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Why isn't the Lagrange equation trivial?
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The Lagrange Equation is at the heart of Lagrangian mechanics, which generate the equations of motion of a system for a given Lagrangian. Let $q^alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time, and $L$ represent the Lagrangian. The Lagrangian is a function of the state of a particle, which is the collection of the particle's position $q^alpha$ and velocity $dot q^alpha$ values. The Lagrange Equation then reads as follows.
$$ fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0$$
My question is this: why is this a fundamental law of physics and not a simple triviality of ANY function $L$ on the variables $q^alpha$ and $dot q^alpha$? The following "proof" (titled below as "THE 'PROOF'") of the Lagrange Equation takes into account no notions of physical properties of moving objects and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.
Obviously, this is not the case. For, if it were, nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning below?
THE "PROOF"
$$beginalign
fracddt fracpartial Lpartial dot q^alpha & = fracpartialpartial dot q^alpha fracdLdt &textcommutativity of derivatives \ \
&= fracpartial dot Lpartial dot q^alpha \ \
&= fracpartial Lpartial q^alpha & textcancellation of dots \ \
&Rightarrow fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0
endalign$$
classical-mechanics lagrangian-formalism
asked 6 hours ago
Trevor Kafka
644
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The Lagrange Equation is at the heart of Lagrangian mechanics, which generate the equations of motion of a system for a given Lagrangian. Let $q^alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time, and $L$ represent the Lagrangian. The Lagrangian is a function of the state of a particle, which is the collection of the particle's position $q^alpha$ and velocity $dot q^alpha$ values. The Lagrange Equation then reads as follows.
$$ fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0$$
My question is this: why is this a fundamental law of physics and not a simple triviality of ANY function $L$ on the variables $q^alpha$ and $dot q^alpha$? The following "proof" (titled below as "THE 'PROOF'") of the Lagrange Equation takes into account no notions of physical properties of moving objects and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.
Obviously, this is not the case. For, if it were, nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning below?
THE "PROOF"
$$beginalign
fracddt fracpartial Lpartial dot q^alpha & = fracpartialpartial dot q^alpha fracdLdt &textcommutativity of derivatives \ \
&= fracpartial dot Lpartial dot q^alpha \ \
&= fracpartial Lpartial q^alpha & textcancellation of dots \ \
&Rightarrow fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0
endalign$$
classical-mechanics lagrangian-formalism
asked 6 hours ago
Trevor Kafka
644
7
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
2
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago
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4
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up vote
4
down vote
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The Lagrange Equation is at the heart of Lagrangian mechanics, which generate the equations of motion of a system for a given Lagrangian. Let $q^alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time, and $L$ represent the Lagrangian. The Lagrangian is a function of the state of a particle, which is the collection of the particle's position $q^alpha$ and velocity $dot q^alpha$ values. The Lagrange Equation then reads as follows.
$$ fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0$$
My question is this: why is this a fundamental law of physics and not a simple triviality of ANY function $L$ on the variables $q^alpha$ and $dot q^alpha$? The following "proof" (titled below as "THE 'PROOF'") of the Lagrange Equation takes into account no notions of physical properties of moving objects and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.
Obviously, this is not the case. For, if it were, nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning below?
THE "PROOF"
$$beginalign
fracddt fracpartial Lpartial dot q^alpha & = fracpartialpartial dot q^alpha fracdLdt &textcommutativity of derivatives \ \
&= fracpartial dot Lpartial dot q^alpha \ \
&= fracpartial Lpartial q^alpha & textcancellation of dots \ \
&Rightarrow fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0
endalign$$
classical-mechanics lagrangian-formalism
asked 6 hours ago
Trevor Kafka
644
The Lagrange Equation is at the heart of Lagrangian mechanics, which generate the equations of motion of a system for a given Lagrangian. Let $q^alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time, and $L$ represent the Lagrangian. The Lagrangian is a function of the state of a particle, which is the collection of the particle's position $q^alpha$ and velocity $dot q^alpha$ values. The Lagrange Equation then reads as follows.
$$ fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0$$
My question is this: why is this a fundamental law of physics and not a simple triviality of ANY function $L$ on the variables $q^alpha$ and $dot q^alpha$? The following "proof" (titled below as "THE 'PROOF'") of the Lagrange Equation takes into account no notions of physical properties of moving objects and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.
