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I am doing a course on Lagrangian mechanics and the instructor mentioned that the numerical value of the Lagrangian is conserved when I shift between two inertial reference frames, even though their functional forms might be different. He then mentioned that this is unlike what happens with the Hamiltonian.

I am having trouble with the first part of this itself. If I have a block of mass $m$ moving with a uniform velocity $v_0 = \dot x$ with respect to the ground, then I can say that it's Lagrangian is given by

$$ L = \frac 1 2 m(\dot x)^2 + mgy$$

However, if I am sitting next to the block and moving along with it, then I would consider it to be at rest. And since it is not moving along $y$, I would write down it's Lagrangian as

$$L_1 = mgy$$

How is the value of the Lagrangian then conserved?

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  • $\begingroup$ Related: physics.stackexchange.com/q/526603/2451 and links therein. $\endgroup$
    – Qmechanic
    May 31, 2020 at 7:06
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    $\begingroup$ The adjective you meant to use is invariant, not conserved. $\endgroup$
    – J.G.
    May 31, 2020 at 18:52
  • $\begingroup$ Yes, I came here to leave the same comment as J.G. Normally "conserved" is used for quantities which do not change as the system evolves over time, whereas "invariant" is used for quantities which are the same regardless of reference frame. Conserved quantities are not necessarily invariant and vice versa! $\endgroup$ May 31, 2020 at 21:33
  • $\begingroup$ I didn't mean to ask if the Lagrangian is invariant. I understand that the Lagrangian is invariant under point transformations. I had this question because the instructor explicitly wrote down that the value of the Lagrangian does not depend on the coordinates. $\endgroup$
    – newtothis
    Jun 1, 2020 at 4:39

4 Answers 4

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If you're dealing with a classical Lagrangian as you have mentioned, then the numerical value is definitely not conserved! For example, let's work with just the kinetic term. If a particle of mass $m$ is moving relative to me at some velocity $\dot{x}$, then it has Lagrangian $$ L_\text{me}= \frac{1}{2} m\dot{x}^2. $$ Now, if I move into the frame of the particle it looks still, so $$ L_\text{particle}= 0.$$ However, this is where special relativity comes into play. It's a theory that we use to make sense of being able to work in different frames, and most importantly define frame invariant quantities. We can define a timelike (particle) 4-vector $x^\mu(\tau)$, where $x^\mu(\tau) = (t, \mathbf{x})$, and each element is parametrised by our proper time $\tau$. From this, we define what we call a Lorentz invariant Lagrangian, which is indeed independent of what inertial frame we're in, since the kinetic part of timelike 4-vectors just evaluates to a constant (usually -1). You can look this up yourself. $$ L = -1 = \frac{1}{2} m^2 \dot{x} ^\mu \dot{x} _ \mu$$.

Basically, you need to upgrade the theory and start reading about some space-time geometry :)

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Equivalent Lagrangians

Two Lagrangians which yield the same equations of motion are called equivalent Lagrangians. It follows that Lagrangians which differ by a constant are equivalent Lagrangians (because the equations of motion derived from them turn out to be the same). In your case, $\mathcal L-\mathcal L_1=mv_0 ^2/2=\rm constant$, thus $\mathcal L$ and $\mathcal L_1$ are equivalent. For a general case, if two Lagrangians ($\mathcal L$ and $\mathcal L'$) are equivalent then

$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(\mathcal L-\mathcal L')}{\partial \dot q_i}\right)=0 \quad \text{and} \quad \frac{\mathrm d (\mathcal L-\mathcal L')}{\mathrm dq_i}=0 \qquad \forall \:\: i\tag{1}$$

Derivation

Let's prove that the Lagrangians in any two inertial reference frames will always be equivalent. Let the Lagrangian in the ground frame be

$$\mathcal L=T-V=\sum_i^n \frac 1 2 m_i |\mathbf v_i|^2-\sum_j^m\left(\sum_i^n V_i(q_j)\right)$$

where $m_i$ and $\mathbf v_i$ are the mass and the velocity of the $i$-th particle, and $V_i$ is the potential of the $i$-th particle dependent of the coordinate $q_j$. Now let's move to a refrence frame moving with a velocity $\mathbf v$, then the new Lagrangian is

$$\mathcal L'=T'-V'=\sum_i^n \frac 1 2 m_i |\mathbf v_i-\mathbf v|^2-\sum_j^m\left(\sum_i^n V_i(q)\right)$$

