# Why a falling object has a constant acceleration?

My question is could we derive the fact that a falling body has constant acceleration from Galilean relativity and only?

There is an answer on this site which claims this (accepted answer on the question Why does kinetic energy increase quadratically, not linearly, with speed?) and I don't understand what it means.

• Hello! It might be helpful to explain what exactly you don't understand in said answer and possibly provide a link to it. Thanks! Feb 23, 2021 at 20:44
• Free-fall acceleration can be derived from a universal gravity law and second Newton law, but you will not derive it from Galilean relativity alone. Feb 23, 2021 at 20:52
• In addition to that, why does kinetic energy increases quadratically - check my answer, and it does not relates to free-fall acceleration too. These are different subjects. Feb 23, 2021 at 20:59
• Constant acceleration is a good approximation in a gravitational field that's nearly uniform, but it's not exactly true - as you fall toward a massive body, you get closer to it, meaning gravity pulls harder and accelerates you faster. Falling from a height of 400km above earth, your acceleration due to gravity will increase from about 9m/s^2 to almost 10m/s^2 by the time you hit the ground. Feb 23, 2021 at 21:03
• Fun related fact: the conic section that best represents frictionless projectile motion is not a parabola, it is an ellipse Mar 2, 2021 at 1:45

A falling body does not have constant acceleration. A body falling toward a planet has an acceleration that is inversely proportional to the square of its distance from the center of the planet, so the acceleration constantly increases as it falls.

When a body falls only for a short distance, the distance from the center of the planet changes very little, so the acceleration is almost constant over that distance.

A body in free fall does not feel acceleration; the acceleration is only apparent to an observer is a separate inertial frame, such as that of the planet.

Something that might confuse some people is "terminal velocity". In air having a constant density, a body falling a relatively small distance like a few thousand feet will accelerate only until it reaches a speed where the air resistance matches the weight of the body. Thereafter, the body will fall at that speed with no further acceleration.

• isn't it inversely proportional to the square of the distance?
– user288901
Feb 28, 2021 at 2:00
• You are of course right, thank you. Should have proof-read before posting! I have corrected the answer. Feb 28, 2021 at 3:38

"Could we derive the fact that a falling body has constant acceleration from Galilean relativity and only?"

No. However, there's a related fact which is true, which I will explain soon.

First, let's define what "Galilean relativity" is. It refers to the fact that the laws of physics are the same for people moving at different velocities.

(By the way, what distinguishes it from "special relativity" is that observers moving at different velocities have the same exact notion of time the time coordinate $$t$$, so there is no maximum speed $$c$$ which all observers agree is constant. So, special relativity says that the 1) the laws of physics are the same for observers moving at any constant velocity, and 2) the speed of light $$c$$ is constant in all reference frames. Gallilean invariance says 1) the laws of physics are the same for observers moving at any constant velocity, and 2) the time coordinate $$t$$ doesn't change for different observers.)

So, if one inertial observer is using coordinates $$(t, \vec{x})$$, another inertial observer would use $$(t', \vec{x}')$$, where \begin{align} t' &= t \\ \vec{x}' &= \vec{x} + \vec{v} t \end{align} where $$\vec{v}$$ is the relative velocity of the observers.

Now, "the laws of physics" in any situation take the form of a differential equation, which can be used to solve for how physical variables, say the position vectors of some objects, evolve in time. A general differential equation can be written as $$F\big(t, \vec{x}(t), \tfrac{d}{dt} \vec{x}(t), \ldots, \tfrac{d^n}{dt^n} \vec{x}(t), \ldots \big) = 0$$ where $$F$$ is just some function. (More generally we could write $$(\vec{x}_1, \ldots, \vec{x}_N)$$ for the position of $$N$$ particles.) Gallilean invariance implies that, if $$\vec{x}(t)$$ is some solution to the path of a particle, then $$\vec{x}'(t)$$ MUST also be a solution, where $$\vec{x}'(t) = \vec{x}(t) + \vec{v} t$$. So Gallilean invariance says that we must have a symmetry of our laws of motion.

Now that we know what Gallilean relativity is, what does it imply? Well, it certainly doesn't mean that everything has a constant acceleration. As has been pointed out, Newtonian $$1/r^2$$ gravity has Gallilean invariance, but things do not accelerate uniformly.

The fact that an object accelerates in a constant gravitational field with uniform acceleration is a consequence of Galilean invariance, and the assumption that a gravitational field is frame invariant to uniform motions up and down with a steady velocity.

There are two important claims being made here: if the gravitational field (i.e., the laws of physics) are invariant under both constant translations and Gallilean transformations, then the object must undergo constant accerlation.

Let's now try to prove this claim!

