Question about inertial frames

[quasar987]

There is something that bugs me about inertial frames.

According to Einstein's definition (chapter 4 of "Relativity"), an inertial frame is one for which Newton's first law holds: "a body far enough removed from other bodies continues in a state of rest or uniform motion in a straight line." He then states that it is with respect to these frames that the Newtonian laws of mechanics are assumed to hold true.

And of course, his postulate of special of relativity concerns these reference frames also, saying that wrt any two such frames of reference, the laws of nature take the same form.

Also in chapter 4, he gives as an example of a non inertial frame of refenrence a frame that is attached to the earth, because wrt such a frame, the stars, which are far removed from other bodies, don't move in straight lines (they move in circles due to the Earth rotating).

Ok, so any frame of reference that we use on Earth to measure stuff is non inertial. So then neither the Newtonian laws of mechanics nor the postulate of SR applies to them. What the heck?

 

[WannabeNewton]

Indeed a frame fixed to the Earth is not a true inertial frame, it is just a very very good approximation of one. Also, slightly unrelated, note that inertial frames in SR are different from inertial frames in Newtonian mechanics.

 

[quasar987]

Also, slightly unrelated, note that inertial frames in SR are different from inertial frames in Newtonian mechanics.

In what way? I don't recall Einstein making any distinction in his book.

 

[WannabeNewton]

In what way? I don't recall Einstein making any distinction in his book.

They are more general in the sense that in SR the inertial frames are related by lorentz transformations whereas in Newtonian mechanics they are related by galilean transformations, which the lorentz transformations reduce to in the slow motion limit. In Newtonian mechanics if you were in an inertial frame and galilean boosted to another frame then you would still call the new frame an inertial frame. In SR, making no assumptions about slow motion, if you are in an inertial frame and lorentz boosted to another frame then you would still call the new frame an inertial frame. The notion of an inertial observer being one who undergoes free motion is the same in both theories of course.

 

[PeterDonis]

Indeed a frame fixed to the Earth is not a true inertial frame, it is just a very very good approximation of one.

I would state this a bit differently: a frame fixed to the Earth is a very, very good approximation to an inertial frame *for horizontal motion* (over distances that are small compared to the radius of the Earth). Such a frame is *not* a good approximation to an inertial frame for vertical motion. (At least, it's not a good approximation to an inertial frame in Einstein's sense. It may be in Newton's sense.)

 

[Dale]

Ok, so any frame of reference that we use on Earth to measure stuff is non inertial. So then neither the Newtonian laws of mechanics nor the postulate of SR applies to them. What the heck?

This is correct. A Foucault pendulum is an example of a crude experiment which was possible around Newton's day* and which demonstrates the non-inertial nature of an earth-fixed reference frame.

*although possible in Newton's day, it wasn't actually performed until the 1850's:
http://en.wikipedia.org/wiki/Foucault_pendulum

 

[Darwin123]

There is something that bugs me about inertial frames.

According to Einstein's definition (chapter 4 of "Relativity"), an inertial frame is one for which Newton's first law holds: "a body far enough removed from other bodies continues in a state of rest or uniform motion in a straight line." He then states that it is with respect to these frames that the Newtonian laws of mechanics are assumed to hold true.

And of course, his postulate of special of relativity concerns these reference frames also, saying that wrt any two such frames of reference, the laws of nature take the same form.

Also in chapter 4, he gives as an example of a non inertial frame of refenrence a frame that is attached to the earth, because wrt such a frame, the stars, which are far removed from other bodies, don't move in straight lines (they move in circles due to the Earth rotating).

Ok, so any frame of reference that we use on Earth to measure stuff is non inertial. So then neither the Newtonian laws of mechanics nor the postulate of SR applies to them. What the heck?

The "distant stars" are not a good reference for defining an inertial frame because a) There is more than one distant star and b) all of them are moving relative to each other. Hence, Einstein gave a detailed protocol in his 1905 article that defines inertial frame. Actually, he called them reference frames in his 1905 article but the phrase got generalized later. In any case, an inertial frame is not defined relative to "the distant stars."

I once tried to summarize the Einstein protocol for myself. I had to read an essay by H. A. Lorentz before I understood it. So there is a little Lorentz in the following description. However, here it goes.

