How to demonstrate the acceleration using spherical coordinates and spherical unit vectors?

Assuming, you have polar coordinates: {r[t],th[t],ph[t]} as functions of time t and you want to calculate the acceleration that is defined as the second derivatives of the cartesian coordinates.

With the polar coordinates: {r,th,ph}, the cartesian coordinates: {x,y,z}, the cartesian unit vectors: {ex,ey,ez} and the polar unit vectors: {er,eth,eph}, the spherical unit vectors are:

er= Sin[th] Cos[ph] ex + Sin[th] Sin[ph] ey + Cos[th] ez; eth= Cos[th] Cos[ph] ex + Cos[th] Sin[ph] ey - Sin[th] ez; eph= -Sin[ph] ex + Cos[ph] ey; 

The position vector: vecr= x ex+y ey+ z ez = r er + th eth + ph eph. From this we get the transformation matrix from polar to cartesian coordinates (where we introduce the time for later on):

pol2cart= {{Sin[th] Cos[ph], Sin[th] Sin[ph], Cos[th] }, {Cos[th] Cos[ph], Cos[th] Sin[ph], - Sin[th]}, {-Sin[ph] ,Cos[ph],0 } 

this is a orthogonal matrix, therefore its inverse is the transposed matrix. With this:

{er,eth,eph}= pol2cart . {ex,ey,ez} {ex,ey,ez}= Transpose[pol2cart ] . {er,eth,eph} 

And as {x,y,z}.{ex,ey,ez} = {r,th,ph}.{er,eth,eph} = {r,th,ph}.pol2cart.{ex,ey,ez} we have for the coordinates:

{x,y,z}= Transpose[pol2cart].{r,th,ph} {r,th,ph}= pol2cart.{x,y,z} 

Now for derivatives relative to time. As the spherical unit vectors are not constant, it is easier to use coordinates relative to cartesian unit vectors to calculate derivatives and subsequently transform the coordinates back to spherical unit vectors.

For this, consider that the position vector is: vecr= r er. That is: vecr= {r,0,0}.{er,eth,eph} = {x,y,z}.{ex,ey,ez}.

{x,y,z}= pol2cart . {r,0,0}; (*coord. of position vector rel. to cart. unit vectors*) {r,th,ph}= Transpose[pol2cart] . {x,y,z}; (*back transformation to spherical unto vectors*) 

To get the acceleration in cartesian coord. we introduce the time and write the position vector in coordinates relative to cart. unit vectors:

coord= Transpose[pol2cart] . {r[t], 0, 0}; {x''[t],y''[t],z''[t]]]}= D[coord,{t,2}] 

And back transformation to polar coordinates:

acceleration= pol2cart.D[Transpose[pol2cart] . {r,0,0},{t,2}] {-r[t] (Sin[th[t]]^2 Derivative[1][ph][t]^2 + Derivative[1][th][t]^2) + (r^\[Prime]\[Prime])[t], 2 Derivative[1][r][t] Derivative[1][th][t] + r[t] (-Cos[th[t]] Sin[th[t]] Derivative[1][ph][t]^2 + ( th^\[Prime]\[Prime])[t]), 2 Derivative[1][ph][ t] (Sin[th[t]] Derivative[1][r][t] + Cos[th[t]] r[t] Derivative[1][th][t]) + r[t] Sin[th[t]] (ph^\[Prime]\[Prime])[t]} 

Perhaps a bit simpler than Daniel Huber's answer:

r = {R[t] Sin[\[Theta][t]] Cos[\[CurlyPhi][t]],R[t] Sin[\[Theta][t]] Sin[\[CurlyPhi][t]], R[t] Cos[\[Theta][t]]}; (* vector in cartesian coordinates*) 

basis bR, b\[Theta], b \[CurlyPhi] in spherical coordinates R[t], \[Theta][t], \[CurlyPhi][t]

basis = D[r, {{R[t], \[Theta][t], \[CurlyPhi][t]}}]; 

acceleartion

acc=D[r,{t,2}] 

components in spherical coordinates

Map[# . acc/# . # &, basis] // FullSimplify