How to show $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$











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0
down vote

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If in ${mathbb R}^3$,



$r=sqrt{x^2+y^2+z^2}text{ and }vec{P}=
dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$



how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?




Edit in response to the answer



It is said in the answer that "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either".



With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin. Why? Why not?







calculus multivariable-calculus vectors vector-fields divergence







share|cite|improve this question




















  • 1




    What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
    – SZN
    Nov 27 at 12:41










  • It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
    – Joe
    Nov 27 at 13:04












  • We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
    – Christian Blatter
    Nov 28 at 13:43

















up vote
0
down vote

favorite












If in ${mathbb R}^3$,



$r=sqrt{x^2+y^2+z^2}text{ and }vec{P}=
dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$



how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?




Edit in response to the answer



It is said in the answer that "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either".



With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin. Why? Why not?







calculus multivariable-calculus vectors vector-fields divergence







share|cite|improve this question




















  • 1




    What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
    – SZN
    Nov 27 at 12:41










  • It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
    – Joe
    Nov 27 at 13:04












  • We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
    – Christian Blatter
    Nov 28 at 13:43















up vote
0
down vote

favorite









up vote
0
down vote

favorite











If in ${mathbb R}^3$,



$r=sqrt{x^2+y^2+z^2}text{ and }vec{P}=
dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$



how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?




Edit in response to the answer



It is said in the answer that "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either".



With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin. Why? Why not?







calculus multivariable-calculus vectors vector-fields divergence







share|cite|improve this question















If in ${mathbb R}^3$,



$r=sqrt{x^2+y^2+z^2}text{ and }vec{P}=
dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$



how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?




Edit in response to the answer



It is said in the answer that "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either".



With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin. Why? Why not?







calculus multivariable-calculus vectors vector-fields divergence




calculus multivariable-calculus vectors vector-fields divergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 12:55

























asked Nov 27 at 12:36









Joe

292113




292113








  • 1




    What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
    – SZN
    Nov 27 at 12:41










  • It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
    – Joe
    Nov 27 at 13:04












  • We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
    – Christian Blatter
    Nov 28 at 13:43
















  • 1




    What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
    – SZN
    Nov 27 at 12:41










  • It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
    – Joe
    Nov 27 at 13:04












  • We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
    – Christian Blatter
    Nov 28 at 13:43










1




1




What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
– SZN
Nov 27 at 12:41




What have you done so far? What is the coordinate expression for $P$? Given a vector field $F$, how do you compute $nablacdot F$?
– SZN
Nov 27 at 12:41












It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
– Joe
Nov 27 at 13:04






It can be very easily seen that $vec{nabla} cdot vec{P}=0$ everywhere except origin. How shall we show $vec{nabla} cdot vec{P}=0$ at origin? This is what I mean by "how can we show that $vec{nabla} cdot vec{P}=0$ over the whole domain of ${mathbb R}^3$?"
– Joe
Nov 27 at 13:04














We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
– Christian Blatter
Nov 28 at 13:43






We had the same question by the same OP yesterday: math.stackexchange.com/questions/3015453/…
– Christian Blatter
Nov 28 at 13:43












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










The problem is that the field $f = 1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either. That being said you could define the field



$$
f_a = frac{1}{r + a} tag{1}
$$



for some constant $a > 0$, and then calculate ${bf P}_a$ as



$$
{bf P}_a = frac{partial}{partial z}f_a ~hat{y} - frac{partial}{partial y}f_a ~hat{z} tag{2}
$$



so that



$$
nabla cdot {bf P}_a = frac{partial^2 }{partial ypartial z}f_a - frac{partial^2 }{partial zpartial y}f_a tag{3}
$$



since $f_a$ has continuous derivatives, you can switch the order, therefore



$$
nabla cdot {bf P}_a = 0 tag{4}
$$



Now take the limit



$$
nabla cdot {bf P} = lim_{ato 0}nabla cdot {bf P}_a = 0
$$






share|cite|improve this answer





















  • "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
    – Joe
    Nov 28 at 12:52













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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










