STRACT: Inertial Frames in Special and General Relativity

[madness]

How are inertial frames defined in Special and General Relativity? In Newtonian physics, an inertial frame is usually defined as one in which N2 holds. Clearly this cannot be the same definition as for SR. In GR an inertial frame is one in which SR holds (I think). However, there is now a distinction between globally and locally inertial, and frames which are apparently accelerating towards each other can still be locally inertial.

 

[espen180]

In both SR og GR, an inertial frame is one which is moving along a geodesic in space-time. In SR this reduces to nonaccelerating frames as inertial.

 

[torquil]

How are inertial frames defined in Special and General Relativity? In Newtonian physics, an inertial frame is usually defined as one in which N2 holds. Clearly this cannot be the same definition as for SR. In GR an inertial frame is one in which SR holds (I think). However, there is now a distinction between globally and locally inertial, and frames which are apparently accelerating towards each other can still be locally inertial.

First of all, I concur with the reply above.

Yes, in SR an intertial frame can be said to fill the entire space. N2 would be valid for all movements of particles as seen from this frame, no matter how far away.

In GR this needn't be the case There could be tidal forces present, if the frame is located in a non-uniform gravitational field. So in GR it could be the case that the inertial frame is only exact in an infinitesimal volume, since deviations from N2 appear as soon as a particle is not in the origin of the inertial frame.

The fact that each planet experiences tidal forces is a manifestation of this. They do not follow N2 because their size is finite.

In GR, two intertial frames can accelerate towards each other. Just consider two frames, each freely falling towards a massive object, from opposite sides.

Torquil

 

[madness]

Isn't the Earth moving on a geodesic? Would someone standing on a non-rotating Earth be in an inertial frame? And the geodesics depend on the frame we're working in, but it still holds that someone traveling along one will be in an inertial frame?

 

[sylas]

Isn't the Earth moving on a geodesic? ...

Yes.

... Would someone standing on a non-rotating Earth be in an inertial frame?

No. They are kept on the surface by normal forces. A geodesic at the surface is the line traced by a falling object which otherwise does not interact with matter. The geodesics proceed downwards, through the surface. You can tell it is not inertial because you can feel weight.

The Earth itself, however, is falling unimpeded around the Sun... a geodesic.

And the geodesics depend on the frame we're working in, but it still holds that someone traveling along one will be in an inertial frame?

In GR, a geodesic is a frame which is freely falling. As I understand it, in GR, a frame is really only local to the observer falling along the geodesic.

Cheers -- sylas

 

[Ich]

Isn't the Earth moving on a geodesic? Would someone standing on a non-rotating Earth be in an inertial frame?

Earth's center is moving approximately on a geodesic.
Geodesics at the surface are accelerated inwards by Earth's gravitation, so someone standing at the surface actually accelerates away from a geodesic.
Further, and maybe more pertaining to your question: a point on the surface is moving alway parallel to Earth's center, but, due to tidal effects, experiences different gravitational "forces" from moon and sun. So even in the absence of Earth's gravitational field, it doesn't move on a geodesic. That's how inertial frames are valid only for small regions in curved spacetime.

And the geodesics depend on the frame we're working in, but it still holds that someone traveling along one will be in an inertial frame?

No, geodesics depend on initial conditions (velocity) and the curvature of spacetime. They are not coordinate dependent.

 

[madness]

"No. They are kept on the surface by normal forces. A geodesic at the surface is the line traced by a falling object which otherwise does not interact with matter. The geodesics proceed downwards, through the surface. You can tell it is not inertial because you can feel weight."

It seems a little strange that the Earth is moving along a geodesic and I'm moving along with the Earth yet I'm not moving along a geodesic. I suppose Ich is right that if I was in the centre of the Earth I would be moving along a geodesic.

"No, geodesics depend on initial conditions (velocity) and the curvature of spacetime. They are not coordinate dependent."

From the point of view of a man in freefall, he sits at the origin the whole time, whereas from the point of view of the Earth he traces out a geodesic. Or from the point of view of someone on a roundabout, a geodesic might be a spiral. They do depend on the frame of reference - that's where the affine connection come's into the geodesic equation.

 

[sylas]

From the point of view of a man in freefall, he sits at the origin the whole time, whereas from the point of view of the Earth he traces out a geodesic. Or from the point of view of someone on a roundabout, a geodesic might be a spiral. They do depend on the frame of reference - that's where the affine connection come's into the geodesic equation.

No; the geodesics are the same in all cases. All that is different is the co-ordinates of the geodesic. The geodesic itself is a set of points in spacetime... a path or a locus. The same set of points is a geodesic independent of what co-ordinate system you use to identify the points.