Therefore,

$$\mathcal L-\mathcal L'=\sum_i^n \frac 1 2 m_i |\mathbf v_i|^2-\sum_i^n \frac 1 2 m_i |\mathbf v_i-\mathbf v|^2$$

Now using the fact that $|\mathbf v_i-\mathbf v|^2=|\mathbf v_i|^2+|\mathbf v|^2-2\mathbf v_i\cdot \mathbf v$, we get

\begin{align} \mathcal L-\mathcal L'&=\sum_i^n \frac 1 2 m_i (|\mathbf v_i|^2-(|\mathbf v_i|^2+|\mathbf v|^2-2\mathbf v_i\cdot \mathbf v))\\ &= \sum_i^n m_i (\mathbf v_i\cdot \mathbf v)-\sum_i^n \frac 1 2 m_i |\mathbf v|^2\\ &= \left(\sum_i^n m_i \mathbf v_i\right)\cdot \mathbf v-\sum_i^n \frac 1 2 m_i |\mathbf v|^2 \end{align}

Now we'll use the fact that $\displaystyle\sum_i^n m_i \mathbf v_i=M\mathbf v_{\rm COM}$, where $\displaystyle M=\sum_i^n m_i$ and $\mathbf v_{\rm COM}$ is the velocity of the center of mass. Thus,

\begin{align} \mathcal L-\mathcal L'&=M(\mathbf v_{\rm COM}\cdot \mathbf v)-\sum_i^n \frac 1 2 m_i |\mathbf v|^2 \end{align}

Let the coordinate along the direction of $\mathbf v$ be $q_k$, then $\mathbf v_{\rm COM}\cdot\mathbf v=\dot{q_k}|\mathbf v|$. Substituting this, we get

\begin{align} \mathcal L-\mathcal L'&=M\dot{q_k}|\mathbf v|-\sum_i^n \frac 1 2 m_i |\mathbf v|^2\\ \mathcal L-\mathcal L'&=M\dot{q_k}|\mathbf v|-\rm constant \end{align}

Now testing the conditions in equation $(1)$, we get:

\begin{align} \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(\mathcal L-\mathcal L')}{\partial \dot q_k}\right)&=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(M\dot{q_k}|\mathbf v|-\rm constant)}{\partial \dot q_k}\right)\\ &=\frac{\mathrm d}{\mathrm dt}(M |\mathbf v|)\\ &=0\\ &\text{and}\\ \frac{\mathrm d(\mathcal L-\mathcal L')}{\mathrm d q_k}&=\frac{\mathrm d(M\dot{q_k}|\mathbf v|-\rm constant)}{\mathrm d q_k}\\ &=0 \end{align}

As you see that the conditions in equation $(1)$ hold true, which implies that both the Lagrangians are equivalent. This could also be trivially proven by showing the invariance of equations of motions under changing from one inertial reference frame to another.

Conclusion

Your instructor's statement (as you stated) is completely incorrect. However, it is correct to say that Lagrangians obtained in different inertial frames of reference are always equivalent.

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For a free particle the potential is zero and the numerical value of the Lagrangian is equal to that of the Hamiltonian. I don't see how they should behave differently under coordinate transformations (like Galilei or Lorentz boosts). Possibly, it was the action that was meant, not the Lagrangian.

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Your professor is totally wrong. The Lagrangian $\mathcal{L}$ can always have an additive constant, or even a non-zero multiplicative constant. So $\mathcal{L}+c$, or $k\mathcal{L}$ also equally satisfy the Euler Lagrange equations, if $\mathcal{L}$ also does so. Even a total time derivative of a function, when added to $\mathcal{L}$, does not alter its ability to satisfy the E-L equations.

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