# What is the four-acceleration for Newtonian gravity?

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

$$A^\mu=\frac{dU^\mu}{d\tau}=\gamma\frac{dU^\mu}{dt}$$

$$\gamma(\frac{dU^0}{dt},\frac{d\textbf{U}}{dt}).$$

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

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

$$(0,\vec{a})$$

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

$$(\gamma^4\vec{\beta}\cdot\vec{a},\gamma^4(\vec{a}+\vec{\beta}\times(\vec{\beta}\times\vec{a}))),$$

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

$$\gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

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.

• Wouldn't the gamma terms cause faster than light travel? Shouldn't the acceleration tend to 0 as approaching the speed of light? Commented Apr 16, 2020 at 20:57
• No and No. Intuition is a poor guide to relativity. Commented Apr 16, 2020 at 21:38
• 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? Commented Apr 16, 2020 at 23:47
• Or am I thinking of coordinate dependent acceleration? Commented Apr 17, 2020 at 0:32
• 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. 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

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

Similarly

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

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

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

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

Thus,

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

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

• How would it look for Newtonian gravity? Commented Apr 15, 2020 at 22:20
• @JoshuaPasa What do you mean by that ? Commented Apr 15, 2020 at 22:30
• 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. Commented Apr 15, 2020 at 22:32
• @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. 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.

• How does this stop acceleration causing faster than light travel Commented Apr 16, 2020 at 9:18
• Doesn't there need to be a gamma term in there somewhere? Commented Apr 16, 2020 at 23:50
• @JoshuaPasa You mean in the first line or the second? Commented Apr 16, 2020 at 23:53
• 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? Commented Apr 16, 2020 at 23:55
• @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. Commented Apr 17, 2020 at 0:05