Lagrangian doesn't change when adding total time derivative

[gionole]

TL;DR Summary

Lagrangian when adding total time derivative

When we have $L(q, \dot q, t)$, The change in action is given by:
$\int_{t_1}^{t_2} dt (\frac{\partial L}{\partial x} f(t) + \frac{\partial L}{\partial \dot x} \dot f(t))$ when we change our true path $x(t)$ by $x(t) + \epsilon f(t)$.

Now, attaching the image, check what Landau says.

I'm wondering whether he means that in the $f(q,t)$, whether q also depends on $t$ or not.

Let's say It doesn't. In this case, when we change $x(t)$ by $x(t) + \epsilon f(t)$, change in lagrangian $dL = \frac{\partial L}{\partial x} dx + \frac{\partial L}{\partial \dot x} d\dot x$ . What I did in there is even when true path changed with variation, $f(q,t)$ stayed the same in the new lagrangian(it didn't change), and we directly got dL to be the same as if we had used variation for only $L(q, \dot q, t)$ and not $L(q, \dot q, t) + \frac{d}{dt}f(q,t)$. In this case, if lagrangian didn't change, action wouldn't change and the derivation of euler lagrange would be for the same action as for the $L(q, \dot q, t)$. Would this logic apply ? but i also realize that this is only valid if $f(q,t)$ is just an ordinary function of some coordinate and time where coordinate doesn't act like in the same way as it does in the L itself. The coordinate in L is a function of t in itself.

Another way would be to just treat $L'(q, \dot q, t)$ as the normal lagrangian(just imagine that it doesn't include $\frac{d}{dt}f(q,t)$ and once I get to $\int_{t1}^{t2} (\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot x}) dt f(t)$, I could just plug in instead of L, the $L +\frac{d}{dt}f(q,t)$ and this still proves the equation of motion will be the same, but this here only would be correct if $f(q,t)$ is 0 at $t2$ and $t1$.

Please note that $\frac{d}{dt}f(q,t)$ is not the same thing as $\epsilon f(t)$ we use for variation of true path(at least this is my undestanding). So would my first way be correct ? is $f(q,t)$ such a function where q is just coordinate and nothing more(not a function of $t$) ? or my 2nd way is correct which assumes that this $f(q,t)$ has to be 0 at t2 and t1 ?

 

[renormalize]

TL;DR Summary: Lagrangian when adding total time derivative

Use the second equation in your snippet from Landau to write the difference:$$S^{\prime}-S=f\left(q^{\left(2\right)},t_{2}\right)-f\left(q^{\left(1\right)},t_{1}\right)\equiv f\left(q\left(t_{2}\right),t_{2}\right)-f\left(q\left(t_{1}\right),t_{1}\right)$$and then simply take the variation of this difference:$$\delta\left(S^{\prime}-S\right)=\delta\left(f\left(q\left(t_{2}\right),t_{2}\right)-f\left(q\left(t_{1}\right),t_{1}\right)\right)=\frac{\partial f\left(q\left(t_{2}\right),t_{2}\right)}{\partial q\left(t_{2}\right)}\delta q\left(t_{2}\right)-\frac{\partial f\left(q\left(t_{1}\right),t_{1}\right)}{\partial q\left(t_{1}\right)}\delta q\left(t_{1}\right)=0$$This vanishes because of condition (2.3) in Landau: $$\delta q\left(t_{1}\right)=\delta q\left(t_{2}\right)=0$$Hence, even though the Lagrangians $L^{\prime}$ and $L$ differ by a total derivative, they both yield precisely the same Euler-Lagrange equation.

 

[gionole]

I wanted to prove it differently.

we got $L' = L + \frac{d}{dt}(f(q,t)$

Let's say true path for this L' is $\hat q(t)$. Then our varied path is $q(t) = \hat q(t) + \epsilon k(t)$ (I brought variation of function named $k(t)$ so we make sure we know it's not the same as $f(q,t)$. Now, when $\epsilon$ changes by $d\epsilon$, our change of lagrangian is:

$dL' = \frac{\partial L}{\partial q} k(t) + \frac{\partial L}{\dot \partial q} \dot k(t) + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

$dL' = dL + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

The change in action would be integral over $dL'$

$dS = \int_{t_1}^{t_2} dL dt + \int_{t_1}^{t_2} \frac{d}{dt} \frac{\partial f}{\partial q} k(t) dt$

$k(t)$ at $t2$ and $t1$ is 0, hence $dS = \int_{t_1}^{t_2} dL dt$ which is the same thing.

