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Question

The Euler-Lagrange equation gives the equations of motion of a system with Lagrangian $L$. Let $q^alpha$ represent the generalized coordinates of a configuration manifold, $t$ represent time. The Lagrangian is a function of the state of a particle, i.e. the particle's position $q^alpha$ and velocity $dot q^alpha$. The Euler-Lagrange equation is

$$ fracddt fracpartial Lpartial dot q^alpha = fracpartial Lpartial q^alpha$$

Why is this a law of physics and not a simple triviality for any function $L$ on the variables $q^alpha$ and $dot q^alpha$? The following "proof" of the Lagrange Equation uses no physics, and seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.

$$beginalign
fracddt fracpartial Lpartial dot q^alpha & = fracpartialpartial dot q^alpha fracdLdt &textcommutativity of derivatives \ \
&= fracpartial dot Lpartial dot q^alpha \ \
&= fracpartial Lpartial q^alpha & textcancellation of dots endalign$$

This can't be right, or else 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 above?

The very first step is wrong. You canÃ¢Â€Â™t commute a total and partial derivative. — 50 mins ago

score 24 · Answer 1 · 2018-09-16 15:03:20Z

Ah, what a tricky mistake you've made there. The problem is that you've simply confused some 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.

score 6 · Answer 2 · 2018-09-16 02:58:03Z

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.

score 6 · Answer 3 · 2018-09-16 11:18:06Z

The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~stackrel(2)=~fracpartialhphantomq^jpartial q^jtag1$$
of a velocity derivative $fracpartialhphantomdotq^jpartial dotq^j$ with the total time derivative
$$fracmathrm dhphantomtmathrm d t
~=~fracpartialhphantomtpartial t
+dotq^jfracpartialhphantomq^jpartial q^j
+ddotq^jfracpartialhphantomdotq^jpartial dotq^j
+dddotq^jfracpartialhphantomddotq^jpartial ddotq^j
+ldots tag2$$
is not zero. See also e.g. this related Math.SE post & this related Phys.SE post.

The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^jtag3$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But a Lagrangian typically depends on velocities. See also this related Phys.SE post.

Note the following algebraic Poincare lemma:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivativetag4$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE posts.

out of curiosity, what's up with all those phantoms? Why don't you centre the $partial$s? — 47 mins ago

Mozibur Ullah 4,37222144 · Answer 4 · 2018-09-16 06:49:32Z

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.

hkBst 1112 · Answer 5 · 2018-09-16 09:10:43Z

First some notation. Following SICM, I write the Lagrange equations like so:

$$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) Ã¢ÂˆÂ’ (partial_1 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q] = 0$$ (where $mathrmD$ is the total derivative and corresponds to the time derivative, and $ÃŽÂ“[q] = (q,mathrmDq)$.)

Your question: ''Why is the Lagrange equation not a triviality? What is wrong with the calculation below?''.

Trying to rewrite your calculation using the unambiguous notation from SICM immediately reveals some problems:

Impossible to simply commute derivatives:
Neither $$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_2((mathrmD L) Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ nor $$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_2mathrmD (L Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ make any sense.

Impossible to cancel dots: $$partial_2mathrmD (L Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_1 (L Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ both left and right look pretty non-sensical.

Then you need to do $$partial_1 (L Ã¢ÂˆÂ˜ ÃŽÂ“[q]) = (partial_1 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]$$ to reconstruct a sane expression.

Thus no step in your proof is warranted.

my2cts 3,3412416 · Answer 6 · 2018-09-16 11:42:19Z

You can require the EL equation for any functional. However, your thesis that it is a general identity is wrong. Where did you get the ideas for step 1 and 3 ?

Aganju 413410 · Answer 7 · 2018-09-16 14:34:01Z

N. Steinle already gave a great answer on the question

why is this a fundamental law of physics and not a simple triviality of ANY function L

but I would like to point out an additional tidbit regarding the part

.. seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.

While the Lagrange equations mathematically really only describe a function/process that is an extremal value of some Lagrangian (or also some energy or action potential), the important part is that the converse is not as simple.

It seems to be a "a fundamental law of physics" that many processes, that we observe in nature even have a Lagrangian, an energy potential. This is actually not trivial, not every multidimensional function has such a potential and is a statement about the symmetry of these processes.

