Surface area of sphere using Dirac delta
$begingroup$
This question is related to this one.
Suppose I want to calculate the surface area $S(R)$ of a sphere of radius $R$. I can express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |-R) d vec x$$
I would then obtain
$$S(R) = 4 pi int_0^infty delta(r-R) r^2 dr = 4 pi R^2$$
which is the correct result.
But it seems to me that I could equivalently express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |^2-R^2) d vec x$$
which gives
$$S(R) = 4 pi int_0^infty delta(r^2-R^2) r^2 dr $$
From the property of composition of the delta with a function,
$$delta(r^2-R^2)=frac{delta(r-R)+delta(r+R)}{2R}$$
but since $r geq 0$ I only have to consider the positive root, so that
$$S(R) = 4 pi int_0^infty frac{delta(r-R)}{2R} r^2 dr = 2 pi R$$
Why do I get two different results? Is something wrong with the second way of expressing $S(R)$?
integration dirac-delta surface-integrals
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
$endgroup$
add a comment |
$begingroup$
This question is related to this one.
Suppose I want to calculate the surface area $S(R)$ of a sphere of radius $R$. I can express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |-R) d vec x$$
I would then obtain
$$S(R) = 4 pi int_0^infty delta(r-R) r^2 dr = 4 pi R^2$$
which is the correct result.
But it seems to me that I could equivalently express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |^2-R^2) d vec x$$
which gives
$$S(R) = 4 pi int_0^infty delta(r^2-R^2) r^2 dr $$
From the property of composition of the delta with a function,
$$delta(r^2-R^2)=frac{delta(r-R)+delta(r+R)}{2R}$$
but since $r geq 0$ I only have to consider the positive root, so that
$$S(R) = 4 pi int_0^infty frac{delta(r-R)}{2R} r^2 dr = 2 pi R$$
Why do I get two different results? Is something wrong with the second way of expressing $S(R)$?
integration dirac-delta surface-integrals
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
$endgroup$
$begingroup$
What exactly is this question asking that the other one isn't?
$endgroup$
– David Z
Jul 19 '16 at 13:14
$begingroup$
@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
$endgroup$
– valerio
Jul 19 '16 at 13:16
1
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
$endgroup$
– David Z
Jul 19 '16 at 13:19
$begingroup$
@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
$endgroup$
– valerio
Jul 19 '16 at 13:24
1
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
$endgroup$
– reuns
Jul 19 '16 at 13:35
add a comment |
$begingroup$
This question is related to this one.
Suppose I want to calculate the surface area $S(R)$ of a sphere of radius $R$. I can express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |-R) d vec x$$
I would then obtain
$$S(R) = 4 pi int_0^infty delta(r-R) r^2 dr = 4 pi R^2$$
which is the correct result.
But it seems to me that I could equivalently express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |^2-R^2) d vec x$$
which gives
$$S(R) = 4 pi int_0^infty delta(r^2-R^2) r^2 dr $$
From the property of composition of the delta with a function,
$$delta(r^2-R^2)=frac{delta(r-R)+delta(r+R)}{2R}$$
but since $r geq 0$ I only have to consider the positive root, so that
$$S(R) = 4 pi int_0^infty frac{delta(r-R)}{2R} r^2 dr = 2 pi R$$
Why do I get two different results? Is something wrong with the second way of expressing $S(R)$?
integration dirac-delta surface-integrals
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
$endgroup$
This question is related to this one.
Suppose I want to calculate the surface area $S(R)$ of a sphere of radius $R$. I can express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |-R) d vec x$$
I would then obtain
$$S(R) = 4 pi int_0^infty delta(r-R) r^2 dr = 4 pi R^2$$
which is the correct result.
