# When a car accelerates relative to earth, why can't we say earth accelerates relative to car?

When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not?

• Velocity is relative, but acceleration is not! The answers make this clear of course, but it's worth pointing out that we don't usually say "the car accelerates relative to Earth", but just "the car accelerates". This is because the Earth is actually irrelevant to whether the car is accelerating or not. Sep 8, 2021 at 8:36
• IMO, Claudio Saspinski's answer deserves to be the most up-voted because it goes right to the heart of your confusion. You absolutely can measure the position of the Earth in the car's reference frame, and you can find the 2nd derivative of position with respect to time. We call that derivative "acceleration;" *BUT* it's only a pale shadow of everything that we usually mean by "acceleration" because none of the laws of physics that depend on or describe acceleration are valid in the car's non-inertial reference frame. Sep 8, 2021 at 17:45
• Note! An Earth-based reference frame isn't truly inertial either, but it's so close to inertial that, for many purposes, we can ignore the slight difference. Sep 8, 2021 at 17:46
• How is velocity any different? If you are in the car, you see the earth accelerating. Sep 9, 2021 at 11:20
• Some of these answers would not stand if the car was much more massive than the earth. ;) Sep 9, 2021 at 22:46

Isn't it equally plausible to say that the Earth has accelerated from the point of view of the reference frame of the car?

Yes it is.

The car accelerates because the static friction force exerted forward on the car by the earth is equal and opposite to the static friction force the car exerts backwards on the earth per Newton's 3rd law.

But the acceleration of the car and the Earth due to the equal and opposite forces is determined by Newton's 2nd law. For the car that acceleration is

$$a_{car}=\frac{F}{m_{car}}$$

And the acceleration of the Earth is

$$a_{earth}=\frac{F}{m_{earth}}$$

But since $$m_{earth}\gg m_{car}$$, $$a_{earth}\ll a_{car}$$.

In other words, the acceleration of the Earth is so small as to be unmeasurable.

Hope this helps.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 9, 2021 at 6:18
• The question is about whether all reference frames are equally valid. This answer does not address that question.
– Noah
Sep 10, 2021 at 3:09
• @Noah I covered that in my other answer when I got clarification from the OP. That answer supplemented this one. See OP’s comment on that Sep 10, 2021 at 3:28

The second Newton's law is valid for inertial frames of reference.

If we are for example in a airplane that is braking after landing, any loose object will accelerate forwards, without any force that can be identified. On the other hand, if we hold the object, we do a force on it and it is at rest in the plane's frame. In the first case, an acceleration without a force, and in the second one, a force without acceleration.

It is the criteria to know for sure that we are in an accelerated frame of reference.

• Sorry, but doesn't "acceleration" merely means non-linearity of function x = f(t)? Sep 9, 2021 at 19:38
• @IgorG yes, when you take measurements of an object from an inertial external observer. But the point here is: it is possible for an observer inside that object, only by doing internal measurements, to know if its frame is inertial or accelerated. Sep 10, 2021 at 10:42
• Isn't that the frictional force in the plane which makes loose object moves?
– user326901
Jul 22, 2022 at 15:36

What breaks the symmetry here?

The accelerations are not symmetric because (proper) acceleration itself is not relative (frame variant). A simple accelerometer can measure the asymmetry. The car’s accelerometer measures a large acceleration. The earth’s does not. The measured asymmetry in the acceleration is due to the asymmetry in the mass under equal (but opposite) forces.

Note, the accelerometer measures proper acceleration. It is possible to discuss coordinate acceleration instead. Coordinate acceleration is relative, but it is also not particularly physical. No physical experiment can depend on coordinate acceleration.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Sep 8, 2021 at 12:50

Yes, we can consider the Earth as accelerating in respect to the car. However, the car in this case will be a non-inertial reference frame - that is, besides the forces entering the Newtons laws for inertial reference frames, we will need to add several fictitious forces.

Working in the inertial reference frames is usually easier mathematically, since they are singled out by the laws of nature.

• Yep, as simple as this. Short and to the point. Sep 9, 2021 at 12:52

Unlike velocity, acceleration is not relative. When you consider the earth accelerating towards the car, you use a noninertial reference frame and you have to take so-called fictional forces into account.

A person inside the car feels the acceleration while a person outside does not. Somewhere the symmetry is broken, as noted by many others.

The symmetry is restored if one considers the gravitational field.

The person inside the car can say «I'm at rest in a gravitational field pushing me to my back, so that I feel pushed on my seat while the guy I see outside is free falling in that gravitational field».

