Physics 374
Fall 2005
Prof. E. F. Redish

MP11

	Path 1 is certainly the easiest, so we'll do that one first.  This path is easiest because is just so happens that it lies entirely along a single coordinate axis if we choose to do the work integral in polar coordinates.  Point A is at (Sqrt[2]) on the radial axis and Point B is at (Sqrt[18]) on the radial axis [since (1^2)+ (1^(2 ))= ((((Sqrt[2])))^2) and (3^2)+ (3^2) = ((((Sqrt[18])))^2) ].  Thus:
	(Subscript[W, 1 - 2]) = \(TraditionalForm\`∫\_A\%B\(\(\(\ \)\(F\ •\)\)\& -> \) d\(r\& -> \)\) = \(TraditionalForm\`∫\_\(\(\ \)\(r\  = \ \@2\)\)\%\(\(\ \)\(r\  = \ \@18\)\)\((\(-\(GMm\/r\^2\)\) 
              r\&∧)\)\ \(•(\(r\&∧\) dr)\)\)
The minus sign in the force of gravity term is important.  The unit vector (Overscript[r, ∧]) points out away from the origin (where M is), but the force on m due to M must be attractive, which is back towards the origin.  Since both vectors in the integrand lie along the (Overscript[r, ∧]) direction, the dot product is easy:
	 			 (Subscript[W, 1 - 2]) = \(TraditionalForm\`∫\_\(\(\ \)\(r\  = \ \@2\)\)\%\(\(\ \)\(r\  = \ \@18\)\)\(-\(GMm\/r\^2\)\) dr\) 
And now the resulting integral is straightforward:
		(Subscript[W, 1 - 2]) = -GMm \(TraditionalForm\`∫\_\(\(\ \)\(r\  = \ \@2\)\)\%\(\(\ \)\(r\  = \ \@18\)\)\(r\^\(-2\)\) dr\) = GMm ((1/(Sqrt[18])) -  (1/(Sqrt[2])))
		
	Path 2 is more complicated.  This complication comes from the fact that gravity depends on the radial coordiate r while we'd need to use the Cartesian coordinates x and y to conveniently trace out the path.  Taking Point C to be at the corner, i.e. the point (3,1):
		(Subscript[W, 1 - 2]) = \(TraditionalForm\`∫\_A\%B\(\(\(\ \)\(F\ •\)\)\& -> \) d\(r\& -> \)\) = \(TraditionalForm\`∫\_A\%C\(\(\(\ \)\(F\ •\)\)\& -> \) d\(r\& -> \)\) + \(TraditionalForm\`∫\_C\%B\(\(\(\ \)\(F\ •\)\)\& -> \) d\(r\& -> \)\)
		
			= \(TraditionalForm\`∫\_\(x\  = \ 1\)\%\(x\  = \ 3\)\((\(-\(GMm\/r\^2\)\) 
              r\&∧)\)\ \(•(\(x\&∧\) dx)\)\) + \(TraditionalForm\`∫\_\(y\  = \ 1\)\%\(y\  = \ 3\)\((\(-\(GMm\/r\^2\)\) 
              r\&∧)\)\ \(•(\(y\&∧\) dy)\)\)
And here we need to pause and think about what exactly (Overscript[r, ∧]) • (Overscript[x, ∧]) and  (Overscript[r, ∧]) • (Overscript[y, ∧]) mean.