But it seems to me that I could equivalently express $S(R)$ as
$$S(R)=int_{mathbb{R}^3} delta (| vec x |^2-R^2) d vec x$$
which gives
$$S(R) = 4 pi int_0^infty delta(r^2-R^2) r^2 dr $$
From the property of composition of the delta with a function,
$$delta(r^2-R^2)=frac{delta(r-R)+delta(r+R)}{2R}$$
but since $r geq 0$ I only have to consider the positive root, so that
$$S(R) = 4 pi int_0^infty frac{delta(r-R)}{2R} r^2 dr = 2 pi R$$
Why do I get two different results? Is something wrong with the second way of expressing $S(R)$?
integration dirac-delta surface-integrals
integration dirac-delta surface-integrals
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Jul 19 '16 at 12:55
valeriovalerio
462210
asked Jul 19 '16 at 12:55
valeriovalerio
462210
asked Jul 19 '16 at 12:55
valeriovalerio
462210
462210
$begingroup$
What exactly is this question asking that the other one isn't?
$endgroup$
– David Z
Jul 19 '16 at 13:14
$begingroup$
@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
$endgroup$
– valerio
Jul 19 '16 at 13:16
1
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
$endgroup$
– David Z
Jul 19 '16 at 13:19
$begingroup$
@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
$endgroup$
– valerio
Jul 19 '16 at 13:24
1
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
$endgroup$
– reuns
Jul 19 '16 at 13:35
add a comment |
$begingroup$
What exactly is this question asking that the other one isn't?
$endgroup$
– David Z
Jul 19 '16 at 13:14
$begingroup$
@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
$endgroup$
– valerio
Jul 19 '16 at 13:16
1
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
$endgroup$
– David Z
Jul 19 '16 at 13:19
$begingroup$
@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
$endgroup$
– valerio
Jul 19 '16 at 13:24
1
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
$endgroup$
– reuns
Jul 19 '16 at 13:35
$begingroup$
What exactly is this question asking that the other one isn't?
$endgroup$
– David Z
Jul 19 '16 at 13:14
$begingroup$
What exactly is this question asking that the other one isn't?
$endgroup$
– David Z
Jul 19 '16 at 13:14
$begingroup$
@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
$endgroup$
– valerio
Jul 19 '16 at 13:16
$begingroup$
@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
$endgroup$
– valerio
Jul 19 '16 at 13:16
1
1
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
$endgroup$
– David Z
Jul 19 '16 at 13:19
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
$endgroup$
– David Z
Jul 19 '16 at 13:19
$begingroup$
@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
$endgroup$
– valerio
Jul 19 '16 at 13:24
$begingroup$
@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
$endgroup$
– valerio
Jul 19 '16 at 13:24
1
1
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
$endgroup$
– reuns
Jul 19 '16 at 13:35
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
$endgroup$
– reuns
Jul 19 '16 at 13:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your result is actually correct even if it may not seem intuitive at first glance. I had a very similar problem here: Compute area of a sphere through a Dirac delta
The key part is that:
$$int_{R^n}{f(x)delta(g(x))|nabla g(x)|,dx} = int_{R^n}{f(x)delta_S(x),dx} = int_{S}{f(x),dsigma(x)}$$
That means that your parametrisation function $g(x)$ is actually very important. If you want to get $4pirho^2$ you will have to multiply your delta function with $|nabla g(r)| = 2r$, $delta_S(r) = 2rdelta(r^2-rho^2)$. Then those would cancel out, and you will get:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{delta(r^2-rho^2)2r^3sintheta,dr},dtheta},dphi} = int_{0}^{2pi}{int_{0}^{pi}{frac{2rho^3sintheta}{2rho},dtheta},dphi} = 4pirho^2$$
The key thing to understand is that you cannot simply substitute if your delta is of the form $delta(g(r))$ where $g(r)$ is not the identity plus/minus a shift.
