Trying to do some basic manipulations with 4-vectors and I have a question about the proper (no pun intended) approach. It's probably easiest if we look at a simple example. So let's define a 4-velocity:

$$V=(\sqrt{t^2x^2 - 1}, tx, 0,0)$$

(Let me know if this isn't a valid 4-velocity. I tried to come up with a simple example that has both t and x in the components of V).

If I were to take the derivative $\partial_0V$ I'm unsure if I can do this the "obvious" way:

$$\frac{\partial}{\partial t}V = (\frac{tx}{\sqrt{t^2x^2-1}}, x, 0, 0)$$

or do I need instead to use the chain rule in the following way:

$$\partial_0V = \frac{\partial V}{\partial t} \frac{\partial t}{\partial \tau} = \gamma (\frac{tx}{\sqrt{t^2x^-1}}, x, 0, 0)$$

If we look at spatial derivatives this approach might look like:

$$\partial_1V = \frac{\partial V}{\partial t} \frac{\partial t}{\partial \tau} \frac{\partial \tau}{\partial x^1} = (\frac{tx}{\sqrt{t^2x^-1}}, x, 0, 0)\frac{\gamma}{tx} $$

using $\frac{\partial x^1}{\partial \tau}$ is the first component of $V = \frac{\partial X}{\partial \tau}$ which is $t$ where X is the 4-vector representing position.

I suppose the question comes down to:

Is $\partial_0V$ equal to $\frac{\partial V}{\partial t}$ or $\frac{\partial V}{\partial \tau}$

  • 4
    $\begingroup$ By definition, $\partial_0 = \partial / \partial t$. $\endgroup$
    – knzhou
    May 11, 2020 at 0:23
  • $\begingroup$ However, it seems likely that this is an XY problem. If you're even considering anything of the form $\partial_\tau V$, where the components of $V$ depend on both $x$ and $t$, it's likely that you're trying to do something else and getting confused. So it might be useful to give more context. $\endgroup$
    – knzhou
    May 11, 2020 at 0:24
  • $\begingroup$ This is an entirely made-up exercise, just trying to verify my understanding. So taking $\partial_0 V$ is just a simple matter of taking the partial by t of each component of the 4-vector. That's what I thought, thanks. Curious about why "∂τV, where the components of V depend on both x and t," is especially unusual just at a glance. $\endgroup$
    – Metropolis
    May 11, 2020 at 1:31
  • 1
    $\begingroup$ If you have a velocity that depends both on $x$ and $t$, then presumably you're talking about a velocity field. (The velocity of a particle should just be a single, time-dependent vector.) But then $\partial_\tau$ doesn't make sense -- if you have a whole velocity field, what $\tau$ is this referring to? $\endgroup$
    – knzhou
    May 11, 2020 at 1:50
  • $\begingroup$ Got it, thanks. $\endgroup$
    – Metropolis
    May 11, 2020 at 2:20

1 Answer 1


What does your four-velocity represent?

If it is a world-line tangent

I would guess, it represents a tagent to some world-line ($x^\mu$). Now this world-line is a 1d-curve, i.e. the position on this line is captured by 1d parameter, so you can only take full derivatives, i.e. with respect to that one parameter.

Having said that, there are several ways you can choose this parameter. You can paramerize the position on the curve with respect to its arc-length, i.e. parametrization in terms of proper time ($x^{\mu}=x^\mu\left(\tau\right)$). Or you could parametrize it in terms of lab-time ($x^{\mu}=x^\mu\left(t\right)$). You could even parametrize it in terms of spatial coordinate, say $z$ ($x^{\mu}=x^\mu\left(z\right)$), if each point of the line corresponds to unique $z$. Technically, what I talked about above are three different functions, but Physicists will often denote them with the same symbol. Also, note that $x^\mu$ is, in general, not a vector. It is simply a 'tuple' of coordinates.

Next lets assume your four-velocity $V$ is the tangent to the world-line $x^\mu=x^\mu\left(t\right)=(t,\,x(t),\, y(t),\, z(t))^\mu$

Then the full derivative is: $$\frac{dV^\mu}{dt}=\frac{d}{dt}\left(\sqrt{t^2 x(t)^2 -1 }, t\,x(t), 0, 0\right)^\mu$$. And one would normally use $\dot{x}=dx/dt$ to denote derivatives of $x$.

Also, note that using the standard metric $g_{\alpha\beta}=diag\left(1,\,-1,\,-1,\,-1\right)$ the 'magnitude' of your four-velocity is $g_{\alpha\beta}V^\alpha V^\beta=-1$, whereas normally, for conventional observers $g_{\alpha\beta}V^\alpha V^\beta=1$ (or $c^2$ depending on normalization).

If it is a set of four-velocities

Perhaps what you meant is that you have a set of world-lines, where different $x$ identifies a different world-line and $t$ denotes the position on the world-line. In this case partial derivatives are possible, but I would strongly suggest using a different label for worldlines, e.g. $x\to\xi$, to avoid confusion with spatial cooridinates $xyz$.


So what did you mean?


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