# Effect of general relativity on Newton's second law

Let's say I have some object, like a long pole, that is 30 km long with non-zero mass $m$. One end is resting on the Earth's surface and it is sticking straight up, perpendicular to the surface.

According to the gravitational time dilation equations, a clock at the far end should gain about 49 microseconds per day when observed from the surface. More generally, the clock at the far end runs $1 + 5.63 \times {10}^{-10}$ times faster.

If I were to apply some force to this object that accelerated it along a tangent (for example) to the earth's surface (so both ends have the same displacement), the acceleration would be $F/m$ according to Newton's second law.

But now I am confused. I read Confused on Newton's second law being invariant under relativity but that actually seemed to boil down to a typo, and I couldn't really understand how to apply the information there in any case. I will attempt to explain why I'm confused but I'm not sure how to express this concisely in words:

Basically, I assume that the object must move the same distance at the surface as it does at the far end, and that an observer at the surface sees the same displacement profile at both ends but also sees more clock ticks pass at the far end. An observer at the far end would also see more clock ticks on their local clock than the ground observer would see at theirs, but the object would move the same distance during this time.

So if you were to work backwards and derive acceleration based on position over time (position → velocity → acceleration) at either end, the acceleration would appear lower at the far end (more clock ticks).

But if the acceleration is not the same on both ends, then in $F = ma$, either $F$ or $m$ must also be different on both ends.

It seems like the force wouldn't change, so does that mean each observer sees the object as having a different mass? Does an observer at the far end have to apply more force compared to an observer at the near end to accelerate and displace the object by the same amount?

What is happening here? How is the time parameter (acceleration) in Newton's Second Law reconciled with general relativity in this example? What do the observers see and what happens to the object?

Here's an example of my confusion in terms of math. As a simple example consider a constant force $F$, so a constant acceleration. After some time $t$, the relative displacement $x$ of an object with mass $m$ would be:

$$x = \frac{1}{2}at^2 = \frac{Ft^2}{2m}$$

So for a ground ($_g$) and a far ($_f$) observer:

$$x_g = \frac{F_gt_g^2}{2m_g}$$ $$x_f = \frac{F_ft_f^2}{2m_f}$$

Now, if displacement, force, and mass are the same from both ends (let $x = x_g = x_f$, $F=F_g=F_f$, $m=m_g=m_f$) and $t_f = kt_g$ where $k$ is a time dilation factor due to gravity ($1 + 5.63 \times {10}^{-10}$ in the above example), then substituting for time, we have:

$$\frac{Ft_g^2}{2m} = \frac{F(kt_g)^2}{2m}$$

Simplifying under the assumption that $m \neq 0$, then, we end up with:

$$t_g = \pm kt_g$$

But we've already calculated that $k \neq 1$, and therefore this isn't possible for $t_g \neq 0$ unless one of the assumptions about the displacement, force, or mass being the same at both ends is incorrect (... or the assumption that $2$ means the same thing at both ends, heh ...).

But I know it must be possible because physics works. So I'm missing something.

• Presumably an observer on the ground does not see the same displacement at both ends. Mar 8, 2016 at 18:34
• @OrangeDog Wouldn't they see the same displacement profile, just lagged in time slightly because of the time it takes for the light to travel to their eyes? So same displacement per time, but just with the perceived start time of the displacement curve at the far end offset a little. So the acceleration curve would still be the same. Like, if there were a ruler at both ends, the object would cover the same distance; and the amount of time between the observed start of the motion and a specific displacement would appear the same at both ends to the ground observer? Mar 8, 2016 at 18:37
• The transmission delay (sometimes called retardation) is separate from time dilation. You still observe time dilation if you calculate back to where and when the observed event actually took place. Mar 8, 2016 at 18:45

$F^{a} = \frac{d }{d\tau}\left(mv^{a}\right)$
Where $\tau$ is the affine parameter of the curve, and $v^{a}$ statisfies $v^{a}v_{a} = -1$. Even in special relativity, this has a bunch of consquences, including that you can no longer use that quadratic form for the position, even for constant force.
But, for your case, the most important thing is that if $F^{a}$ is ONLY in the $x$ direction in an unboosted frame, it will pick up a $t$ component in the boosted frame, whihc means that you can't just simply do the type of multiplication you do in the OP.