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Consider a particle moving freely, where $\vec{r}(t)$ is the position of the particle. Suppose I move into a frame with $$\vec{r}' =\vec{r} + \epsilon \vec{F}(\vec{r}, t)\tag{1},$$ where $\epsilon$ is an infinitesimal variation and $\vec{F}(\vec{r}, t)$ is an arbitrary function (e.g. $\vec{F}(\vec{r}, t) = 1 \vec{u}$ for translation, $F(\vec{r}, t) = t \vec{u}$ for Galilean boosts, $F(\vec{r}, t) = \vec{u} t^2/2$ for accelerating frames, where $\vec{u}$ is a unit vector in an arbitrary direction etc.). The Lagrangian in the new frame is given by (to first order):

\begin{align} L(\vec{r}', \dot{\vec{r}'}, t) &\equiv L'(\vec{r}, \dot{\vec{r}}, t) = L(\vec{r} + \epsilon \vec{F}, \dot{\vec{r}}+ \epsilon \dot{\vec{F}}, t) \Leftrightarrow \end{align}

\begin{align} L' = L + \frac{\partial L}{\partial \vec{r}} \cdot \epsilon \vec{F} + \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \dot{\vec{F}} \tag{2}\end{align}

Using the chain rule (or integration by parts):

\begin{align} L' = L + \frac{\partial L}{\partial \vec{r}} \cdot \epsilon \vec{F} + \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) - \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}}) \cdot \epsilon \vec{F} \Leftrightarrow \end{align}

\begin{align} L' = L + \Big(\frac{\partial L}{\partial \vec{r}} - \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}}) \Big) \cdot \epsilon \vec{F}+ \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) \tag{3}\end{align}

The term after $L$ cancels out due to the Euler-Lagrange equation, therefore we are left with:

\begin{align} L' = L + \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) \tag{4}\end{align}

Now, we know that the Equation of Motion is the same if we add a total time derivative to the Lagrangian (since the action changes by a constant), so this equation would imply that the equation of motion is unchanged, independent of $\vec{F}$. Now, this is obviously not correct, because for example the equation of motion is not the same in an accelerating frame of reference ($F(\vec{r}, t) = \vec{u} t^2/2$).

Even though it feels wrong, I can't seem to pinpoint where the mistake is. Perhaps someone could help me understand what went wrong.

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  • $\begingroup$ Comment to the post (v3): $\vec{F}$ seems to depend on $\vec{u}$ not $\vec{r}$. $\endgroup$
    – Qmechanic
    Feb 25 at 14:38

4 Answers 4

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Well here minimizing the action is the same demanding zero variations both on $L'$ and $L$. That's good. But now when you use write the Euler-Lagrange equations you'll have one equation in $\overrightarrow{r}$ and one with $\overrightarrow{r}'$ but expressing one in terms of the other will give you a different equation. In practice you can't tell if your system is accelerating or not (not going into spacetime curvature here) if you're in a vacuum. So whatever generalized coordinates you choose to work with will give you the same expressions.

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You have very nicely proven Noether's Theorem here. If a given infinitesimal transformation, $q_i \rightarrow q_i^\prime = q_i + \varepsilon \, Q_i\left(\vec{q},t\right)$, is a symmetry of the Lagrangian, i.e., $$ L^\prime(\vec{q}^\prime, \dot{\vec{q}}^\prime, t) = L(\vec{q}^\prime, \dot{\vec{q}}^\prime, t) = L(\vec{q}, \dot{\vec{q}}, t) + O\left(\varepsilon^2\right)\,, $$ then there exists a constant of motion $$ \sum_i Q_i \frac{\partial L}{\partial \dot{q}_i}\,. $$ In other words, that time derivative must be zero if the transformation is to be a symmetry.


