# Is there an analog to polar coordinates for Minkowski 4-vectors?

Is there a way to represent a Minkowski 4-vector using a 4D polar coordinate system, i.e. a single radial coordinate and 3 angular coordinates?

I know this can be done in Euclidean 4-space with spherical coordinates. And I know that 3D spherical coordinates can be used for the spacelike part of a Minkowski 4-vector. But I'm not sure how to combine the spacelike and timelike parts into one set of polar coordinates.

• At the most basic level a four-vector is just a list of four real numbers, so you can use the Euclidean spherical coordinates if you want. Is there some additional special property you want the coordinates to have? – Luke Pritchett May 3 '17 at 12:06
• Well, I'm simulating elastic collisions, which I'm presently doing by Lorentz-transforming 4-momentums in/out of the center-of-momentum frame. So, presumably I need 4-vectors that can be Lorentz-transformed, and that allow me to find the CoM frame easily. I'm wondering if polar coordinates might help with numerical stability. – jedediah May 3 '17 at 12:25
• web.mit.edu/edbert/Alexandria/notes2.pdf – mmesser314 May 3 '17 at 13:16

## 1 Answer

One sometimes useful decomposition of four-vectors uses two Euclidean angles and what is called the rapidity, a sort of unitless measure of energy.

You can imagine starting with a four-vector in its rest frame, $p=(m,0,0,0)$, where $p^2 = m^2$. Then you can boost in the $z$ direction with rapidity parameter $\eta \in [0,\infty)$, giving you $p = m(\cosh\eta,0,0,\sinh\eta)$. You can check that this left the Minkowski norm invariant. Finally, you can do a spatial rotation to make the three-vector part point any direction you want. $p = m(\cosh\eta,\sinh\eta \sin\theta\cos\phi,\sinh\eta\sin\theta\sin\phi,\sinh\eta\cos\theta)$.

For physical interpretation, note that $E = m\cosh\eta$, so $\cosh\eta$ is the same as the Lorentz factor $\gamma$, and $\sinh\eta = \beta\gamma$.

Note this form is only for time-like vectors. For lightlike vectors or spacelike vectors you have to start with different initial four-vectors, like $m(1,0,0,1)$ or $m(0,0,0,1)$.