Having trouble taking derivative of a cross product when finding Lagrangian to find force equation for rotating non-inertial frame

Let $G:Bbb{R}^3to Bbb{R}^3$ be defined as $G(r) = Omega times r$. Note that $G$ is a linear transformation in the sense of linear-algebra ($G(cr_1 + r_2) = c G(r_1) + G(r_2)$ for all $cin Bbb{R}, r_1,r_2in Bbb{R}^3$). So, at any point $ainBbb{R}^3$, its derivative (or total derivative, or Frechet derivative whatever terminology you prefer most) is simply $dG_a(cdot) = G(cdot)$. In other words, at every point, $G$ is its own derivative.

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Really Explicit and Systematic Approach To Differentiation.

But, this by itself is probably not very helpful for you in simplifying the math. I'll first do this the systematic way, and then later you can see the "quick way" (which is still rigorous if everything is interpreted properly). You have a Lagrangian $L:Bbb{R}^3times Bbb{R}^3to Bbb{R}$ defined as begin{align} L(a,b) &= dfrac{m}{2}lVert b + Omega times arVert^2 - U(a) tag{$*$} end{align}

Now, to be precise, we shall define the following functions:

• $r:Bbb{R}^3times Bbb{R}^3to Bbb{R}^3$, $r(a,b) := a$. In words, $r$ is that function which eats a pair of vectors and spits out the first vector

• $v:Bbb{R}^3times Bbb{R}^3to Bbb{R}^3$, $v(a,b) := b$. So, $v$ is that function which eats a pair of vectors and spits out the second vector.

With this, we can rewrite $(*)$ as follows: begin{align} L &= frac{m}{2}lVert v + Omega times r rVert^2 - Ucirc r end{align} (The only reason I explicitly defined these functions is so that we have a proper, mathematically correct equality between functions; where both the LHS and RHS are functions $Bbb{R}^3times Bbb{R}^3to Bbb{R}$, and they agree at every point of their domain).

Now, here are the "rules" of differentiation you need to know:

• If $T:Vto W$ is a linear transformation between finite-dimensional spaces, then for every point $ain V$, we have $dT_a = T$. In words, this says a linear transformation is its own derivative.

• Chain Rule: if $f:Vto W$ and $g:Wto X$ are differentiable functions, then for every $ain V$, we have $d(gcirc f)_a = dg_{f(a)} circ df_a$.

• A "generalized product rule". Essentially, it amounts to "keep the first, differentiate the second plus differentiate the first keep the second"; but see the link for the precise statement.

So, for example, since $r(a,b) = a$ is a linear transformation, we have $dr_{(a,b)} = r$; or even more explicitly, for all $(a,b), (xi,eta)in Bbb{R}^3times Bbb{R}^3$, we have begin{align} dr_{(a,b)}(xi,eta) = r(xi,eta) = xi. end{align} Similarly, $dv_{(a,b)} = v$, or more explcitly, $dv_{(a,b)}(xi,eta) = v(xi,eta) = eta$. Next, by combining this with the chain rule, we have begin{align} d(Ucirc r)_{(a,b)} = dU_{r(a,b)} circ dr_{(a,b)} = dU_a circ r end{align}

Finally, by recalling that $lVert xirVert^2 = langle xi,xirangle$, we can calculate the derivative of the function $lVert v + Omega times rrVert^2$, and this is for any $(a,b)in Bbb{R}^3times Bbb{R}^3$, begin{align} dleft(lVert v + Omega times rrVert^2right)_{(a,b)}(xi,eta) &= 2 langle (v + Omega times r)(a,b), d(v + Omega times r)_{(a,b)}(xi,eta)rangle &= 2langle b + Omega times a, dv_{(a,b)}(xi,eta) + Omega times dr_{(a,b)}(xi,eta) rangle &= 2langle b + Omega times a, eta + Omega times xi rangle &= 2 bigg(langle b + Omega times a, etarangle + langle b + Omega times a, Omega times xirangle bigg) end{align} Now, we use the well-known identity $langle alpha, betatimes gammarangle = langle alphatimes beta, gamma rangle$ (i.e $alpha cdot (betatimes gamma) = (alphatimes beta)cdot gamma$) to the last term to get begin{align} dleft(lVert v + Omega times rrVert^2right)_{(a,b)}(xi,eta) &= 2 bigg( langle (b + Omega times a)times Omega, xirangle + langle b + Omega times a, etarangle bigg) end{align}