First let's begin with the translational symmetery, that if we take a solution and shift it by a constant vector $$\vec{a} = (a_1, a_2, a_3)$$ $$\vec{x}' = \vec{x} + \vec{a}$$ then we will still have a solution. This means that if $$0 = F(t, \vec{x}(t), \tfrac{d}{dt} \vec{x}(t), \ldots)$$ then we must have \begin{align} 0 &= F(t, \vec{x}'(t), \tfrac{d}{dt} \vec{x}'(t), \ldots) \\ &= F(t, \vec{x}(t) + \vec{a}, \tfrac{d}{dt} \vec{x}(t), \ldots). \end{align} Now, the great thing about symmetry is general is that it constrains the laws of physics! We can see that, if we differentiate the above expression by $$a_i$$ (for one of $$i = 1,2,3$$) we get \begin{align} 0 &= \frac{\partial}{\partial a_i} F(t, \vec{x}(t) + \vec{a}, \tfrac{d}{dt} \vec{x}(t), \ldots) \\ &= \frac{\partial F}{\partial x_i}. \end{align} This means that for the laws of physics to respect translational invariance, i.e. if the gravitational field is constant in space, we must have $$\frac{\partial F}{\partial x_i} = 0$$, which means that $$F$$ cannot depend on the position coordinate $$\vec{x}$$!

Now, what if we ALSO demand that the laws of physics respect Gallilean invarince too? Well, that means that if $$0 = F(t, \vec{x}(t), \tfrac{d}{dt} \vec{x}(t), \ldots)$$ then \begin{align} 0 &= F(t, \vec{x}'(t), \tfrac{d}{dt} \vec{x}'(t), \ldots) \\ &= F(t, \vec{x}(t) + \vec{v} t, \tfrac{d}{dt} \vec{x}(t) + \vec{v}, \ldots ). \end{align} Just as before, we can differentiate by $$v_i$$ to find that \begin{align} 0 &= \frac{\partial}{\partial v_i} F(t, \vec{x}(t) + \vec{v} t, \tfrac{d}{dt} \vec{x}(t) + \vec{v}, \ldots) \\ &= \frac{\partial F}{\partial x_i} t + \frac{\partial F}{\partial \dot{x_i} } \\ &= \frac{\partial F}{\partial \dot{x_i} } \end{align} where in the final step we used the equation we got from translational invariance. The equation $$\frac{\partial F}{\partial \dot{x_i}} = 0$$ means that $$F$$ cannot depend on the velocity! Therefore, taking both of these symmetries into account, $$F$$ can only depend on $$t$$ and derivatives of position of order $$2$$ or higher. $$F = F(t, \frac{d^2}{d t^2} \vec{x}(t), \frac{d^3}{d t^3} \vec{x}(t), \ldots)$$

Now, have we proved that acceleration must be constant? Well, sadly, no. We've just showed that the acceleration can't depend on position or velocity. However, if we add a few assumptions, that

1. The laws of physics don't depend explicitly on the time coordinate $$t$$
2. The laws of physics don't depend on derivatives of position to order $$3$$ or more

then we can conclude that the laws of physics must only depend on the second derivative, i.e. $$F = F( \frac{d^2}{d t^2} \vec{x}(t) )$$ for which the only possibilities amount to the acceleration being constant.

Now, you might argue that, while assuming that the laws of physics don't depend explicitly on $$t$$ might be somewhat reasonable, assuming that they don't depend on derivatives higher than $$3$$ is a big assumption. I agree with you. However, it happens to be an empirically accepted fact, and the truth of the matter is that having a law of physics of the form $$\frac{d^3}{dt^3} \vec{x}(t) = \vec{b}$$ is perfectly consistent with Gallilean (and position) invariance. It just so happens that physics is not of this form. (If you are interested in why physics doesn't depend on the third derivative, there are some pretty interesting answers on this site, but I won't go into it here.)

So, TLDR: If you assume that if the laws of physics are invariant under time translations, space translations, and Gallilean translation, AND that they don't involve derivatives higher than two, then the only possibility for the laws is that the acceleration must be constant.

## No, we cannot.

In fact, the claim is not even true. Bodies do not fall with constant acceleration. The gravitational force decreases with the distance squared, so as the body falls its acceleration increases, if only a bit. Plus, there is air friction, etc.

If the falling distance is short enough that you can neglect the dependence of the gravitational force on distance, and the velocity is small enough that you can neglect friction, then the body does fall with approximately constant acceleration. But these are dynamical conditions. Galilean transformations are a kinematical statement. You can never solve dynamics using only kinematics.

Galilean transformations are universal. Constant acceleration is specific to very concrete situations. The former has little to say about the latter.