To address your main point: yes, it is true that if we put the primed Lagrangian back into an action (ignoring second-order terms in $\varepsilon$): $$ S^\prime = \int_{t_A}^{t_B} L^\prime(\vec{q}^\prime, \dot{\vec{q}}^\prime, t) \, dt \approx \int_{t_A}^{t_B} L(\vec{q}, \dot{\vec{q}}, t) + \varepsilon \frac{d}{dt} \left[ \sum_i Q_i \frac{\partial L}{\partial \dot{q}_i} \right] \, dt $$ then we find that: $$ S^\prime = \int_{t_A}^{t_B} L(\vec{q}, \dot{\vec{q}}, t) \, dt + \varepsilon \left[ \sum_i Q_i \frac{\partial L}{\partial \dot{q}_i} \right]_{t_A}^{t_B}\,. $$ Since the second term depends only on the end points of the path, variation there will be zero and therefore the Euler-Lagrange equations will be exactly what you would find from: $$ S = \int_{t_A}^{t_B} L(\vec{q}, \dot{\vec{q}}, t) \, dt $$ But note that we have, by making the Taylor expansion, written $L^{\prime}$ in the unprimed coordinates. The Euler-Lagrange equations we find will be in the unprimed coordinates: $$ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i}\, . $$ To find the equations in terms of $\vec{q}^\prime$, you would need to transform the Euler-Lagrange equations using your originally specified transformation (or, find the Euler-Lagrange equation associated to $L^\prime$ in its primed coordinates). Only when $$ \sum_i Q_i \frac{\partial L}{\partial \dot{q}_i} = \text{ const.} $$ will the equations of motion have the same form in both the primed and unprimed coordinates. That only occurs when your transformation of coordinates is a symmetry of the Lagrangian.


For example, let's take: $$ L(x,y, v_x, v_y) = \frac{1}{2} m \left( v_x^2 + v_y^2 \right) + U(x)\, . $$ Then the transformation of coordinates: $$ \begin{align} x^\prime &= x \\ y^\prime &= y + \varepsilon a \end{align} $$ yields: $$ L^\prime = L(x^\prime, y^\prime, v_x^\prime, v_y^\prime) = \frac{1}{2} m \left({v_x^\prime}^2 + {v_y^\prime}^2 \right) + U(x^{\prime}) = L(x, y, v_x, v_y) $$ and $m v_y$ is the conserved quantity.

A less trivial example is the two-body problem: $$ L(\vec{r}, \dot{\vec{r}}) = \frac{1}{2} \mu | \vec{r} |^2 + U(|\vec{r}|) $$ you can show that the transformation given by a small rotation $\Delta \theta$ about an arbitrary axis in the $\hat{n}$ direction: $$ \vec{r} \rightarrow \vec{r}^\prime = \vec{r} + \Delta \theta \,\hat{n} \times \vec{r} $$ leaves the Lagrangian unchanged to second order in $\Delta \theta$, and the angular momentum component along $\hat{n}$ is the conserved quantity.

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Briefly speaking, OP's mistake in eq. (4) [as compared to eq. (3)] is to prematurely use the equations of motion (EOM) in the Lagrangian, i.e. $L^{\prime}$ is not necessarily $L$ (modulo a total time derivative) off-shell. This mean that the corresponding Euler-Lagrange (EL) equations for $L^{\prime}$ and $L$ could be different.

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  • $\begingroup$ I don't see what you're saying Qmechanic. He uses the EoM properly in (3), right, since those are the E-L equations for $L$. In (4) he doesn't use the EoM, he just says that if you put that $L^\prime$ into an action you would get the "same EoM" (I think that's what @Andrei Cosmin is saying). $\endgroup$
    – Ben H
    Feb 25 at 15:28
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Feb 25 at 15:37
  • $\begingroup$ Are you just saying that it's not an equality in Eq (4) because of higher-order terms? Sorry, I don't really understand. Eq (4) is true up to those higher-order corrections. Or, if it is not true, could you explain why? $\endgroup$
    – Ben H
    Feb 25 at 15:57
  • $\begingroup$ Oh, maybe I understand now. Sadly, I don't think I knew what "off-shell" meant. You're saying that the analysis in Eq (2)-(4) --- in particular, assuming the E-L equation for $L$ in going from (3) to (4) --- only holds when $\vec{r}(t)$ is the actual path, that which extremizes the action. But the comparison of $L$ and $L^\prime$ should be made for arbitrary paths $\vec{r}$, prior to their insertion in the action? $\endgroup$
    – Ben H
    Feb 25 at 16:04
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Thank you for the great responses, this makes way more sense to me now. I also want to add my understanding of the answers and a clarification of the subtlety that confused me initially.

