I know that Newtonian gravity is not Lorentz invariant, however, I'm just interested in the result.

Let's say we have a four-velocity, $U^\mu$, to get four-acceleration we differentiate with respect to proper time



However, I do not know how to continue, and how to input $$\textbf{a}=\frac{GM}{r^2}$$ Into the equation. I know intuitively that the acelleration must tend to $0$ if the velocity of the particle is close to $c$ , so there must be a factor of $\frac{1}{\gamma}$ somewhere. Can someone help?


3 Answers 3


An object’s four-acceleration, being a Lorentz four-vector, depends on the Lorentz frame in which it is measured and thus on the three-velocity of the object as well as its three-acceleration. This is somewhat counterintuitive because in Newtonian physics three-acceleration is the same in all inertial frames and not dependent on three-velocity.

An object with three-acceleration $\vec{a}$ — for example, one falling under Earth’s gravity — has the simple four-acceleration


in a frame in which the object is instantaneously at rest.

But in a general Lorentz frame where the object has three-velocity $\vec{\beta}c$ in addition to its three-acceleration $\vec{a}$, the four acceleration is


which is derived in Wikipedia. Here $\gamma$ is the usual Lorentz factor,


As for “I know intuitively that the acceleration must tend to $0$ if the velocity of the particle is close to $c$”, you can see from the formula that your intuition is wrong in this case.

  • $\begingroup$ Wouldn't the gamma terms cause faster than light travel? Shouldn't the acceleration tend to 0 as approaching the speed of light? $\endgroup$ Commented Apr 16, 2020 at 20:57
  • $\begingroup$ No and No. Intuition is a poor guide to relativity. $\endgroup$
    – G. Smith
    Commented Apr 16, 2020 at 21:38
  • $\begingroup$ But if gamma goes to infinity whistled v goes to c, wouldn't the acceleration blow up to infinity. If that's not the case can you explain why? $\endgroup$ Commented Apr 16, 2020 at 23:47
  • $\begingroup$ Or am I thinking of coordinate dependent acceleration? $\endgroup$ Commented Apr 17, 2020 at 0:32
  • $\begingroup$ It shouldn't be surprising that components of the four-acceleration go to infinity as $v\to c$, because $d\tau$ is going to zero due to time dilation. The components of the four-velocity go to infinity for the same reason. $\endgroup$
    – G. Smith
    Commented Apr 18, 2020 at 0:02

So let me write,

$A^{\alpha} = \frac{dU^{\alpha}}{d\tau}$ where $A$:four acceleration, $U$:four-velocity

For $\alpha = 0$,

$A^{0} = \frac{dU^{0}}{d\tau}$, $U^0 = \gamma$ and $\frac{dt}{d\tau}=\gamma$ so we can write

$$A^{0} = \frac{d}{d\tau}(\gamma) = \frac{dt}{d\tau}\frac{d}{dt}(\gamma) = \gamma \dot{\gamma}$$


$A^{1} = \frac{dU^{1}}{d\tau}$, $U^1 = u^1\gamma$

$$A^{1} = \frac{d}{d\tau}(u^1\gamma) = \frac{du^1}{d\tau}\gamma + u^1\frac{d\gamma}{d\tau}$$

$$A^{1} = \frac{dt}{d\tau}\frac{d}{dt}(u^1)\gamma + u^1\gamma\dot{\gamma} $$

let me denote $\frac{du^1}{dt}=a^1$


$$A^{1} = a^1\gamma^2 + u^1\gamma\dot{\gamma} $$

The same goes for $A^2$ and $A^3$ so we have,

$$\vec{A} = (\gamma\dot{\gamma}, \vec{a}\gamma^2 + \vec{u}\gamma\dot{\gamma})$$

where $\vec{a}$ three-acceleration and $\vec{u}$ three-velocity

  • $\begingroup$ How would it look for Newtonian gravity? $\endgroup$ Commented Apr 15, 2020 at 22:20
  • $\begingroup$ @JoshuaPasa What do you mean by that ? $\endgroup$
    – seVenVo1d
    Commented Apr 15, 2020 at 22:30
  • $\begingroup$ Using the Newtonian formula for the acceleration how would you write the four acceleration. Because there's also a four-velocity in the equation which slightly complicates things. $\endgroup$ Commented Apr 15, 2020 at 22:32
  • $\begingroup$ @JoshuaPasa I am not sure but I dont think that is possible. Also four acceleration is a general thing the particle does not need to be in the gravitational field to accelerate. Newtonian theory works for $v <<c$ so you can try to take a limit for the four-acceleration in order to get something. $\endgroup$
    – seVenVo1d
    Commented Apr 15, 2020 at 22:53

It's a lot simpler than you think. The four-acceleration in the presence of a Newtonian potential $\phi$, in the limits where this notion makes sense at all, is just $$a^\mu = \frac{\partial^\mu \phi}{m}$$ in the $(+---)$ metric convention. In the nonrelativistic limit, this simply reduces to $$m \mathbf{a} = - \nabla \phi, \quad \frac{dE}{dt} = m \dot{\phi}$$ which are exactly what you expect.

  • $\begingroup$ How does this stop acceleration causing faster than light travel $\endgroup$ Commented Apr 16, 2020 at 9:18
  • $\begingroup$ Doesn't there need to be a gamma term in there somewhere? $\endgroup$ Commented Apr 16, 2020 at 23:50
  • $\begingroup$ @JoshuaPasa You mean in the first line or the second? $\endgroup$
    – knzhou
    Commented Apr 16, 2020 at 23:53
  • $\begingroup$ Since nothing can travel faster than light, doesn't the acceleration need to approach 0 as v goes to c? So doesn't there need a gamma in the second line? $\endgroup$ Commented Apr 16, 2020 at 23:55
  • $\begingroup$ @JoshuaPasa Ah, in the second line I took the nonrelativistic limit. Of course, in general there are messy factors of $\gamma$ all over the place. $\endgroup$
    – knzhou
    Commented Apr 17, 2020 at 0:05

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