I think you're asking about this circuit:

simulate this circuit – Schematic created using CircuitLab
In order for this circuit to work, the current source simply needs to pull the MOSFET's source node below ground enough that \$V_d-V_s > V_{th}\$. Since it's an ideal current source, it has no problem doing this.
On the other hand, many real-world circuits that we model as current sources can't really change their output voltages arbitrarily. That's why I connected the other node of the current source to "VSS" instead of ground. If VSS is a negative voltage, then the current source could actually be implemented with another transistor (NMOS or NPN BJT).
What is happening physically that leads the source to that potential?
That depends entirely what real circuit is being used to make your current source. An "ideal current source" like we often draw in circuit theory class can't be realized perfectly, so there's no physical explanation for how it works. There are numerous ways to implement an approximation to a current source, but each one operates on its own principles.
You might want to research the cascode circuit, which is a very common real-world example of this idea.