$.

This all goes back to the problem with our contemporary calculus notation. Specifically, it's crap, or in academic language, unsatisfactory. We mean what we say, but we do not always say what we mean. No one has come up with anything better however, so we're stuck with what we have unfortunately.

16. Mar 30, 2007

### cesiumfrog

$$\iiint f(x) d^3 x = \int \left( \int \left( \int f(x) dx \right) dx \right) dx$$

Obsessive, you can't just insert different (subscripted) dummy variables, because the function is of "x", not of "x_2" etc, therefore what you wrote was not strictly equivalent.

leright, if you're content with differentials as ratios, I don't see the intuitive problem with this notation representing an infinitisimal volume element.

One reason to put the exponent on the operator rather than the dummy variable is that the operator is a special symbol (difficult to confuse), whereas the alternative would be ambiguous: sometimes $d(x^3)$ really does mean to integrate (once) with respect to (the dummy variable) $x^3$ itself. In fact, this is the kind of notation I personally use most, because I think it is clearer than introducing new variables into my algebra (say, having to write out "let $q=x^3$", operating over q, then never using that variable name as anything else.. and it gets hard finding enough distinct symbols that aren't associated with particular meanings in a physics context).

Last edited: Mar 30, 2007
17. Mar 30, 2007

### masudr

Surely the function is of x, y, z. i.e. say charge density:

$$\rho(x,y,z)$$

18. Mar 30, 2007

### Swapnil

But instead of $$d\mathbf{x}$$ wouldn't a better notation be $$d^3\vec{s}$$? I say this for three reasons:

1) First, it is easier than writing a boldface letter;

2) Second, since $$d\vec{s}$$ is an infinitesimal displacement in three dimentions, it would remind readers to integrate with respect to all three spacial coordinates.

3) Third, since $$d\vec{s}$$ has a different form in different coordinate systems, the reader would be reminded of the Jocobian scaling factor. For example, in spherical coordinates, $$d\vec{s} = dr\hat{r} + rd\phi\hat{\phi} + r\sin(\phi)d\theta\hat{\theta}$$ and thus it somewhat makes sense why $$d^3\vec{s}$$ should be defined to stand for $$r^2\sin(\phi)dr d\phi d\theta$$

1) Reader might thing that $$d^3\vec{s}$$ is a vector eventhough it is a scalar volume.

Last edited: Mar 31, 2007
19. Mar 31, 2007

### Crosson

An appropriate notation would be:

D (func, var, order)

So the second derivative of f wrt to x is:

D(f,x,2)

where the order can be any real etc.