# Equation of Motion Invariance in Galilean Mechanics

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.

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

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.

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.

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.

• 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). Feb 25 at 15:28
• I updated the answer. Feb 25 at 15:37
• 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? Feb 25 at 15:57
• 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? Feb 25 at 16:04

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).