Verifying that Newton's Equations are equivalent to the EL equations

  • #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.
 
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  • #2
The equation, $$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$ is incorrect. The indices don't match.
 
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  • #3
Hill said:
The equation, $$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$ is incorrect. The indices don't match.
they do in the ##2\times 2## case right?
$$\sum_{a=1}^{2} g_{1a}\dot{\phi}^a = g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$
or are you trying to say that this pattern changes, and becomes more complicated when we deal with a higher-order matrix(##g_{ab}##) so that's why this equation is incorrect?
 
  • #4
Hamiltonian said:
they do in the ##2\times 2## case right?
$$\sum_{a=1}^{2} g_{1a}\dot{\phi}^a = g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$
or are you trying to say that this pattern changes, and becomes more complicated when we deal with a higher-order matrix(##g_{ab}##) so that's why this equation is incorrect?
They don't match because the index ##a## is a free index on the left, but it is a dummy index on the right:
$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$
 
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  • #5
Hill said:
They don't match because the index ##a## is a free index on the left, but it is a dummy index on the right:
$$\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_a g_{1a}\dot{\phi}^a$$
ok, I see the mistake.
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}_2+g_{21}\dot{\phi}_1$$
so how could I combine these to write a more general formula for the ##n\times n## metric?
I am not too familiar with this kind of manipulation of indices, is this supposed to be relatively straightforward, or are there any resources I could refer to that will make my life easier?
 
  • #6
Hamiltonian said:
ok, I see the mistake.
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}_2+g_{21}\dot{\phi}_1$$
so how could I combine these to write a more general formula for the ##n\times n## metric?
I am not too familiar with this kind of manipulation of indices, is this supposed to be relatively straightforward, or are there any resources I could refer to that will make my life easier?
I think it is straightforward. Take the first equation: $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$ There is index ##1## on the left. Where does it appear on the right?
Take the second: $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}_2+g_{21}\dot{\phi}_1$$ The index on the left is ##2##. Where does it appear on the right? See the pattern?

Another mistake: on the right, the coordinates should be ##\dot{\phi}^1## and ##\dot{\phi}^2## rather than ##\dot{\phi}_1## and ##\dot{\phi}_2##.
 
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  • #7
Hill said:
I think it is straightforward. Take the first equation: $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}_1+g_{12}\dot{\phi}_2$$ There is index ##1## on the left. Where does it appear on the right?
Take the second: $$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}_2+g_{21}\dot{\phi}_1$$ The index on the left is ##2##. Where does it appear on the right? See the pattern?

Another mistake: on the right, the coordinates should be ##\dot{\phi}^1## and ##\dot{\phi}^2## rather than ##\dot{\phi}_1## and ##\dot{\phi}_2##.
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^1}=g_{11}\dot{\phi}^1+g_{12}\dot{\phi}^2$$
$$\frac{\partial\mathcal{L}}{\partial\dot{\phi}^2}=g_{22}\dot{\phi}^2+g_{21}\dot{\phi}^1$$
$$\implies \frac{\partial\mathcal{L}}{\partial \dot{\phi}^a}=\sum_{b=1}^n g_{ab}\dot{\phi}^b$$
now if we plug this into our EL equation we get,
$$\frac{d}{dt}\left(\sum_{b=1}^n g_{ab}\dot{\phi}^b \right)=\sum_{b=1}^n g_{ab}\ddot{\phi}^b$$
$$\sum_{b=1}^n g_{ab}\ddot{\phi}^b = -\frac{\partial}{\partial\phi^a}(V(\phi^1,...,\phi^n))$$
so now I go back to my original question,
Hamiltonian said:
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,
$$g^{ab}\sum_{b=1}^n g_{ab}\ddot{\phi}^b = -g^{ab}\frac{\partial}{\partial\phi^a}(V(\phi^1,...,\phi^n))$$
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.
would it make more sense to multiply both sides by ##\sum g^{ab}## because then on the LHS if you can absorb both sums into one(although I am not sure if you can do that because in general ## \sum a_i \sum b_i \ne \sum a_i b_i ##) leading to getting ##g^{ab}g_{ab}=I##?
 