This means:

1. our $f(q,t)$ could have been anything and not important whether at t2, t1, it's 0, even if it's not zero, we still yield at the same result due to $k(t)$ being in there.
2. in the $f(q,t)$, $q$ is not just a simple coordinate, but it's an equation of motion, remember, how I arrived with $dL'$. I arrived at $\frac{d}{dt} \frac{\partial f}{\partial q} k(t)$ because changing $\epsilon$ by $d\epsilon$, I wrote the change in $f(q,t)$ which means $q$ definitely is the same equation of motion as it's in L(q, \dot q, t) and not just the simple coordinate.

Would these 2 be correct ?

 

[vanhees71]

TL;DR Summary: Lagrangian when adding total time derivative

When we have $L(q, \dot q, t)$, The change in action is given by:
$\int_{t_1}^{t_2} dt (\frac{\partial L}{\partial x} f(t) + \frac{\partial L}{\partial \dot x} \dot f(t))$ when we change our true path $x(t)$ by $x(t) + \epsilon f(t)$.

Now, attaching the image, check what Landau says.

I'm wondering whether he means that in the $f(q,t)$, whether q also depends on $t$ or not.

Let's say It doesn't. In this case, when we change $x(t)$ by $x(t) + \epsilon f(t)$, change in lagrangian $dL = \frac{\partial L}{\partial x} dx + \frac{\partial L}{\partial \dot x} d\dot x$ . What I did in there is even when true path changed with variation, $f(q,t)$ stayed the same in the new lagrangian(it didn't change), and we directly got dL to be the same as if we had used variation for only $L(q, \dot q, t)$ and not $L(q, \dot q, t) + \frac{d}{dt}f(q,t)$. In this case, if lagrangian didn't change, action wouldn't change and the derivation of euler lagrange would be for the same action as for the $L(q, \dot q, t)$. Would this logic apply ? but i also realize that this is only valid if $f(q,t)$ is just an ordinary function of some coordinate and time where coordinate doesn't act like in the same way as it does in the L itself. The coordinate in L is a function of t in itself.

Another way would be to just treat $L'(q, \dot q, t)$ as the normal lagrangian(just imagine that it doesn't include $\frac{d}{dt}f(q,t)$ and once I get to $\int_{t1}^{t2} (\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot x}) dt f(t)$, I could just plug in instead of L, the $L +\frac{d}{dt}f(q,t)$ and this still proves the equation of motion will be the same, but this here only would be correct if $f(q,t)$ is 0 at $t2$ and $t1$.

Please note that $\frac{d}{dt}f(q,t)$ is not the same thing as $\epsilon f(t)$ we use for variation of true path(at least this is my undestanding). So would my first way be correct ? is $f(q,t)$ such a function where q is just coordinate and nothing more(not a function of $t$) ? or my 2nd way is correct which assumes that this $f(q,t)$ has to be 0 at t2 and t1 ?

Two Lagrangians are equivalent if they differ by a total time derivative of a function $\Omega(q,t)$, but $\Omega$ must not depend on the $\dot{q}$. That's, because the action has boundary constraints for the variations of the generalized coordinates but not for the generalized velocities. Only than the variation of the action and thus the equations of motion are the same using the two different Lagrangians.

You can see this directly from the Euler-Lagrange equations too. So let's set
$$L'(q,\dot{q},t)=L(q,\dot{q},t) + \mathrm{d}_t \Omega(q,t) = L(q,\dot{q},t) + \dot{q}^k \frac{\partial \Omega}{\partial q^k} + \partial_t \Omega(q,t).$$
Note that here and the following Einstein's summation convention is used, i.e., you have to sum over indices appearing twice in an equation.