The problem is that the field $f = 1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either. That being said you could define the field



$$
f_a = frac{1}{r + a} tag{1}
$$



for some constant $a > 0$, and then calculate ${bf P}_a$ as



$$
{bf P}_a = frac{partial}{partial z}f_a ~hat{y} - frac{partial}{partial y}f_a ~hat{z} tag{2}
$$



so that



$$
nabla cdot {bf P}_a = frac{partial^2 }{partial ypartial z}f_a - frac{partial^2 }{partial zpartial y}f_a tag{3}
$$



since $f_a$ has continuous derivatives, you can switch the order, therefore



$$
nabla cdot {bf P}_a = 0 tag{4}
$$



Now take the limit



$$
nabla cdot {bf P} = lim_{ato 0}nabla cdot {bf P}_a = 0
$$






share|cite|improve this answer





















  • "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
    – Joe
    Nov 28 at 12:52

















up vote
0
down vote



accepted










The problem is that the field $f = 1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either. That being said you could define the field



$$
f_a = frac{1}{r + a} tag{1}
$$



for some constant $a > 0$, and then calculate ${bf P}_a$ as



$$
{bf P}_a = frac{partial}{partial z}f_a ~hat{y} - frac{partial}{partial y}f_a ~hat{z} tag{2}
$$



so that



$$
nabla cdot {bf P}_a = frac{partial^2 }{partial ypartial z}f_a - frac{partial^2 }{partial zpartial y}f_a tag{3}
$$



since $f_a$ has continuous derivatives, you can switch the order, therefore



$$
nabla cdot {bf P}_a = 0 tag{4}
$$



Now take the limit



$$
nabla cdot {bf P} = lim_{ato 0}nabla cdot {bf P}_a = 0
$$






share|cite|improve this answer





















  • "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
    – Joe
    Nov 28 at 12:52















up vote
0
down vote



accepted







up vote
0
down vote



accepted






The problem is that the field $f = 1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either. That being said you could define the field



$$
f_a = frac{1}{r + a} tag{1}
$$



for some constant $a > 0$, and then calculate ${bf P}_a$ as



$$
{bf P}_a = frac{partial}{partial z}f_a ~hat{y} - frac{partial}{partial y}f_a ~hat{z} tag{2}
$$



so that



$$
nabla cdot {bf P}_a = frac{partial^2 }{partial ypartial z}f_a - frac{partial^2 }{partial zpartial y}f_a tag{3}
$$



since $f_a$ has continuous derivatives, you can switch the order, therefore



$$
nabla cdot {bf P}_a = 0 tag{4}
$$



Now take the limit



$$
nabla cdot {bf P} = lim_{ato 0}nabla cdot {bf P}_a = 0
$$






share|cite|improve this answer












The problem is that the field $f = 1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either. That being said you could define the field



$$
f_a = frac{1}{r + a} tag{1}
$$



for some constant $a > 0$, and then calculate ${bf P}_a$ as



$$
{bf P}_a = frac{partial}{partial z}f_a ~hat{y} - frac{partial}{partial y}f_a ~hat{z} tag{2}
$$



so that



$$
nabla cdot {bf P}_a = frac{partial^2 }{partial ypartial z}f_a - frac{partial^2 }{partial zpartial y}f_a tag{3}
$$



since $f_a$ has continuous derivatives, you can switch the order, therefore



$$
nabla cdot {bf P}_a = 0 tag{4}
$$



Now take the limit



$$
nabla cdot {bf P} = lim_{ato 0}nabla cdot {bf P}_a = 0
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 14:12









caverac

12.7k21027




12.7k21027












  • "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
    – Joe
    Nov 28 at 12:52




















  • "The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
    – Joe
    Nov 28 at 12:52


















"The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
– Joe
Nov 28 at 12:52






"The problem is that the field $f=1/r$ itself is not even defined at the origin, so its derivatives are not going to be defined either". With this information, is it correct to say that $vec{P}=nabla times vec{A}$ everywhere except the origin.
– Joe
Nov 28 at 12:52




















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