It is a mathematical fact that once we make a transformation of the form $\vec{r}' =\vec{r} + \epsilon \vec{F}(\vec{r}, t)$, we can always write

\begin{align} L' = L + \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) \tag{1}\end{align}

if $L(\vec{r}(t), \dot{\vec{r}(t)}, t)$ satisfies the Euler-Lagrange equation (I think this is what is meant by on-shell in @Qmechanic's answer).

Due to the total time derivative, this implies that the Euler-Lagrange equation will always have the same form with respect to $\vec{r}$:

\begin{align} \frac{\partial L}{\partial \vec{r}} - \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}}) = \frac{\partial L'}{\partial \vec{r}} - \frac{d}{dt}( \frac{\partial L'}{\partial \dot{\vec{r}}}) \tag{2}\end{align}

Now there are two separate cases that need to be considered for the Lagrangian $L'$.

Case 1: $L'$ is invariant under the transformation, meaning $L' = L$. This implies the well known result that (roughly speaking) any continuous symmetry leads to a conserved quantity with respect to time.

\begin{align} \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) = 0 \tag{3}\end{align}

Case 2: $L'$ is NOT invariant under the transformation. Hence,

\begin{align} \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) \neq 0 \tag{4}\end{align}

However, it seems to me that we can $\underline{\text{always}}$ (this is where the confusion started, see end of page Edit) write the Lagrangian $L'$ as:

\begin{align} L' = L + \frac{d}{dt}(\Lambda(t)) \tag{5}\end{align}

where $\Lambda(t)$ is some function of time. Therefore, Case 2 can also be written as:

\begin{align} \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F} - \Lambda(t)) = 0 \tag{6}\end{align}

For some examples, Case 1 is very well illustrated by @Ben H's answer. For Case 2, let's consider a slight tweak on @Ben H's first example.

\begin{align} x' &= x + \epsilon a \\ y' &= y \end{align}

\begin{align} L(x, y, v_x, v_y) = \frac{m}{2}(v_x^2 + v_y^2) + U(x) \tag{7}\end{align}

which yields:

\begin{align} L' = L(x', y', v_x', v_y') = \frac{m}{2}(v_x^2 + v_y^2) + U(x + \epsilon a) \tag{8}\end{align}

\begin{align} L' - L = U(x + \epsilon a) - U(x) = \frac{\partial U}{\partial x} \epsilon a \tag{9}\end{align}

where we identify $\Lambda(t) = \int \frac{\partial U}{\partial x} dt$. Therefore, equation (6) becomes:

\begin{align} \frac{d}{dt}( \frac{\partial L}{\partial \dot{\vec{r}}} \cdot \epsilon \vec{F}) = \frac{\partial U}{\partial x} \epsilon a \tag{10}\end{align}

using $\vec{F} = \hat{x} a$, (10) simplifies to:

\begin{align} \frac{d}{dt}( \frac{\partial L}{\partial \dot{x}}) = \frac{\partial U}{\partial x} \tag{10}\end{align}

which is just the equation of motion for the $x$ direction, as given by the Euler-Lagrange equation.

In conclusion, it did not occur to me until recently that Noether's theorem and the Euler-Lagrange equation are equivalent statements. What Noether's theorem seems to "explicitly" bring to the table is the link between conserved quantities and symmetries.

Edit:

Inspired by a similar post, I noticed that it is not technically correct to write $\Lambda(t) = \int \frac{\partial U}{\partial x} dt$ since $\Lambda(t)$ must be a $\underline{\text{local}}$ function. Therefore, equation (10) is not a consequence of equation (5).

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