# Why is the scalar product of four-velocity with itself -1?

My GR book Hartle says the scalar product of four-velocity with itself $-1$?

Consider the definition of four velocity $\mathbf{u} = \frac{dx^{\alpha}}{d\tau}$. Suppose I take the scalar product of four-velocity with itself. I then get $$\mathbf{u}\cdot \mathbf{u} = \eta_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau}$$ But these aren't numbers. So why would it yield $-1$?

Granted, I suppose I could consider this a unit four-velocity. But I still don't see how that would yield $-1$. It should be

$$\mathbf{u}\cdot \mathbf{u} = -1(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1) = 2$$

using $$\eta_{\alpha\beta}= \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$

• What happens when you take $\mathbf u=(1,0,0,0)^T$? I.e. an observer that is standing still. Or $\mathbf u=(0,1,0,0)^T$ Mar 2, 2023 at 15:59
• It's $+1$ in the West Coast metric. Mar 2, 2023 at 17:21

First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices.

In this parameterization, the four velocity takes the form

$$u^\mu = \left(\gamma, \gamma v_x, \gamma v_y, \gamma v_z\right)$$ where $\gamma=(1-v^2)^{-1/2}$. Then it's easy to explicitly check that $$u_\mu u^\mu = -\gamma^2 + \gamma^2 \vec{v} \cdot \vec{v} = -\gamma^2\left(1-v^2\right)=-1$$

Note that this assumes $|\vec{v}| < 1$, since otherwise $\gamma$ is not a finite real number. So only timelike paths can be paramaterized so that $\eta_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$.

• +1 Thank you for pointing this out. I wasn't clear on the fact that it was a convention. And your derivation was excellent as well. I haven't seen that anywhere else yet. Mar 1, 2015 at 4:35

Why is the scalar product of four-velocity with itself -1

• The scalar product is invariant
• In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component.

See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have

$$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0$$

Thus

$$\eta_{\alpha \beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt} = -1$$

• OH! I get it. Thanks! That's so simple. I thought of the rest frame idea, but I couldn't justify it in my mind. The invariant point makes it obvious. Grazie. Mar 1, 2015 at 4:02

The derivatives with respect to $\tau$ very much are numbers, but they are not all $1$.

Consider your worldline as a curve $\gamma$ parameterized by $\lambda$. We have \begin{align} \gamma : \mathbb{R} & \to \mathbb{R}^4 \\ \lambda & \mapsto (x^0, x^1, x^2, x^3). \end{align} At any point in your worldline you have a position $(x^0, x^1, x^2, x^3)$, where all components are scalar functions of $\lambda$. You can differentiate these to find your velocity components: $$\dot{x}^\alpha \equiv \frac{\mathrm{d}x^\alpha}{\mathrm{d}\lambda}.$$ Again, these are functions of $\lambda$, and each of the four evaluates to a real number at every point on the line.

In the special case where your parameter $\lambda$ is the proper time, then we call $\dot{x}^\alpha$ instead $u^\alpha$, the $\alpha$ component of $4$-velocity.

For an example, if you were not moving with respect to the coordinates, then each spatial $x^i$ would be constant as $\lambda$ varied, so we would have $\dot{x}^i = 0$. We also know $$\mathrm{d}\tau = \sqrt{-\mathrm{d}s^2} = \sqrt{-g_{\alpha\beta} \mathrm{d}x^\alpha \mathrm{d}x^\beta} = \sqrt{-g_{\alpha\beta} \dot{x}^\alpha \dot{x}^\beta} \mathrm{d}\lambda = \sqrt{-g_{00}} \, \dot{x}^0 \, \mathrm{d}\lambda,$$ so $$u^\alpha \equiv \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} = \frac{\mathrm{d}x^\alpha}{\mathrm{d}\lambda} \frac{\mathrm{d}\lambda}{\mathrm{d}\tau} = \dot{x}^\alpha \frac{1}{\sqrt{-g_{00}} \dot{x}^0} = \big((-g_{00})^{-1/2}, 0, 0, 0\big).$$ You can easily check that $g_{\alpha\beta} u^\alpha u^\beta = -1$ in this case.

1 In general relativiy, replace $\mathbb{R}^4$ with some open set $U \subseteq \mathbb{R}^4$ covering the appropriate part of the manifold.

• +1 Really like this, connecting it to the metric tensor. Btw, re chat, you are of course right. I was kind of sloppy in my statement "they aren't numbers". What I should have said was "they are not specific numbers, but instead arbitrary ones so I don't see how we can turn out a -1 from a bunch of arbitrary numbers." That's also really interesting about replacing $\Bbb{R}^4$ with some open set covering the appropriate part of the manifold. I just ordered Wald. I look forward to reading more specifically about the geometry there. Mar 1, 2015 at 4:46

I will consider $$c=1$$.

We define the interval by $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$

and the proper time by $$d\tau^2=-ds^2=-g_{\mu\nu}dx^\mu dx^\nu$$ From here, it's straightforward to show that $$g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$

Now, let's take the inner product of the four-velocity $$U^\mu=dx^\mu/d\tau$$, $$U=g_{\mu\nu}U^\mu U^\nu=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$ for what was said before.

In conclusion, the four-velocity always will be normalized $$-1$$.

• $u_{i}u^{j}=-1\;\;,\;\;u^{i}u_{j}=1\;\;\;$ Mar 2, 2023 at 17:55
• @TheTiler why do you say this? I think that's incorrect because both expressions are the same, so both are equal to -1. Mar 2, 2023 at 18:37

From the definition of the spacetime invariant and a 4-displacement vector, we know:

$$\mathrm ds^2 = \mathrm d\mathbf x\cdot \mathrm d\mathbf x$$

We also know proper time is defined as:

$$\mathrm d\tau = \sqrt{-\mathrm ds^2}$$

Thus:

$$\mathrm d\tau^2 = -\mathrm ds^2$$

Now:

$$\mathrm d\tau^2 = - \mathrm d\mathbf x\cdot \mathrm d\mathbf x$$

$$\frac{\mathrm d\mathbf x}{\mathrm d\tau} \cdot \frac{\mathrm d\mathbf x}{\mathrm d\tau} = - \mathbf U\cdot \mathbf U = -1$$