  • #8
  • #9
Orodruin said:
##g_{ab}## is not generally constant. You cannot move it in and out of a time derivative freely.

I also strongly suggest that you adopt the Einstein summation convention (indices repeated twice imply a sum over the range of the index, sums are not explicitly written out) and read this: https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/
ok so in Einstein summation convention, my final equation reads,
$$g_{ab}\ddot{\phi}^b=-\frac{\partial}{\partial \phi^a}\left (V(\phi^1,...,\phi^n) \right )$$
is it fair to say that I have solved the given problem?
Have I shown this equation that I have obtained by plugging the Lagrangian into the EL Lagrange equation is equivalent to Newton's equation, $$\ddot{\mathbf{q}}=-\nabla V(\mathbf{q})$$ or do I still have more work to do?
The equation kind of resembles this, but again I am not familiar with the matric and only know it's a matrix I don't know how to interpret it here that's why I am having a hard time telling if I am done or not.
 
  • #10
Hamiltonian said:
ok so in Einstein summation convention, my final equation reads,
$$g_{ab}\ddot{\phi}^b=-\frac{\partial}{\partial \phi^a}\left (V(\phi^1,...,\phi^n) \right )$$
is it fair to say that I have solved the given problem?

No, it is incorrect unless ##g_{ab}## are constants as per my first comment. To connect to your other thread, if you wrote this down in polar coordinates with ##V = 0## you would obtain simply ##\ddot r = \ddot \phi = 0##, which is not the case for a particle moving in a straight line.

Hamiltonian said:
Have I shown this equation that I have obtained by plugging the Lagrangian into the EL Lagrange equation is equivalent to Newton's equation, $$\ddot{\mathbf{q}}=-\nabla V(\mathbf{q})$$ or do I still have more work to do?
In the case of Cartesian coordinates, ##g_{ab} = \delta_{ab}## are the components of the identity matrix and it is then directly what you have. In the general case, it is more delicate.
 
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  • #11
A missing piece of information in this thread is the relationship between ##\mathbf q## and the ##\phi^a##.
 
  • #12
Orodruin said:
No, it is incorrect unless ##g_{ab}## are constants as per my first comment. To connect to your other thread, if you wrote this down in polar coordinates with ##V = 0## you would obtain simply ##\ddot r = \ddot \phi = 0##, which is not the case for a particle moving in a straight line.
Ok, so since ##g_{ab}## is not generally constant I will have on the LHS of my EL equation,
$$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial \dot{\phi}^a} \right )=\frac{d}{dt}(g_{ab}\dot{\phi}^b)$$
from here would it make sense to use the product rule? although I wouldn't really know what to do with ##\frac{d}{dt}(g_{ab})## and since it's a matrix I would have to apply ##\frac{d}{dt}## to every single one of its entries, but I don't know what the matrix is except for a bunch of ##g_{ij}##'s which may or may not be functions of time.
Nothing is given about the specifics of the metric, could you help me on how I could arrive at Newtons equations from what I have, if that's even possible?
 
  • #13
In general, the metric components are functions of the coordinates ##\phi^a## so
$$
\frac{d}{dt} g_{ab} = \frac{d\phi^c}{dt} \frac{\partial g_{ab}}{\partial \phi^c} \equiv \dot \phi^c \partial_c g_{ab}
$$
by the chain rule of derivatives.

The generalized version of Newton’s second law is on the form
$$
A_a \equiv \dot \phi^c \nabla_c U_a = - \partial_a V
$$
which is basically what you have modulo some definitions you are likely unfamiliar with based on what we have discussed so far.

Edit: One can write entire textbook chapters on this (I have done it). Containing everything in a single discussion forum thread is a bit much.
 
