I was curious, why do they specify the differential of Work using an lowercase delta symbol $\delta$ as in "$\delta W$", instead of using a $d$, as in $dW$. For example:
$$\delta W=\vec{F} \cdot d\vec{s}$$
Does that have anything to do with the idea of absolute differentiation from tensor calculus? Example:
$$\frac{\delta T^i}{\delta t} = \frac{dT^i}{dt} + \Gamma^i_{rs} T^r \frac{dx^s}{dt}$$
If I convert this formula into a absolute differential, by multiplying both sides by $\delta t$, then I get:
$$\delta T^i = dT^i + \Gamma^i_{rs} T^r dx^s$$
I would interpret this to mean that if you are using Cartesian coordinates then $\delta T^i = dT^i$ since the christoffel symbols of the second kind $\Gamma^i_{rs}$ are all zero for Cartesian coordinates.
I was just curious if this has a similar means for the physics formula:
$$\delta W=\vec{F} \cdot d\vec{s}$$
how would you apply this formula in cylindrical coordinates instead of Cartesian coordinates for example.
(see https://en.wikipedia.org/wiki/Work_(physics) where i found the formula for work differential)
If I look in my tensor book i find that the christoffel symbols of the second kind for cylinderical coordinates:
$\Gamma^r = \begin{bmatrix}0&0&0 \\ 0&-r&0\\0&0&0\end{bmatrix}$
$\Gamma^\theta = \begin{bmatrix}0& 1/r&0 \\ 1/r &0&0 \\ 0&0&0 \end{bmatrix}$
$\Gamma^{z} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0 \end{bmatrix}$
so in this case the absolute differential would have a none zero second term.
Another possibility that using under case $\delta$ as in $\delta W$ has nothing to do with absolute differentiation because work is a scalar?
I just was wondering if it is somehow linked to $\delta$ absolution differentiation operator in tensors class...
The textbook i'm looking at says nothing about tensors... but it does tell me that if you are using cylindrical coordinates then the formula for a differential is different which sounds very tensor like to me. Example:
$ds = dr \frac{ds}{dr} + r~d\theta \frac{ds}{d\theta} + dz \frac{ds}{dz}$
Normally, in Cartesian coordinates the formula for a differential would be:
$ds = \frac{\partial s}{\partial x}dx + \frac{\partial s}{\partial y}dy + \frac{\partial s}{\partial z}dy$