Measuring devices are defined as materials that exist in a local state of quasi-equilibrium. That I got from Lorentz. Rulers and clocks have to be built in such a way that left to themselves, they do not change significantly with time. They are systems that don't change phases. They are systems that don't absorb or emit significant amounts of energy. They are systems held together by forces which vary very little with time.

The measuring device may be able to oscillate without emitting or absorbing significant amounts of energy. Then you can call such a device a clock. It may have observable features that can be observed without emitting or absorbing significant amounts of energy. Then you would call such a device a ruler.

A hypothetical measuring device that does not have an external force or external torque acting on it defines the inertial frame. The position of that measuring device is the origin of the coordinate system of that inertial frame. If the measuring device oscillates, any single oscillation can be the initial time.

The internal parts of that device will be interacting through matching forces that satisfy Newton's third law of motion. Every part in that device can have a nonzero force acting on it. However, that nonzero force has to be matched in another object by a force in the opposite direction. If there is any internal part that is acted on by an unpaired force, then it can't be used to define an inertial frame. The unpaired force is referred to as a pseudoforce.

All hypothetical measuring devices that are stationary with respect to that "origin" are part of the inertial frame. All such devices won't have external forces acting upon them.

Consider a clock "resting" on the surface of the earth. The contact force of the ground is holding it up. The force of gravity is pulling it down. The sum of contact force and gravity is not zero. The gravity has to be a little larger than the contact force because the clock is accelerating toward the center.

The sum of contact force and gravity is the centripetal force. The clock is being acted on by a nonzero centripetal force. Therefore, the clock can't be considered part of an inertial frame if it is fixed to the surface of the earth.

This protocol isn't much different from the protocol that Newton describes in Principia. Newton describes an "absolute space" which also has to be defined in terms of "measuring devices". That is why Principia is so painfully long. If all Newton had to do was write down the three laws, the book would have been a pamphlet.

Einstein's "inertial frame" is only slightly weirder than Newton's "absolute space". In both cases, there are certain quantities that are relative. What defines both are "measuring instruments" made of parts which are subject to paired forces. Every part in the device has to have a matching part with an equal and opposite force acting on it.

That is why before I do any mechanics problem, relativistic or Newtonian, I write down an inventory of the forces and parts. Even in some of these "paradox" problems (especially in "paradox" problems), I make sure that I know where all significant forces are pointing. One can't a priori assume that a force is insignificant. Because if the force is unpaired, then the object it acts on can't be part of an "inertial frame" or an "absolute space."

 

[PeterDonis]

Consider a clock "resting" on the surface of the earth. The contact force of the ground is holding it up. The force of gravity is pulling it down.

Ok, so you're using a Newtonian definition of "force", not a relativistic one. (By the relativistic definition of "force", the only force on the clock is the ground pushing it up.)

The sum of contact force and gravity is not zero. The gravity has to be a little larger than the contact force

This is the right answer (in the Newtonian view), but for the wrong reason:

because the clock is accelerating toward the center.

Huh? This doesn't make sense in either the Newtonian or the relativistic view. In the Newtonian view, the clock is at rest; it's not accelerating in any direction. In the relativistic view, the clock is accelerating *away* from the center; its proper acceleration is outward, which is what keeps it at the same radial coordinate (a freely falling frame would decrease its radial coordinate with time).

The sum of contact force and gravity is the centripetal force. The clock is being acted on by a nonzero centripetal force.

No, you've got it backwards. In the Newtonian view, the clock is being acted on by a nonzero *centrifugal* force (i.e., *away* from the center) because it's in a rotating frame. The sum of the upward contact force from the ground, and the upward centrifugal force, equals the downward force of gravity, so the net force on the clock is zero and it is at rest.

 

[zn5252]

There is something that bugs me about inertial frames.
Ok, so any frame of reference that we use on Earth to measure stuff is non inertial. So then neither the Newtonian laws of mechanics nor the postulate of SR applies to them. What the heck?

If you were in a lift in free fall on Earth (discard air drag ), and then you let go of an apple, it will follow a straight line. This is a local inertial frame, limited in time and space so that tidal effects are neglected...

