Is there any true inertial reference frame in the universe?

Is there any true inertial reference frame in the universe?

Newton's first law states that an object at rest remains at rest, and an object performing uniform motion performs uniform motion, until and unless acted upon by an external force, if viewed from an inertial frame. It is the definition of an inertial frame of reference. And Newton's second law states that the net external force acting on a particle equals its mass times its acceleration. Thus we need to have an inertial frame in order for Newton's 1st and 2nd laws to be applicable.

Scientists claim that Earth is an inertial frame, either the ECEF frame or the ECI frame. Why is that? Earth's different parts have different accelerations when it performs rotational motion about its axis. Now you may say that Earth's axis is inertial, but the Earth is also revolving around the Sun. Thus Earth's frame should be non-inertial.

Even supposing that earth's frame is inertial, then it means that rest of the universe is non-inertial, because only a frame that moves with constant velocity with respect to an inertial frame is also an inertial frame and according to various scientific experiments there is no other matter in this universe which satisfies that criterion. Now you may say that earth is an approximate inertial frame, but still, my question is: is there any perfect inertial frame in the universe where Newton's laws are exactly applicable?

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manishi, I edited your question to bring it up to our quality standards. As I've told you, we ask that questions (and answers) use reasonably standard spelling and grammar, and avoid ALL CAPS or unnecessary formatting. Please use this and other questions on the site as examples of how to write your posts in the future - you can't count on someone editing your posts into shape every time. – David Zaslavsky Nov 26 '11 at 6:01
I do not think that there are any scientists that claims that the earth is an inertial frame. They sure would be bad scientists. – Hans-Peter E. Kristiansen Nov 26 '11 at 7:08

When you ask for a "perfect" or "true" inertial reference frame you are asking for something that cannot be answered in physics. Perfection is only possible in mathematics, not physics.

So in physics, what can asked is whether or not the reference frame is an inertial to a certain level of accuracy. The surface of the earth is not inertial because of the gravitational field of the earth - not because the earth is moving around the sun and the sun is moving around the galaxy. But if you consider motion only in a horizontal plane on the surface of the earth and if you are only doing the typical high school physics tabletop experiments, the earth is an inertial reference frame as far as the accuracy of the measurements performed is concerned. If you do more accurate measurements, then it would not be an acceptable inertial reference frame.

Consider a satellite in orbit around the earth and examine a relatively small volume near the center of mass of the satellite. That small volume over a suitably small period of time will be an inertial reference frame to a very high level of accuracy. For example, two small masses that are 1 inch apart (radially) in orbit around the earth that start out "exactly" at rest relative to each other will over a time period of 10 seconds come to have a relative speed of 0.006 inches/second due to the differences in orbital velocity for two orbits that differ by 1 inch. So it depends on the level of accuracy needed for an experiment that you want to perform in an inertial reference frame.

To get a reference frame that is more accurately inertial it would necessary to be orbiting much further from all gravitating objects. Thus it is all about the level of accuracy you require of the inertial reference frame.

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you did not get my question... – manishi Nov 26 '11 at 8:17
@manishi - first of all welcome to Physics.SE. 2nd please explain what I did not get about your question. To say that a certain frame is an inertial frame, you have to specify what level of accuracy is being requested. For example, on the surface of the earth, if you only consider horizontal motion in the typical high school physics tabletop experiments, the earth is an inertial reference frame as far as the accuracy of the measurements performed is concerned. If you do more accurate measurements, then it would not be an acceptable inertial reference frame. – FrankH Nov 26 '11 at 8:46
Gravity does not make Earth non-inertial in Newton physics. It is just a force like any other forces. -1 – Anixx Apr 19 at 14:01
@Anixx It is gravitational attraction between the Earth and sun that causes the Earth to orbit the sun (i.e. accelerate). Without gravity this acceleration would not be present. Reference frames must be referenced to something (typically matter). I think FrankH means that you need to be very far from all gravitating objects otherwise gravity would cause this matter to accelerate ruining your inertial reference frame. – OSE Apr 19 at 17:19
@Anixx - According to en.wikipedia.org/wiki/Inertial_frame_of_reference - "All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration." So since an accelerometer would have non zero acceleration on the Earth's surface, it would not be an inertial reference frame. Remember that gravitation and acceleration are exactly the same thing according to the principle of equivalence and General Relativity. Only if you are in FREE FALL in gravity then it would be an inertial frame. – FrankH Apr 21 at 5:52
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OK, you got a little confused. Earth is approximately inertial, not exactly. When we do an experiment in the lab, we treat it like an inertial frame because it is nearly one. Actually, there's gravity but an experiment in two dimensions on a table top can ignore that, and an experiment with light or very light particles can also ignore it since it's so small. Ditto for centrifugal force of rotation caused by spin of Earth, orbit of Earth around sun, pull of moon, pull of sun, pull of jupiter, etc.