Then the canonical momenta of the new Lagrangian are
$$p_j'=\frac{\partial L'}{\partial \dot{q}^j}=p_j + \frac{\partial \Omega}{\partial q^j}.$$
and the Euler-Lagrange equations
$$\dot{p}_k' = \dot{p}_j + \dot{q}^k \frac{\partial^2 \Omega}{\partial q^k \partial q^j} + \frac{\partial^2 \Omega}{\partial q^j \partial t}=\frac{\partial L'}{\partial q^k} = \frac{\partial L}{\partial q^k} + \dot{q}^k \frac{\partial^2 \Omega}{\partial q^j \partial t} + \frac{\partial^2 \Omega}{\partial q^j \partial t},$$
which are obviously fulfilled if and only if the Euler-Lagrange equations of the original Lagrangian are fulfilled,
$$\dot{p}_j=\frac{\partial L}{\partial q^j}.$$

 

[gionole]

I wanted to prove it differently.

we got $L' = L + \frac{d}{dt}(f(q,t)$

Let's say true path for this L' is $\hat q(t)$. Then our varied path is $q(t) = \hat q(t) + \epsilon k(t)$ (I brought variation of function named $k(t)$ so we make sure we know it's not the same as $f(q,t)$. Now, when $\epsilon$ changes by $d\epsilon$, our change of lagrangian is:

$dL' = \frac{\partial L}{\partial q} k(t) + \frac{\partial L}{\dot \partial q} \dot k(t) + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

$dL' = dL + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

The change in action would be integral over $dL'$

$dS = \int_{t_1}^{t_2} dL dt + \int_{t_1}^{t_2} \frac{d}{dt} \frac{\partial f}{\partial q} k(t) dt$

$k(t)$ at $t2$ and $t1$ is 0, hence $dS = \int_{t_1}^{t_2} dL dt$ which is the same thing.

This means:

1. our $f(q,t)$ could have been anything and not important whether at t2, t1, it's 0, even if it's not zero, we still yield at the same result due to $k(t)$ being in there.
2. in the $f(q,t)$, $q$ is not just a simple coordinate, but it's an equation of motion, remember, how I arrived with $dL'$. I arrived at $\frac{d}{dt} \frac{\partial f}{\partial q} k(t)$ because changing $\epsilon$ by $d\epsilon$, I wrote the change in $f(q,t)$ which means $q$ definitely is the same equation of motion as it's in L(q, \dot q, t) and not just the simple coordinate.

Would these 2 be correct ?

@vanhees71 what don't you like in this approach ?

 

[vanhees71]

What I didn't like is that in the 1st line it looked as if $f$ would also depend on $\dot{q}$. Maybe I should have read on more carefully. What you calculated is in fact not $d S$ but $\delta S$ and that this variation of the action is invariant when changing the Lagrangian by a total derivative of an arbitrary function of $q$ and $t$. Put in this way, it's of course correct. It must not depend on $\dot{q}$ though.

 

[gionole]

What I didn't like is that in the 1st line it looked as if $f$ would also depend on $\dot{q}$. Maybe I should have read on more carefully. What you calculated is in fact not $d S$ but $\delta S$ and that this variation of the action is invariant when changing the Lagrangian by a total derivative of an arbitrary function of $q$ and $t$. Put in this way, it's of course correct. It must not depend on $\dot{q}$ though.

Then all good. Thanks a lot.

 

[gionole]

Here is the tricky part I found. I will repeat my above proof first in this message for you to read easily. Thank you in advance for taking a look at this, I know it's not pleasant looking at this huge text below! so much appreciated.

$L' = L + \frac{d}{dt}(f(q,t)$

True path for this L' is $\hat q(t)$. Then our varied path is $q(t) = \hat q(t) + \epsilon k(t)$ (I brought variation of function named $k(t)$ so we make sure we know it's not the same as $f(q,t)$. Now, when $\epsilon$ changes by $d\epsilon$, our change of lagrangian is:

$dL' = \frac{\partial L}{\partial q} k(t) + \frac{\partial L}{\dot \partial q} \dot k(t) + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

$dL' = dL + \frac{d}{dt} \frac{\partial f}{\partial q} k(t)$

The change in action would be integral over $dL'$

$dS = \int_{t_1}^{t_2} dL dt + \int_{t_1}^{t_2} \frac{d}{dt} \frac{\partial f}{\partial q} k(t) dt$

$k(t)$ at $t2$ and $t1$ is 0, hence $dS = \int_{t_1}^{t_2} dL dt$ which is the same thing. So we showed $S' = S$

Now, Landau wants to calculate Lagrangian in two frames where speed is $v$ and $v+\epsilon$. In the frame where speed is $v$, we got $L(v^2)$. Let's write this in taylor around point $v$ and we get: $L(v^2) + \frac{dL}{dv^2}2v(x-v)$ and since $\epsilon$ is infinitely small, we can plug in $v+\epsilon$ instead of $x$ and we will get L' in $v+\epsilon$ frame.