Obviously, this is not the case. For, if it were, nobody would give a hoot about this equation and it would be totally useless to solve any problem. What is wrong with the logical reasoning below?
THE "PROOF"
$$beginalign
fracddt fracpartial Lpartial dot q^alpha & = fracpartialpartial dot q^alpha fracdLdt &textcommutativity of derivatives \ \
&= fracpartial dot Lpartial dot q^alpha \ \
&= fracpartial Lpartial q^alpha & textcancellation of dots \ \
&Rightarrow fracddt fracpartial Lpartial dot q^alpha - fracpartial Lpartial q^alpha = 0
endalign$$
classical-mechanics lagrangian-formalism
classical-mechanics lagrangian-formalism
asked 6 hours ago
Trevor Kafka
644
asked 6 hours ago
Trevor Kafka
644
asked 6 hours ago
Trevor Kafka
644
asked 6 hours ago
Trevor Kafka
644
asked 6 hours ago
Trevor Kafka
644
644
7
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
2
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago
add a comment |Â
7
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
2
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago
7
7
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
2
2
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago
add a comment |Â
4 Answers
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Ah, what a tricky mistake you've made there. The problem is that you've simply confused some confusions notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle in moving in 1 dimension in an external potential energy $V(q)$ is
$$
L(q, dot q) = frac12m dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clear $q$ and $dot q$ are not independent variables. Once $q$ is specified for all times, $dot q$ is also specified for all times.
A better way to write the above Lagrangian might be
$$
L(a, b) = frac12m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
fracpartial Lpartial a = -V'(a) hspace1cm fracpartial Lpartial b = m b.
$$
Usually, most people write this as
$$
fracpartial Lpartial q = -V'(q) hspace1cm fracpartial Lpartial dot q = m dot q.
$$
However, $q$ and $dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $dot q$ are too when they're being put into the Lagrangian. In otherwords, we could put any two numbers into $L$, we just decided to put in $q$ and $dot q$.
Furthermore, let's look at the total time derivative $fracddt$. How should we understand the following expression?
$$
fracddt L(q, dot q)
$$
Both $q$ and $dot q$ are functions of time. Therefore, $L(q, dot q)$ depends on time simply because $q$ and $dot q$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
fracddt L(q, dot q) = fracdqdt fracpartial Lpartial a(q, dot q) + fracd dot qdt fracpartial Lpartial b(q, dot q) = dot q fracpartial Lpartial a(q, dot q) + ddot q fracpartial Lpartial b(q, dot q)
$$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $partial L / partial a$ and $partial L / partial b$ by plugging in $(q, dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
answered 2 hours ago
user1379857
992515
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So, in principle one can choose essentially $itany$ lagrangian $mathcalL$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that dont correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $dotq$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $mathcalL$. But nowadays, almost everyone that works with GR or modified gravity start from $mathcalL$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.
answered 4 hours ago
N. Steinle
4666
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The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~=~fracpartialhphantomq^jpartial q^j$$ is not zero. See also e.g. this related Math.SE post.
The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^j$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But the Lagrangian typically depends on velocities.
Fact:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivative$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE post.
answered 1 hour ago
Qmechanic♦
96.8k121631028
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That's an interesting sequence of symbolic manipulations!
It's because of the lack of rigour that it's easy to fall into these pitfalls and typically physics text don't go into where these are and why and how to avoid them. It's a skill that one picks up by doing problems, going through the theory and reading around.
Similar problems are associated with the path integral which has no rigorous definition. However, the variational calculus can be made rigourous. However, this is difficult. It's typically not touched upon in an undergraduate mathematics course where they will rigourously define calculus for one real variable, for one complex variable and many real variables - either calculus on a manifold or more typically, multi-variable calculus, which is calculus in a (finite-dimensional) vector space.
To make the mathematics of this rigorous requires apparatus of jet bundles. You can find an exposition of Saunders Jet Bundles and Michors Natural Operations. It's takes quite some development.
answered 35 mins ago
Mozibur Ullah
4,33522144
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Ah, what a tricky mistake you've made there. The problem is that you've simply confused some confusions notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle in moving in 1 dimension in an external potential energy $V(q)$ is
$$
L(q, dot q) = frac12m dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clear $q$ and $dot q$ are not independent variables. Once $q$ is specified for all times, $dot q$ is also specified for all times.
