Transverse and longitudinal component of a photon propagator

I'm studying the QFT with Peskin and Schroeder's book. There is a point which is mentioned several times in the book and I don't quiet understand which is the transverse and longitudinal component of a photon propagator.

For example, the paragraph below the Feynman gauge photon propagator (12.56) $$D^{\mu v}(q)=D(q)(g^{\mu v}-\frac{q^\mu q^{v}}{q^2})+\frac{-i}{q^2} \frac{q^\mu q^v}{q^2}$$says that the first term here corresponding to the transverse components of the propagator. I think he means the $$D(q)(g^{\mu v}-\frac{q^\mu q^{v}}{q^2})$$. What I don't understand is that why this term is the transverse component of the propagator and the second is the longitudinal component of the propagator?

For example, the paragraph below (12.56) $$D^{\mu v}(q)=D(q)(g^{\mu v}-\frac{q^\mu q^{v}}{q^2})+\frac{-i}{q^2} \frac{q^\mu q^v}{q^2}$$says that the first term here corresponding to the transverse components of the propagator. I think he means the $$D(q)(g^{\mu v}-\frac{q^\mu q^{v}}{q^2})$$.

What I don't understand is that why this term is the transverse component of the propagator and the second is the longitudinal component of the propagator?

It is "transverse" because it is orthogonal to $$q$$.

$$q_\mu \left(q^2 g^{\mu\nu} - q^{\mu}q^{\nu} \right)D(q)/q^2 = 0$$

Similarly, "longitudinal" means "in the same direction as $$q$$."

• But I think the transverse component means that the polarization vector is orthogonal to the momentum vector, how could you derive this from the equation in your answer. Jul 12, 2022 at 17:03
• No, there's no polarization in your expression. It turns out that the polarization is orthogonal to the momentum for light. But here "longitudinal" means along the momentum and "transverse" means orthogonal to the momentum.
– hft
Jul 12, 2022 at 17:18