Mixing
Why $L(x,u,p)=sqrt{frac{1+p^2}{u}}$ does not depend on $x$, even when $u=u(x)$?
$begingroup$
I'm reading an article which writes that a Lagrangian:
$L(x,u,p)=sqrt{frac{1+p^2}{u}}$ does not depend on $x$, even when $u$ is function of $x$.
So how does it not depend on $x$? Is it some increase in abstraction (thinking that it could work for any kind of similar $u$ so the specific nature of $x$ doesn't matter)?
http://www-users.math.umn.edu/~olver/ln_/cv.pdf, p. 15-16.
functions euler-lagrange-equation
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
$endgroup$
|
show 2 more comments
$begingroup$
I'm reading an article which writes that a Lagrangian:
$L(x,u,p)=sqrt{frac{1+p^2}{u}}$ does not depend on $x$, even when $u$ is function of $x$.
So how does it not depend on $x$? Is it some increase in abstraction (thinking that it could work for any kind of similar $u$ so the specific nature of $x$ doesn't matter)?
http://www-users.math.umn.edu/~olver/ln_/cv.pdf, p. 15-16.
functions euler-lagrange-equation
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
$endgroup$
$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18
|
show 2 more comments
$begingroup$
I'm reading an article which writes that a Lagrangian:
$L(x,u,p)=sqrt{frac{1+p^2}{u}}$ does not depend on $x$, even when $u$ is function of $x$.
So how does it not depend on $x$? Is it some increase in abstraction (thinking that it could work for any kind of similar $u$ so the specific nature of $x$ doesn't matter)?
http://www-users.math.umn.edu/~olver/ln_/cv.pdf, p. 15-16.
functions euler-lagrange-equation
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
$endgroup$
I'm reading an article which writes that a Lagrangian:
$L(x,u,p)=sqrt{frac{1+p^2}{u}}$ does not depend on $x$, even when $u$ is function of $x$.
So how does it not depend on $x$? Is it some increase in abstraction (thinking that it could work for any kind of similar $u$ so the specific nature of $x$ doesn't matter)?
http://www-users.math.umn.edu/~olver/ln_/cv.pdf, p. 15-16.
functions euler-lagrange-equation
functions euler-lagrange-equation
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
edited Jan 20 at 17:24
mavavilj
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
asked Jan 20 at 17:04
mavaviljmavavilj
2,81411137
2,81411137
$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18
|
show 2 more comments
$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18
$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18
|
show 2 more comments
1 Answer
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Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.
answered Jan 20 at 17:13
GReyesGReyes
1,71515
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$begingroup$
Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.
answered Jan 20 at 17:13
GReyesGReyes
1,71515
$endgroup$
add a comment |
$begingroup$
Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.
answered Jan 20 at 17:13
GReyesGReyes
1,71515
$endgroup$
add a comment |
$begingroup$
Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.
answered Jan 20 at 17:13
GReyesGReyes
1,71515
$endgroup$
Formally, your Lagrangian does not depend on $x$. It is only by solving the variational problem that you get the dependence of $u$ on $x$ required to minimize the action along all possible trajectories $u(x)$.
answered Jan 20 at 17:13
GReyesGReyes
1,71515
answered Jan 20 at 17:13
GReyesGReyes
1,71515
answered Jan 20 at 17:13
GReyesGReyes
1,71515
answered Jan 20 at 17:13
GReyesGReyes
1,71515
1,71515
add a comment |
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$begingroup$
If $L$ is dependent on $u$ which is dependent on $x$ then $L$ must be dependent on $x$ surely?
$endgroup$
– Henry Lee
Jan 20 at 17:13
$begingroup$
They are simply making the remark that $L(x,u,p)=ell(u,p)$ for some given function $ell$.
$endgroup$
– Did
Jan 20 at 17:13
$begingroup$
@Did So it's an increase in abstraction? That for that result the $x$ doesn't matter.
$endgroup$
– mavavilj
Jan 20 at 17:15
$begingroup$
"So it's an increase in abstraction?" Yeah, I tried to avoid this one... because what do you mean by an increase in abstraction here?
$endgroup$
– Did
Jan 20 at 17:16
$begingroup$
@Did That one formulates the result by thinking that "this will work for any $u,p$, regardless of what the $x$ inside $u(x)$ does". So $u$ is an abstraction on $u(x)$ where one doesn't consider $x$ or $u(x)$, but only $u$ as a function.
$endgroup$
– mavavilj
Jan 20 at 17:18