2
$\begingroup$

Earth is an accelerating frame. However, all experiments done on earth behave as if Earth is an inertial frame. How is this possible? Is there an experiment which can be done on earth to show it really is accelerating?

$\endgroup$
3
  • 6
    $\begingroup$ Your second sentence is very, very wrong. Can you explain how you came to believe that all experiments show a result that you know to be false? I am interested to learn how people come to believe false things, particularly things they know to be false. $\endgroup$ Commented Sep 19, 2017 at 11:11
  • 2
    $\begingroup$ Quite apart from the relativity point of view as given by John Rennie (which is the correct one), if you are interested in predicting the weather then you emphatically can not assume the Earth's frame is inertial. Indeed, where I work, we have a road called 'Coriolis way' whose name is a reference to just this. $\endgroup$
    – user107153
    Commented Sep 19, 2017 at 11:15
  • 4
    $\begingroup$ Possible duplicate of Earth-Centered Inertial (ECI) reference frame as approximate inertial frame of reference $\endgroup$ Commented Sep 19, 2017 at 11:19

6 Answers 6

21
$\begingroup$

Scientists doing experiments in labs on the surface of the Earth are not in an inertial frame. If we were in an inertial frame then if you dropped an object it would remain stationary beside you, as happens if you're an astronaut in the International Space Station. What actually happens is that if we drop an object it accelerates away from us at $9.81$ m/s², and this proves we don't inhabit an inertial frame.

$\endgroup$
11
  • $\begingroup$ Which definition of inertial frame are you using? $\endgroup$
    – Farcher
    Commented Sep 19, 2017 at 7:56
  • 10
    $\begingroup$ @Farcher: yes in the sense that I implicitly treat gravity as a fictitious force. My perspective would be that you cannot understand what an inertial frame is unless you take the relativistic view. Suppose you are in deep space inside a rocket accelerating at 9.81 m/s². I'm sure most of us would say you were in a non-inertial frame. But how would you distinguish this from being stationary on the surface of the Earth? The trajectories of dropped or thrown objects would be the same for both. $\endgroup$ Commented Sep 19, 2017 at 10:01
  • 3
    $\begingroup$ @JR it might be a good idea to include that last comment as part of the answer, particularly since this is now HNQ. $\endgroup$ Commented Sep 19, 2017 at 12:27
  • 3
    $\begingroup$ I think this answer might be more accessible if it wasn't a relativistic view. Modeling gravity as an attractive force between objects rather than an effect of a non-inertial frame is much easier, until you need to talk about objects accelerating to near the speed of light, or black holes. $\endgroup$
    – Cort Ammon
    Commented Sep 19, 2017 at 15:41
  • 3
    $\begingroup$ The question has a tag "newtonian-mechanics". The concept of an intertial reference frame is well-defined in Newtonian mechanics (a frame where Newton's laws hold), and doesn't include treating gravity as fictitious force. I'm afraid this really doesn't answer the question. $\endgroup$
    – JiK
    Commented Sep 19, 2017 at 16:42
16
$\begingroup$

The surface of the earth is not an inertial frame. All experiments show it to be a rotating frame (rotating at basically one rotation every 24 hours). The Focault Pendulum is the classic demonstration that the environment on the surface of Earth is not an inertial frame.

Even if you canceled out this rotation, as is done in the ECI coordinate system, there is still the rotation about the sun. However, this effect is dwarfed by that of the rotation of the earth, so it is very hard to detect without going into space. The rotation about the axis of the Earth is 14.76 deg/hr, while the rotation of the Earth around the sun is 0.042 deg/hr. However, the simplest proof that the rotation about the sun does matter is, in fact, that the Earth rotates around the sun! That, in and of itself, is proof that we are not in an inertial frame on Earth!

That all being said, for human scale problems, the effects of the rotation of the Earth are small enough that we can model things as though we were in an inertial frame. If you throw a ball, the effect of the Earth's rotation is so slight compared to the errors you might have in the trajectory from throwing it slightly differently that you'd never notice. However, once we start looking at things with scientific instruments, we can quickly discern the effect of these rotating frames.