Cheers -- sylas

 

[Ich]

I suppose Ich is right that if I was in the centre of the Earth I would be moving along a geodesic.

Approximately. An extended object is always subject to tides, so no point of it moves exactly on a gedesic. But the center is as close as possible.

And sylas beat me to your second statement. There is only one geodesic, but infinitely many ways to assign coordinates to it.

 

[George Jones]

No; the geodesics are the same in all cases.

Yes.

The geodesic itself is a set of points in spacetime... a path or a locus.

This confuses a function with the image of a function, but I'm probably being overly pedantic here.

 

[sylas]

This confuses a function with the image of a function, but I'm probably being overly pedantic here.

Not at all! I'd much prefer to be as correct as I can manage, and I don't always get the terminology right. But I'm not sure of the issue here. A pedantic explanation of the best way to express it would be much appreciated!

Cheers -- sylas

 

[George Jones]

A geodesic depends not only on a set of points in spacetime, but also on how ("fast") the points are traversed. A curve in a spacetime M is a mapping

\gamma : I \rightarrow M,

where I is an interval (often closed) of the real line. A curve \gamma is a geodesic if the tangent vector \dot{\gamma} is parallel transport along the curve, that is, if

\nabla_{\dot{\gamma}} \dot{\gamma} = 0.[/itex]<br /> <br /> An observer is a future-direct timelike curve \gamma parametrized by proper time. This means that<br /> <br /> g \left( \dot{\gamma} , \dot{\gamma} \right) = 1<br /> <br /> all along the curve. In this case, the tangent vector &quot;field&quot; is the observer&#039;s 4-velocity, often denoted by \mathbf{u}, the observer&#039;s 4-acceleration is<br /> <br /> \mathbf{a} = \nabla_{\mathbf{u}} \mathbf{u},<br /> <br /> and a freely falling observer is a geodesic observer curve.<br /> <br /> Sometimes pinning down things so precisely confuses, and sometimes it helps with understanding.

 

[sylas]

A geodesic depends not only on a set of points in spacetime, but also on how ("fast") the points are traversed.

Ah! Thank you. I think I have been thinking of a geodesic as what should more properly be called the image of a timelike geodesic. Given the image of a timelike geodesic, I guess there is only one unique way to give a geodesic that is parameterized by proper time, and so I didn't worry about the map.

Sometimes pinning down things so precisely confuses, and sometimes it helps with understanding.

Quite so! But if I understand the definitions better, I may be able to improve my own descriptions, even if not given in full detail, so that they remain correct when made more precise; which would be nice. Not to mention that I simply wanted to know it better myself.

Thanks -- sylas

 

[Altabeh]

A geodesic depends not only on a set of points in spacetime, but also on how ("fast") the points are traversed. A curve in a spacetime M is a mapping

\gamma : I \rightarrow M,

where I is an interval (often closed) of the real line. A curve \gamma is a geodesic if the tangent vector \dot{\gamma} is parallel transport along the curve, that is, if

\nabla_{\dot{\gamma}} \dot{\gamma} = 0.[/itex]<br /> <br /> An observer is a future-direct timelike curve \gamma parametrized by proper time. This means that<br /> <br /> g \left( \dot{\gamma} , \dot{\gamma} \right) = 1<br /> <br /> all along the curve. In this case, the tangent vector &quot;field&quot; is the observer&#039;s 4-velocity, often denoted by \mathbf{u}, the observer&#039;s 4-acceleration is<br /> <br /> \mathbf{a} = \nabla_{\mathbf{u}} \mathbf{u},<br /> <br /> and a freely falling observer is a geodesic observer curve.<br /> <br /> Sometimes pinning down things so precisely confuses, and sometimes it helps with understanding.

<br /> <br /> Mathematicians mostly use M as a n-dimensional embedded manifold in a (n+1)-space and they of course clarify what topology the manifold is equipped with. In this case (basically in GR), since M is a smooth manifold and of course equipped with the metric topology in a Riemannian space(time), i.e. the local coordinates are curvilinear, one has to notice that<br /> <br /> \gamma : I \rightarrow M,<br /> <br /> depends, above all else, on the metric itself and then when it comes to the dynamical properties of the moving particles, mass points, on geodesics, considering the fact that \nabla_{\dot{\gamma}} \dot{\gamma} = 0[/itex], it is said to be &lt;i&gt;a geodesic depends on how fast the points are traversed&lt;/i&gt;. This means that the geodesics are split into various categories just by studying the dynamics of particles moving on them. This is the reason why Mathematicians are not that much acquainted with, for instance, light-like geodesics. &lt;br /&gt; &lt;br /&gt; So I would see that both sylas and George look at one thing from two different angles which each calls for its own viewpoints.&lt;br /&gt; &lt;br /&gt; AB

 

[George Jones]

As usual, Altabeh, I think I disagree with what you have written, but I am sufficiently confused by what you have written to be sure that I disagree.

