Mixing
Laplacian as limit of Integral Identity
$begingroup$
Let $psi(vec{r})$ be a scalar field,
show that:
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega psi(vec{r}')-psi(vec{r})dOmega'$$
where $rho=|vec{r}-vec{r'}|$, $Omega$ is a sphere of radius $rho$ with it's center at $vec{r}$ (and $dOmega'$ symbolises the solid angle)
what I have tried it a Taylor expansion of $psi(vec{r}')$ around $vec{r}$
that is:
$$psi(vec{r}')=psi(vec{r})+nablapsi(vec{r})cdot (vec{r}-vec{r}')+..$$
which gives me (because the limith should kill all the higher terms in the expansion, I think..)
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
now I know that divergence is defined as
$$nabla cdot vec{F}=lim_{rho to 0} frac{1}{4pi rho^2} int_Omega vec{F}cdot vec{dA}$$
so I get:
$$lim_{rho to 0} frac{1}{4pi rho^2} int_Omeganablapsi(vec{r})cdot vec{dA}=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
I just don't know how to show that these two limits are the same.
can someone help (if what I have done so far is correct of course)
limits multivariable-calculus vector-analysis vector-fields electromagnetism
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
$endgroup$
add a comment |
$begingroup$
Let $psi(vec{r})$ be a scalar field,
show that:
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega psi(vec{r}')-psi(vec{r})dOmega'$$
where $rho=|vec{r}-vec{r'}|$, $Omega$ is a sphere of radius $rho$ with it's center at $vec{r}$ (and $dOmega'$ symbolises the solid angle)
what I have tried it a Taylor expansion of $psi(vec{r}')$ around $vec{r}$
that is:
$$psi(vec{r}')=psi(vec{r})+nablapsi(vec{r})cdot (vec{r}-vec{r}')+..$$
which gives me (because the limith should kill all the higher terms in the expansion, I think..)
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
now I know that divergence is defined as
$$nabla cdot vec{F}=lim_{rho to 0} frac{1}{4pi rho^2} int_Omega vec{F}cdot vec{dA}$$
so I get:
$$lim_{rho to 0} frac{1}{4pi rho^2} int_Omeganablapsi(vec{r})cdot vec{dA}=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
I just don't know how to show that these two limits are the same.
can someone help (if what I have done so far is correct of course)
limits multivariable-calculus vector-analysis vector-fields electromagnetism
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
$endgroup$
$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06
add a comment |
$begingroup$
Let $psi(vec{r})$ be a scalar field,
show that:
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega psi(vec{r}')-psi(vec{r})dOmega'$$
where $rho=|vec{r}-vec{r'}|$, $Omega$ is a sphere of radius $rho$ with it's center at $vec{r}$ (and $dOmega'$ symbolises the solid angle)
what I have tried it a Taylor expansion of $psi(vec{r}')$ around $vec{r}$
that is:
$$psi(vec{r}')=psi(vec{r})+nablapsi(vec{r})cdot (vec{r}-vec{r}')+..$$
which gives me (because the limith should kill all the higher terms in the expansion, I think..)
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
now I know that divergence is defined as
$$nabla cdot vec{F}=lim_{rho to 0} frac{1}{4pi rho^2} int_Omega vec{F}cdot vec{dA}$$
so I get:
$$lim_{rho to 0} frac{1}{4pi rho^2} int_Omeganablapsi(vec{r})cdot vec{dA}=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
I just don't know how to show that these two limits are the same.
can someone help (if what I have done so far is correct of course)
limits multivariable-calculus vector-analysis vector-fields electromagnetism
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
$endgroup$
Let $psi(vec{r})$ be a scalar field,
show that:
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega psi(vec{r}')-psi(vec{r})dOmega'$$
where $rho=|vec{r}-vec{r'}|$, $Omega$ is a sphere of radius $rho$ with it's center at $vec{r}$ (and $dOmega'$ symbolises the solid angle)
what I have tried it a Taylor expansion of $psi(vec{r}')$ around $vec{r}$
that is:
$$psi(vec{r}')=psi(vec{r})+nablapsi(vec{r})cdot (vec{r}-vec{r}')+..$$
which gives me (because the limith should kill all the higher terms in the expansion, I think..)
$$nabla^2 psi(vec{r})=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
now I know that divergence is defined as
$$nabla cdot vec{F}=lim_{rho to 0} frac{1}{4pi rho^2} int_Omega vec{F}cdot vec{dA}$$
so I get:
$$lim_{rho to 0} frac{1}{4pi rho^2} int_Omeganablapsi(vec{r})cdot vec{dA}=lim_{rho to 0} frac{3}{pi rho^2} int_Omega nablapsi(vec{r})cdotvec{rho} dOmega'$$
I just don't know how to show that these two limits are the same.
can someone help (if what I have done so far is correct of course)
limits multivariable-calculus vector-analysis vector-fields electromagnetism
limits multivariable-calculus vector-analysis vector-fields electromagnetism
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
asked Jan 4 at 19:28
Alexandar SolženjicinAlexandar Solženjicin
858
858
$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06
add a comment |
$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06
$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06
add a comment |
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$begingroup$
All sorts of confusing things here. Are you talking about a sphere in 3-D? From the mention of solid angle, I deduce that you are. But your formula for divergence is wrong. The factor should be the reciprocal of the volume of the ball; so the factor should be $dfrac3{4pirho^3}$. Now let me try to sort out the rest ...
$endgroup$
– Ted Shifrin
Jan 4 at 22:51
$begingroup$
You're right that the higher order terms go away in the limit. However, the correct factor in the original formula should be $dfrac 3{4pirho^2}$. As I said, you have a $dfrac 3{4pirho^3}$; this gets multiplied by a factor of $1/rho$ because the unit normal is $(vec r-vec r')/rho$. But then the solid angle takes care of a factor of $1/rho^2$, since $dA/rho^2 = dOmega'$.
$endgroup$
– Ted Shifrin
Jan 4 at 23:06