$L = L(v^2)$
$L' = L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$

Since we're looking at two inertial frames, equation of motion should be the same in both, which means our lagrangians above should both yield the same result.

We know that adding total time derivative of $f(q,t)$ doesn't change it, so Landau assumes that $\frac{\partial L}{\partial v^2}2v\epsilon$ must be of such function(total time derivative of something). That's where I need your help. He says that this whole thing must be a linear function of v, which means since it already contains $v$, $\frac{\partial L}{\partial v^2}$ can't contain $v$ anymore and if so, his next point is great. But what I tried is I assumed $\frac{\partial L}{\partial v^2}$ contains $v$(it contains some other constant for sure, but not important as constants won't change anything) and if so, we will get:

$\frac{\partial L}{\partial v^2}2v\epsilon = v2v\epsilon = \frac{d}{dt}r2\epsilon\frac{dr}{dt} = \frac{d}{dt}2\epsilon r \frac{dr}{dt}$

I don't like using $r$, we can generalize and use $q$ since we need to be on the same coordinate system. So $\frac{d}{dt}2\epsilon q \frac{dq}{dt}$

If you remember, in the beginning we proved that for $L' = L + \frac{d}{dt}(f(q,t)$, we showed $L' = L$ in terms of solutions. So in our latest derivation, we can treat $f(q,t) = 2\epsilon q$ So we get:

$L' = L + \frac{d}{dt}(f(q,t)) \frac{dq}{dt}$. Now, we start with the same proof: our varied path is $q(t) = \hat q(t) + \epsilon k(t)$.

We're curious now when $\epsilon$ changes, how $L'$ changes. When $\epsilon$ changes, $dq = k(t)$, so we end up with:

$dL' = dL + \frac{d}{dt}\frac{\partial f}{\partial q} k(t) \frac{d}{dt} k(t)$

Integrating over this, we get:

$S' = S + \int_{t_1}^{t_2} \frac{d}{dt}\frac{\partial f}{\partial q} k(t) \frac{d}{dt} k(t) dt$ (note now that we got $k(t)$ inside the integral and it's 0 at t2 and t1 due to our variation principle. Hence, we proved $S' = S$.

So Landau only arrived to his final derivation of $1/2mv^2$ because he based his logic that $\frac{\partial L}{\partial v^2}$ couldn't contain $v$ because the right part of the $L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$ must have been a linear function of $v$, but I just showed that even if it's not linear($\frac{\partial L}{\partial v^2}$ contains $v$), he would still arrive at the same idea that actions don't change. I'm definitely thinking that I'm making a mistake somehow in the proof where I still arrive at $S' = S$ when $\frac{\partial L}{\partial v^2} = v$, while I shouldn't arrive here, because if so, Landau's logic is not strong.

 

[vanhees71]

If you know change your $dS$ to $\delta S$, everything looks good!

 

[gionole]

If you know change your $dS$ to $\delta S$, everything looks good!

I'm not sure what you mean. In the last message, Landau requires $\frac{\partial L}{\partial v^2}2v\epsilon$ to be linear function of $v$, which means $\frac{\partial L}{\partial v^2}$ can't contain $v$ because he says if it does, then total time derivative added to Lagrangian won't yield the same action, but I just showed that even if $\frac{\partial L}{\partial v^2}$ contains $v$ which in turn makes $\frac{\partial L}{\partial v^2}2v\epsilon$ non-linear of $v$, I still got $S' = S$ and this is what I don't get, because Landau's logic seems wrong then to base his logic that it must be a linear function of $v$. Attaching the image below.

 

[vanhees71]

It's just a question of the correct nomenclature. You are calulating variations of a functional and not differentials of a function.

The point of Landau is to derive the correct Lagrangian for a free particle from full Galilei invariance. You can just use all the convenient subgroups one at a time. Time translation, space translation, and rotation invariance are unproblematic, because for them you don't need an $\Omega$. Using these symmetries Landau comes to the conclusion that $L(\vec{x},\dot{\vec{x}},t)$ must in fact be of the form $L(\dot{\vec{x}}^2)$.

