- #1

Hamiltonian

- 296

- 190

- Homework Statement
- Suppose that ##K=\frac{1}{2}\sum g_{ab}\dot{\phi}^a\dot{\phi}^b## and that ##V=V(\phi^1,...,\phi^n)##. Verify that Newtons equations: ##\ddot{\mathbf{q}}=-\nabla V(\mathbf{q})##, are equivalent to the Euler-Lagrange equations with respect to the coordinates ##\phi^a##

- Relevant Equations
- The Euler-Lagrange Equation:

##\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial \dot{\phi}}\right )=\frac{\partial \mathcal{L}}{\partial \phi}##

In the past, I have shown relatively easily that if we have a lagrangian of the form ##\mathcal{L}=\frac{1}{2}\dot{\mathbf{q}}^2-V(\mathbf{q})## simply plugging this into the EL equation gives us newtons second law: ##\ddot{\mathbf{q}}=-\frac{\partial V}{\partial \mathbf{q}}##. I am unfamiliar with the metric(##g_{ab}##), my Professor told me that I could still do this problem if I simply assume that the metric is simply a ##n\times n## matrix.

Plugging the Lagrangian which in this case is, $$\mathcal{L}= K - V = \frac{1}{2}\sum g_{ab}\dot{\phi}^a \dot{\phi}^b - V(\phi^1,...,\phi^n)$$

into the EL equation should still be equivalent to Newton's equation.

The RHS of the EL equation is,

$$\frac{\partial\mathcal{L}}{\partial\phi^a}=-\frac{\partial V}{\partial \phi^a}(\phi^1,...,\phi^n)$$

the LHS is,

$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^a}=\frac{1}{2}\frac{\partial}{\partial \dot{\phi}^a}(\sum g_{ab}\dot{\phi}^a\dot{\phi}^b)$$

A hint my professor gave me to make things a little simpler was to consider a ##2\times 2## ##g_{ab}## and then later generalize to the ##n\times n## case,

$$g_{ab}=\begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{pmatrix}$$

$$\sum g_{ab}\dot{\phi}^a\dot{\phi}^b=g_{11}\dot{\phi}_1\dot{\phi}_1 + g_{12}\dot{\phi}_1\dot{\phi}_2+g_{21}\dot{\phi}_1\dot{\phi}_2+g_{22}\dot{\phi}_2\dot{\phi}_2$$

so then,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\frac{1}{2}\frac{\partial }{\partial \dot{\phi}_1}(g_{11}\dot{\phi}_1\dot{\phi}_1 + g_{12}\dot{\phi}_1\dot{\phi}_2+g_{21}\dot{\phi}_1\dot{\phi}_2+g_{22}\dot{\phi}_2\dot{\phi}_2)$$

and because ##g_{12} = g_{21}##, everything simplifies to,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\frac{1}{2}(2g_{11}\dot{\phi}_1+2g_{12}\dot{\phi}_2)$$

then for a general case, we can conclude by looking at the pattern that,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$

Putting the EL equation together we get,

$$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}\right ) = \sum_a g_{1a}\ddot{\phi}^a$$

$$\sum_a g_{1a}\ddot{\phi}^a = - \frac{\partial}{\partial \phi^a}(V(\phi^1,...,\phi^n))$$

from here I am a little confused my professor told me that since Newton's 2nd law in the problem statement is given in terms of the mass metric, by which he means that the mass term is absorbed into the ##\nabla##, we must multiply both sides of our equation by ##g^{ab}## which is defined as the inverse of the metric in tensor notation. so then we get,

$$g^{ab}\sum_{a}g_{1a}\ddot{\phi}^a=-g^{ab}\frac{\partial}{\partial \phi^a}(V(\phi^1,..,\phi^n))$$

from here I am not sure if I am done or if there is more simplification I need to do so that this looks even more like(or exactly like) Newton's equation. The LHS kind of looks like the acceleration as there is a second derivative and the right-hand side kind of looks like the gradient of the potential as we are differentiating ##V## but I don't know if this is it or there is more to be done.

Plugging the Lagrangian which in this case is, $$\mathcal{L}= K - V = \frac{1}{2}\sum g_{ab}\dot{\phi}^a \dot{\phi}^b - V(\phi^1,...,\phi^n)$$

into the EL equation should still be equivalent to Newton's equation.

The RHS of the EL equation is,

$$\frac{\partial\mathcal{L}}{\partial\phi^a}=-\frac{\partial V}{\partial \phi^a}(\phi^1,...,\phi^n)$$

the LHS is,

$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^a}=\frac{1}{2}\frac{\partial}{\partial \dot{\phi}^a}(\sum g_{ab}\dot{\phi}^a\dot{\phi}^b)$$

A hint my professor gave me to make things a little simpler was to consider a ##2\times 2## ##g_{ab}## and then later generalize to the ##n\times n## case,

$$g_{ab}=\begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{pmatrix}$$

$$\sum g_{ab}\dot{\phi}^a\dot{\phi}^b=g_{11}\dot{\phi}_1\dot{\phi}_1 + g_{12}\dot{\phi}_1\dot{\phi}_2+g_{21}\dot{\phi}_1\dot{\phi}_2+g_{22}\dot{\phi}_2\dot{\phi}_2$$

so then,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\frac{1}{2}\frac{\partial }{\partial \dot{\phi}_1}(g_{11}\dot{\phi}_1\dot{\phi}_1 + g_{12}\dot{\phi}_1\dot{\phi}_2+g_{21}\dot{\phi}_1\dot{\phi}_2+g_{22}\dot{\phi}_2\dot{\phi}_2)$$

and because ##g_{12} = g_{21}##, everything simplifies to,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\frac{1}{2}(2g_{11}\dot{\phi}_1+2g_{12}\dot{\phi}_2)$$

then for a general case, we can conclude by looking at the pattern that,

$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$

Putting the EL equation together we get,

$$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}\right ) = \sum_a g_{1a}\ddot{\phi}^a$$

$$\sum_a g_{1a}\ddot{\phi}^a = - \frac{\partial}{\partial \phi^a}(V(\phi^1,...,\phi^n))$$

from here I am a little confused my professor told me that since Newton's 2nd law in the problem statement is given in terms of the mass metric, by which he means that the mass term is absorbed into the ##\nabla##, we must multiply both sides of our equation by ##g^{ab}## which is defined as the inverse of the metric in tensor notation. so then we get,

$$g^{ab}\sum_{a}g_{1a}\ddot{\phi}^a=-g^{ab}\frac{\partial}{\partial \phi^a}(V(\phi^1,..,\phi^n))$$

from here I am not sure if I am done or if there is more simplification I need to do so that this looks even more like(or exactly like) Newton's equation. The LHS kind of looks like the acceleration as there is a second derivative and the right-hand side kind of looks like the gradient of the potential as we are differentiating ##V## but I don't know if this is it or there is more to be done.

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