Here's a nice and intuitive explanation of the problem: https://www.mathpages.com/home/kmath663/kmath663.htm
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
$endgroup$
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
add a comment |
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1 Answer
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oldest
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$begingroup$
Your result is actually correct even if it may not seem intuitive at first glance. I had a very similar problem here: Compute area of a sphere through a Dirac delta
The key part is that:
$$int_{R^n}{f(x)delta(g(x))|nabla g(x)|,dx} = int_{R^n}{f(x)delta_S(x),dx} = int_{S}{f(x),dsigma(x)}$$
That means that your parametrisation function $g(x)$ is actually very important. If you want to get $4pirho^2$ you will have to multiply your delta function with $|nabla g(r)| = 2r$, $delta_S(r) = 2rdelta(r^2-rho^2)$. Then those would cancel out, and you will get:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{delta(r^2-rho^2)2r^3sintheta,dr},dtheta},dphi} = int_{0}^{2pi}{int_{0}^{pi}{frac{2rho^3sintheta}{2rho},dtheta},dphi} = 4pirho^2$$
The key thing to understand is that you cannot simply substitute if your delta is of the form $delta(g(r))$ where $g(r)$ is not the identity plus/minus a shift.
Here's a nice and intuitive explanation of the problem: https://www.mathpages.com/home/kmath663/kmath663.htm
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
$endgroup$
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
add a comment |
$begingroup$
Your result is actually correct even if it may not seem intuitive at first glance. I had a very similar problem here: Compute area of a sphere through a Dirac delta
The key part is that:
$$int_{R^n}{f(x)delta(g(x))|nabla g(x)|,dx} = int_{R^n}{f(x)delta_S(x),dx} = int_{S}{f(x),dsigma(x)}$$
That means that your parametrisation function $g(x)$ is actually very important. If you want to get $4pirho^2$ you will have to multiply your delta function with $|nabla g(r)| = 2r$, $delta_S(r) = 2rdelta(r^2-rho^2)$. Then those would cancel out, and you will get:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{delta(r^2-rho^2)2r^3sintheta,dr},dtheta},dphi} = int_{0}^{2pi}{int_{0}^{pi}{frac{2rho^3sintheta}{2rho},dtheta},dphi} = 4pirho^2$$
The key thing to understand is that you cannot simply substitute if your delta is of the form $delta(g(r))$ where $g(r)$ is not the identity plus/minus a shift.
Here's a nice and intuitive explanation of the problem: https://www.mathpages.com/home/kmath663/kmath663.htm
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
$endgroup$
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
add a comment |
$begingroup$
Your result is actually correct even if it may not seem intuitive at first glance. I had a very similar problem here: Compute area of a sphere through a Dirac delta
The key part is that:
$$int_{R^n}{f(x)delta(g(x))|nabla g(x)|,dx} = int_{R^n}{f(x)delta_S(x),dx} = int_{S}{f(x),dsigma(x)}$$
That means that your parametrisation function $g(x)$ is actually very important. If you want to get $4pirho^2$ you will have to multiply your delta function with $|nabla g(r)| = 2r$, $delta_S(r) = 2rdelta(r^2-rho^2)$. Then those would cancel out, and you will get:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{delta(r^2-rho^2)2r^3sintheta,dr},dtheta},dphi} = int_{0}^{2pi}{int_{0}^{pi}{frac{2rho^3sintheta}{2rho},dtheta},dphi} = 4pirho^2$$
The key thing to understand is that you cannot simply substitute if your delta is of the form $delta(g(r))$ where $g(r)$ is not the identity plus/minus a shift.
Here's a nice and intuitive explanation of the problem: https://www.mathpages.com/home/kmath663/kmath663.htm
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
$endgroup$
Your result is actually correct even if it may not seem intuitive at first glance. I had a very similar problem here: Compute area of a sphere through a Dirac delta
The key part is that:
$$int_{R^n}{f(x)delta(g(x))|nabla g(x)|,dx} = int_{R^n}{f(x)delta_S(x),dx} = int_{S}{f(x),dsigma(x)}$$
That means that your parametrisation function $g(x)$ is actually very important. If you want to get $4pirho^2$ you will have to multiply your delta function with $|nabla g(r)| = 2r$, $delta_S(r) = 2rdelta(r^2-rho^2)$. Then those would cancel out, and you will get:
$$int_{0}^{2pi}{int_{0}^{pi}{int_{0}^{infty}{delta(r^2-rho^2)2r^3sintheta,dr},dtheta},dphi} = int_{0}^{2pi}{int_{0}^{pi}{frac{2rho^3sintheta}{2rho},dtheta},dphi} = 4pirho^2$$
The key thing to understand is that you cannot simply substitute if your delta is of the form $delta(g(r))$ where $g(r)$ is not the identity plus/minus a shift.
