Transverse Comoving distance

Question

I don't understand the link between the comoving distance and transverse comoving distance : how can they be equal ?

Here an example of definition that I have found :

Angular Diameter Distance : The angular diameter distance $D_{A}$ is defined as the ratio of an object's physical transverse size to its angular size (in radians). It is used to convert angular separations in telescope images into proper separations at the source. It is famous for not increasing indefinitely as $z$ -> infinity; it turns over at $z \sim 1$ and thereafter more distant objects actually appear larger in angular size. Angular diameter distance is related to the transverse comoving distance by $D_{A}=\dfrac{D_{M}}{1+z}(17)$

Some clarifications would be fine.

EDIT 1: @benrg If I have well understood, the transverse comoving distance is simply equal to the comoving distance between object that has emitted at redshift "z" and us that we reveive now this light, isn't it ? If yes, why complicate the things by using the qualified word "transverse" into "transverse comoving distance" ?

Answer 1

Angular diameter distance is the reduced circumference of the circle, centered at our location, on which the object was located when it emitted the light (or the reduced area of the sphere if you prefer). If the universe is spatially flat then this is the same as the metric radius of the circle. In general it's related to the radius by the function $S_k$ defined here, where $k$ is the Gaussian curvature at that time.

Transverse comoving distance is the same thing scaled up by $1{+}z$, i.e., the reduced circumference of the circle on which the same object is located at the present cosmological time, if it had no net peculiar motion in the mean time. It's therefore equal to (radial) comoving distance if the universe is spatially flat.