Using reference frames for derivation of Lagrangian

[gionole]

I had an interesting thought and it just might be me, but I'm looking forward to hearing your thoughts, but not like the thoughts - "yeah, Landau is messy, complicated, don't read that, e.t.c". Just think about what you think about my thoughts.

Landau first starts to mention the variational calculus technique and how euler lagrange equation is given. He uses $q(t)$ such as for the path, but his proof is the general one we all know. So he derives EL equation and writes it down. Then starts and builds his logic why $L = K - U$.

His first argument is that equations of motions from different reference frames(inertial and non-inertial) have the different EOM form and in that case, he decides that we can always find inertial frame and stick with it. Ok, clear till now. Then starts to derive $L = K - U$ for the freely moving particle. He says that in inertial frame, the freely moving particle would only change speed if external force acts on it and he says that in this frame, space and time are homogeneous. i.e in this frame, freely moving particle's equation of motion would be the same(wouldn't change) at all times and at all points in this inertial frame. So because of this, he says that Lagrangian can not depend on $q$ and $t$.

Question 1: Well, Everything was clear, till when he said that Lagrangian can not depend on $q$ and $t$. He bases this logic on the homogenous time and space argument, but how homogeneous time and space implies anything about Lagrangian ? they imply that equations of motions exactly stay the same, but not sure why it implies anything about $L$ not containing the $q$ and $t$. It seems to me that because Landau already has derived the EL before this moment, in his head, he imagines what if $L$ contained $q$ ? then for sure equation of motion would end up to be $m\ddot x \neq 0$(due to EL) which is wrong due to Newton. Then what if $L$ contained $t$? then EOM wouldn't be the same as of newton's. So if you think as well, that Landau imagines $L$ to be containing $q$ and why it would be wrong and thats why he says, $L$ couldn't contain $q$, then why in the first place does he mention homogenity, inertial frames, .e.t.c ?
Question 2: I like the logic of how he gets to the point(excluding Question 1's thoughts) that L must be $cv^2$ where $c$ is something he doesn't know yet. Then he out of nowhere says that $c$ must be $\frac{1}{2}m$ and that's it. It seems to me that he might have used the Euler lagrange on $cv^2$ and tries to find $c$ such as he arrives to Newton's equations and he would only arrive at newton's equations if $c == \frac{1}{2}m$. What was the whole point of using: inertial, non inertial frames, "adding total time derivative of Lagrangian doesn't change equations of motion" if he in the end still relies on Euler Lagrange and specifically newton's already derived equations for free particle. He could have just never mention the whole explanation and just wrote something like this: "We're looking for the Lagrangian for the free particle, we know newton arrives with a constant speed equation for such particle, so what should our L be such that using Euler lagrange on it would get us $m\ddot =0$ and it's super easy to come up with $L = \frac{1}{2}mv^2$".

Question 3: Not only that, I think he does the same to arrive at $K - U$ for the system of particles and still relies on Euler lagrange equation.

If Landau uses Euler Lagrange + Newton's already derived EOM's, then what's the need at all to be mentioning such huge explanations about inertial frames, homogenity, lagrangian total time derivatives, e.t.c. It's pointless.

Here is the book link. (from page 4)

 

[vanhees71]

I had an interesting thought and it just might be me, but I'm looking forward to hearing your thoughts, but not like the thoughts - "yeah, Landau is messy, complicated, don't read that, e.t.c". Just think about what you think about my thoughts.

To the contrary the Landau Physics Course is among the best theoretical-physics books ever written. It's sometimes a bit brief and thus not suited as an introductory physics book, but it's very clear to the point. For me the highlights of the series are volumes 2, 5, and 10, but also volume 1 on analytical mechanics is great.

Landau first starts to mention the variational calculus technique and how euler lagrange equation is given. He uses $q(t)$ such as for the path, but his proof is the general one we all know. So he derives EL equation and writes it down. Then starts and builds his logic why $L = K - U$.