$\endgroup$
3
  • $\begingroup$ "However, this effect is dwarfed by that of the rotation of the earth"Why is it so?How do rotation and revolution end up cancelling each other? $\endgroup$ Commented Sep 19, 2017 at 5:44
  • 4
    $\begingroup$ @spatialdelusion The earth's rotation about it's axis is 0.0041 deg/s. The earth's rotation about the sun is .0000117 deg/s. Thus, you see roughly 400 times more pronounced of an effect from the rotation about its axis than you see caused by its rotation around the sun. $\endgroup$
    – Cort Ammon
    Commented Sep 19, 2017 at 6:44
  • 2
    $\begingroup$ Rotation about the sun would be almost inertial. The only difference from an inertial frame that gravity from the sun causes is tidal forces. $\endgroup$
    – Taemyr
    Commented Sep 19, 2017 at 11:48
9
$\begingroup$

There is a Foucault pendulum in the physics building of my university which does precisely that. In addition, the Caribbean and the southern US just got pounded by a couple of hurricanes - the fact that they all circulate counterclockwise is a result of the Coriolis force, which is an artifact of our non-inertial frame.

The mathematical framework which describes the non-inertial forces which arise in rotating references frames goes as follows (if you're not interested in the math, this part can be skipped - the actual answer to your question is at the end).


First, we need to consider how vectors $\hat e_i$ change when we're working in a rotating coordinate system. In an inertial frame, our unit vectors are constant in both magnitude (by definition) and direction. Therefore, for some general vector $\vec x = \sum x_i \hat e_i$,

$$\vec v \equiv \frac{d}{dt} \vec x = \sum_i \frac{dx_i}{dt} \hat e_i $$ and $$ \vec a \equiv \frac{d^2}{dt^2} \vec x = \sum_i \frac{d^2x_i}{dt^2} \hat e_i$$

On the other hand, if our coordinate system is rotating with angular velocity $\vec \omega$, then we have that

$$ \frac{d}{dt} \hat e_i = \vec \omega \times \hat e_i$$ and $$ \frac{d^2}{dt^2} \hat e_i = \left(\frac{d\vec \omega}{dt}\right) \times \hat e_i + \vec\omega \times (\vec \omega \times \hat e_i) $$

And so for a general vector $\vec x = \sum x_i \hat e_i$, we therefore have that

$$\vec v \equiv \frac{d}{dt}\vec x = \sum_i \frac{dx_i}{dt} \hat e_i + \sum_i x_i(\vec \omega \times \hat e_i) = \left(\sum_i \frac{dx_i}{dt} \hat e_i\right) + \vec \omega \times \vec x$$

and

$$ \frac{d^2}{dt^2}\vec x = \sum_i \frac{d^2x_i}{dt^2} \hat e_i + \sum_i \frac{dx_i}{dt}(\vec \omega \times \hat e_i) + \vec \omega \times \left(\sum_i \frac{dx_i}{dt} \hat e_i\right) + \sum_i x_i\left(\frac{d\vec \omega}{dt}\right)\times \hat e_i + \vec \omega \times (\vec \omega \times \vec x)$$ so $$ \vec a \equiv \frac{d^2}{dt^2} \vec x = \sum_i \frac{d^2x_i}{dt^2} \hat e_i + 2\vec \omega \times \left(\sum_i \frac{dx_i}{dt} \hat e_i \right) + \frac{d\vec \omega}{dt} \times \vec x + \vec \omega \times (\vec \omega \times \vec x)$$


Comparing the two cases, we see that at the moment the two frames coincide, the position, velocity, and acceleration as measured in the rotating frame are related to their inertial counterparts as follows:

$$ \vec x = \vec x_{rot}$$

$$ \vec v = \vec v_{rot} + \vec \omega \times \vec x_{rot}$$

$$ \vec a = \vec a_{rot} + 2\vec \omega \times \vec v_{rot} + \frac{d\vec \omega}{dt} \times \vec x_{rot} + \vec \omega \times (\vec \omega \times \vec x_{rot})$$