The person outside the car says «there is no gravitational field, I'm at rest and the guy in the car accelerates.»

The very point here is that there are no physical means to distinguish «being accelerating» from «being at rest in a gravitational field».

The general relativity makes the whole story precise and relates each non inertial frame (i.e. each manner of accelerating) to a gravitational field that makes you feel the same.

EDIT As pointed out by gs in the comment, every such "gravitationa field" equivalent to an acceleration cannot be created by a physical distribution of mass. So the person in the car has to say «I'm at rest in a curved space-time» while the person outside the car says «I'm at rest in a flat space-time».

• I suspect that the OP's question can be answered perfectly well in the context of Galilean relativity, without invoking GR.
– Kai
Sep 8, 2021 at 14:42
• The observer in the accelerating car, however, will measure that the apparent acceleration of objects outside the car does not follow an inverse-square law, and cannot therefore be a gravitational field. The principle of equivalence only holds for measurements taken of objects inside the accelerated frame.
– g s
Sep 8, 2021 at 15:00
• To devils-advocate myself, we could suppose an infinite flat planar distribution of mass located somewhere between us and the far side of the observable universe. This would produce a homogenous gravitational field, which would explain the motion of everything up to nearby galaxies. But the observer in the car can still experimentally disprove the existence of such a plane mass, since the far side of the observable universe is observable, and will still be redshifted by our acceleration.
– g s
Sep 8, 2021 at 15:27
• Even though the OP asked in the frame of reference (if I may say so) of Newtonian physics, this is a good answer. That, in Newtonian physics, accelerated frames of reference are distinguishable from inertial systems is quite clear, and I believe it is (or became very quickly) clear to the OP as well. Therefore it is fair to assume that they were, maybe, also asking: "The interaction between car and Earth appears symmetrical; why then is the effect not symmetrical?" Or: "Why can I not at will switch the frames of reference?" The answer is "Good thinking. You can, in GR." Sep 8, 2021 at 19:22

Thanks. Yes the earth will have a miniscule acceleration as well in this case due to the opposite force. But I was thinking acceleration as more in terms of what the observer perceives as the rate of change with which objects in the other frame were receding/approaching.

This will supplement my other answer in an attempt to answer your above comment and the related ones that followed.

What the observer perceives will depend on whether the observer is in an inertial (non-accelerating) or non-inertial (accelerating) reference frame. Although the velocity of an object will depend on the inertial frame of an observer, an observer in any inertial frame will perceive the acceleration of an object to be the same because acceleration is the rate of change in velocity.

Although the acceleration of the earth due to the car will be the same (albeit infinitesimal) to an observer in any inertial frame, an observer in the car, because the observer is in a non-inertial (accelerating forwards) reference frame, will perceive the ground to have an acceleration backwards equal in magnitude to the forward acceleration of the car. In order to explain the earth's backwards acceleration in accordance with Newton's second law, the observer in the car needs to invoke a fictitious (pseudo) force that acts on the road since Newton's laws of motion only apply to inertial frames.

Although an observer on the ground is also in an non-inertial frame, its acceleration due to the car is so infinitesimal that, at least locally, the ground can be approximated as an inertial frame. Thus the observer on the ground will perceive an acceleration of the car approximating the true acceleration of the car, even though the observer on the ground is technically not in an inertial frame.

Hope this helps.

• Thanks! Also found an interesting quote about inertial frames: Anthony French (physicist who also worked on the Manhattan project) writes in his book - "Our ultimate definition of an inertial frame may indeed be that it is a frame having zero acceleration with respect to the matter of the universe at large". Sep 9, 2021 at 0:36

Isn't it equally plausible to say that the earth has accelerated from the point of view of the reference frame of the car?

It is! Just as the motion of the planets has once been described by the epicycle theory (which was very complicated compared to heliocentric theory), you can consider the earth accelerating towards the car. Choose whatever reference frame is suitable for you, and be prepared that other people have other, more suitable descriptions, which allow them to make predictions in less time with less words than you.

Actually, from an economic point of view, neither the car accelerates towards the resting earth, nor does the earth accelerate towards the resting car. It is that car and earth are accelerating relative to one another, inversely proportional to their mass. This is because that description allows for the usual angular momentum and linear momentum of the isolated system "earth+car" to be conserved, which answers

Are the forces being applied to move the car have a special significance in deciding which frame is inertial and which one is not?

The forces that act between earth and car are only significant with respect to inertia, in that the sum of all forces acting on the system "earth+car" are zero (implying conservation of linear momentum) and the sum of all torques are also zero (implying conservation of angular momentum). This defines the inertial system in classical mechanics.

