12 
This evidently corresponds with argument (A) above. 
Adding M, and M,, we have a total momentum forwards of 
msa°U and this agrees with the permanent forward displace- 
ment. 
On the other hand, we have 
M, = 4ua?s de f fay =0. 
This is the argument (B), and it gives a total momentum 
of wsa?u backwards. 
We may write the integral M, as a limit in the form 
M, = Liga dive s ae [fdy. 
In this integral with 8, ¢ finite and a not zero, the order of 
integration may be inverted without changing the value; we 
have in either case 
M, = 4ua?s Lim (tan? = tan"), 
2 SS ¢ c 
This form brings out the indeterminateness of the prob- 
lem, for the limit can have any value we please between 
7/2 and O according to the limiting value of the ratio b/c. 
The argument (A) above supposes that 6 and ¢ are both 
infinite in such a way that 6 is infinitely greater than ¢; 
while we obtain the result of argument (B) by supposing 
c/b infinite in the limit. Another special case would be to 
suppose 6 and ¢ to become indefinite in a ratio of equality. 
Then M, is zsa?u and the total momentum of the fluid is 
zero. In this case we picture the fluid as of equally infinite 
extent in and at right angles to the line of motion. Up to 
the present we have taken the solution of the fluid motion 
without considering the conditions under which it was 
obtained. These included the condition that the fluid should 
be at rest at infinity, that is, the velocity should become 
infinitesimal as the distance from the cylinder increased 
indefinitely. If we could imagine the fluid to be contained 
in a fixed boundary at infinity, the condition to be satisfied 
there would be the vanishing of the normal component of 
velocity. At first sight, there would not seem to be much 
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