According to Newton's 2nd Law,

$$ \sum_i \vec F_i = m \vec a$$

So using our earlier work, we can re-arrange this to give

$$ m \vec a_{rot} = \left(\sum_i \vec F_i\right) - 2m\vec \omega \times \vec v_{rot} - m\vec \omega\times(\vec \omega \times \vec x_{rot}) - m \frac{d\vec \omega}{dt} \times \vec x_{rot} $$

This is Newton's 2nd Law applied to a rotating reference frame (positions are measured from the center of the rotating frame). The first term on the right is genuine, and encodes the "real" forces acting on the object. The rest of the terms are inertial forces (sometimes called "fictitious forces", though I somewhat dislike that name) and reflect the fact that our reference frame is accelerating.

  • $\vec F_{cor} \equiv -2m \vec \omega \times \vec v$ is the aforementioned Coriolis force, which drives hurricane formation and causes the plane of oscillation of a Foucault pendulum to precess
  • $\vec F_{cent} \equiv -m \vec \omega \times(\vec \omega \times \vec x)$ is the familiar Centrifugal force, which pushes particles away from the axis of rotation
  • $\vec F_{Euler} \equiv -m \frac{d\vec \omega}{dt} \times \vec x$ is the lesser-discussed Euler force, which affects systems with time-dependent rates of rotation, and acts in the tangential direction

Now, to answer your actual question. We often perform experiments in the lab under the assumption that the "lab frame" is inertial. We can quantify how good this assumption is by comparing the acceleration due to the "genuine" forces to the acceleration due to the non-inertial forces.

The magnitudes of the inertial accelerations are

  • $|\vec a_{cor}| = 2\omega v_\perp$ (where $v_\perp$ is the East-West velocity of the particle)
  • $|\vec a_{cent}| = \omega^2 r_\perp$ (where $r_\perp$ is the distance from the axis of rotation)
  • $|\vec a_{Euler}| = 0$ (because $\omega$ is constant)

On Earth, $\omega = \frac{2\pi}{24\text{ hours}} \approx 10^{-4}$ Hz, and the mean equatorial radius of the earth is approximately $6.5 \times 10^6$ m. Therefore,

  • $|\vec a_{cor}| \approx 2\times 10^{-4} v_\perp$ m/s$^2$ (where $v_\perp$ is measured in m/s)
  • $|\vec a_{cent}| \approx 6.5 \times 10^{-2}$ m/s$^2$ (at the equator)

As you can see, the accelerations due to the inertial forces are on the order of $10^{-4} - 10^{-2}$ m/s$^2$. They are small and easily "washed out" by gravity, air resistance, and the like - but they are certainly not trivial, and careful experiments like the Foucault pendulum are capable of measuring them. Of course, the effects of these small accelerations are much more visible over long times and large distances, so in that sense they can be tough to see in a closed laboratory - but you can see their effects on everything from weather patterns to the ocean currents.

$\endgroup$
2
  • 2
    $\begingroup$ This contains an excellent answer, namely it shows the relatively tiny effect of forces that the OP seems to be envisaging, but this information is buried a bit in the middle. The answer would be better if it gave a summary first rather than diving into differential equations, at least to start with, in case the OP gets put off and misses the key points. $\endgroup$
    – Stilez
    Commented Sep 19, 2017 at 11:55
  • $\begingroup$ @Stilez Good point - I added a note at the top. $\endgroup$
    – J. Murray
    Commented Sep 19, 2017 at 12:26
4
$\begingroup$

Although other answers explain the physics of the earth's frame, the OP seems to be looking for a different kind of answer, namely, why don't we seem to notice these in everyday life, more than why it isn't seen "in experiments".

The answer is that you can and do see it, on a large scale, over non-brief time periods, and in specific specialised situations. Weather (especially spiralling winds and fluids) is one case, so are the Foucault pendulum and large guns/missile trajectories. Other answers cover these.

But in everyday life our frame can be masked by other effects or "hidden in plain sight":

  • Simply not falling (moving as gravity dictates) on a massive planet suggests a non inertial frame, because a force must be constantly acting on us. It is - the ground pushes us and all our houses etc up. We just don't see this as a non inertial frame because we are so used to it, we don't think "some force must be acting on us constantly to offset the effect of gravity".