Now comes invariance under Galilei boosts, which is a bit more tricky, because here you need an $\Omega$, i.e., the Lagrangian is not invariant but only equivalent under Galilei boosts:
$$t'=t, \quad \vec{x}'=\vec{x}-\epsilon \vec{n} t.$$
The symmetry condition reads
$$-\vec{n} \cdot \frac{\partial L}{\partial \cdot \dot{\vec{x}}} + \dot{\vec{x}} \cdot \frac{\partial \Omega}{\partial \vec{x}} + \frac{\partial \Omega}{\partial t}=0.$$
Given the form of the Lagrangian as $L(\dot{\vec{x}}^2)$ this reads
$$-2 \vec{n} \cdot \dot{\vec{x}} L'(\dot{\vec{x}}^2) + \dot{\vec{x}} \cdot \frac{\partial \Omega}{\partial \vec{x}} + \frac{\partial \Omega}{\partial t}=0.$$
Obviously that's fulfilled only for
$$\Omega(\vec{x},t)=m \vec{x} \cdot \vec{n}, \quad m=\text{const}.$$
This implies
$$L'(\dot{\vec{x}}^2)=\frac{m}{2} \; \Rightarrow \; L=\frac{m}{2} \dot{\vec{x}}^2.$$

 

[gionole]

It really depends how you look at it. Landau more likely uses my approach - see #8. And i was curious where I made a mistake here. This would be better for me to learn where I made a mistake in there.

 

[vanhees71]

The mistake is that you have to add an arbitrary $\mathrm{d}/\mathrm{d} t f(q,t)$ to your $L'$ and then demand that $L'=L$. This determines, how $f$ must look and this then how $L$ must look. That's what Landau explains a bit hand-wavy with words ;-).

 

[gionole]

The mistake is that you have to add an arbitrary $\mathrm{d}/\mathrm{d} t f(q,t)$ to your $L'$ and then demand that $L'=L$. This determines, how $f$ must look and this then how $L$ must look. That's what Landau explains a bit hand-wavy with words ;-).

And what makes you think that I don’t add it ? Which equation from my reply do you mean ?

 

[vanhees71]

Maybe, I simply don't understand your calculation. Just compare it to what I wrote (my $\Omega$ is of course your $f$).

 

[gionole]

Maybe, I simply don't understand your calculation. Just compare it to what I wrote (my $\Omega$ is of course your $f$).

I think I dont understand your symmetry condition. Is it how L’ changes when you change $\epsilon$ ?

 

[vanhees71]

That's a bit lengthy, i.e., it's the derivation of Noether's theorem. Let's look at the special case, where time is not involved in the transformation, which is sufficient for the Galilei symmetry (except time translations, but these are also very easy to treat as a special case).

Then in the Lagrangian formulation you consider an infinitesimal transformation acting on the configuration variables $q^j$ only:
$$t'=t, \quad q_j'=q_j +\epsilon Q_j(q,\dot{q},t).$$
Then this is a symmetry if there exists a function $\Omega(q,t)$ such that
$$L(q',\dot{q}',t)=L(q,\dot{q},t) + \frac{\mathrm{d}}{\mathrm{d}t} \Omega(q,t).$$
Now we evaluate the left-hand side up to 1st order in $\epsilon$. For this we need
$$\dot{q}_j' = \dot{q}_j + \epsilon \frac{\mathrm{d}}{\mathrm{d} t} Q_j(q,\dot{q},t).$$
Thus, setting $\Omega=-\epsilon \tilde{\Omega}$, we have as the "symmetry condition"
$$L(q',\dot{q}',t)-L(q,\dot{q},t)=-\epsilon \tilde{\Omega} + \mathcal{O}(\epsilon^2).$$
So expanding the left-hand side to 1st order in $\epsilon$:
$$Q_k \frac{\partial L}{\partial q_k} + \frac{\mathrm{d} Q_k}{\mathrm{d} t} + \frac{\mathrm{d} \tilde{\Omega}}{\mathrm{d} t}=0. \qquad (*)$$
This must hold true for any $q(t)$, not only for the solutions of the Euler-Lagrange equations,
$$\frac{\partial L}{\partial q_k}=\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}_k}=\frac{\mathrm{d} p_k}{\mathrm{d} t}.$$
Now let's see, what (*) means for solutions of these equations of motion. Just using the Euler-Lagrange equations in the first term leads to
$$\frac{\mathrm{d} p_k}{\mathrm{d} t} Q_k + p_k \frac{\mathrm{d} Q_k}{\mathrm{d} t} + \frac{\mathrm{d} \tilde{\Omega}}{\mathrm{d} t}=0,$$
but that's obviously a total time derivative, i.e.,
$$\frac{\mathrm{d}}{\mathrm{d} t} \left ( p_k Q_k+\tilde{\Omega} \right)=0.$$
Thus the quantity in the brackets is conserved. This is Noether's theorem: For any symmetry of the variation of the action there's a conserved quantity.

