Notes to a video lecture on UNIZOR.COM

Lagrangian Math+

This lecture contains additional mathematical material needed to introduce Lagrangian Mechanics in generalized coordinates and, in particular, to prove the uniformity of Euler-Lagrange equation for non-Cartesian coordinates.

Assume, we have a smooth real function S(...) of n real arguments q_i (i∈[1,n]), each of which, in turn, is a function of some unique for all functions parameter t that we can call "time" for definitiveness.
That is, function S(...), ultimately, is a function of time t and can be expressed as
S(t) = S(q₁(t),q₂(t),...q_n(t))

We will use apostrophe (') to indicate a derivative of any function by time t.
So, S'(t)=dS(t)/dt and
q_k'(t)=dq_k(t)/dt.

When taking partial derivatives, we treat q_k and q_k' as independent variables, even though both originate from the same time-dependent functions.

Lemma 1
∂S' /∂q_k' = ∂S /∂q_k
for any k∈[1,n]

PROOF:
From the rules of differentiation follows that
S'(t) = dS(t)/dt =
= dS(q₁(t),...q_n(t))/dt =
= Σ_i{[∂S(q₁,...q_n)/∂q_i]·q_i'(t)} =
= Σ_i [h_i(q₁,...q_n) · q_i'(t)]
where for each i∈[1,n]
h_i(q₁,...q_n) = ∂S(q₁,...q_n)/∂q_i

Now S'(t), the derivative of function S(t) by time, is represented as a function of 2n arguments - all q_i and all q_i':
S' = S'(q₁,...q_n,q₁',...q_n')

This function is a sum of n components, and only one of them, the k^th one, contains q_k' multiplied by h_k(q₁,...q_n).
Other components are independent of q_k', and differentiating by q_k' would nullify them.

Partially differentiating this sum by q_k' and dropping all nullified members except the k^th one, we get
∂S' /∂q_k' = h_k(q₁,...q_n) =
= ∂S(q₁,...q_n)/∂q_k = ∂S /∂q_k
End of proof.

Lemma 2
∂/∂q_k[dS/dt] = d/dt[∂S/∂q_k]
for any k∈[1,n]

PROOF:
Consider separately left and right sides of the equation we have to prove.
From the rules of differentiation follows that the expression on the left side equals to
∂/∂q_k[dS/dt] =
= ∂/∂q_k[dS(q₁(t),...q_n(t))/dt] =
= ∂/∂q_kΣ_i{[∂S(...)/∂q_i]·q_i'(t)} =
since each q_i'(t) in the sum above is independent of q_k(t), it can be considered as a constant for partial differentiation by q_k(t)
= Σ_i{∂/∂q_k[∂S(...)/∂q_i]}·q_i'(t) =
= Σ_i[∂²S(...)/∂q_k∂q_i]·q_i'(t)

On the other hand, the right side of the expression we have to prove is
d/dt[∂S/∂q_k] =
= d/dt[∂S(q₁(t),...q_n(t))/∂q_k] =
= Σ_i[∂²S(...)/∂q_i∂q_k]·q_i'(t)

Comparing the two final expressions for left and right sides of the original equation, we see that the only difference is in the order of differentiation of S(...) by two arguments:
∂²S(...)/∂q_k∂q_i on the left side,
∂²S(...)/∂q_i∂q_k on the right.

According to the known properties of partial differentiation, the order of differentiation by different arguments is irrelevant (Schwarz Theorem, not covered in this course, states that the sufficient condition for this is that both mixed derivatives exist and are continuous),
∂²S(...)/∂q_i∂q_k = ∂²S(...)/∂q_k∂q_i
That proves the equality between left and right sides of the original equation we have to prove.

Uniform Cartesian Coordinates

Consider a mechanical system of N objects in three-dimensional space.
Let's introduce the uniform Cartesian coordinates {s₁,...s_n} (where n=3N) to replace familiar symbolics {x₁,y₁,z₁,...x_N,y_N,z_N}.

This is just a change of symbolics we use, not a change in a concept. These uniform Cartesian coordinates are still the same Cartesian coordinates, just with another name.
This will help us to more conveniently express the transformation from Cartesian coordinates (in their uniform symbolics) to non-Cartesian (generalized) and back.

So, we will use
s₁ for x₁
s₂ for y₁
s₃ for z₁
etc. up to
s_n−2 for x_N
s_n−1 for y_N
s_n for z_N

In princip...