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Mathematical bases for which $q$ and $dotq$ could be treated as independent variable in $L(q,dotq,t)$
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Mathematical bases for which $q$ and $dotq$ could be treated as independent variable in $L(q,dotq,t)$.
In Lagrangian mechanics with single degree of freedome $q(t)$ and it's first degree deritative $dotq(t)$ were considered as independent variable.
$displaystyle fracpartial dotqpartial q=fracpartial qpartial tpartial q=fracpartial qpartial qpartial t=fracpartial Cpartial t=0$ and its general form $displaystyle fracpartial q^(n)partial q=0$ could be easily proven.
However, $displaystyle fracpartial qpartial dotq=0$ was not so straight forward.
My question was that:
Prove the general form $displaystyle fracpartial qpartial q^(n)=0$ in Lagrangian equation $L(q,dotq,t)$ where $displaystylefracpartial Lpartial q=fracddt(fracpartial Lpartial dotq)$ and $q^(n)$ was the $n$ th deritative of $q$.
calculus analysis euler-lagrange-equation
asked Jul 30 at 9:00
J C
399314
 |Â
show 6 more comments
up vote
2
down vote
favorite
Mathematical bases for which $q$ and $dotq$ could be treated as independent variable in $L(q,dotq,t)$.
In Lagrangian mechanics with single degree of freedome $q(t)$ and it's first degree deritative $dotq(t)$ were considered as independent variable.
$displaystyle fracpartial dotqpartial q=fracpartial qpartial tpartial q=fracpartial qpartial qpartial t=fracpartial Cpartial t=0$ and its general form $displaystyle fracpartial q^(n)partial q=0$ could be easily proven.
However, $displaystyle fracpartial qpartial dotq=0$ was not so straight forward.
My question was that:
Prove the general form $displaystyle fracpartial qpartial q^(n)=0$ in Lagrangian equation $L(q,dotq,t)$ where $displaystylefracpartial Lpartial q=fracddt(fracpartial Lpartial dotq)$ and $q^(n)$ was the $n$ th deritative of $q$.
calculus analysis euler-lagrange-equation
asked Jul 30 at 9:00
J C
399314
2
It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
1
In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
1
Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
2
"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
1
Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Mathematical bases for which $q$ and $dotq$ could be treated as independent variable in $L(q,dotq,t)$.
In Lagrangian mechanics with single degree of freedome $q(t)$ and it's first degree deritative $dotq(t)$ were considered as independent variable.
$displaystyle fracpartial dotqpartial q=fracpartial qpartial tpartial q=fracpartial qpartial qpartial t=fracpartial Cpartial t=0$ and its general form $displaystyle fracpartial q^(n)partial q=0$ could be easily proven.
However, $displaystyle fracpartial qpartial dotq=0$ was not so straight forward.
My question was that:
Prove the general form $displaystyle fracpartial qpartial q^(n)=0$ in Lagrangian equation $L(q,dotq,t)$ where $displaystylefracpartial Lpartial q=fracddt(fracpartial Lpartial dotq)$ and $q^(n)$ was the $n$ th deritative of $q$.
calculus analysis euler-lagrange-equation
asked Jul 30 at 9:00
J C
399314
Mathematical bases for which $q$ and $dotq$ could be treated as independent variable in $L(q,dotq,t)$.
In Lagrangian mechanics with single degree of freedome $q(t)$ and it's first degree deritative $dotq(t)$ were considered as independent variable.
$displaystyle fracpartial dotqpartial q=fracpartial qpartial tpartial q=fracpartial qpartial qpartial t=fracpartial Cpartial t=0$ and its general form $displaystyle fracpartial q^(n)partial q=0$ could be easily proven.
However, $displaystyle fracpartial qpartial dotq=0$ was not so straight forward.
My question was that:
Prove the general form $displaystyle fracpartial qpartial q^(n)=0$ in Lagrangian equation $L(q,dotq,t)$ where $displaystylefracpartial Lpartial q=fracddt(fracpartial Lpartial dotq)$ and $q^(n)$ was the $n$ th deritative of $q$.
calculus analysis euler-lagrange-equation
asked Jul 30 at 9:00
J C
399314
asked Jul 30 at 9:00
J C
399314
asked Jul 30 at 9:00
J C
399314
asked Jul 30 at 9:00
J C
399314
399314
2
It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
1
In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
1
Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
2
"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
1
Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08
 |Â
show 6 more comments
2
It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
1
In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
1
Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
2
"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
1
Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08
2
2
It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
1
1
In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
1
1
Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
2
2
"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
1
1
Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08
Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08
 |Â
show 6 more comments
1 Answer
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As I commented on an answer to a related question, I think part of the problem is conflation of functions with variables.