In general relativity (gravitational field theory), things are getting more complicated, but you wanted a simple explanation.

• Thanks for the answer. As you said that we "can consider the earth accelerating towards the car", if I am understanding correctly then it seems to imply that in that case the usual formulation of newton's second law won't work from the point of view of the observer inside the car and he would have to come up with a more complicated description to describe the motion of objects . Seems to suggest that acceleration is somehow less arbitrary of a quantity than velocity or distance. Sep 7, 2021 at 23:26
• @cometrraza: Exactly! Usually there is consensus about what is the most economical description/the best one for some given field of physics. With the advent of the heliocentric theory of the planets, the epicycles became unpopular (besides some religious issues...). But sometimes it is less clear what is most economical. For example, in quantum mechanics, there is the Schrödinger picture, the Heisenberg picture, and the interaction picture. All three are mathematically equivalent but very different in their details, and they are almost equal in their effort to practice. Sep 8, 2021 at 14:45

It isn't really plausible. You could, if you wished, take your car to be stationary and everything else to be accelerating relative to it. The problem is that you would find it hard to develop a set of rules for modelling what actually happens if you adopt such a frame of reference.

Consider, for example, Netwon's second law. Suppose you push a trolley laden with 100kg with a certain force to accelerate it to move at 1m/s in one second- if you took the trolley to be stationary, you would have to suppose you had applied a force to accelerate the Earth to 1m/s in the opposite direction. Fair enough, you might say, let's do that. But now suppose you apply a force to accelerate a trolley laden with 200kg to accelerate to 1m/s in the same time. You have applied twice the force to make that happen to the heavier trolley, but if you consider that trolley to be stationary, that force has still only made the Earth accelerate to 1m/s in the other direction. So you now find that two different forces cause the Earth to move backwards at the same rate, which seems nonsense.

So, the rules of physics make a clear distinction between the two possible viewpoints, showing that acceleration is not a relative phenomenon in the way that motion is.

You can absolutely say that the Earth is accelerating in the reference frame of the car. It is equally valid to saying the car is accelerating in the Earth's reference frame.

We tend not to do it as a pragmatic matter. In these situations with accelerating frames, each frame has different equations of motion. Some are easier to work through mathematically than others.

Consider a hypothetical case where you have a very light weight vehicle, and are carrying a very heavy object. From the perspective of the earth, it's easy to see that accelerating the vehicle plus the heavy object takes roughly the same energy as if you were to just throw the heavy object, leaving the vehicle at the same speed. From the perspective of the car, it is equally easy to to accelerate the entire Earth as it is to accelerate the object being carried. This is despite the fact that the weight is much lower in mass than the Earth.

The equations of motion will bear this out. But it isn't very intuitive in this frame. It's hard to calculate. Its easier to calculate from a reference frame that is more inertial. So that's what we tend to do. While it's equally valid to be accelerating the Earth behind you, the calculations are easier if we choose to view this as a fixed Earth and an accelerating car.

From a physics perspective, everything you say is correct. But human language is considerably more subjective and messy when it comes to definitions.

It's not that we can't say that the Earth accelerates in this scenario, it's that we don't say it because we don't observe it.

Even if we did observe it, we tend to say that the object whose velocity changes the most is the thing that accelerates. We discuss how fast a tennis player served a ball. We do not discuss how much the racket decelerated during the serve. Both things happened, with equal momentum, but we only care about one of them.

A gun is another a great example here. The bullet is considered to be the projectile, even though the gun recoiling has the same momentum as the bullet being propelled forwards.

To summarize, you are completely correct from a physics point of view, but you're not accounting for the fact that human language is much messier, subjective and ambiguous compared to scientific and mathematical principles.

• Exactly! Plus there is the matter of causality. I put gas in the tank (or electrons in the battery), press the accelerator (another bit of language for you :-)), the engine spins, transmitting force to the wheels, and the car begins to move. I do nothing to the Earth except push against it with the tires. Ergo, the car accelerates: I don't exert my willpower to shove the Earth around in order to bring my destination to me. Sep 8, 2021 at 16:43

Here is my favorite view of what a person in the car would observe. I don't see it represented very often, for some reason.

By the equivalence principles (that e.g. general relativity builds on), the moment the car engine starts applying torque to the wheels, a person inside the car will feel a change in gravity. It will get a new component in the backward direction of the car.

