I Longitudinal DIS Structure functions

[CAF123]

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]

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.

 

[RGevo]

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: $C_L(x,Q,y)$