If variables $x$ and $y$ are related by $y=f(x)$, that doesn't mean that $y$ is the function $f$; the same variable could be a different function of another variable, like $y=f(x)=g(z)$. A function is a relation, while a variable is... (I haven't seen a satisfactory definition.)
If we insist that the Lagrangian is a function, not a variable, then $fracpartial Lpartial q$ is an abuse of notation, and the Lagrangian equation should be
$$L^(1,0,0)(q(t),dot q(t),t) = fracddtL^(0,1,0)(q(t),dot q(t),t)$$
assuming that $t$ is a variable. Note that $L^(1,0,0)$ is a function in itself, regardless of what we call its inputs.
If the Lagrangian is a variable, then the partial derivatives must specify what other variables are held constant for the differentiation, as joriki said.
But how can we "hold constant" a dependent variable? (Not only is $dot q$ dependent on the function $q$, but also the variable $q$ is dependent on $t$.) Perhaps the idea of "dependent/independent variables" doesn't make sense, and should be replaced with "domain and codomain of functions". Other views are welcome.
answered Jul 30 at 10:38
mr_e_man
748219
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As I commented on an answer to a related question, I think part of the problem is conflation of functions with variables.
If variables $x$ and $y$ are related by $y=f(x)$, that doesn't mean that $y$ is the function $f$; the same variable could be a different function of another variable, like $y=f(x)=g(z)$. A function is a relation, while a variable is... (I haven't seen a satisfactory definition.)
If we insist that the Lagrangian is a function, not a variable, then $fracpartial Lpartial q$ is an abuse of notation, and the Lagrangian equation should be
$$L^(1,0,0)(q(t),dot q(t),t) = fracddtL^(0,1,0)(q(t),dot q(t),t)$$
assuming that $t$ is a variable. Note that $L^(1,0,0)$ is a function in itself, regardless of what we call its inputs.
If the Lagrangian is a variable, then the partial derivatives must specify what other variables are held constant for the differentiation, as joriki said.
But how can we "hold constant" a dependent variable? (Not only is $dot q$ dependent on the function $q$, but also the variable $q$ is dependent on $t$.) Perhaps the idea of "dependent/independent variables" doesn't make sense, and should be replaced with "domain and codomain of functions". Other views are welcome.
answered Jul 30 at 10:38
mr_e_man
748219
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
add a comment |Â
up vote
0
down vote
As I commented on an answer to a related question, I think part of the problem is conflation of functions with variables.
If variables $x$ and $y$ are related by $y=f(x)$, that doesn't mean that $y$ is the function $f$; the same variable could be a different function of another variable, like $y=f(x)=g(z)$. A function is a relation, while a variable is... (I haven't seen a satisfactory definition.)
If we insist that the Lagrangian is a function, not a variable, then $fracpartial Lpartial q$ is an abuse of notation, and the Lagrangian equation should be
$$L^(1,0,0)(q(t),dot q(t),t) = fracddtL^(0,1,0)(q(t),dot q(t),t)$$
assuming that $t$ is a variable. Note that $L^(1,0,0)$ is a function in itself, regardless of what we call its inputs.
If the Lagrangian is a variable, then the partial derivatives must specify what other variables are held constant for the differentiation, as joriki said.
But how can we "hold constant" a dependent variable? (Not only is $dot q$ dependent on the function $q$, but also the variable $q$ is dependent on $t$.) Perhaps the idea of "dependent/independent variables" doesn't make sense, and should be replaced with "domain and codomain of functions". Other views are welcome.
answered Jul 30 at 10:38
mr_e_man
748219
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As I commented on an answer to a related question, I think part of the problem is conflation of functions with variables.
If variables $x$ and $y$ are related by $y=f(x)$, that doesn't mean that $y$ is the function $f$; the same variable could be a different function of another variable, like $y=f(x)=g(z)$. A function is a relation, while a variable is... (I haven't seen a satisfactory definition.)
If we insist that the Lagrangian is a function, not a variable, then $fracpartial Lpartial q$ is an abuse of notation, and the Lagrangian equation should be
$$L^(1,0,0)(q(t),dot q(t),t) = fracddtL^(0,1,0)(q(t),dot q(t),t)$$
assuming that $t$ is a variable. Note that $L^(1,0,0)$ is a function in itself, regardless of what we call its inputs.
If the Lagrangian is a variable, then the partial derivatives must specify what other variables are held constant for the differentiation, as joriki said.
