what is the physical meaning of the 4-vector inner product? Given two 4-vectors, $P^\mu, Q_\mu$, what is the physical meaning of the inner product, $P^\mu Q_\mu = a$?  Specifically I would like to know what it means when the vectors are orthogonal ($a=0$) with respect to this inner product.  
I know that this product is Lorentz invariant. I also know when the inner product of a four-vector with itself is zero, then the momentum is on the light/null cone.
 A: If you have a way of defining the norm (squared magnitude) of vectors, then you automatically get an inner product from the relation $(P+Q)^2=P^2+Q^2+2P\cdot Q$. It works that way for the relativistic inner product, and it also works that way for the quantum-mechanical inner product.
The norm of a timelike displacement vector is the square of the time measured by a clock that travels inertially along that vector. The norm of a spacelike displacement vector is minus the square of the proper length along that vector (i.e., the length of a ruler that's at rest in the frame of an observer who says the vector connects simultaneous points). (This is for the $+---$ metric. You can also flip the signs if you prefer $-+++$.)
Orthogonality can be interpreted in different ways, depending on the spacelike, null, or timelike characters of the vectors. If they're both spacelike, then orthogonality means that an observer could pick the two vectors as their orthogonal x and y axes, i.e., an observer who says they're vectors of simultaneity will then say that they're orthogonal in the Euclidean sense.
If one is timelike and one is spacelike, then it means that an observer could use them as their t and x axes. In other words, an inertial observer whose world-line coincides with the timelike vector will say that the spacelike one is a vector of simultaneity.
A null vector is orthogonal to itself, and can also be orthogonal to a spacelike vector or to another null vector.