Now let's use it for spatial translations and a single particle. It's most simple to express in Cartesian coordinates, i.e, we set $q_j=x_j$ ($j \in \{1,2,3 \}$).
$$t'=t, \quad \vec{x}'=\vec{x}+\epsilon \vec{n}, \quad \vec{Q}=\vec{n}=\text{const}.$$
The symmetry condition (*) reads
$$\vec{n} \cdot \frac{\partial L}{\partial \vec{x}} + \frac{\mathrm{d}}{\mathrm{d} t} \tilde{\Omega}=0$$
or
$$\vec{n} \cdot \frac{\partial L}{\partial \vec{x}} + \dot{\vec{x}} \frac{\partial \tilde{\Omega}}{\partial \vec{x}} + \frac{\partial \tilde{\Omega}}{\partial t}=0.$$
This can only be fulfilled, if $\tilde{\Omega}=\tilde{\Omega}(t)$, but if we assume that $L=L(\vec{x},\dot{\vec{x}})$ (which follows from time-translation invariance anyway) then we can just set $\tilde{\Omega}=0$. Since then for all $\vec{n}$
$$\vec{n} \cdot \frac{\partial L}{\partial \vec{x}}=0,$$
we must have
$$\frac{\partial L}{\partial \vec{x}}=0 \; \Rightarrow \; L=L(\dot{\vec{x}}).$$
Now consider an arbitary infinitesimal rotation
$$\vec{x}'=\vec{x}-\epsilon \vec{n} \times \vec{x} \; \Rightarrow \; \vec{Q}=-\vec{n} \times \vec{x}.$$
Then the symmetry condition for our case reads (again with $\tilde{\Omega}=0$)
$$(\vec{n} \times \dot{\vec{x}}) \cdot \frac{\partial L}{\partial{\dot{x}}}=\vec{n} \cdot \left (\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}} \right)=0.$$
Since this should hold for all $\vec{n}$ we must have
$$\dot{\vec{x}} \times \frac{\partial L}{\partial \dot{\vec{x}}}=0,$$
which means that
$$\frac{\partial L}{\partial \dot{\vec{x}}} \propto \dot{\vec{x}},$$
and that's the case if and only if
$$L=L(\dot{\vec{x}}^2).$$
Finally using the formalism for the boost, you get to what's written in #11.

 

[gionole]

Thanks so much for the amazing analysis. Though, I still want to say that you overcomplicate it. It still seems there's something i don't get. Could you exactly follow my approach ? I made it easier for you to follow my logic, I did my best so you don't have to think too much what I'm doing below. I described every single step. Unless you follow this exactly, you mightn't understand my logic. Hope you do that. I tried following your logic, but I already had started thinking about my way and even when I followed yours, I couldn't figure out what was wrong with mine.

Imagine we got our lagrangian = $L$. From this, we get: $\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0$.

Now, imagine we got our lagrangian = $L + \frac{d}{dt}f(q,t)$. By using the principle of variations(calculus), we arrive at $\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0$ (This means that when we solve this, ofc in the end, the equation of motion will be the exact same thing - not different by a bit, which is what we set out to prove)

Everything is clear till now. Now, follow me. Landau also now has 2 lagrangians.

Lagrangian 1: $L(v^2)$
Lagrangian 2: $L(v^2) + \frac{\partial L}{\partial v^2} 2v\epsilon$

He wants to prove that solving both of these will in the end get us the same equation of motion(i.e - same function - for example $x(t) = 2t^2 + 5t$ in both cases).

Well, The only way to prove this is somehow if he sees the exact resemblance between $\frac{\partial L}{\partial v^2} 2v\epsilon$ and $\frac{d}{dt}f(q,t)$, bingo, then he proved that their solutions will be the same, hence how it should be in inertial frames.

