Hello All,(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Limit of zeo mass for a dynamical system

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**