    (If you're interested, the actual forces doing this are mainly electromagnetic - including electrostatic - and degeneracy pressure of electrons and other particles. These forces tend to keep particles separate when other forces such as gravity would otherwise pull them closer. Technically, degeneracy pressure isn't a force, in the sense that quantum exclusion isn't an 'interaction' in the way that gravity and electromagnetism are, but it has a similar effect, so it's often referred to as one.)

  • On everyday small scales, other forces and effects are more prominent. A pool ball seems to move as it would in an inertial frame because the lateral forces due to smoothness and level of the table have so much more effect than the non-inertial frame effects can have, over the few seconds and couple of metres of a pool ball being hit.

  • An object dropped or thrown through the air is much more affected by wind and air resistance (and random Brownian Motion in some cases). Also because the air is already moving in a way that would mask non-inertial effects, and on a small scale the air's motion and resistance are more related to viscosity and flow, so they carry and impart a motion that doesn't show obvious signs of being non-inertial.

  • Even if you remove the air and fluid effects above, an object moving in a vacuum on earth would still start with the inertial frame of the hand (or whatever else) projected it. So even without air, if you threw a tennis ball across a tennis-court-sized vacuum chamber, for the few seconds and metres it is in motion, the divergence from a straight line is masked by short time, short distance, and by the fact it started with the same motion your hand did, so any divergence is measured from that and not from a static non-inertial frame. It's also still in the non-inertial frame of earth's gravity which isn't affected by vacuum.

    However..... If you could create a huge vacuum chamber that was much larger, you would see the non-inertial frame somewhat, as the ball would appear to slightly diverge from a straight path or not move at constant velocity as measured along its path. But this wouldn't just be because of the earth's frame or other non-inertial frame effects. Much/most of it would probably be due to the earth's shape as a sphere, gravity acting unopposed on it (but not on the chamber walls/floors/instruments), and so on.

Summary

In general, non inertial effects are therefore often masked by: the medium imparting motion to the objects we are watching; friction/viscosity/turbulence effects of the medium that our objects are in, or in contact with; forces we don't see because we are so used to them; short distances or timescales where the effect us too small for us to see it clearly, and so on.

In experiments and in many specific real-world situations, we can and do see the effects of our frame.

Just look at water swirling down the plug-hole of your next bath, and it's right there, hidden in plain view because we're so used to it.

$\endgroup$
6
  • $\begingroup$ I am not entirely convinced of your first bullet point regarding "simply falling". The ground pushes us up, yes, but the weight caused by gravity pulls us down. Thus is an equilibrium and the result is that we neither fall nor fly. How is this an example of the rotating centrifugal effect? $\endgroup$
    – Steeven
    Commented Sep 20, 2017 at 10:44
  • $\begingroup$ The OP didn't specifically state centrifugal or any other. It asked about any/all kinds of non-inertial frame effects. Without analysing every frame we're in, the fact that we are carried on the earth rather than at its centre, and the forces applying to us which we don't see because of familiarity, are going to be a factor in these, although I agree I may have stated it simplistically. My aim in that point was to show the general kinds of ideas which might explain "why we don't seem (in daily life) such effects, or not obviously", because that seemed to be the thrust of the question. $\endgroup$
    – Stilez
    Commented Sep 20, 2017 at 14:35
  • $\begingroup$ Ok. But "the fact that we are carried on the earth rather than at its centre is going to be a factor in these" is not really fulfilling for me. Imagine a not-rotating and not-revolving planet. Gravity will still pull down and we will still be held up by the ground. But this is an inertial frame (unless we take the relativistic view). Am I misunderstanding your point? $\endgroup$
    – Steeven
    Commented Sep 20, 2017 at 14:38
  • $\begingroup$ Not really :) But if answering technically, I'd also extend the answer - earth also keeps us in a non-inertial frame with regard to motion round the sun, within the galaxy, and as we partake in the motion of our galaxy and cluster etc. There are many non-inertial frames we could consider ourselves part of, but we stick with earth's surface movement due to 1) predominance locally of earths gravity, and 2) being stuck 4000km from earths COG on our rotating and wildly moving points. I just didn't want to get into deep analysis on it, preferring to keep it simple as the OP seemed to prefer. $\endgroup$
    – Stilez
    Commented Sep 20, 2017 at 14:43
  • $\begingroup$ "Just look at water swirling down the plug-hole of your next bath, and it's right there, hidden in plain view because we're so used to it." How is this related to reference frames? In an inetrial frame, water would not go straight down either. $\endgroup$
    – JiK
    Commented Sep 21, 2017 at 8:38
4
$\begingroup$