A better way to write the above Lagrangian might be
$$
L(a, b) = frac12m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
fracpartial Lpartial a = -V'(a) hspace1cm fracpartial Lpartial b = m b.
$$
Usually, most people write this as
$$
fracpartial Lpartial q = -V'(q) hspace1cm fracpartial Lpartial dot q = m dot q.
$$
However, $q$ and $dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $dot q$ are too when they're being put into the Lagrangian. In otherwords, we could put any two numbers into $L$, we just decided to put in $q$ and $dot q$.
Furthermore, let's look at the total time derivative $fracddt$. How should we understand the following expression?
$$
fracddt L(q, dot q)
$$
Both $q$ and $dot q$ are functions of time. Therefore, $L(q, dot q)$ depends on time simply because $q$ and $dot q$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
fracddt L(q, dot q) = fracdqdt fracpartial Lpartial a(q, dot q) + fracd dot qdt fracpartial Lpartial b(q, dot q) = dot q fracpartial Lpartial a(q, dot q) + ddot q fracpartial Lpartial b(q, dot q)
$$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $partial L / partial a$ and $partial L / partial b$ by plugging in $(q, dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
answered 2 hours ago
user1379857
992515
add a comment |Â
up vote
3
down vote
Ah, what a tricky mistake you've made there. The problem is that you've simply confused some confusions notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle in moving in 1 dimension in an external potential energy $V(q)$ is
$$
L(q, dot q) = frac12m dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clear $q$ and $dot q$ are not independent variables. Once $q$ is specified for all times, $dot q$ is also specified for all times.
A better way to write the above Lagrangian might be
$$
L(a, b) = frac12m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
fracpartial Lpartial a = -V'(a) hspace1cm fracpartial Lpartial b = m b.
$$
Usually, most people write this as
$$
fracpartial Lpartial q = -V'(q) hspace1cm fracpartial Lpartial dot q = m dot q.
$$
However, $q$ and $dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $dot q$ are too when they're being put into the Lagrangian. In otherwords, we could put any two numbers into $L$, we just decided to put in $q$ and $dot q$.
Furthermore, let's look at the total time derivative $fracddt$. How should we understand the following expression?
$$
fracddt L(q, dot q)
$$
Both $q$ and $dot q$ are functions of time. Therefore, $L(q, dot q)$ depends on time simply because $q$ and $dot q$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
fracddt L(q, dot q) = fracdqdt fracpartial Lpartial a(q, dot q) + fracd dot qdt fracpartial Lpartial b(q, dot q) = dot q fracpartial Lpartial a(q, dot q) + ddot q fracpartial Lpartial b(q, dot q)
$$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $partial L / partial a$ and $partial L / partial b$ by plugging in $(q, dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
answered 2 hours ago
user1379857
992515
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Ah, what a tricky mistake you've made there. The problem is that you've simply confused some confusions notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle in moving in 1 dimension in an external potential energy $V(q)$ is
$$
L(q, dot q) = frac12m dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clear $q$ and $dot q$ are not independent variables. Once $q$ is specified for all times, $dot q$ is also specified for all times.
A better way to write the above Lagrangian might be
$$
L(a, b) = frac12m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
fracpartial Lpartial a = -V'(a) hspace1cm fracpartial Lpartial b = m b.
$$
Usually, most people write this as
$$
fracpartial Lpartial q = -V'(q) hspace1cm fracpartial Lpartial dot q = m dot q.
$$
However, $q$ and $dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $dot q$ are too when they're being put into the Lagrangian. In otherwords, we could put any two numbers into $L$, we just decided to put in $q$ and $dot q$.
Furthermore, let's look at the total time derivative $fracddt$. How should we understand the following expression?
$$
fracddt L(q, dot q)
$$
Both $q$ and $dot q$ are functions of time. Therefore, $L(q, dot q)$ depends on time simply because $q$ and $dot q$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
fracddt L(q, dot q) = fracdqdt fracpartial Lpartial a(q, dot q) + fracd dot qdt fracpartial Lpartial b(q, dot q) = dot q fracpartial Lpartial a(q, dot q) + ddot q fracpartial Lpartial b(q, dot q)
$$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $partial L / partial a$ and $partial L / partial b$ by plugging in $(q, dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
answered 2 hours ago
user1379857
992515
Ah, what a tricky mistake you've made there. The problem is that you've simply confused some confusions notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle in moving in 1 dimension in an external potential energy $V(q)$ is
$$
L(q, dot q) = frac12m dot q^2 - V(q).
$$
This is how most people write it. However, this is very confusing, because clear $q$ and $dot q$ are not independent variables. Once $q$ is specified for all times, $dot q$ is also specified for all times.