So, if you combine everything I've said so far, we have that begin{align} dL_{(a,b)}(xi,eta) &= m bigg( langle (b + Omega times a)times Omega, xirangle + langle b + Omega times a, etarangle bigg) - dU_a(xi) end{align}

Now, recall that the gradient is defined as the vector $nabla U(a)$ such that for all $xi in Bbb{R}^3$, begin{align} langle nabla U(a), xirangle = dU_a(xi). end{align}

Hence, the equation becomes begin{align} dL_{(a,b)}(xi,eta) &= langle m [(b + Omega times a)times Omega] - nabla U(a),,, xirangle + mlangle b + Omega times a, eta rangle tag{$ddot{smile}$} end{align}

In other words, the way to interpret this result is that begin{align} begin{cases} frac{partial L}{partial r}(a,b) &= m [(b + Omega times a)times Omega] - nabla U(a) frac{partial L}{partial v}(a,b) &= m(b+ Omega times a) tag{i} end{cases} end{align}

Or if you don't plug in the points of evaluation, then we can rewrite this result as:

begin{align} begin{cases} frac{partial L}{partial r} &= m [(v + Omega times r)times Omega] - (nabla U) circ r frac{partial L}{partial v} &= mleft(v+ Omega times rright) tag{ii} end{cases} end{align} and in this form it's a proper equality of functions $Bbb{R}^3times Bbb{R}^3 to Bbb{R}^3$ (and in this notation, things are also probably more familiar to you). Now, I'm sure you can apply the Euler-Lagrange equations, and rearrange terms appropriately to get the desired equations of motion.

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"Quicker" Way of Performing the Differentiation

Now, the "quick way" of deriving the result is to simply not include the point of evaluation $(a,b)$ of the derivatives, and not include the vectors $(xi,eta)$ on which the derivative is evaluated. Also, let's write the inner product using a $cdot$ rather than $langle cdot, cdotrangle$. Then, from begin{align} L&= frac{m}{2}lVert v+ Omega times rrVert^2 - Ucirc r, end{align} we get begin{align} dL &= frac{m}{2} cdot 2 (v+ Omega times r)cdot (dv + Omega times dr) - d(Ucirc r) &= m bigg( (v+ Omega times r) cdot (Omega times dr) + (v+ Omega times r) cdot dvbigg) - [(nabla U)circ r]cdot dr &= m bigg( [(v+ Omega times r)times Omega] cdot dr + (v+ Omega times r) cdot dvbigg) - [(nabla U)circ r]cdot dr &= bigg(m [(v+ Omega times r)times Omega] - nabla U circ rbigg) cdot dr + m (v+ Omega times r) cdot dv tag{$ddot{smile}ddot{smile}$} end{align}

Now, note that $(ddot{smile})$ and $(ddot{smile}ddot{smile})$ say the exact same thing because if you evaluate $(ddot{smile}ddot{smile})$ on $(a,b)$ and then on $(xi,eta)$ (and you revert back to the $langle cdot, cdot rangle$ notation) we get the exact same thing. Hence, we can easily read off the derivatives in (i) and (ii).

(By now you should be able to justify each equal sign carefully, and also interpret each equality as a proper equality of functions with appropriate domain and target space, and know where various things need to be plugged in to make an evaluation. By the way, this pretty much what Landau and Lifshitz do in Volume $1$, page $128$).