a differential is something is independent of the coordinates. i.e. it is rigged up to give the same answer for a path integral no matter how you parametrize the path integral.
i.e. given a path, how would you integrate a function over it? you would subdivide the path into smaller pieces, then multiply the value of the function at an endpoint of one piece by the "length" of that piece, add up and take limits,..
but what if there is no measure of "length" for paths (as on a manifold)? then you would not know what to multiply the value of your function by.
so instead we integrate "differentials". these assign a number not to a point on the path, but to a point and a tangebntnvector at that point.
now if we parametrize the path by running along it at a certain speed, we can say that the length of the path between two points is the time it took to traverse that piece.
but that changes when we go faster, i.e. going faster makes the path look shorter. but since we are integrating a differential, the differential sees the velocity vector of the parametrization, and gives a bigger number on a longer velocity vector, so it comes out even.
now if we already have a coordinate system in our space then we could use that to measure lengths of tangenht svectors. i.e. we call dx the differential that projects a vector on the x-axis and takes that length.
but some other differential might act diferently, so we multiply dx by some coefficicne function like f(x)dx to get our mroe general differential.
but then suppose someone changes the coordiantes from x to t. then dt is a different measure for vectors. if dt sees a vector ir does not give the same lnegth as dx does.
rather the product (dx/dt) dt does give the same length as dx does on eacxh vector.;
so injtegratinhg (dx/dt) dt over any path will give the same result as integarting dx will.
sorr this isn't so great, but these little devils are a little confusing.
e.g. consider the path to be the unit circle in the x, y plane.
If I map the unit t interval onto the circle by sending t to (cos(2pi t), sin(2pi t)), then at the point t = 0, the velocity vector is the vertical vector (0, 2pi) in the x,y plane.
this means the parametrization by polar coordinates maps the unit vector from the t interval onto this vector of length 2pi in the x, y plane.
on the other hand, if we paramnewtrize the circle by the map sending u to (cos(u),sin(u)) we get velocity vectors of length 1, but the time interval requires u=0 to u=2pi to get all the way around.
if we integrate the differential (-ydx + xdy)/(x^2 + y^2), in these two cases, it transforms into either 2pi dt, integrated from t=0 to t=1, or into du, integrated from u=0 to u = 2pi.
we get the same thing both ways, namely 2pi.
anyway, the reason dx = (dx/dt) dt, is that these two give the same answer on all velocity vectors.