Unlike Cartesian coordinates, where (Overscript[x, ∧]) and (Overscript[y, ∧]) are fixed, the radial unit vector (Overscript[r, ∧]) depends on the coordiate θ !  Think about it... (Overscript[r, ∧]) points right if θ = 0 degrees and up if θ = 90 degrees.  Convince yourself with a quick geometric argument that:
				 (Overscript[r, ∧]) = cos θ (Overscript[x, ∧])  +  sin θ (Overscript[y, ∧])
And then it's easy to see that  (Overscript[r, ∧]) • (Overscript[x, ∧]) = cos θ and (Overscript[r, ∧]) • (Overscript[y, ∧]) = sin θ.  So now our integrals are:
		(Subscript[W, 1 - 2]) = \(TraditionalForm\`∫\_\(x\  = \ 1\)\%\(x\  = \ 3\)\(-\(GMm\/r\^2\)\) cos\ θ\ dx\) + \(TraditionalForm\`∫\_\(y\  = \ 1\)\%\(y\  = \ 3\)\(-\(GMm\/r\^2\)\) sin\ θ\ dy\)
That's progress, but there's still a problem.  We're trying to integrate with respect to x and y, but the integrands are in terms of r and θ.  Obviously r and θ depend on x and y, but we'll need to write out that dependence explicitly if we're going to do these integrals.  Again, convince yourself with a short geometric argument that:
					r = (Sqrt[(x^2)  +  (y^2)])
					cos θ = (x/r) = (x/(Sqrt[(x^2)  +  (y^2)]))
					sin θ = (y/r) = (y/(Sqrt[(x^2)  +  (y^2)]))
Which makes the work:
	(Subscript[W, 1 - 2]) = \(TraditionalForm\`∫\_\(x\  = \ 1\)\%\(x\  = \ 3\)\((\(-\(GMm\/\(x\^2 + 
                    y\^2\)\)\))\) \((x\/\@\(x\^2\  + \ y\^2\))\)\ dx\) + \(TraditionalForm\`∫\_\(y\  = \ 1\)\%\(y\  = \ 3\)\((\(-\(GMm\/\(x\^2 + y\^2\)\)\))\) \((\ 
            y\/\@\(x\^2\  + \ y\^2\))\) dy\)
	
		= -GMm\(TraditionalForm\`∫\_\(x\  = \ 1\)\%\(x\  = \ 3\)\(x\/\((x\^2 + y\^2)\)\^\(3\/2\)\) dx\) - GMm\(TraditionalForm\`∫\_\(y\  = \ 1\)\%\(y\  = \ 3\)\(y\/\((x\^2 + y\^2)\)\^\(3\/2\)\) dy\)
And now we must make one final observation.  Along the first half of path B, as x varies from 1 to 3, y is constant (y = 1).  Along the second half of path B, as y varies from 1 to 3, x is constant (x = 3).  Putting this in gives:

(Subscript[W, 1 - 2]) = -GMm\(TraditionalForm\`∫\_\(x\  = \ 1\)\%\(x\  = \ 3\)\(x\/\((x\^2 + 1)\)\^\(3\/2\)\) dx\) - GMm\(TraditionalForm\`∫\_\(y\  = \ 1\)\%\(y\  = \ 3\)\(y\/\((\ 9 + y\^2)\)\^\(3\/2\)\) dy\)
These integrals can be done fairly easily once we notice that the derivative of (x^2) goes as x, which suggests the substitutions u = (x^2)  +  1and v = (y^2)+ 9  will likely simplify the complicated looking integrands.  After some math, we get
			(Subscript[W, 1 - 2]) = Gmm ((1/(Sqrt[18])) -  (1/(Sqrt[2])))
Which is (comfortingly) the same as we got for the Path 1 case.  At least for this case, the claim that the work done by a conservative force depends only upon the starting and ending points of the motion and is independent of the particular path taken between them seems to be true.  
	As a final thought, note that (1/(Sqrt[18])) <  (1/(Sqrt[2])), and it makes sense that the work done by the force of gravity on the little mass due to the big mass is negative.  The big mass wants to pull the little mass towards it.  If the little mass moves away from the big mass (like from Point A to Point B), the gravitational pull of the big mass will be trying to slow the little mass down (decrease its kinetic energy).  If (Subscript[KE, final])  <  (Subscript[KE, initial]) then by conservation of energy, something must have done negative work on the little mass.  Sure enough, our result is negative.

(Solution by T .  Bing)