A better way to write the above Lagrangian might be
$$
L(a, b) = frac12m b^2 - V(a).
$$
Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that
$$
fracpartial Lpartial a = -V'(a) hspace1cm fracpartial Lpartial b = m b.
$$
Usually, most people write this as
$$
fracpartial Lpartial q = -V'(q) hspace1cm fracpartial Lpartial dot q = m dot q.
$$
However, $q$ and $dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $dot q$ are too when they're being put into the Lagrangian. In otherwords, we could put any two numbers into $L$, we just decided to put in $q$ and $dot q$.
Furthermore, let's look at the total time derivative $fracddt$. How should we understand the following expression?
$$
fracddt L(q, dot q)
$$
Both $q$ and $dot q$ are functions of time. Therefore, $L(q, dot q)$ depends on time simply because $q$ and $dot q$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus.
$$
fracddt L(q, dot q) = fracdqdt fracpartial Lpartial a(q, dot q) + fracd dot qdt fracpartial Lpartial b(q, dot q) = dot q fracpartial Lpartial a(q, dot q) + ddot q fracpartial Lpartial b(q, dot q)
$$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $partial L / partial a$ and $partial L / partial b$ by plugging in $(q, dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have
$$
f(x) = x^2
$$
and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
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So, in principle one can choose essentially $itany$ lagrangian $mathcalL$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that dont correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $dotq$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $mathcalL$. But nowadays, almost everyone that works with GR or modified gravity start from $mathcalL$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.
answered 4 hours ago
N. Steinle
4666
add a comment |Â
up vote
1
down vote
So, in principle one can choose essentially $itany$ lagrangian $mathcalL$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that dont correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $dotq$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $mathcalL$. But nowadays, almost everyone that works with GR or modified gravity start from $mathcalL$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.
answered 4 hours ago
N. Steinle
4666
add a comment |Â
up vote
1
down vote
up vote
1
down vote
So, in principle one can choose essentially $itany$ lagrangian $mathcalL$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that dont correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $dotq$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $mathcalL$. But nowadays, almost everyone that works with GR or modified gravity start from $mathcalL$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.
answered 4 hours ago
N. Steinle
4666
So, in principle one can choose essentially $itany$ lagrangian $mathcalL$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that dont correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $dotq$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $mathcalL$. But nowadays, almost everyone that works with GR or modified gravity start from $mathcalL$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.
answered 4 hours ago
N. Steinle
4666
edited 4 hours ago
answered 4 hours ago
N. Steinle
4666
answered 4 hours ago
N. Steinle
4666
answered 4 hours ago
N. Steinle
4666
4666
add a comment |Â
add a comment |Â
up vote
1
down vote
The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~=~fracpartialhphantomq^jpartial q^j$$ is not zero. See also e.g. this related Math.SE post.
The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^j$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But the Lagrangian typically depends on velocities.
Fact:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivative$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE post.
answered 1 hour ago
Qmechanic♦
96.8k121631028
add a comment |Â
up vote
1
down vote
The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~=~fracpartialhphantomq^jpartial q^j$$ is not zero. See also e.g. this related Math.SE post.
The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^j$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But the Lagrangian typically depends on velocities.
Fact:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivative$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE post.
answered 1 hour ago
Qmechanic♦
96.8k121631028
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~=~fracpartialhphantomq^jpartial q^j$$ is not zero. See also e.g. this related Math.SE post.
The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^j$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But the Lagrangian typically depends on velocities.
Fact:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivative$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE post.
answered 1 hour ago
Qmechanic♦
96.8k121631028
The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~=~fracpartialhphantomq^jpartial q^j$$ is not zero. See also e.g. this related Math.SE post.
The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^j$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But the Lagrangian typically depends on velocities.
Fact:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivative$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE post.
answered 1 hour ago
Qmechanic♦
96.8k121631028
edited 1 hour ago
answered 1 hour ago
Qmechanic♦
96.8k121631028
answered 1 hour ago
Qmechanic♦
96.8k121631028
answered 1 hour ago
Qmechanic♦
96.8k121631028
96.8k121631028
add a comment |Â
add a comment |Â
up vote
0
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That's an interesting sequence of symbolic manipulations!