His first argument is that equations of motions from different reference frames(inertial and non-inertial) have the different EOM form and in that case, he decides that we can always find inertial frame and stick with it. Ok, clear till now. Then starts to derive $L = K - U$ for the freely moving particle. He says that in inertial frame, the freely moving particle would only change speed if external force acts on it and he says that in this frame, space and time are homogeneous. i.e in this frame, freely moving particle's equation of motion would be the same(wouldn't change) at all times and at all points in this inertial frame. So because of this, he says that Lagrangian can not depend on $q$ and $t$.

That's a very modern approach, and for me it was one of the great motivations to study physics, i.e., it uses symmetry considerations to explain, why the physical laws look the way they do. The rationale is that you take the Galilei-Newtonian spacetime model as given, and this spacetime model implies a large symmetry group, the Galilei group, i.e., it implies that the physical laws for closed systems must be symmetric under the Galilei group and derives, assuming that the physical laws can be formulated with the action principle, which form the Lagrangian can take to obey the symmetries of Galilei-Newton spacetime.

Question 1: Well, Everything was clear, till when he said that Lagrangian can not depend on $q$ and $t$. He bases this logic on the homogenous time and space argument, but how homogeneous time and space implies anything about Lagrangian ? they imply that equations of motions exactly stay the same, but not sure why it implies anything about $L$ not containing the $q$ and $t$. It seems to me that because Landau already has derived the EL before this moment, in his head, he imagines what if $L$ contained $q$ ? then for sure equation of motion would end up to be $m\ddot x \neq 0$(due to EL) which is wrong due to Newton. Then what if $L$ contained $t$? then EOM wouldn't be the same as of newton's. So if you think as well, that Landau imagines $L$ to be containing $q$ and why it would be wrong and thats why he says, $L$ couldn't contain $q$, then why in the first place does he mention homogenity, inertial frames, .e.t.c ?

Take the homogeneity of space, i.e., the symmetry under spatial translations, and obviously he considers a single particle as the closed system. So it's clear that the Lagrangian can be written in terms of Cartesian coordinates $\vec{x}$, $L=L(\vec{x},\dot{\vec{x}},t)$. Now the translation symmetry of space implies that the transformation
$$\vec{x} \rightarrow \vec{x}'=\vec{x}+\vec{a}$$
is a symmetry for $\vec{a}=\text{const}$.

Now a symmetry is defined by the demand that the Lagrangian $L(\vec{x}',\dot{\vec{x}}',t)$, written in terms of the old coordinates is equivalent to the original Lagrangian, i.e., there must be a function $\Omega(\vec{x},t)$ such that
$$L(\vec{x}+\vec{a},\dot{\vec{x}},t)=L(\vec{x},\dot{\vec{x}},t) + \frac{\mathrm{d}}{\mathrm{d} t} \Omega(\vec{x},t).$$
That must hold true also to the approximation that $|\vec{a}|$ is very small, and we can expand the left-hand side to linear order in $\vec{a}$, which leads to
$$\vec{a} \cdot \frac{\partial L}{\partial \vec{x}} = \frac{\mathrm{d}}{\mathrm{d} t} \Omega(\vec{x},t)=\dot{\vec{x}} \cdot \frac{\partial \Omega}{\partial \vec{x}} + \frac{\partial \Omega}{\partial t}.$$
On the left-hand side you have no expression proportional to $\dot{\vec{x}}$. So it's clear that you can make $\partial \Omega/\partial \vec{x}=0$, i.e., $\Omega=\Omega(t)$. But then $L$ would also be only a function of $t$, which doesn't lead to any equations of motion. So the only way to fulfill the symmetry is
$$\vec{a} \cdot \frac{\partial L}{\partial \vec{x}} =0,$$
and this must hold for all vectors $\vec{a}$, but this implies that
$$\frac{\partial L}{\partial \vec{x}}=0 \; \Rightarrow \; L=L(\dot{\vec{x}},t).$$
In the same way you can go through all the other symmetries.

 

[anuttarasammyak]

What Landau wrote in the textbook is usually reffered as Noether's therorem which explains the relation between symmetry and conservation of the system. See e.g. https://en.wikipedia.org/wiki/Noether's_theorem for your study.

 

[gionole]

@vanhees71 your answer is actually great, but here're my worries about it.

1. shouldn't it be $L = L(\dot x)$ instead of $L = L(\dot x, t)$ ?

