Mathematical Tools for Physics - Department of Physics - University

9—Vector Calculus 1 229

V dV/dr d^2 V/dr^2

The second derivative on the left side of Eq. (9.44) has a double spike that does not appear on

the right side. It can’t be there, so my assumption thatV is discontinuous is false andV must be

continuous.

Assume next thatV is continuous but its derivative is not. The graphs ofV, ofdV/dr, and of

d^2 V/dr^2 then look like

V dV/dr d^2 V/dr^2

The second derivative on the left side of Eq. (9.44) still has a spike in it and there is no such

spike in theρon the right side. This is impossible, sodV/drtoo must be continuous.

Back to the Problem

Of the four constants that appear in Eqs. (9.45) and (9.46), one is already known,C′. For the rest,

V(R−) =V(R+) is 4 πGρ 0

R^2

6

+D′=−

C

R

+D

dV

dr

(R−) =

dV

dr

(R+) is 8 πGρ 0

R

6

= +

C

R^2

These two equations determine two of the constants.

C= 4πGρ 0

R^3

3

, then D−D′= 4πGρ 0

R^2

6

+ 4πGρ 0

R^2

3

= 2πGρ 0 R^2

Put this together and you have

V R

V(r) =

{ 2

3 πGρ^0 r

(^2) − 2 πGρ 0 R (^2) +D (r < R)

−^43 πGρ 0 R^3

/

r+D (r > R)

(9.48)

Did I say that the use of potentials is supposed to simplify the problems? Yes, but only the harder

problems. The negative gradient of Eq. (9.48) should be~g. Is it? The constantDcan’t be determined

and is arbitrary. You may choose it to be zero.