The Student Room Group

Volume Differential

Hello TSR,

I am told that the volume differential dV=dx dy dzdV = dx\ dy\ dz is not the same as another differential I had seen and thought to be the same: d3rd^3 \mathbf{r}.

Basically, is this true? And if not, why not? And what would be true?

RΨ(r,t)2 d3r=zyxΨ(r,t)2 dxdydz\displaystyle \int_R |\Psi (\mathbf{r}, t)|^2 \ d^3 \mathbf{r} = \int_{z} \int_{y} \int_{x} |\Psi(\mathbf{r}, t)|^2 \ dx dy dz
(edited 10 years ago)
Reply 1
Original post by Astronomical
Hello TSR,

I am told that the volume differential dV=dx dy dzdV = dx\ dy\ dz is not the same as another differential I had seen and thought to be the same: d3rd^3 \mathbf{r}.

Basically, is this true? And if not, why not? And what would be true?

RΨ(r,t) d3r=zyxΨ(r,t) dxdydz\displaystyle \int_R |\Psi (\mathbf{r}, t)| \ d^3 \mathbf{r} = \int_{z} \int_{y} \int_{x} |\Psi(\mathbf{r}, t)| \ dx dy dz


I guess it may make people uncomfortable to see  d3r\ d^3 \mathbf{r} referred to as a volume differential/element as you need to scale it by the modulus of the Jacobian determinant for the transformation between the two coord systems

For example, whereas dxdydzdx dy dz is a good volume element in the usual Cartesian coords, drdϕdθdr d\phi d\theta is not a good volume element in spherical polars, even though the spherical polar coord system is orthogonal.
Reply 2
Original post by atsruser
I guess it may make people uncomfortable to see  d3r\ d^3 \mathbf{r} referred to as a volume differential/element as you need to scale it by the modulus of the Jacobian determinant for the transformation between the two coord systems

For example, whereas dxdydzdx dy dz is a good volume element in the usual Cartesian coords, drdϕdθdr d\phi d\theta is not a good volume element in spherical polars, even though the spherical polar coord system is orthogonal.


So if we had, say, Ψ(x,y,z,t)\Psi (x,y,z,t) would it be true?

So you're saying that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 where the subscripts denote arbitrary orthogonal coordinates?

Isn't that just dVdV, which is why I am confused.
Reply 3
Original post by Astronomical
So if we had, say, Ψ(x,y,z,t)\Psi (x,y,z,t) would it be true?

So you're saying that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 where the subscripts denote arbitrary orthogonal coordinates?

Isn't that just dVdV, which is why I am confused.


No, I'm saying that if someone thinks that d3r=dx1dx2dx3d^3\mathbf{r} = dx_1 dx_2 dx_3, then they may not be happy to refer to it as a volume element, since it is only a volume element in Cartesian coordinates.

If someone thinks that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 then I suspect that they would also think that it makes perfect sense to refer to it as a volume element.

It's a matter of notation, and I don't seem to have any books that use this specific notation, so I can't check how it is usually defined. Your usage of it in the integral with what I guess is a wave function (missing a square?) looks familiar to me however, and if I saw that written down without any other explanation, then I would have no problem in assuming that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 in that context, since that would be required for the integral to make sense.

See this usage of the notation, for example.
Reply 4
Original post by atsruser
No, I'm saying that if someone thinks that d3r=dx1dx2dx3d^3\mathbf{r} = dx_1 dx_2 dx_3, then they may not be happy to refer to it as a volume element, since it is only a volume element in Cartesian coordinates.

If someone thinks that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 then I suspect that they would also think that it makes perfect sense to refer to it as a volume element.

It's a matter of notation, and I don't seem to have any books that use this specific notation, so I can't check how it is usually defined. Your usage of it in the integral with what I guess is a wave function (missing a square?) looks familiar to me however, and if I saw that written down without any other explanation, then I would have no problem in assuming that d3r=Jdx1dx2dx3d^3\mathbf{r} = |J|dx_1 dx_2 dx_3 in that context, since that would be required for the integral to make sense.

See this usage of the notation, for example.


Indeed, I missed the square! Oops!

I'm glad to hear you believe it to be a notational thing, as that's what I'd thought as well. The lecturer who said it was more than just a notational thing has a habit of being wrong, so I wouldn't be surprised if he was wrong on this as well. +repped!
Reply 5
Original post by Astronomical
Indeed, I missed the square! Oops!

I'm glad to hear you believe it to be a notational thing, as that's what I'd thought as well. The lecturer who said it was more than just a notational thing has a habit of being wrong, so I wouldn't be surprised if he was wrong on this as well. +repped!


I don't really understand what "more than just a notational thing" means here.

I think that you may be misinterpreting what I said. My response was trying to suggest that there are two "obvious" interpretations of the notation, and I'm not sure which, if either, is standard. You really need to dig out a few text books and see how it tends to be used if you want to be sure.
Reply 6
Original post by atsruser
I don't really understand what "more than just a notational thing" means here.

I think that you may be misinterpreting what I said. My response was trying to suggest that there are two "obvious" interpretations of the notation, and I'm not sure which, if either, is standard. You really need to dig out a few text books and see how it tends to be used if you want to be sure.


I've only seen it used by our quantum physics lecturer when doing the integrals I mentioned above, and like you said it only makes sense in that scenario if it's effectively a volume element. Right?

Looking around online, I've just seen d3xd^3 x used as well doing a divergence integral; it seems that it is just notational.
Reply 7
Original post by Astronomical
I've only seen it used by our quantum physics lecturer when doing the integrals I mentioned above, and like you said it only makes sense in that scenario if it's effectively a volume element. Right?


Right. The only sensible interpretation in my mind is that d3rd^3{\bold{r}} is a volume element, and is equivalent to dVdV. In that case, you are mentally supposed to think of it with the Jacobian, and the correct coords, but since it's too much effort to write that out each time, and it looks messier, then this notation is used. I think.


Looking around online, I've just seen d3xd^3 x used as well doing a divergence integral; it seems that it is just notational.


If I knew what you meant by "just notational" then I could agree or disagree.
Reply 8
Original post by atsruser
Right. The only sensible interpretation in my mind is that d3rd^3{\bold{r}} is a volume element, and is equivalent to dVdV. In that case, you are mentally supposed to think of it with the Jacobian, and the correct coords, but since it's too much effort to write that out each time, and it looks messier, then this notation is used. I think.



If I knew what you meant by "just notational" then I could agree or disagree.


By just notational I merely mean that it's another way of writing dVdV and not some profound new beast.

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