There are also instantaneous inertial reference frames which can be used for a brief moment, but of course it will change abruptly from one moment to another...

 

[WannabeNewton]

Such a frame is *not* a good approximation to an inertial frame for vertical motion

Ah yes that is a good point Peter.

 

[Darwin123]

Ok, so you're using a Newtonian definition of "force", not a relativistic one. (By the relativistic definition of "force", the only force on the clock is the ground pushing it up.)

There are no "inertial frames" in general relativity. The phrase "inertial frames" implies that the question involves special relativity only.

I am explaining things in terms of special relativity, not general relativity. I am using an approximation, consistent with special relativity, where gravity is just a force like any other.

I used the wrong example, maybe. Others have pointed out that there are some first-order problems with mixing "gravitational force" with "special relativity. I was thinking of the Hafele-Keating experiment, which can be analyzed with some accuracy by that approximation. However, this may be a special case.

I stick to my answer because you did not mention "gravity" and because you used the phrase "inertial frame". As long as there is no "source" of gravity", special relativity is accurate in an inertial frame. The inertial frame is what I described.

This is the right answer (in the Newtonian view), but for the wrong reason:

This "absolutely" makes sense in the "absolute space" of Newton's Laws for the reason that I detailed. The clock is accelerating toward the axis of the Earth in the "absolute space" defined by Newton. This acceleration is referred to as "centripetal acceleration."

Velocity is a vector, both in Newton's Laws and in special relativity. Vectors include both magnitude and direction. Although the "speed" of the clock may not be changing, the direction of motion is changing. The change in direction is part of acceleration.

The acceleration as measured in the "absolute space" is a nonzero quantity. The acceleration may be zero if you use the surface of the Earth as a reference. However, you then have to explain two "unpaired <i.e., pseudo> forces." They are the centrifugal force and the Coriolis force

Vector analysis isn't that different in special relativity. The formula for centripetal acceleration is used in Einstein's article, although he doesn't call attention to it. He assumes that his reader knows basic vector analysis.

Huh? This doesn't make sense in either the Newtonian or the relativistic view. In the Newtonian view, the clock is at rest; it's not accelerating in any direction.

In the "absolute space" of Newton, the clock is accelerating inward. It is not at rest in "absolute space".

This is a rather basic part of Newton's mechanics. Newton's Three Laws of motion are only valid in "absolute space." If you try to apply Newton's Laws in any frame that isn't an "absolute space", you are liable to make errors. One can use frames other than "absolute space", but then one has to hypothesize "pseudoforces". As I said, a pseudoforce is an "unpaired force".

In the relativistic view, the clock is accelerating *away* from the center; its proper acceleration is outward, which is what keeps it at the same radial coordinate (a freely falling frame would decrease its radial coordinate with time).

You have one idea that is right for "general relativity". Gravity is not a force. However, there is no "acceleration" in general relativity either.

My knowledge of general relativity is very limited. However, I think I know the issue you are referring to. Therefore, I will address this issue at the risk of incurring the wrath of the moderators.

What substitutes for "inertial frame" in general relativity is the "geodesic". What is often called "acceleration" in special relativity is a "deviation from the geodesic" in general relativity. A body which is not acted on by external forces is on a world-line which is a geodesic.

"Free fall" in general relativity mean being on a geodesic. The shortest path in four dimensional space between any two events is a geodesic. A clock in orbit (i.e., free fall) would be on a geodesic. If any non-gravitational force acts on the clock, then the clock would not be on a geodesic.

A clock fixed to the surface of the Earth would not be on a geodesic. It is acted on by the contact forces of the ground. An airplane in the air is not on a geodesic. It is being acted on by aerodynamic forces (i.e., aero=related to air). What ever non-gravitational forces are acting on the measuring device make it transform from one geodesic to another. This very roughly corresponds to a difference in inertial frames, but is slightly different.

In any case, the clock fixed to the surface of the Earth is in no way on an inertial frame. The surface of the Earth exerts a non-gravitational force on the clock. To use relativity, you have to compare the ticking of that clock to the ticking of a clock that is NOT being acted on by a non-gravitational force. The hypothetical clock can be in orbit around the Earth or even suspended at the center of the earth. End of digression on general relativity.