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The answer to your question is no. There is no exact Newtonian inertial frame of rereference in the universe. (And by the way, it's not true that knowledgeable physicists claim the earth's frame to be exactly inertial in the Newtonian sense.)

In Newtonian mechanics, an inertial frame is one in which Newton's second law holds for a test particle, without the need to introduce fictitious forces. Suppose we observe a test particle to have an acceleration $a$ in a certain frame. To determine whether it's obeying the second law, we need to find the sum $F$ all the forces acting on it. This sum includes the gravitational forces of all objects, no matter how distant. Once we've determined $a$ and $F$, we can say whether the second law is obeyed, and if it's not, we can always determine a frame in which it would have been obeyed.

But this assumes we can determine $F$. As we take into account matter farther and farther away, the sum will settle down fairly rapidly. By truncating the sum at some distance $r$, we expect that the error in $F$ will probably shrink like some power of $r$. But in order to define an exact Newtonian frame, we would have to take the limit as $r\rightarrow\infty$. The trouble is that beyond a certain length scale, Newtonian mechanics itself starts to become a poorer and poorer approximation. There will certainly be values of $r$ such as, say, a million light years, for which Newtonian mechanics is a very good approximation, and we can get a very good approximation to an inertial frame, but we can never have a perfect inertial frame.

Note that in general relativity, we simply define an inertial frame as the frame of a free-falling particle, which is much simpler to verify -- it can be verified locally.

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Yup. Earth is not an inertial frame, but we make it inertial by including additional forces so that we can write down Newton's Laws. These are called "fictitious forces" for obvious reasons and you can read about it on Wikipedia http://en.wikipedia.org/wiki/Fictitious_force . You might have already heard about the Coriolis effect, which is one of the [three] additional forces, due to Earth spinning/moving.

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psuedo force is an imaginary force which is applied on the body if the frame is non inertial,in order to use the equation F=ma.The direction of psuedo force is opposite to the direction of acceleration of non inertial observer w.r.t inertial frame.the magnitude of psuedo force=mass of the body on which it is applied*acceleration of non inertial observer w.r.t inertial frame......thus if earth is non inertial and you want to make it inertial by applying fictitious forces then again you require an inertial frame to do so.. – manishi Nov 26 '11 at 8:28
An inertial frame is a mathematical construct... by adding in the fictitious forces, we have have obtained an inertial frame of reference. – Chris Gerig Nov 26 '11 at 8:37
Comment to the answer (v1): If, when describing Newton's second law in some reference frame, fictitious forces are needed, then that reference frame is by definition not an inertial frame. – Qmechanic Apr 19 at 14:35

Wiki writes "In general relativity, an inertial reference frame is only an approximation that applies in a region that is small enough for the curvature of space to be negligible." - since the gravity force is infinite(and a frame anyway needs to be referenced to an object(with mass)), there is no really true reference frame anywhere. The earth is a good everyday approximation on the small scale(buildings and smaller). The fixed stars(http://en.wikipedia.org/wiki/Fixed_stars) are very good approximations.

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For a parcel of atmosphere (or any object) that is moving in the Earth's reference frame with constant speed, another fictitious force, the Coriolis force appears in the equation of motion

0 = (omega * (omega*r))centrifugal force - (2omega*v)Coriolis force - ((1/density)*∇pressure) pressure gradient + (g)gravity + (v(∇^2*v)) frictional force)

And this different from the equation of inertial frame which is

Omega * (omega*r) = - ((1/density)*∇pressure) pressure gradient + (g)gravity + (v(∇^2*v)) frictional force)

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