Your question is a deep one. Since, as we shall see in a minute, what stands in the way of being an inertial frame is gravity in practice, the answer is provided by General Relativity, which is our best theory of gravity. It stems from the following essential equivalence: an observer in a lab without windows cannot distinguish between the two following situations:

  • (a) the lab is in free fall in a gravity field;
  • (b) the lab is in motion at constant velocity in vacuum.

Since (b) is the very definition of inertial frame, it means that an inertial frame is a free falling lab. Then this explains why there are, qualitatively, two ways to depart from being an inertial frame, which are two ways to mess up situation (a).

Not being in free fall

Well, obviously! This is the case of a lab on the surface of the Earth. Were it in free fall, it would move toward the centre of the Earth. Instead it is carried along by the Earth spinning about its North-South axis. This results in the Coriolis force, and in the speed of light not being isotropic. The former results in the peculiar motion of the Foucault pendulum, in the vortices of hurricanes. The latter results in the Sagnac effect (a light signal traveling along a closed rectangular path takes a different time in one direction than in the other).

But, as you have surely heard, the Coriolis force needs spatially large system and long periods of time to have a measurable effect. Similarly, for the Sagnac effect, if we collapse the rectangle onto a line, the effect disappears. The beauty of General Relativity is that it can be quantitative here. Given the motion of the lab, its spatial extension and the duration of the experiment, one can compute the deviation from inertial frame (it is given by the curvature of the world line of the lab for the record). In any case, the lesson to take home is that for small enough experiments performed for small enough times, an observer at the surface of the Earth is almost inertial, and the smaller and the shorter, the better [*]. That's why your little nephew playing marbles in the schoolyard assumes his marble will go straight, and he is right about that, to within the precision that matters for his game! Where he to play with 1 ton marble hurled several kilometres away, he would need to correct for the Coriolis force, as naval officers operating the good ol' 50-caliber guns on battleships knew.

Tidal effects

This is the second way to mess up the situation (a) above. This is totally negligible for a lab at the surface of Earth: the dominant effect is that of the previous section. But let's imagine we magically stop the Earth spinning. A lab at the surface of Earth is still not in free fall, so let's make it so: imagine we have a large container that we drop from 20,000 feet. Neglecting air drag, it free falls toward the Earth and by the argument of the introduction, we have created an inertial frame. However the gravitational pull of the Earth on any object inside the container is toward a different direction at one end A of the container compared to the other end B: this is measurable if the container is so large that the angle between the straight line from the center O of the Earth to A and the straight line from O to B is large enough. As a result, an observer inside that big container will measure a slight different of motion for objects at A and at B. This is a tidal effect. Again, it can be made as small as we want by making the container small enough and/or measuring the motion of those objects for short enough a duration. And again, General Relativity provides a way to quantitatively predict his effect (the deviation from inertiality will be related to the curvature of spacetime in this case).

[*] Moreover, the slower the Earth spins, the larger and the longer the experiment can be for a given quality of inertial-ness of the lab.

$\endgroup$
3
$\begingroup$

Surely, the Foucault pendulum experiment can easily prove earth's rotational motion and hence its acceleration. Also Coriolis effect proves earth's rotational motion. Most experiments conducted in Earth's frame (lab frame) are correct upto and order. You surely don't require to consider Earth's rotational effect for experiments that are small scale. But long range experiments require that correction. e.g. long range ballistic missiles and even snipers at times require to consider Earth's rotation.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.