2. I get your point, but note that at the time when Landau explains all this, he hasn't explained translation of space and symmetries. So can we look at this differently ? When you have a free particle, in the inertial frame, space is homogeneous - this means that it shouldn't matter whether particle is at $x=2$ or $x=10$, the equation of motion must be exactly the SAME in both places. If that's true, what would that mean ? for free particle, at $x=2$, equation of motion is $vt$ and at $x=10$, it should be the same $vt$ right ? if so, can you give an example where at different places in non-inertial frame($x=2$ and $x=10$, the same object would have different EOM - let's stick with non-rotational example for now) ?

3. when arrived at $L = cq^2$, how does he arrive that $c=\frac{1}{2}m$ ? he seems to be using euler lagrange and bases his logic on newton in order to get the same result as newton. In my opinion, if he bases all the logic on newton, he could have easily said: what would $L$ to be so that applying EL to it will get me the exact same newton's results ? he would easily arrive at the correct lagrangian and wouldn't have to deal with explanations at all of frames. My point is either you go all in and derive lagrangian based on your own logic all the way and derive everything mathematically just by yourself or just assume something and that's it.

 

[vanhees71]

@vanhees71 your answer is actually great, but here're my worries about it.

1. shouldn't it be $L = L(\dot x)$ instead of $L = L(\dot x, t)$ ?

If you don't know anything about how $L$ looks like, it could well be that $L$ also depends explicitly on $t$. This, however, is then ruled out by time-translation invariance, which leads to $\partial_t L=0$ as spatial translation invariance lead to $\partial_{\vec{x}}L=0$.

2. I get your point, but note that at the time when Landau explains all this, he hasn't explained translation of space and symmetries. So can we look at this differently ? When you have a free particle, in the inertial frame, space is homogeneous - this means that it shouldn't matter whether particle is at $x=2$ or $x=10$, the equation of motion must be exactly the SAME in both places. If that's true, what would that mean ? for free particle, at $x=2$, equation of motion is $vt$ and at $x=10$, it should be the same $vt$ right ? if so, can you give an example where at different places in non-inertial frame($x=2$ and $x=10$, the same object would have different EOM - let's stick with non-rotational example for now) ?

No. The equation of motion for the free particle turns out to be at the end $m \ddot{\vec{x}}=0$, and that's indeed invariant under translations.

Nowhere in the entire discussion are non-inertial frames under debate, because the full symmetries hold of course only in inertial frames.

3. when arrived at $L = cq^2$, how does he arrive that $c=\frac{1}{2}m$ ? he seems to be using euler lagrange and bases his logic on newton in order to get the same result as newton. In my opinion, if he bases all the logic on newton, he could have easily said: what would $L$ to be so that applying EL to it will get me the exact same newton's results ? he would easily arrive at the correct lagrangian and wouldn't have to deal with explanations at all of frames. My point is either you go all in and derive lagrangian based on your own logic all the way and derive everything mathematically just by yourself or just assume something and that's it.

Of course, with one free particle alone you cannot determine the overall factor physically, i.e., you cannot even measure mass.

 

[Ibix]

Nowhere in the entire discussion are non-inertial frames under debate, because the full symmetries hold of course only in inertial frames.

Would it be better to say that the symmetries hold in all frames of any kind, but are only obvious when one picks coordinate systems that are well adapted to the symmetry in question? Or am I missing something?

 

[vanhees71]

Sure, you can express Galilei symmetry also in arbitrary coordinates, but what sense should such a complication make?

 

[gionole]

To the contrary the Landau Physics Course is among the best theoretical-physics books ever written. It's sometimes a bit brief and thus not suited as an introductory physics book, but it's very clear to the point. For me the highlights of the series are volumes 2, 5, and 10, but also volume 1 on analytical mechanics is great.

That's a very modern approach, and for me it was one of the great motivations to study physics, i.e., it uses symmetry considerations to explain, why the physical laws look the way they do. The rationale is that you take the Galilei-Newtonian spacetime model as given, and this spacetime model implies a large symmetry group, the Galilei group, i.e., it implies that the physical laws for closed systems must be symmetric under the Galilei group and derives, assuming that the physical laws can be formulated with the action principle, which form the Lagrangian can take to obey the symmetries of Galilei-Newton spacetime.