Also, as usual, work intervenes, and I don't know when I'll get back to this thread. IF I do get back to this thread, I want to address some technical issues, but first a simple example. Take R^4 to be Minkowski spacetime, with g suitably defined using the standard global chart as an inertial coordinate system.

Let \gamma : [0, 1] -> R^4 be given by \gamma(t) = (t, 0, 0, 0), and let \alpha : [0, 1] -> R^4 be given by \alpha(t) = (t^2, 0, 0, 0).

What is the relevance to this example of what you have written.

 

[Altabeh]

As usual, Altabeh, I think I disagree with what you have written, but I am sufficiently confused by what you have written to be sure that I disagree.

Also, as usual, work intervenes, and I don't know when I'll get back to this thread. IF I do get back to this thread, I want to address some technical issues, but first a simple example. Take R^4 to be Minkowski spacetime, with g suitably defined using the standard global chart as an inertial coordinate system.

Let \gamma : [0, 1] -> R^4 be given by \gamma(t) = (t, 0, 0, 0), and let \alpha : [0, 1] -> R^4 be given by \alpha(t) = (t^2, 0, 0, 0).

What is the relevance to this example of what you have written.

I'm so happy that you disagree again with me this time on this issue and I hope you are not going to not be back to this thread as you once left my post https://www.physicsforums.com/showthread.php?t=371614" without an answer.

Let's say you wanted to declare that geodesics are realized by two qualities:

1- A geodesic is a set of points in spacetime that is defined to be locally the shortest path between points in the spacetime. (From a mathematical point of view)

2- The instantaneous derivative of a curve wrt the proper time s at each point and the change in the direction of this derivative along the curve must be taken into account so a little analysis on these two things is required when studying geodesics. For example, \dot{\gamma}=a with a being a constant, represents a straight line or the shortest path between two points in a Minkowski spacetime and thus \dot{\dot{\gamma}}=0 for every line; this is the shortest because the curvature vector of \gamma is null.

In mathematics, the second part of the "quality" 2 is indeed a manifestation of the first formula of Frenet formulae,

R=|\dot{\dot{\gamma}}|, (1)

where R is the magnitude of curvature vector.

Which in physics and specially GR plays a significant role as the equation of geodesic. So that one can see that it just falls upon the equation of geodesic to determine how fast geodesics are traversed. All these things happen to exist on metric spaces so the first thing that a geodesic depends on is the metric of that space(time) not anything else. You say it is the first part of "quality" 2 which can along with "quality" 1 distinguish a geodesic from other curves. But I'm saying that it

depends, above all else, on the metric itself

not how fast geodesics are traversed. The difference between my idea and yours is that I don't talk about a special M but you take M for granted to be a given metric (I think). So you can see that it is so simple-minded to even think that we are in a disagreement with each other.

Let \gamma : [0, 1] -> R^4 be given by \gamma(t) = (t, 0, 0, 0), and let \alpha : [0, 1] -> R^4 be given by \alpha(t) = (t^2, 0, 0, 0).

What is the relevance to this example of what you have written.

See that you specify R^4 or the Minkowski spacetime. Now you must get my idea: through the use of equation (1), |\frac{d^2}{dt^2}\alpha(t)|=R_\alpha=2, which means \alpha is not a geodesic in R^4. But this could be if we considered another spacetime with another metric.

AB

 

[George Jones]

Okay, now I have some idea what you mean, but in this thread, I think that is best to take spacetime to be an arbitary but fixed pair \left( M , \mathbf{g} \right), where M is a 4-dimensional differentiable manifold and \mathbf{g} is a Lorentzian "metric", so that a metric-compatible torsion-free connection is fixed.

I think that box 10.2 form Misner, Thorne, and Wheeler hints at a physicist's version of what you mean. I don't seem to have any math references that talk about this. Could you give some?

Like always, I am hard-pressed to find time for this, as I'm preparing to go out of town today, and much of my weekend will be spent doing stuff with my wife and three-year-old daughter. Maybe Monday. I want to discuss some of the technical details.

For example, while it is true in a Riemannian mainfold that shortest distance (if it exists) between two points is a geodesic, it not true that

A geodesic is a set of points in spacetime that is defined to be locally the shortest path between points

Easy counterexamples to this statement exist! For example, consider S^2 with its metric inherited by considering it to be a surface in \mathbb{R}^3. Think about geosdesics on the surface of the Earth!