But how can we "hold constant" a dependent variable? (Not only is $dot q$ dependent on the function $q$, but also the variable $q$ is dependent on $t$.) Perhaps the idea of "dependent/independent variables" doesn't make sense, and should be replaced with "domain and codomain of functions". Other views are welcome.
answered Jul 30 at 10:38
mr_e_man
748219
As I commented on an answer to a related question, I think part of the problem is conflation of functions with variables.
If variables $x$ and $y$ are related by $y=f(x)$, that doesn't mean that $y$ is the function $f$; the same variable could be a different function of another variable, like $y=f(x)=g(z)$. A function is a relation, while a variable is... (I haven't seen a satisfactory definition.)
If we insist that the Lagrangian is a function, not a variable, then $fracpartial Lpartial q$ is an abuse of notation, and the Lagrangian equation should be
$$L^(1,0,0)(q(t),dot q(t),t) = fracddtL^(0,1,0)(q(t),dot q(t),t)$$
assuming that $t$ is a variable. Note that $L^(1,0,0)$ is a function in itself, regardless of what we call its inputs.
If the Lagrangian is a variable, then the partial derivatives must specify what other variables are held constant for the differentiation, as joriki said.
But how can we "hold constant" a dependent variable? (Not only is $dot q$ dependent on the function $q$, but also the variable $q$ is dependent on $t$.) Perhaps the idea of "dependent/independent variables" doesn't make sense, and should be replaced with "domain and codomain of functions". Other views are welcome.
answered Jul 30 at 10:38
mr_e_man
748219
edited Jul 30 at 12:07
answered Jul 30 at 10:38
mr_e_man
748219
answered Jul 30 at 10:38
mr_e_man
748219
answered Jul 30 at 10:38
mr_e_man
748219
748219
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
add a comment |Â
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
A tricky way for $dotq$ (which I was not sure if it's correct) for Lagrange's equations of the second kind $displaystyle fracddt ( fracpartial qpartial dotq ) = fracddt ( fracpartial qpartial q partial t ) = fracddt ( partial q fracpartial tpartial q ) = fracd q cdot 0dt=0$, so it's simplest to write $q$ and $dotq$ as independent variables to avoid calculations.(No assumption) There is a difference between generalized equations of motion and Lagrange's equations of the second kind. I was wondering if there was a formal proof.
– J C
Jul 30 at 12:37
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
Acturally, the equation before lagrange could be derived from variation as a function of dependent variables, then one might introduce one of the $u_i$ as ($t$) independent variables. Although the "initial assumption" that $q$ and $dotq$ were independent seemed to be doable, but was not an initial assumption for the lagrange's equation. Lagrange's equation of motion actually didn't specify or require $p(t)$ and $dotp(t)$ to have any relation, so it's a mathematical reason.
– J C
Jul 30 at 16:38
add a comment |Â
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It's not clear to me what you want to prove starting from what. In Lagrangian mechanics, we treat $q$ and $dot q$ as independent variables on which $L$ depends; thus $fracpartial qpartialdot q=0$ is true by definition simply because of the way we introduce these variables. Your "calculation" of $fracpartialdot qpartial q$ makes no sense; that derivative is also zero simply by definition.
– joriki
Jul 30 at 10:27
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In case you're confused by the terrible notation we unfortunately use for partial derivatives, it might help to explicitly include all the variables in the notation. Then in Lagrangian mechanics we would usually mean $left.fracpartial q(q,dot q)partialdot qright|_q$ when we write $fracpartial qpartialdot q$ (where the subscript indicates the variable held fixed). In this notational form, it's clear that the derivative vanishes. If you mean something else by $fracpartial qpartialdot q$, I'd suggest that you write it in this form to clarify what you mean.
– joriki
Jul 30 at 10:31
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Did you read my comments above? None of that makes any sense; I suspect you'll notice that yourself if you try to transcribe it into the more explicit notation I suggested above.
– joriki
Jul 30 at 12:38
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"I was wondering if there was a formal proof." For a formal proof, we need a formal question. So far you haven't stated anything about what functions and variables you're considering to depend on each other in which ways. The only thing we can formally say so far is how things are usually done in Lagrangian mechanics. This I stated above, and there's nothing more to say since it's just a matter of definitions. If you want to operate in a different picture, you'll have to formally define that picture before we can start providing formal proofs in it.
– joriki
Jul 30 at 12:40
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Related: math.stackexchange.com/questions/1963640/…, math.stackexchange.com/questions/580858/…, physics.stackexchange.com/questions/885/…
– Hans Lundmark
Jul 30 at 13:08