Now I confess some limitations. I don't get the mathematics of space-time completely. Therefore, I generally try to stay away from general relativity. When asked a question about relativity, I generally answer using special relativity. I use the approximation that a gravity is a force like any other, which is consistent with special relativity.

Furthermore, I know Newtonian physics very well. So I really have to object to the statement, "the surface of the Earth is at rest in Newtonian physics". Newtonian physics is not as simple as some non-scientists believe.

If I am asked about the Hefel-Keating experiment, or any other relativistic experiment, I use the special relativity/gravitational force approximation. It is not always accurate, but for many issues it is close enough. If you insist on an explanation that is totally consistent with general relativity, then there are other people who would be better able to help you.

No, you've got it backwards. In the Newtonian view, the clock is being acted on by a nonzero *centrifugal* force (i.e., *away* from the center) because it's in a rotating frame. The sum of the upward contact force from the ground, and the upward centrifugal force, equals the downward force of gravity, so the net force on the clock is zero and it is at rest.

There is no centrifugal force in either Newton's absolute space or Einstein's inertial frame. The centrifugal force is a pseudoforce. The centrifugal force only exists in two types of places. The centrifugal force can exist in a space that is accelerating relative to an absolute space. The centrifugal force can exist in a reference frame that is accelerating relative to an inertial frame.

In frames where the centrifugal force "exists", it has physical effects. However, there are tests to determine whether or not it is a pseudo-force. If the force is a pseudo-force, than the objects that it is associated with are unpaired. If the centrifugal force acts on a body, there is no matching body on which an opposite and equal force is being exerted. The issue that you are discussing is just as important in Newtonian physics as in relativity. Determination of the proper frame for using mechanics has to be done regardless of the mechanics one is using.

 

[PeterDonis]

There are no "inertial frames" in general relativity.

First of all, who said anything about general relativity? I just said "relativistic". What I said would apply just as well to someone standing inside a rocket whose engine was accelerating it at 1 g, i.e., in a context where SR is sufficient.

Second, yes, there are inertial frames in GR. They just don't cover all of spacetime; they only cover small patches of it.

That said, if you want to limit discussion here to SR, that's fine. But see further comments below.

I am explaining things in terms of special relativity, not general relativity. I am using an approximation, consistent with special relativity, where gravity is just a force like any other.

There is no general "approximation" in SR where "gravity is just a force like any other", because the "force" of gravity is not felt, and other forces are--i.e., objects moving solely under the "force" of gravity are in free fall, but objects moving under other forces are not. In other words, gravity is best thought of, in the "SR approximation" you are talking about, as a pseudoforce, not a "real" force. See further comments below.

You can approximate being at rest in a gravity field in SR by treating it as being at rest in an accelerated frame, in accordance with the equivalence principle. This works as long as tidal gravity is negligible. I'm not sure what you are trying to do fits within that limitation; see further comments below.

Also, I'm not sure that you're actually treating the force on an object rotating with the Earth as "gravity" anyway. See further comments below.

I used the wrong example, maybe. Others have pointed out that there are some first-order problems with mixing "gravitational force" with "special relativity.

Yes, there are. See above.

I was thinking of the Hafele-Keating experiment, which can be analyzed with some accuracy by that approximation.

And which doesn't require treating gravity as a "force" at all. It only requires knowing about gravitational time dilation with altitude (and combining it with SR time dilation due to relative speed).

The clock is accelerating toward the axis of the Earth in the "absolute space" defined by Newton. This acceleration is referred to as "centripetal acceleration."

Ah, I see. You are not using a rotating frame; you are using an inertial frame in which the Earth's center of mass is at rest, but in which the Earth itself is rotating, so an object on the surface of the Earth is not at rest. Yes, what you say is correct in that frame, as far as it goes (but see further comments below); I just didn't realize that was the frame you were using.

In this frame, however, the object on the surface of the Earth is *not* "at rest in the gravity field", because the gravity field is *not* rotating with the Earth. (At least, it isn't in your approximation; the Earth's rotation does have a small effect on the "gravitational field" in its vicinity, but it's much smaller than the effects you're considering.) So the application of the equivalence principle is not as simple as I described above, and in fact I don't think that's really what your model is doing.

