muzialis
Dec4-11, 03:46 PM
Hello All,
from considering the system of equations (differentiation is mean with respect to time for the function v = v(t) )
G' = a - v
v' = (1/m) (G - p (v))
Very briefly, they characterize the speed of a crack with an "effective mass" m under a generalized force G.
Dividing the top equation by the lower oneone can conclude that
dG / dV = m (a - v) / (G- p(v))
Considering the massless limit m-> 0 one then could obtain
dG(G- p(v)) = 0.
After a long preambe, my question : should one not be able to get to the massless limit right from the start by ignoring the term mv' (inertial term) from the start? If I try i do not recover the relationship dG(G- p(v)) = 0
Any help would really be the most appreciated
thanks
from considering the system of equations (differentiation is mean with respect to time for the function v = v(t) )
G' = a - v
v' = (1/m) (G - p (v))
Very briefly, they characterize the speed of a crack with an "effective mass" m under a generalized force G.
Dividing the top equation by the lower oneone can conclude that
dG / dV = m (a - v) / (G- p(v))
Considering the massless limit m-> 0 one then could obtain
dG(G- p(v)) = 0.
After a long preambe, my question : should one not be able to get to the massless limit right from the start by ignoring the term mv' (inertial term) from the start? If I try i do not recover the relationship dG(G- p(v)) = 0
Any help would really be the most appreciated
thanks