# Deriving the number of loops and number of lines meeting at a vertex in a diagram

When reading about Superficial Degrees of Divergence (SDOD) I have seen in An Introduction to Field Theory (Chapter 10.1) that for Lagrangians with a $$\phi^n$$ interaction term we know that:

$$\tag{1} L=P -V +1,$$ $$\tag{2} nV = N + 2P,$$

Where $$L$$ represents the number of loops in a diagram, $$P$$ the number of internal propagators, $$N$$ the number of external lines and $$V$$ the number of vertices.

The degree of divergence of the diagram is given by:

$$\tag{3}D=dL-2P.$$

Why is this so? Is there a way to derive these equations though calculations or is it just a known fact?

if you have something like this : $$\sim \int \dfrac{d^4 k_1 d^4 k_2 \dots d^4 k_L}{(\bar{k}_i-m) \dots k^2_j \dots }$$ that for each propagator you have power of 2 of momentum in denominator and for each loop you have power of $$d-1$$ of momentum for measure of integral in numerator that after integrating become $$d$$.so for degree of divergence you should calculate(Show that with ND) :
$$ND = (power\ of\$$k$$\ in\ numerator - power\ of\$$k$$\ in\ denominator)$$.(after integrating that plus 1 to measure of numerator)
because we have $$L$$ measures that should be integrating, for power of $$k$$ in numerator we have $$dL$$(after integrating) , and because we have $$P$$ propagator each have power of $$2$$ in denominator so power of $$k$$ in denominator should be $$2P$$. so final result for degree of divergence is $$dL-2P$$.
• For a connected Feynman diagram the superficial degree of (UV) divergence $$D$$ is defined as $$D~:=~ \#\{\text{\mathrm{d}p in int. measure}\} ~+~ \#\{\text{p in numerator}\}~-~ \#\{\text{p in denominator}\}.$$ The 1st term on the RHS is $$Ld$$. The 2nd term on the RHS is 0 if there are no derivative couplings. In the simplest cases, the 3rd term on the RHS comes from $$P$$ internal scalar propagators. This leads to OP's eq. (3). A generalization of eq. (3) is discussed in my Phys.SE answer here.