Here's a nice and intuitive explanation of the problem: https://www.mathpages.com/home/kmath663/kmath663.htm
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
edited Jan 25 at 21:59
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
answered Jan 25 at 1:36
lightxbulblightxbulb
1,115311
1,115311
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
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– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
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– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
add a comment |
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
It looks like my question is basically a duplicate of yours! I have to admit that it is not completely clear to me why the first relation should hold, even if I have (maybe too quickly) read the extensive explanations on the linked answer. However, since I am a physicist, maybe I can bring the point of view of a physicist: it's clear that there is something wrong with my attempt, because the delta function has the physical dimensions of the inverse of its argument. Therefore, it's clear that integrating $delta(r-R)$ and $delta(r^2-R^2)$ cannot physically give you the same result.
$endgroup$
– valerio
Jan 25 at 9:54
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
@valerio The first relation is a result from geometric measure theory known as the coarea formula (just applied to the Dirac delta in this case, you can check the linked threads and references in my original post to see derivations for the Dirac delta) which lets you integrate over the level sets of a function (exactly what you want since $g^{-1}(0)$ is a level set). I didn't know that the Dirac delta has the physical dimensions of its argument (if you have references do link them since I am very interested in reading more) but I guess that makes sense and explains why you get $2pirho$.
$endgroup$
– lightxbulb
Jan 25 at 11:01
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
$begingroup$
It has dimensions of the inverse of its argument. It has to be, for it to make physical sense. Here is a Physics SE post about it: physics.stackexchange.com/questions/33760/…. I think the second answer is especially useful to understand why this must be the case (basically it follows from the property $delta(alpha x) = delta(x)/|alpha|$ if you think of $alpha$ as a physical unit, for example "meters").
$endgroup$
– valerio
Jan 25 at 12:58
1
1
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
@valerio I always thought of it from a mathematical point of view, so the physical perspective definitely makes sense. Note that the identity you gave is a special case to the one I presented (the $|a|$ in the denom is due to the derivative). I found a very good and intuitive explanation concerning your problem specifically: mathpages.com/home/kmath663/kmath663.htm give it a read.
$endgroup$
– lightxbulb
Jan 25 at 17:21
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
$begingroup$
Thanks, this page looks really useful. I think you could also add it to the answer (like "see also this page"), since comments could be deleted in the future but that page could be helpful for future visitors ;)
$endgroup$
– valerio
Jan 25 at 21:39
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$begingroup$
What exactly is this question asking that the other one isn't?
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– David Z
Jul 19 '16 at 13:14
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@DavidZ The other question was based on an error in the calculation of $int delta(r^2-R^2) r^2 dr$. In this question I'm asking why two expression that seem to be equivalent to me hold different results. I know that the two questions look the same, but they aren't.
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– valerio
Jul 19 '16 at 13:16
1
$begingroup$
Well, but that seems like the same thing the other question was asking. In the other question, you calculated the same two expressions and obtained the same two different results, and you asked which one is right (and implicitly why). I don't see what more is being asked here.
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– David Z
Jul 19 '16 at 13:19
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@DavidZ No, in the other question I calculated $int delta(r^2-R^2) r^2 dr$ and got two different results. The first was wrong, the second was correct. Now, I am calculating two different integrals which to me should hold the same result. I think I am calculating both of them correctly, and I would like to know why they hold different results because I would expect them to hold the same result. It is very similar but different.
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– valerio
Jul 19 '16 at 13:24
1
$begingroup$
you have to define first $delta(r^2 - R^2)$ (and $delta(|x|- R)$ and $delta(|x|^2- R^2)$). change of variables can work with $delta$ if you are careful and think to $delta(x)$ in $mathbb{R}^3$ as the limit of $f_a(x) = displaystylefrac{1_{ |x| < a}}{4 pi a^3/3}$ as $a to 0^+$
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– reuns
Jul 19 '16 at 13:35