You have some duplicated paragraphs there...
â€“Â Pedro A
2 hours ago — 2 hours ago

score 24 · Answer 8 · 2018-09-16 15:03:20Z

Ah, what a tricky mistake you've made there. The problem is that you've simply confused some 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.

score 6 · Answer 9 · 2018-09-16 02:58:03Z

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.

score 6 · Answer 10 · 2018-09-16 11:18:06Z

The commutator $$biggl[fracpartialhphantomdotq^jpartial dotq^j,fracmathrm dhphantomtmathrm d tbiggr]~stackrel(2)=~fracpartialhphantomq^jpartial q^jtag1$$
of a velocity derivative $fracpartialhphantomdotq^jpartial dotq^j$ with the total time derivative
$$fracmathrm dhphantomtmathrm d t
~=~fracpartialhphantomtpartial t
+dotq^jfracpartialhphantomq^jpartial q^j
+ddotq^jfracpartialhphantomdotq^jpartial dotq^j
+dddotq^jfracpartialhphantomddotq^jpartial ddotq^j
+ldots tag2$$
is not zero. See also e.g. this related Math.SE post & this related Phys.SE post.

The cancellation of dots $$fracpartial dotLpartial dotq^j~=~fracpartial Lpartial q^jtag3$$ works for functions $L(q,t)$ that don't depend on velocities $dotq^k$. But a Lagrangian typically depends on velocities. See also this related Phys.SE post.

Note the following algebraic Poincare lemma:
$$Ltext satisfies the Euler-Lagrange (EL) eqs. identically quadLeftrightarrowquad Ltext is a total time derivativetag4$$
(modulo possible topological obstructions). For details, see e.g. this & this Phys.SE posts.

out of curiosity, what's up with all those phantoms? Why don't you centre the $partial$s? — 47 mins ago

Mozibur Ullah 4,37222144 · Answer 11 · 2018-09-16 06:49:32Z

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.

hkBst 1112 · Answer 12 · 2018-09-16 09:10:43Z

First some notation. Following SICM, I write the Lagrange equations like so:

$$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) Ã¢ÂˆÂ’ (partial_1 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q] = 0$$ (where $mathrmD$ is the total derivative and corresponds to the time derivative, and $ÃŽÂ“[q] = (q,mathrmDq)$.)

Your question: ''Why is the Lagrange equation not a triviality? What is wrong with the calculation below?''.

Trying to rewrite your calculation using the unambiguous notation from SICM immediately reveals some problems:

Impossible to simply commute derivatives:
Neither $$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_2((mathrmD L) Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ nor $$mathrmD((partial_2 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_2mathrmD (L Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ make any sense.

Impossible to cancel dots: $$partial_2mathrmD (L Ã¢ÂˆÂ˜ ÃŽÂ“[q]) neq partial_1 (L Ã¢ÂˆÂ˜ ÃŽÂ“[q])$$ both left and right look pretty non-sensical.

Then you need to do $$partial_1 (L Ã¢ÂˆÂ˜ ÃŽÂ“[q]) = (partial_1 L) Ã¢ÂˆÂ˜ ÃŽÂ“[q]$$ to reconstruct a sane expression.

Thus no step in your proof is warranted.

my2cts 3,3412416 · Answer 13 · 2018-09-16 11:42:19Z

You can require the EL equation for any functional. However, your thesis that it is a general identity is wrong. Where did you get the ideas for step 1 and 3 ?

Aganju 413410 · Answer 14 · 2018-09-16 14:34:01Z

N. Steinle already gave a great answer on the question

why is this a fundamental law of physics and not a simple triviality of ANY function L

but I would like to point out an additional tidbit regarding the part

.. seems to suggest that the Lagrange Equation is simply a mathematical fact that works for every function.

While the Lagrange equations mathematically really only describe a function/process that is an extremal value of some Lagrangian (or also some energy or action potential), the important part is that the converse is not as simple.

It seems to be a "a fundamental law of physics" that many processes, that we observe in nature even have a Lagrangian, an energy potential. This is actually not trivial, not every multidimensional function has such a potential and is a statement about the symmetry of these processes.

You have some duplicated paragraphs there...
â€“Â Pedro A
2 hours ago — 2 hours ago

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