Take the homogeneity of space, i.e., the symmetry under spatial translations, and obviously he considers a single particle as the closed system. So it's clear that the Lagrangian can be written in terms of Cartesian coordinates $\vec{x}$, $L=L(\vec{x},\dot{\vec{x}},t)$. Now the translation symmetry of space implies that the transformation
$$\vec{x} \rightarrow \vec{x}'=\vec{x}+\vec{a}$$
is a symmetry for $\vec{a}=\text{const}$.

Now a symmetry is defined by the demand that the Lagrangian $L(\vec{x}',\dot{\vec{x}}',t)$, written in terms of the old coordinates is equivalent to the original Lagrangian, i.e., there must be a function $\Omega(\vec{x},t)$ such that
$$L(\vec{x}+\vec{a},\dot{\vec{x}},t)=L(\vec{x},\dot{\vec{x}},t) + \frac{\mathrm{d}}{\mathrm{d} t} \Omega(\vec{x},t).$$
That must hold true also to the approximation that $|\vec{a}|$ is very small, and we can expand the left-hand side to linear order in $\vec{a}$, which leads to
$$\vec{a} \cdot \frac{\partial L}{\partial \vec{x}} = \frac{\mathrm{d}}{\mathrm{d} t} \Omega(\vec{x},t)=\dot{\vec{x}} \cdot \frac{\partial \Omega}{\partial \vec{x}} + \frac{\partial \Omega}{\partial t}.$$
On the left-hand side you have no expression proportional to $\dot{\vec{x}}$. So it's clear that you can make $\partial \Omega/\partial \vec{x}=0$, i.e., $\Omega=\Omega(t)$. But then $L$ would also be only a function of $t$, which doesn't lead to any equations of motion. So the only way to fulfill the symmetry is
$$\vec{a} \cdot \frac{\partial L}{\partial \vec{x}} =0,$$
and this must hold for all vectors $\vec{a}$, but this implies that
$$\frac{\partial L}{\partial \vec{x}}=0 \; \Rightarrow \; L=L(\dot{\vec{x}},t).$$
In the same way you can go through all the other symmetries.

I think I realize now how great this answer this. @vanhees71

Just wanted to ask something

I understand the following from you:

$\vec a \frac{\partial L}{\partial \vec x} = \frac{\partial \Omega}{\partial t}$.

Since right side after derivating it has dependence on $t$, in order for it to be equal to left side, $\frac{\partial L}{\partial \vec x}$ must also be the same way dependent on $t$, though for any $L$, it won't hold true and therefore $\frac{\partial \Omega}{\partial t}$ must be 0. I think i get this point, but what confused me is your statement:

But then $L$ would also be only a function of $t$,

Why does $\vec a \frac{\partial L}{\partial \vec x} = \frac{\partial \Omega}{\partial t}$ state that $L$ must be only a function of $t$ ? because if it is, left side $\frac{\partial L}{\partial \vec x}$ would be 0 and the equation would always fail even if $L$ is only a function of $t$.

 

[vanhees71]

Hm, thinking about it, I'd say this is nonsense. Of course you can always add an arbitrary function $\Omega(t)$ to the Lagrangian without changing the equations of motion. So a better formulation is:

Because $\partial \Omega/\partial \vec{x}=0$ you must have $\Omega=\Omega(t)$ and thus you can set it to 0, because adding such a function to $L$ doesn't change the equations of motion at all.

 

[gionole]

Hm, thinking about it, I'd say this is nonsense. Of course you can always add an arbitrary function $\Omega(t)$ to the Lagrangian without changing the equations of motion. So a better formulation is:

Because $\partial \Omega/\partial \vec{x}=0$ you must have $\Omega=\Omega(t)$ and thus you can set it to 0, because adding such a function to $L$ doesn't change the equations of motion at all.

Well, then you didn't even need to mention

$\partial \Omega/\partial \vec{x}=0$

and directly say that: "hey, let's make $\Omega(\vec x, t)$ to be 0" in which case adding $\frac{d}{dt}0$ to lagrangian doesn't change EOM. . I am not sure what would be different.