He thinks how can I give $\frac{\partial L}{\partial v^2} 2v\epsilon$ the same exact representation ? well, we definitely need $\frac{d}{dt}$ , so he goes ahead and changes $v$ by $\frac{dr}{dt}$, so we get: $\frac{d}{dt}\frac{\partial L}{\partial v^2} 2r\epsilon$. Now, is the trickiest part ever. He somehow thinks that if $\frac{\partial L}{\partial v^2}$ contains $v$, the exact resemblance between $\frac{\partial L}{\partial v^2} 2v\epsilon$ and $\frac{d}{dt}$ dies and for sure, if it dies, then those 2 lagrangians won't yield the same equation of motions. There's no way I might be wrong till this point.

Well, now, I step in and say to Landau: "hey, you definitely say if $\frac{\partial L}{\partial v^2}$ contains $v$, the exact resemblance between $\frac{\partial L}{\partial v^2} 2v\epsilon$ and $\frac{d}{dt}$ dies and you won't get the same equations for the Lagrangians. Landau says: "yes".

Ok, let's say $\frac{\partial L}{\partial v^2} = v$, so we got:

Lagrangian 1: $L(v^2)$
Lagrangian 2: $L(v^2) + v 2v\epsilon = L(v^2) + \frac{d}{dt}2r \frac{d}{dt}r\epsilon$ (changed both $v$ into dr/dt).

Now, are you really saying that these 2 lagrangians won't have the same equation of motion ? It seems like we can still resemble that. Look: $L(v^2) + \frac{d}{dt}2r \frac{d}{dt}r\epsilon = L(v^2) + \frac{d}{dt}2\epsilon r \frac{d}{dt}r$ It pretty much feels like to me that $\frac{d}{dt}2\epsilon r \frac{d}{dt}r$ has a form of $\frac{d}{dt}f(q,t)$ - well, the only difference is one more $\frac{d}{dt}r$ is included into it, but would it make any difference such as this Lagrangian wouldn't have the same equation of motion than $L(v^2)$ ? Let's see.

$\frac{d}{dt}2\epsilon r \frac{d}{dt}r$ From this, we can treat $f(q,t) = r2\epsilon \frac{dr}{dt}$ because it's still a function of q and t. (Note that we're using generalized coordinates, so $r$ can be treated as $q$). $\frac{dr}{dt}$ would still end up being the function of t, so the total $f(q,t)$ is correctly said to be treated as $r2\epsilon \frac{dr}{dt}$ and in front of it, we got our main $\frac{d}{dt}$. So this will get us the same $\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0$.

So, in the end, what logic I derive is that, it doesn't really matter whether $\frac{\partial L}{\partial v^2}$ contains $v$ or not. Landau insists that it shouldn't contain $v$ in order to show lagrangians would end up being the same equation of motion, but I insist, even if it contains $v$, it still ends up being the same.

 

[vanhees71]

I still don't understand what you are doing. The whole point is that if two Lagrangians are equivalent, their difference must be a total derivative of a function depending only on the $q$ and $t$. Here you want to enforce that Galilei boosts are symmetries, i.e., from Landau's (and oviously also yours) calculation it follows that for any Lagrangian of the form $L=L(\dot{\vec{q}}^2)$ (which follows from the symmetries under temporal and spatial translation and spatical rotation) that there must be a function $f(\vec{x},t)$ such that
$$2 \dot{\vec{x}} \cdot \vec{n} L'(\dot{\vec{x}}^2) =-\frac{\mathrm{d}}{\mathrm{d} t} f =-\dot{\vec{x}} \cdot \frac{\partial f}{\partial \vec{x}}-\partial_t f.$$
Now on the left-hand side nothing depends on $\vec{x}$, which implies that $\frac{\partial f}{\partial \vec{x}}=-\vec{n}$ or $f=-\vec{n} \cdot \vec{x}$.

Finally you get the conserved quantity
$$\vec{R}=\vec{x}-\frac{\vec{p}}{m} t=\text{const}$$
as it should be. Together with $\vec{p}=m \dot{\vec{x}}=\text{const}$, following from translation invariance. That's just that the particle moves with constant velocity in accordance with Newton's 1st postulate.

 

[gionole]

All good now, Finally clear.

Thanks so much.