# Partial Integration and the Levi-Civita Symbol

I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential.

I understand the generalized force that they derive from a Lagrangian of the form $$L = \frac{1}{2}m|\vec v|^2 - V(\vec r,\vec v,t)$$

$$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$

Through a series of steps that I still do not quite understand, the author derives the identity for the mixed velocity derivatives of the force:

$$\frac{\partial F_i}{\partial v_j\partial v_k} = 0$$

At this point, "by integrating this equation once" with respect to $$v_k$$ , they obtain the equation:

$$\frac{\partial F_i}{\partial v_j} = \sum_k \varepsilon_{ijk}B_k(\vec r,t)$$

where $$B_k$$ is the $$k^{th}$$ component of a vector function $$\vec B$$ that does not depend on velocity.

I'm having trouble understanding where this expression for the integral comes into play. The left-hand side clearly comes from the FTC. Were I to perform the integration myself I would do the same and include an arbitrary function
$$\frac{\partial F_i}{\partial v_j}=g(\vec r, v_1,...,v_{k-1}, v_{k+1},..., t)$$

where $$g$$ is a function that does not depend on $$v_k$$ explicitly. In this way $$\frac{\partial g}{\partial v_k} =0$$ as we need.

I've tried to work out how this function is related to the expression with $$B_k$$, but I cannot find any source that could point me in the right direction, especially because my best guess for $$g$$ depends on the other $$n-1$$ components of the velocity while the author's $$\vec B$$ vector is a function of position and time only.

Could I have some help understanding what's being done here?

## Edit: Additional important context

Additionally, Razavy goes a step further and assumes that the generalized force is independent of acceleration, just like the Lagrangian. Using this assumption, we can take the second condition listed in another related question I asked to form the anti-symmetry relation

$$\frac{\partial F_i}{\partial v_j} =- \frac{\partial F_j}{\partial v_i}$$

And then we can start taking partial derivatives, assuming all these derivatives are continuous. Taking the left side first:

$$\frac{\partial}{\partial v_k}(LHS)=\frac{\partial^2 F_i}{\partial v_j\partial v_k} = \frac{\partial^2 F_i}{\partial v_k\partial v_j} = \frac{\partial}{\partial v_j}\frac{\partial F_i}{\partial v_k}= \frac{\partial}{\partial v_j}\left(-\frac{\partial F_k}{\partial v_i}\right) = -\frac{\partial^2 F_k}{\partial v_i\partial v_j}$$

So, we can differentiate and swap the top index and a bottom index at the cost of a negative sign. In a similar way, the right hand side can be differentiated

$$\frac{\partial}{\partial v_k}(RHS)=-\frac{\partial^2 F_j}{\partial v_i\partial v_k}=\frac{\partial^2 F_k}{\partial v_i\partial v_j}$$

Thus, We can write: $$\frac{\partial}{\partial v_k}(LHS) = -\frac{\partial}{\partial v_k}(RHS)$$.

Because $$LHS=-RHS$$, we have

$$\frac{\partial}{\partial v_k}(LHS) = \frac{\partial^2 F_i}{\partial v_j\partial v_k} = 0$$

## 1 Answer

Yes, integrating with respect to one $$v_k$$ will give a right-hand side that depends on every $$v_l$$ with $$l\neq k$$. But notice we have one equation for every possible value of $$k$$, and in the end, the left hand side of the equation is always the same, and so should be the right hand side. Thus, we conclude that the function at the right hand side cannot depend on any of the variables $$v_l$$.

More precisely:

$$\frac{\partial F_i}{\partial v_j}=g(\vec r, v_1,...,v_{k-1}, v_{k+1},..., t) \\ = h_{i,j}(\vec r, v_1,...,v_{k}, v_{k+2},..., t) \\ = m_{i,j}(\vec r, v_1,...,v_{k-2}, v_{k},..., t) \\ = n_{i,j}(\vec r, t)$$

• I'm not quite sure that I understand fully. So you're saying that the original equation with $\frac{\partial F_i}{\partial v_j\partial v_k} = 0$ represents $n$ equations, with one for each $k$, correct? I hadn't realized this point before, so that helps a lot. And because this holds for each $v_k$ individually, the partial derivative $\frac{\partial F_i}{\partial v_j}$ is equal to each of these individually. Because each excludes one of the velocity components, $\frac{\partial F_i}{\partial v_j}$ cannot depend on any of them. Is this understanding correct? – spanishinquisitor Jul 4 at 19:05
• My other question was whether you named the functions with the subscripts $ij$ just to emphasize the relationship to the specific $v_j$ derivative we are taking for the $F_i$ component. Yes? – spanishinquisitor Jul 4 at 19:07
• @spanishinquisitor To your first question, yes, that's exactly it. To your second question, also yes. Notice that the left hand side depends on {i,j}, that is, for every pair, there is going to be a coresponding function $n_{i,j}$. I'm just a little bit confused as to why the function has to have the shape it has above, $\epsilon_{ijk} B_k$. – Lucas Baldo Jul 4 at 20:03
• For this shape to be needed we have to require that $\frac{\partial F_i}{\partial v_i} = 0$ and $\frac{\partial F_i}{\partial v_j} = -\frac{\partial F_j}{\partial v_i}$, which I can't see where it comes from. – Lucas Baldo Jul 4 at 20:13
• Great! I'm glad I followed you. I'm also confused as to where those symmetries come from. I suspect it may come from the "series of steps that I still do not quite understand" that I mentioned. I asked another question about this step in this question here. I'm trying to work out those antisymmetry relations from that. – spanishinquisitor Jul 4 at 23:09