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  • #14
Hamiltonian said:
The equation kind of resembles this, but again I am not familiar with the matric and only know it's a matrix I don't know how to interpret it here that's why I am having a hard time telling if I am done or not.
You didn't follow through with the idea of doing the proof in the 2 x 2 case. You only did the first step. I suspect the idea was to prove the whole thing in the simplest case.
 
  • #15
Orodruin said:
In general, the metric components are functions of the coordinates ##\phi^a## so
$$
\frac{d}{dt} g_{ab} = \frac{d\phi^c}{dt} \frac{\partial g_{ab}}{\partial \phi^c} \equiv \dot \phi^c \partial_c g_{ab}
$$
by the chain rule of derivatives.

The generalized version of Newton’s second law is on the form
$$
A_a \equiv \dot \phi^c \nabla_c U_a = - \partial_a V
$$
which is basically what you have modulo some definitions you are likely unfamiliar with based on what we have discussed so far.

Edit: One can write entire textbook chapters on this (I have done it). Containing everything in a single discussion forum thread is a bit much.
ok so for the LHS of the EL equation, we have,
$$\frac{d}{dt}\left (\frac{\partial\mathcal{L}}{\partial\dot{\phi}^a} \right)=\frac{d}{dt}(g_{ab}\dot{\phi}^b)=\frac{d}{dt}(g_{ab})\dot{\phi}^b+g_{ab}\ddot{\phi}^b$$
$$
\frac{d}{dt} g_{ab} = \frac{d\phi^c}{dt} \frac{\partial g_{ab}}{\partial \phi^c} \equiv \dot \phi^c \partial_c g_{ab}
$$
$$
\frac{d}{dt}\left ( \frac{\partial\mathcal{L}}{\partial\dot{\phi}^a} \right)= \dot{\phi^c}\dot{\phi}^b \partial_c g_{ab}+g_{ab}\ddot{\phi}^b=\frac{\partial\mathcal{L}}{\partial\phi^a} = -\partial_a V
$$
so really the equation is,
$$
\dot{\phi}^c\dot{\phi}^b \partial_c g_{ab}+g_{ab}\ddot{\phi}^b=-\partial_a V
$$
and you are saying that this is equivalent to the generalized Version of Newton's second law,
$$
A_a \equiv \dot \phi^c \nabla_c U_a = - \partial_a V
$$
The LHS looks a bit dissimilar but I'll take your word for it.

I actually happen to have your Math methods textbook, would going over chapters 1 & 2 suffice to understand exactly everything that's needed for this problem?
 
  • #16
Hamiltonian said:
I actually happen to have your Math methods textbook, would going over chapters 1 & 2 suffice to understand exactly everything that's needed for this problem?
Yes, in essence. As long as you are doing curvilinear coordinates on a Euclidean space only reading up to 2.3.2 will suffice for what you want to know here. If you want to do more general curved spaces chapter 9, but I’d consider that too advanced for your purposes here. Particularly relevant would be Equations (2.82) and (2.86) with ##v_b = g_{ba} \dot \phi^a## in the former and the latter for the particular form of the Christoffel symbols.

I postponed referring to the generalisation of Newton's second law to quite late in the applications chapter (10), to write it in its "full glory"
$$
M(\nabla_{\dot\gamma} \dot\gamma) = - dV,
$$
but fully understanding the components of this expression requires most of chapter 9.

Hamiltonian said:
$$
\dot{\phi}^c\dot{\phi}^b \partial_c g_{ab}+g_{ab}\ddot{\phi}^b=-\partial_a V
$$
and you are saying that this is equivalent to the generalized Version of Newton's second law,
$$
A_a \equiv \dot \phi^c \nabla_c U_a = - \partial_a V
$$
The LHS looks a bit dissimilar but I'll take your word for it.

Note that ##U_a = g_{ab} \dot\phi^b## and things should fall somewhat into place along with
$$
\nabla_c U_a = \partial_c U_a - \Gamma_{ca}^b U_b,
$$
where you find the expression for the Christoffel symbol in Equation (2.86).
 
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