The person in the car is being held up against this new component of gravity by the back of their seat, and the car is being held up against this gravity by the friction between the wheels and the ground. But nothing holds the ground up, so it begins to fall backwards. Which is to say, yes, the ground is accelerating backwards, just as much as an apple falling to the ground is accelerating downwards.

Assuming the acceleration of the earth is immeasurably small due to its vastly larger mass, I think your confusion actually stems from this:

Isn't it equally plausible to say that the Earth has accelerated from the point of view of the reference frame of the car?

It depends what you mean by the reference frame of the car. In the inertial reference frame in which the car was stationary, the car is now moving, whereas previously it wasn't. It is the car that has accelerated.

In the reference frame in which the car is now stationary, and the earth is moving backwards, the car was previously moving backwards, so it is still the car that has accelerated.

The observer who sees the earth apparently accelerate, does so because she has changed reference frames, or to put it another way she is in (or has been in) an accelerating reference frame. This is where the asymmetry comes from, because this observer is accelerating with the car and not viewing the system from any inertial reference frame.

To be pedantically correct, one should perhaps say that the friction of the wheels on the ground serves to both accelerate the car by a macroscopic amount in one direction, and accelerate the combined center of mass of everything else on Earth by a microscopic (or maybe "femtoscopic") amount in the opposite direction. On the other hand, many factors are causing all locations on the surface of Earth to be accelerated in unpredictable directions by unpredictable directions, and these are so many orders of magnitude larger than the acceleration resulting from the friction of the car's tires on the road as to render the latter meaningless. If a 1,000kg car were to accelerate at 10m/s², the acceleration of the Earth would be about 0.0000000000000016 μm/s².

If instead of looking at a car on the Earth one were to instead consider a 1,000kg car driving around the deck of a 10,000,000kg ferry floating in calm water with no wind, then a 10m/s² acceleration around the deck would cause a roughly 1mm/s² acceleration of the ferry. While it would probably be unusual for the ferry and the water to be sufficiently still that acceleration of the ferry due to friction with the car's tires on the deck would be more significant than acceleration due to waves or friction with the water, the acceleration would be within bounds of what commonplace equipment could measure. On the other hand, the acceleration of the ferry in that scenario is about 592,000,000,000,000,000 times as large as the acceleration of the planet.

[...] deciding which frame is inertial and which one is not?

Strictly speaking, neither the car's frame nor the Earth's frame is inertial. The only inertial frame here is the reference frame tied to the center of mass of the Earth-car system.

Considering that inertial frame:

1. Initially, both the car and the Earth were at rest;
2. Then the engine started rotating the tires, bringing the friction forces.
3. The friction force applied from Earth to the car makes the car accelerate in some direction.
4. The friction force applied from car's tires to the Earth makes the Earth accelerate (and spin) in the opposite direction.

Thus far, everything is symmetric. But then:

What breaks the symmetry here?

Our desire to simplify the calculation does. We don't really want to solve a full two-body problem with friction forces. Ignoring the movement of the Earth (due to its acceleration being so small compared to the acceleration of the car) is the most simple way to reduce it to a one-body problem. In other words, we effectively assume that the Earth stays at rest in the center-of-mass reference frame, which literally reads: "it doesn't accelerate". And that makes the Earth a good enough inertial frame.

Similar reasoning can't be applied to the car: its acceleration is not negligible, and thus there's no way to call it an inertial frame. Hence the asymmetry.

You are correct when we throw a ball down so the ball is accelerating towards earth's center with an acceleration equal to $$-g\vec{j}$$ with respect to earth and so if we are sitting inside the ball we will see everything at rest with respect to earth to have an acceleration $$g\vec{j}$$ with respect to us.

Acceleration is indeed relative because if the first body accelerates with respect to the second body, say with $$\vec{a}$$ then that second body also accelerates with respect to the first body with acceleration $$-\vec{a}$$

In your case, the car (assuming it is moving along the x-axis and some observer is standing at the origin) has three forces $$\displaystyle\vec{f_s}(\text{static friction}), m\vec{g}(\text{weight}), \vec{N}\text{(Normal)},$$

The net acceleration of the car with respect to the observer will be $$\displaystyle\vec{a_c}=\frac{\vec{f_s}}{m}$$

Now consider earth in this inertial frame of reference.

Forces on it are $${-m\vec{g}},-\vec{N},-\vec{f_s}$$ and therefore its acceleration will be $$\displaystyle\vec{a_e}=-\frac{\vec{f_s}}{m_e}$$

Now $$\vec{a_{e/c}}=\vec{a_e}-\vec{a_c}=-\vec{a_{c/e}}$$