 

[gionole]

One thing I got curious about as well is in inertial frame, space is always homogeneous(not only for free particle, but for things that forces act on). Imagine throwing a ball from position x and position x+dx. Definitely, for both positions, gravity force acts on them, but EOM must still stay the same due to homogeneous frame. Why doesn't your logic now work now ? with your logic, we would still get $L$ not to be dependent on $x$, in which case what we get is $L$ never depends on $x$ and not just for free particle which seems wrong as we know when there's potential energy, we got $L$ dependent on $x$.

Does this mean that in inertial frame, EOM's can still be different if there's potential energy present ? in this case, then inertial frame is not homogeneous as they say.

 

[vanhees71]

This logic doesn't work here for the simple fact that now you have an "external potential", $V(x)=-mg x$, and thus the symmetry condition for homogeneity is not fulfilled since now $L$ depends on $x$, and it should, because now we have the gravitational force of the Earth acting on the object under consideration, and the EoM comes correctly out to be
$$m \ddot{\vec{x}}=-\vec{\nabla} V=m g \vec{e}_x,$$
where I've chosen the direction of the $x$ axis "down" (in direction of the gravitational force). In this description of gravity you break the translation invariance.

Of course, you can also treat the problem as a closed system of the Earth and the object under consideration. Then the full Galilei symmetry is again fulfilled. Now the Lagrangian reads
$$L=\frac{m}{2} \dot{\vec{x}}^2 + \frac{M}{2} \dot{\vec{y}}^2 +\frac{G m M}{|\vec{x}-\vec{y}|}.$$
Now you have again translation invariance, i.e., the Lagrangian doesn't change under the transformation
$$\vec{x}'=\vec{x}+\vec{a}, \quad \vec{y}'=\vec{y}+\vec{a},$$
and total momentum
$$\vec{P}=m \dot{\vec{x}}+M\dot{\vec{y}}$$
is conserved.

It's also symmetric under arbitrary rotations, i.e.,
$$\vec{x}'=\hat{D} \vec{x}, \quad \vec{y}'=\hat{D} \vec{y}$$
is also a symmetry, which leads to the conservation of total angular momentum,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}} + M \vec{y} \times \dot{\vec{y}}.$$
And also the Lagrangian doesn't depend explicitly on $t$ (i.e., it's invariant under time translations) and thus the Hamiltonian, which here is the total energy, is conserved,
$$E=\frac{m}{2} \dot{\vec{x}}^2 + \frac{M}{2} \dot{\vec{y}}^2 - \frac{G m M}{|\vec{x}-\vec{y}|}=\text{const}.$$

 

[gionole]

If you assume that L is a function of $x$, then yes, but follow me on this logic and remember that we're trying to derive Lagrangian when there's a non-free particle, but don't yet assume anything and let's stricly follow the homogeneity definition only.

We first start saying that in inertial frame, space is homogeneous, i.e e.o.m doesn't change when the experiment is done from position $x$ and from position $x+a$. I think you agree with me on this, right ?

Because of this, Lagrangian represented from $x+a$ and Lagrangian represented from $x$ definitely must yield the same e.o.m in inertial frame due to homogeneity. I think you also agree with me on this or are you saying this is where I make a mistake ?

If so, we can write:

$L(x+a, \dot x, t) = L(x, \dot x, t) + \frac{d}{dt} \Omega(x, t)$.

$a \frac{\partial L}{\partial x} = \frac{d}{dt} \Omega(x, t)$ (Equation 1)

and if so, we get to the same point. In which exact paragraph of this message have I made a mistake ?

 

[vanhees71]

Of course, Lagrangians that describe some effective model need not have the full Galilei symmetry, i.e., they describe "open systems". I tried to explain this with the example above: If you treat the motion of a body on Earth taking into account the gravitational interaction by using the fact that the Earth is much more massive than the body and you neglect the motio of the Earth due to this interaction and describe thus gravitation on the body just by the fixed gravitational field due to the mass of the then assumed to be resting Earth, you have an open system consisting only of the body of interest. Close to Earth you can approximate the field as constant, and you get the potential $V=-m \vec{g} \cdot \vec{x}$ with $\vec{g}$ the constant gravitational acceleration of the Earth $|\vec{g} \simeq 9.81 \text{m}/\text{s}^2$. This breaks of course translational invariance, because the fixed position of the Earth's center of mass distinguishes this as a "special point". Also you have the vector $\vec{g}$ in the game and thus also rotational invariance is broken to rotations around the direction of this vector.

