# Longitudinal DIS Structure functions

• I
Gold Member
DIS observables can be expressed in terms of structure functions F1, F2 and FL. There exists the relation ##F_L = F_2 - 2xF_1##.

We can write $$F_L = \sum_a x \int_x^1 \frac{dy}{y} C_{a,L}(y,Q) f_a (\frac{x}{y},Q)$$ and similarly for ##F_1## and ##F_2##:

$$F_1 = \sum_a x \int_x^1 \frac{dy}{y} C_{a,1}(y,Q) f_a (\frac{x}{y},Q)$$

$$F_2 = \sum_a x \int_x^1 \frac{dy}{y} C_{a,2}(y,Q) f_a (\frac{x}{y},Q)$$

Then ##F_L = F_2 - 2xF_1## means that also

$$F_L = \sum_a x \int_x^1 \frac{dy}{y} \left( C_{a,2}(y,Q) - 2x C_{a,1}(y,Q) \right) f_a (\frac{x}{y},Q).$$

Comparing this with above eqn for ##F_L## means that ##C_{a,L}## is not just a function of y and Q. Is it possible to extract the longitudinal coefficient function for ##F_L## from knowledge of the coefficient function for ##F_1## and ##F_2##?

MathematicalPhysicist
Gold Member
As far as I can tell it really depends on what the function f_a's and constants C_{a,i}'s are, to tell if C_{a,L} depends on x.

DIS observables can be expressed in terms of structure functions F1, F2 and FL. There exists the relation ##F_L = F_2 - 2xF_1##.

We can write $$F_L = \sum_a x \int_x^1 \frac{dy}{y} C_{a,L}(y,Q) f_a (\frac{x}{y},Q)$$ and similarly for ##F_1## and ##F_2##:

$$F_1 = \sum_a x \int_x^1 \frac{dy}{y} C_{a,1}(y,Q) f_a (\frac{x}{y},Q)$$

$$F_2 = \sum_a x \int_x^1 \frac{dy}{y} C_{a,2}(y,Q) f_a (\frac{x}{y},Q)$$

Then ##F_L = F_2 - 2xF_1## means that also

$$F_L = \sum_a x \int_x^1 \frac{dy}{y} \left( C_{a,2}(y,Q) - 2x C_{a,1}(y,Q) \right) f_a (\frac{x}{y},Q).$$

Comparing this with above eqn for ##F_L## means that ##C_{a,L}## is not just a function of y and Q. Is it possible to extract the longitudinal coefficient function for ##F_L## from knowledge of the coefficient function for ##F_1## and ##F_2##?
You write the relation for ##C_L## there, which depends also on the hadronic variable ##x## (in addition to ##y## and ##Q^2##).

Maybe it would make sense to write: ##C_L(x,Q,y)##