Timelike, spacelike, and lightlike normalization conditions

Question

Let $u^\mu=\frac{dx^\mu}{d\lambda}$ be particle's "4-velocity" where $\lambda$ is affine parameter. If I am not mistaken, we have, for the different cases:

• timelike (massive particles): $u_{\nu}u^{\nu} = g_{\mu \nu}u^{\mu}u^{\nu} = -c^2$
• lightlike (massless particles): $u_{\nu}u^{\nu} = g_{\mu \nu}u^{\mu}u^{\nu} = 0$
• spacelike (hypothetical tachyons?): $u_{\nu}u^{\nu} = g_{\mu \nu}u^{\mu}u^{\nu} = +c^2$

Here the metric signature is $(-,+,+,+)$. I have found in the literature that this is called a normalization condition. Does that mean that other normalization are possible? Where does this normalization even come from?

Questions:

• Can you confirm the formula are correct?
• Are other normalization possible?
• Can these normalization conditions taken as axioms or they can be derived?
• Could you discuss where these normalization are coming from and derive them from first principles?

Answer 1

Depending on the signature, the difference between timelike/spacelike is just that they are of opposite sign. In $(-,+,+,+)$, common to textbooks like MTW Gravitation and such, "timelike" terms in the metric, like $\mathrm{d}t$ and $\mathrm{d}\tau$, usually have a negative sign, while "spacelike" terms in the metric, like $\mathrm{d}x$, $\mathrm{d}y$, $\mathrm{d}z$, etc. have a positive sign.

I'm going to use flat spacetime as an example here; the Minkowski metric in 3+1-dimensional spacetime is

$$\mathrm{d}s^2=-c^2\mathrm{d}\tau^2=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2=g_{\mu\nu}u^\mu u^\nu$$

in your signature. Here $\mathrm{d}t$ is coordinate time, $\mathrm{d}x$, $\mathrm{d}y$, $\mathrm{d}z$ are coordinate distances, and $\mathrm{d}\tau$ is the proper time, the time experienced by an observer whose worldline has a tangent vector $u$ (which may of course be different from one whose tangent vector in any given frame is $u=0$, which would give $\mathrm{d}\tau=\mathrm{d}t$).

If you define the four-velocity as $u^\mu=\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}$ (which is how it is often defined), i.e. if your affine parameter is $\tau$ (or, different by a factor of $c^2\approx1$, $s$), then yes, the norm of the four-velocity is what you say it should be, at least in flat spacetime.

Why choose this normalization? Because it's convenient. In fact to be even more convenient we often set $c=1$ which then makes all your normalizations you mentioned $\pm1$. You can pick whatever normalization you want; the only important thing is that you're consistent with it. $c$ is just a speed, and speeds can be whatever you want with your choice of unit systems.

Can this be taken as an axiom? Sure, if you're working in a spacetime where it holds true. I can't immediately think of a spacetime where it doesn't, but then again I can't read the future; maybe one exists and we haven't found it yet. I am not directly aware of a rigorous general proof that this always holds true but it does in all spacetimes I've ever worked with. Generally I would stay away from holding this as an axiom but if I can prove it in my particular spacetime I'm happy to use that as a starting point for other conjectures.

Want to see the proof why $u_\mu u^\nu=c^2$? Simple: using the Minkowski metric, find $\partial_\tau t$, $\partial_\tau x$, $\partial_\tau y$, and $\partial_\tau z$, and then make them into a four-vector and take its norm. Unless I am also terribly mistaken you should get $c^2$.

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A comment asks how to prove that $g_{\mu\nu}u^\mu u^\nu=u_\mu u^\nu=c^2$ for spacelike trajectories, so I'll prove that here, starting with timelike trajectories which are infinitesimally more intuitive.

In physical terms, the affine parameter usually used is $\mathrm{d}\tau$ since that makes $u^\mu=\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}$, which is what is widely used for the definition of 4-velocity as Qmechanic notes. By definition, $-c^2\mathrm{d}\tau^2=g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$, you can then just divide $\mathrm{d}\tau^2$ and get $-c^2=g_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}=u_\mu u^\nu$ for a timelike vector $u$. Trivially you use this process to also arrive at the results that $u_\nu u^\mu=0$ for lightlike vectors (since for such vectors $\mathrm{d}\tau=0$) and spacelike vectors (where $\mathrm{d}\tau^2<0$).

Answer 2

Ok here is my tentiative answer based on what I know so far. I can't vouch for the correctness.

Trajectory of massive particles can be parametrized by an affine parameter known as proper time $\tau$ which defined by $ds^2=-c^2d\tau^2$. In a curved spacetime described by a metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, the normalization condition of 4-velocity $u^\alpha=\frac{dx^\alpha}{d\tau}$ of a massive particle is given by $ds^2=-c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$ from which we get $g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-c^2$.

On the other hand photon trajectories cannot be parametrized by proper time because $ds=0$, so it doesn't make sense to define 4-velocity of a photon as $u^\alpha=\frac{dx^\alpha}{d\tau}$. But we can choose some other affine parameter $\lambda$ and define photon's 4-velocity as $u^\alpha=\frac{dx^\alpha}{d\lambda}$. In a curved spacetime described by a metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, the normalization condition for this 4-velocity is given by $ds^2=g_{\mu\nu}dx^\mu dx^\nu=0$ from which we obtain $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0$.

Finally, tachyonic trajectories can be parametrized by proper distance $\sigma$ which is defined as $ds^2=d\sigma^2$. 4-velocity of a tachyon can be defined as $u^\alpha=\frac{dx^\alpha}{d\sigma}$. In a curved spacetime described by a metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, the normalization condition for this 4-velocity is given by $ds^2=d\sigma^2=g_{\mu\nu}dx^\mu dx^\nu$ from which we obtain $g_{\mu\nu}\frac{dx^\mu}{d\sigma}\frac{dx^\nu}{d\sigma}=1$ (it is not $g_{\mu\nu}\frac{dx^\mu}{d\sigma}\frac{dx^\nu}{d\sigma}=c^2$).