Pseudovectors and transformations
Posted on Thu 25 September 2025 in Physics
Here are some notes for my Fluid Dynamics II students about vectors, pseudovectors and how they transform under reflections.
Let n
be a unit vector and consider the reflection R
in the plane perpendicular to n
.
Quantities such as the position r
and velocity v
are regular vectors and transform in the obvious way under the reflection. Specifically, we write
r = rn + rp
v = vn + vp
where rn
is the component of r
parallel to n
, rp
is the perpendicular component; and similar for vn
and vp
. Then under the transformation R
we have
r' = R r = -rn + rp
v' = R v = -vn + vp
However, now consider the angular momentum per unit mass am = r * v
, using *
to denote the cross product. Or indeed the angular velocity, which is the above divided by |r|^2
.
We can expand out am
explicitly in terms of rn
, etc.:
am = (rn + rp) * (vn + vp)
= rn * vn + rp * vp + rn * vp + rp * vn
= 0 + rp * vp + (rn * vp + rp * vn)
= amn + amp
Here, the term amn = rp * vp
is parallel to n
since it is the cross product of two vectors that are perpendicular to n
. And the term amp = rn * vp + rp * vn
is perpendicular to n
since each product involves a vector that is parallel to n
.
When r
and v
are transformed, then am
transforms as:
am' = r' * v'
= (R r) * (R v)
= (-rn + rp) * (-vn * vp)
= ...
= rp * vp - (rn * vp + rp * vn)
= amn - amp
= amn' + amp'
Thus the component of am
that is parallel to n
stays the same, while the component of am
that is perpendicular to n
is flipped. Thus the components of am
transform in the opposite direction to how those of r
and v
transform – for it is a pseudovector.
In summary, under a reflection in a plane perpendicular to n
...
of a vector...
the component parallel to n is flipped
perpendicular stays the same
of a pseudovector...
the component parallel to n stays the same
perpendicular is flipped
(If you don't believe me, take a screw and a screwdriver and hold it up against a mirror.)