It's because of the lack of rigour that it's easy to fall into these pitfalls and typically physics text don't go into where these are and why and how to avoid them. It's a skill that one picks up by doing problems, going through the theory and reading around.
Similar problems are associated with the path integral which has no rigorous definition. However, the variational calculus can be made rigourous. However, this is difficult. It's typically not touched upon in an undergraduate mathematics course where they will rigourously define calculus for one real variable, for one complex variable and many real variables - either calculus on a manifold or more typically, multi-variable calculus, which is calculus in a (finite-dimensional) vector space.
To make the mathematics of this rigorous requires apparatus of jet bundles. You can find an exposition of Saunders Jet Bundles and Michors Natural Operations. It's takes quite some development.
answered 35 mins ago
Mozibur Ullah
4,33522144
add a comment |Â
up vote
0
down vote
That's an interesting sequence of symbolic manipulations!
It's because of the lack of rigour that it's easy to fall into these pitfalls and typically physics text don't go into where these are and why and how to avoid them. It's a skill that one picks up by doing problems, going through the theory and reading around.
Similar problems are associated with the path integral which has no rigorous definition. However, the variational calculus can be made rigourous. However, this is difficult. It's typically not touched upon in an undergraduate mathematics course where they will rigourously define calculus for one real variable, for one complex variable and many real variables - either calculus on a manifold or more typically, multi-variable calculus, which is calculus in a (finite-dimensional) vector space.
To make the mathematics of this rigorous requires apparatus of jet bundles. You can find an exposition of Saunders Jet Bundles and Michors Natural Operations. It's takes quite some development.
answered 35 mins ago
Mozibur Ullah
4,33522144
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That's an interesting sequence of symbolic manipulations!
It's because of the lack of rigour that it's easy to fall into these pitfalls and typically physics text don't go into where these are and why and how to avoid them. It's a skill that one picks up by doing problems, going through the theory and reading around.
Similar problems are associated with the path integral which has no rigorous definition. However, the variational calculus can be made rigourous. However, this is difficult. It's typically not touched upon in an undergraduate mathematics course where they will rigourously define calculus for one real variable, for one complex variable and many real variables - either calculus on a manifold or more typically, multi-variable calculus, which is calculus in a (finite-dimensional) vector space.
To make the mathematics of this rigorous requires apparatus of jet bundles. You can find an exposition of Saunders Jet Bundles and Michors Natural Operations. It's takes quite some development.
answered 35 mins ago
Mozibur Ullah
4,33522144
That's an interesting sequence of symbolic manipulations!
It's because of the lack of rigour that it's easy to fall into these pitfalls and typically physics text don't go into where these are and why and how to avoid them. It's a skill that one picks up by doing problems, going through the theory and reading around.
Similar problems are associated with the path integral which has no rigorous definition. However, the variational calculus can be made rigourous. However, this is difficult. It's typically not touched upon in an undergraduate mathematics course where they will rigourously define calculus for one real variable, for one complex variable and many real variables - either calculus on a manifold or more typically, multi-variable calculus, which is calculus in a (finite-dimensional) vector space.
To make the mathematics of this rigorous requires apparatus of jet bundles. You can find an exposition of Saunders Jet Bundles and Michors Natural Operations. It's takes quite some development.
answered 35 mins ago
Mozibur Ullah
4,33522144
answered 35 mins ago
Mozibur Ullah
4,33522144
answered 35 mins ago
Mozibur Ullah
4,33522144
answered 35 mins ago
Mozibur Ullah
4,33522144
4,33522144
add a comment |Â
add a comment |Â
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7
"commutativity of derivatives" doesn't work here ($t$ and $dot q^alpha$ are not independent variables), and there is no such a thing as "cancellation of dots". Try explicit examples (e.g., $L=qdot q$ or simple functions like that) to convince yourself that those manipulations are unjustified.
– AccidentalFourierTransform
6 hours ago
2
@AccidentalFourierTransform That would make a decent answer, albeit a short one.
– StephenG
5 hours ago
@AccidentalFourierTransform There actually is something called “cancellation of dotsâ€Â; see slide 7 of web.physics.ucsb.edu/~dfolsom/CS32/lagrangian_mechanics.pdf. However it’s not so trivial as the OP suggests.
– ZeroTheHero
3 hours ago