 

[gionole]

@vanhees71

You're amazing. Thank you so much. If you got time, would you be able to show why $L$ can't depend on direction of $v$ ? i know that space is isotropic, but i'm asking if you can show why it can't depend on direction, but use only the same approach as you did in reply #2 on this post.

 

[gionole]

Would the following be true ?

$L(x, \dot{\vec x} + \epsilon \vec n, t) - L(x, \dot{\vec x}, t) = \frac{d}{dt} f(x,t)$
$\epsilon\vec n \frac{\partial L}{\partial \vec x} = \dot x \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t}$

from which it's the same argument why L can't depend on $\vec v$ ?

 

[vanhees71]

I'm not sure, how this symmetry should work. If you shift the velocity, this should also affect the position. You can try to check, what follows from invariance under "Galilei boosts":
$$\vec{x} \rightarrow \vec{x}+\delta \vec{v} t, \quad \delta \vec{v}=\text{const}.$$
For rotations, the right ansatz is
$$\vec{x} \rightarrow \vec{x}+\delta \vec{\phi} \times \vec{x}, \quad \delta \vec{\phi}=\text{const}.$$
because an infinitesimal rotation is described by this vector product with the infinitesimal rotation vector. Here, $|\delta \vec{\phi}|$ is the infinitesimal rotation angle and the direction of the vector gives the rotation axis. The rotation direction is according to the right-hand rule: Pointing with the thumb of your right hand in direction of $\delta \vec{\phi}$, the fingers give the rotation direction.

Perhaps with this you can try yourself, what follows for $L(\vec{x},\dot{\vec{x}})$ if you want to have (all rotations) as a symmetry.

 

[gionole]

1. I'm not fully sure that I get your last message. My point was that with your proof about homogeneity, I thought we could use the same logic to prove that $L$ can't depend on the velocity vector. Is there a easy book that explains why infinetisemal rotation is given by $\vec x -> \vec x + d\vec \theta \times \vec x$ ?

2. While I'm at it, are we sure that $L(x+a) - L(x) = a\frac{\partial L}{\partial x}$ ?

First we do:
$L(x+a) - L(x) = a\frac{\partial L}{\partial x} + \frac{a^2}{2!}\frac{\partial^2 L}{\partial x^2} + \frac{a^3}{3!}\frac{\partial^3 L}{\partial x^3} + ...$

My goal is to find the difference -> $L(x+a) - L(x)$ such that it holds true for any and every $a$. So what you do is make all higher orders vanish. I don't get why they must vanish.

I understand that when $a$ is small, approximation is exact and if $a$ is large, approximation is NOT exact, but since we represented the difference with infinite series, even if $a$ is large, approximation will be exact because of "infinite" series of taylor. So what's going on ? why can we proudly say that $L(x+a) - L(x) = a\frac{\partial f}{\partial x}$ ? -(note that we don't know if $L$ is linear in $x$ but we still make it equal to $a\frac{\partial L}{\partial x}$

 

[vanhees71]

Noether's theorem is derived from "infinitesimal" symmetry transformations, i.e., it's sufficient to hold at first order in the expansion of the parameter.

 

[gionole]

In my opinion, what you're doing with this proof is that you're assuming that $L$ can't depend on $\dot x$., because if it does depend, then your proof fails, I get the idea why you're assuming that, because of free particle case, but the same way you assume that, you also could have assumed that L can't depend on $x$. Why is the assumption that $L$ can't depend on $\dot x$ easy to see, but the assumption that $L$ can't depend on $x$ much harder ?

 

[gionole]

@vanhees71
I must have written that you're assuming that L can't depend on the product of $x\dot x$, because if it does, your proof fails, because $\frac{\partial L}{\partial x}$ would now return the member containing $\dot x$ and argument that right side contains $\dot x$ while left side doesn't fails.

