9
$\begingroup$

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 \begin{equation} \mathbf{u}\cdot \mathbf{u} = \eta_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau}\end{equation} 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

\begin{equation} \mathbf{u}\cdot \mathbf{u} = -1(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1) = 2 \end{equation}

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

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

5 Answers 5

17
$\begingroup$

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

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

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 $\eta_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$.

$\endgroup$
1
  • $\begingroup$ +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. $\endgroup$ Mar 1, 2015 at 4:35
15
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Mar 1, 2015 at 4:02
6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ +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. $\endgroup$ Mar 1, 2015 at 4:46
1
$\begingroup$

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 $-1$.

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

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:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.