Your model is basically treating an object on the surface of the Earth as being swung around the Earth's center of mass by a long rope, which pulls it inward just enough to counteract its inertia and keep it in a circular path. There's nothing specific to gravity in the way you're modeling this force that I can see. The only reason "gravity" needs to be brought into it at all is to explain why an object "sticks" to the surface of the Earth and moves in a circular path with it, instead of flying off into space; but the net result, at least for the object's trajectory (see below) is the same as if the object were just on a long rope, as I said.

Note, however, that this model does *not* correctly predict the force *felt* by the object. The "swinging on a rope" model predicts that the object feels an inward force; but in fact it feels an outward force. However, you *cannot* fix this by "adding a force of gravity" to the SR inertial frame you are using, because the direction of that "force of gravity" is not constant; it points in different directions as the object moves around its circular path. That is, tidal effects are not negligible, so the "SR plus gravity as a force" approximation breaks down (see above).

In other words, I do not think the approximation you claim to be using is entirely valid. It predicts the object's trajectory OK, but it does not correctly predict the force felt by the object (i.e., its proper acceleration).

You have one idea that is right for "general relativity". Gravity is not a force. However, there is no "acceleration" in general relativity either.

Yes, there is; there are two kinds, just as there are in SR. There is coordinate acceleration, which is frame-dependent; and there is proper acceleration, which is the invariant path curvature of an object's worldline.

What substitutes for "inertial frame" in general relativity is the "geodesic". What is often called "acceleration" in special relativity is a "deviation from the geodesic" in general relativity. A body which is not acted on by external forces is on a world-line which is a geodesic.

This is OK except for your implication that proper acceleration (which is what you're describing) is somehow not "really" acceleration but is just "called" acceleration. I don't see why you would think that.

A clock fixed to the surface of the Earth would not be on a geodesic.

Correct.

An airplane in the air is not on a geodesic.

Also correct.

What ever non-gravitational forces are acting on the measuring device make it transform from one geodesic to another. This very roughly corresponds to a difference in inertial frames, but is slightly different.

You're getting a bit garbled here. Just saying "is not moving on a geodesic" is sufficient; I'm not sure what "transform from one geodesic to another" means, but it doesn't describe moving on a non-geodesic curve. Also, I'm not sure what "very roughly corresponds to a difference inertial frames, but is slightly different" means, but I think it is connected to your misconception (see above) that GR doesn't have inertial frames. It has local inertial frames.

In any case, the clock fixed to the surface of the Earth is in no way on an inertial frame. The surface of the Earth exerts a non-gravitational force on the clock.

Yes, this is true.

To use relativity, you have to compare the ticking of that clock to the ticking of a clock that is NOT being acted on by a non-gravitational force.

No, you don't. You can compute the proper time along the clock's non-geodesic worldline without having to compare it with proper times along any geodesics. Of course if you want to *compare* the accelerated clock's elapsed time with the elapsed time on a non-accelerated clock between the same two events, you have to know both times. But that's by no means the only way to "use relativity".

I really have to object to the statement, "the surface of the Earth is at rest in Newtonian physics".

Why? Can you be more explicit?

If you insist on an explanation that is totally consistent with general relativity, then there are other people who would be better able to help you.

I didn't ask for an explanation; I already understand how inertial frames work. I was just not clear about which inertial frame you were using in your analysis (see above).

There is no centrifugal force in either Newton's absolute space

I think what you mean is that there is no centrifugal force in Newton's inertial frame, correct? Are you trying to claim that Newtonian physics can't be done in rotating frames? I wish I'd known that when I took physics exams in college; I could have gotten the professors to throw out some of the harder questions.

or Einstein's inertial frame...The centrifugal force can exist in a reference frame that is accelerating relative to an inertial frame.

Yes, agreed, but note that this applies to Newtonian as well as relativistic physics (see above).

Determination of the proper frame for using mechanics has to be done regardless of the mechanics one is using.

I agree if by "the proper frame" you mean "the proper frame for analyzing a given problem". But different problems might have different "proper frames" in which they are most easily analyzed. Inertial frames are not the only possibility.