So you assume that $L$ can't depend on such product of $x \dot x$, but are you able to make that assumption even for free particle ? If you're able to do this assumption without backing it up, why at all are we trying to show that $L$ can't depend on $x$ as it's the same type of assumption. Would be good to also show mathematically why $L$ can't depend on product of $\dot x x$

 

[vanhees71]

Ok, let's do the analysis for rotational invariance. The infinitesimal transformation reads
$$\delta \vec{x}=\vec{\phi} \times \vec{x}, \Rightarrow \; \delta \dot{\vec{x}}=\vec{\phi} \times \dot{\vec{x}}.$$
Inserting this into the Lagrangian the symmetry condition reads
$$\delta L= \delta \vec{x} \cdot \partial_{\vec{x}} L + \delta \dot{\vec{x}} \cdot \partial_{\dot{\vec{x}}} L+ \mathrm{d}_t \delta \Omega(\vec{x},t)=0.$$
Now
$$\mathrm{d}_t \delta \Omega(\vec{x},t) = \dot{\vec{x}} \cdot \delta \partial_{\vec{x}} \Omega+ \partial_t \delta \Omega.$$
So we finally get
$$\delta L = (\delta \phi \times \vec{x}) \cdot \partial_{\vec{x}} L + (\delta \phi \times \dot{\vec{x}}) \cdot \partial_{\dot{\vec{x}}} L + \dot{\vec{x}} \cdot \partial_{\vec{x}} \delta \Omega+ \partial_t \delta \Omega=0.$$
This you can rewrite as
$$\delta \phi \cdot \left ( \vec{x} \times \partial_{\vec{x}} L + \dot{\vec{x}} \times \partial_{\dot{\vec{x}}} L\right ) + \dot{\vec{x}} \cdot \partial_{\vec{x}} \delta \Omega+ \partial_t \delta \Omega=0.$$
Obviously this can be fulfilled by setting $\delta \Omega=0$. Since then the remaining expression must vanish for all vectors $\delta \vec{\phi}$ you must have
$$\vec{x} \times \partial_{\vec{x}} L=0, \quad \dot{\vec{x}} \times \partial_{\dot{\vec{x}}}L=0,$$
This implies that
$$\partial_{\vec{x}} L \propto \vec{x}, \quad \partial_{\dot{\vec{x}}}L \propto \dot{\vec{x}},$$
and this is fulfilled, if
$$L=L(|\vec{x}|,|\dot{\vec{x}}|,t).$$

 

[gionole]

@vanhees71

hey. Have you read #21 reply ? It seems like your proof in #2 only works if $\frac{\partial L}{\partial x}$ doesn't contain $\dot x$, but it would if $L$ is containing product of $x$ and $\dot x$.

It seems like you're assuming that L won't have product of $x\dot x$, but are you able to make that assumption even for free particle ? If you're able to do this assumption without backing it up, why at all are we trying to show that L can't depend on x as it's the same type of assumption. Would be good to also show mathematically why L can't depend on product of $x\dot x$.

 

[vanhees71]

It can't have such a product, because for any function $f$
$$\partial_{\vec{x}} f(\vec{x} \cdot \dot{\vec{x}})=\dot{\vec{x}} f'(\vec{x} \cdot \vec{x}),$$
but $\vec{x} \times \dot{\vec{x}}$ doesn't vanish, because you can always choose arbitrary initial conditions, i.e., at least for this initial time you can always direct $\vec{x}$ and ##\dot{\vec{x}}# in arbitrary directions.

 

[gionole]

@vanhees71
Hm, now, if the $L = \frac{1}{2}mv^2 + x$ , your proof still holds in a sense that $L' - L$ can be represented by total time derivative ($\frac{d}{dt} at$) and not only that, and product terms of $x\dot x$ are absent. If you got such Lagrangian, your proof says that $\frac{\partial L}{\partial x} = 0$ while it's not.

What's your take on this ? Saying that you're considering a free particle doesn't make it true that we can't have $x$ in there just because of that. The whole idea is to prove why $x$ can't be there.

 

[vanhees71]

But $x$, a component of a vector, is not rotationally invariant, as should be clear from the analysis of #22.