• I feel like the last 5 answers in a row I'm seen from you have started with a big bold declaration at the top, like: "No. This object does not exist. It can't be mathematically defined. You can never do this. Don't even try. That doesn't make sense. See Weinberg for details." Consequence of becoming a mathematical physicist? ;) Mar 5, 2021 at 17:48
• @knzhou eh, in a question with eight answers you have to stand out one way or another lol. Also, "See Weinberg" really does answer 90% of the questions on this site, so it is technically always the correct answer ;-) Mar 7, 2021 at 22:57

There are many good and insightful answers here, but I will just add one thing. As stated, the answer to this question is, of course, emphatically no. But I want to come at it a little more from the angle of what relativity actually tells us. First of all, Galilean relativity is not capable of telling us anything about the acceleration at all, except that the minimally non-trivial equations of motion of a body must include the acceleration. Newton's 2nd law is where we go from this minimal constraint on the equations of motion to talking about forces which actually cause the motion of a body to change. By defining the acceleration of a body as proportional to an applied force (or vice versa), we essentially define the motion of bodies using the simplest possible equations of motion consistent with Galilean relativity. We need to know about the specific forces experienced by our system in order to know anything about the acceleration. Even were we to assume that Newton's law of gravitation was the force acting on a body, and nothing else, we would find that in a state of free fall we could at best have an almost constant acceleration due to the fact that the distance between bodies is changing over time (however relatively minutely). This gets even worse when we add in the effects of air resistance, the non-uniform distribution of the matter of the planet as it rotates, wind, etc. ad nauseam.

according to Newton second law and the gravitation law you obtain

$$m\,\ddot R=-\frac {m\,M\,G}{R^2}\tag 1$$

where R is die distance from the earth center to the height $$~h~$$ of the object.

for $$~R\mapsto R_0+h(t)~$$ where $$~R_0~$$ is the earth radius.

Eq. (1)

$$m\,\ddot h=-\frac {m\,M\,G}{(R_0+h)^2}=-\frac {m\,M\,G}{(R_0(1+\frac{h}{R_0}))^2}$$

and with $$~\frac{h}{R_0}\ll 1$$ the solution the the differential equation with the initial conditions $$~h(0)=h_0~,\dot{h}(0)=0~$$

$$h(t)=h_0-\frac 12 g\,t^2$$

where $$~g=\frac{M\,G}{R_0^2}$$

thus g is only constant if $$~\frac{h}{R_0}\ll 1$$ not because the Galilean transformation.

A force as an external influence or action on an object that causes the object to change velocity, that is, to accelerate relative to an inertial reference frame.

Newton’s second-law statement, like Newton’s first–law statement, can be applied only in inertial reference frames (Galilean invariance).

• Newton's second law states: The acceleration of an object is directly proportional to the net force acting on it, and the reciprocal of the mass of the object is the constant of proportionality. Thus,

$$\begin{equation} \vec{a}=\frac{\vec{F}_{\text {net }}}{m}, \quad \text{where} \quad \vec{F}_{\text {net }}=\sum \vec{F} \end{equation}$$

When you say falling object you are supposing a reference frame attached to the ground, that frame is not quite an inertial reference frame because of the small acceleration of the ground due to the rotation of Earth and the small acceleration of Earth itself due to its revolution around the Sun.

So, the question must be reformulated:

Could we derive Newton's second law from Galilean relativity?

The Galilean transformation states:

$$\begin{equation} x=x^{\prime}+v t^{\prime}, \quad y=y^{\prime}, \quad z=z^{\prime}, \quad t=t^{\prime} \end{equation}$$

These equations are consistent with experimental observations as long as $$v$$ is much less than $$c$$. They lead to the familiar classical rules for velocities. If a particle has velocity $$u_{x}=d x / d t$$ in frame $$S$$, its velocity in frame $$S^{\prime}$$ is $$u_{x}^{\prime}=\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x^{\prime}}{d t}=\frac{d}{d t}(x-v t)=u_{x}-v$$

If we differentiate this equation again, we find that the acceleration of the particle is the same in both frames: $$a_{x}=\frac{d u_{x}}{d t}=\frac{d u_{x}^{\prime}}{d t^{\prime}}=a_{x}^{\prime}$$ It should be clear that the Galilean transformation is not consistent with Einstein's postulates of special relativity. If light moves along the $$x$$ axis with speed $$u_{x}^{\prime}=c$$ in $$S^{\prime},$$ these equations imply that the speed in $$S^{\prime}$$ is $$u_{x}=c+v$$ rather than $$u_{x}=c,$$ which is consistent with Einstein's postulates and with experiment.

All this information is taken from:

Tipler, P. A., Mosca. Physics for scientists and engineers - (2003) - New York: W.H. Freeman.

An interesting question that deserves an interesting answer! Galilean relativity: the laws of motion are the same in all inertial frames. The trick here is in defining a limited reference frame. Practically speaking within a limited reference frame we cannot determine whether we are moving or staying at rest. For example, two skydivers exit an aircraft at the same time and observe each other in freefall. Their relative positions with respect to each other remain unchanged during acceleration to terminal velocity. Upon exit they feel weightless, then feel buoyed by upward (relatively) rushing air as they reach terminal velocity. The usefulness of their reference frame expires when the ground approaches. Unfortunately artificially defined limited reference frames do not accurately represent the full picture. A larger universal reference frame dictates pulling the ripcord is imperative! Don't let your limited reference frame thinking limit you. Acceleration due to gravity increases as we approach any object, even the earth, so it is impossible to derive the fact that a falling body has constant acceleration from Galilean relativity and only, since the fact is that it is not a fact!