Mixing
Central Force Fields
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Why does it suffice to show that $V$ is constant on each sphere if we are trying to show that $F(x) = - text{grad} V(x)$ and $V(x) = g(|x|)$ implies that F is central?
Wouldn't a constant $V$ simply mean that $F$ is 0? That would satisfy the condition for being central ($F(x) = lambda(x) x)$ if $lambda = 0$, but what about when $lambda neq 0$?
vector-analysis
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
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add a comment |
$begingroup$
Why does it suffice to show that $V$ is constant on each sphere if we are trying to show that $F(x) = - text{grad} V(x)$ and $V(x) = g(|x|)$ implies that F is central?
Wouldn't a constant $V$ simply mean that $F$ is 0? That would satisfy the condition for being central ($F(x) = lambda(x) x)$ if $lambda = 0$, but what about when $lambda neq 0$?
vector-analysis
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
$endgroup$
add a comment |
$begingroup$
Why does it suffice to show that $V$ is constant on each sphere if we are trying to show that $F(x) = - text{grad} V(x)$ and $V(x) = g(|x|)$ implies that F is central?
Wouldn't a constant $V$ simply mean that $F$ is 0? That would satisfy the condition for being central ($F(x) = lambda(x) x)$ if $lambda = 0$, but what about when $lambda neq 0$?
vector-analysis
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
$endgroup$
Why does it suffice to show that $V$ is constant on each sphere if we are trying to show that $F(x) = - text{grad} V(x)$ and $V(x) = g(|x|)$ implies that F is central?
Wouldn't a constant $V$ simply mean that $F$ is 0? That would satisfy the condition for being central ($F(x) = lambda(x) x)$ if $lambda = 0$, but what about when $lambda neq 0$?
vector-analysis
vector-analysis
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
asked Jan 21 at 20:52
Tarang SalujaTarang Saluja
261
261
add a comment |
add a comment |
1 Answer
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If the level sets of $V$ are spheres centered at the origin, then $text{grad }V(x)$ is normal to the sphere at $x$, hence a scalar multiple of $x$. (The equation $V(x)=g(|x|)$ says, of course, that $V$ is constant on spheres $|x|=text{constant}$.)
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
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$begingroup$
If the level sets of $V$ are spheres centered at the origin, then $text{grad }V(x)$ is normal to the sphere at $x$, hence a scalar multiple of $x$. (The equation $V(x)=g(|x|)$ says, of course, that $V$ is constant on spheres $|x|=text{constant}$.)
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
add a comment |
$begingroup$
If the level sets of $V$ are spheres centered at the origin, then $text{grad }V(x)$ is normal to the sphere at $x$, hence a scalar multiple of $x$. (The equation $V(x)=g(|x|)$ says, of course, that $V$ is constant on spheres $|x|=text{constant}$.)
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
add a comment |
$begingroup$
If the level sets of $V$ are spheres centered at the origin, then $text{grad }V(x)$ is normal to the sphere at $x$, hence a scalar multiple of $x$. (The equation $V(x)=g(|x|)$ says, of course, that $V$ is constant on spheres $|x|=text{constant}$.)
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
If the level sets of $V$ are spheres centered at the origin, then $text{grad }V(x)$ is normal to the sphere at $x$, hence a scalar multiple of $x$. (The equation $V(x)=g(|x|)$ says, of course, that $V$ is constant on spheres $|x|=text{constant}$.)
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 21 at 20:56
Ted ShifrinTed Shifrin
64.2k44692
64.2k44692
add a comment |
add a comment |
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