 

[atyy]

Could the difference between George's and Altabeh's definitions (the acceleration def, not the length def) is that for George, a geodesic is an affinely parameterized geodesic, while for Altabeh, a geodesic can be a non-affinely parameterized curve traversing the same path?

 

[Dale]

How are inertial frames defined in Special and General Relativity?

The concept of inertial frames doesn't really apply to GR. An inertial frame is a coordinate system in which the laws of physics take the standard form, and in GR the laws of physics are expressed in terms of tensor quantities in order to ensure that they take the same form in all coordinate systems.

The comments about geodesics describe inertial objects, which is not the same as describing inertial reference frames.

 

[Fredrik]

The concept of inertial frames doesn't really apply to GR. An inertial frame is a coordinate system in which the laws of physics take the standard form, and in GR the laws of physics are expressed in terms of tensor quantities in order to ensure that they take the same form in all coordinate systems.

The comments about geodesics describe inertial objects, which is not the same as describing inertial reference frames.

You can define an inertial frame in GR too. Consider a point p on the world line of an object. We take p to be the origin. The tangent of the world line at p is by definition a geodesic through p. We take that to be the time axis. The time coordinate at a point q on the time axis is defined to be the proper time along the time axis from p to q. The x,y,z axes are chosen from a hypersurface through p that's orthogonal to the world line. They're either chosen arbitrarily or chosen to match how the object is oriented in space. And, uh, I realize that I don't fully understand the whole construction myself (in curved spacetimes). I'm going to have to think about it, or look it up.

If spacetime is flat, the result is the co-moving inertial frame at p. If spacetime is curved, the procedure may not define a coordinate system that covers the entire manifold. Hence the term "local inertial frame".

 

[Dale]

If spacetime is curved, the procedure may not define a coordinate system that covers the entire manifold

Exactly.

 

[madness]

"The concept of inertial frames doesn't really apply to GR. An inertial frame is a coordinate system in which the laws of physics take the standard form, and in GR the laws of physics are expressed in terms of tensor quantities in order to ensure that they take the same form in all coordinate systems"

Can't you say that an inertial frame is one in which SR applies locally?

 

[atyy]

Can't you say that an inertial frame is one in which SR applies locally?

Yes. BTW, in Newtonian physics, I usually think of an inertial frame as one in which N2 and N3 apply.

 

[Altabeh]

Okay, now I have some idea what you mean, but in this thread, I think that is best to take spacetime to be an arbitary but fixed pair \left( M , \mathbf{g} \right), where M is a 4-dimensional differentiable manifold and \mathbf{g} is a Lorentzian "metric", so that a metric-compatible torsion-free connection is fixed.

I think that is best here to take spacetime to be a "Pseudo-Riemannian" manifold M equipped with a non-degenerate, smooth, symmetric metric tensor g. So you can see that I rule out the part demanding the Minkowskian signature for metric just because we can talk about beautiful and new-brand things like geodesically equivalent metrics on a such a manifold which could bring more motivation into the discussion.

I think that box 10.2 form Misner, Thorne, and Wheeler hints at a physicist's version of what you mean. I don't seem to have any math references that talk about this. Could you give some?

I'm really uncomfortable with the notations M, T and W use in their book to describe things. I'll study whatever I see in that box and come to you and say if I am able to find any math ref. on that!

Like always, I am hard-pressed to find time for this, as I'm preparing to go out of town today, and much of my weekend will be spent doing stuff with my wife and three-year-old daughter. Maybe Monday. I want to discuss some of the technical details.

Say my special hello to your beloved family! Hope to have a nice weekend and of course a challenging monday here in this thread! :)

For example, while it is true in a Riemannian mainfold that shortest distance (if it exists) between two points is a geodesic, it not true that

...

Easy counterexamples to this statement exist! For example, consider S^2 with its metric inherited by considering it to be a surface in \mathbb{R}^3. Think about geosdesics on the surface of the Earth!

I don't see anything here that can be in contrast to my statement. Geodesics can generally be defined for any (Pseudo-) Riemannian manifold. Every shortest path from a point on the manifold to another one gives rise to a geodesic, but the converse is not always true if the discussion does not focus on the local paths or the paths berween sufficiently neighbouring points in which case your counterexample fails to work! This is because the shortest paths between two sufficiently adjacent points must concide with one of the geodesics. Otherwise you are completely right!

Besides, it is possible that there are no shortest paths from a point in M to another point in M, but there are geodesics connecting these two. An example of this is the sphere with a point between two points removed. But remember that this case is not globally smooth, so the idea of counterexamples fails to be valid in any way.

AB

 

[bcrowell]

"The concept of inertial frames doesn't really apply to GR. An inertial frame is a coordinate system in which the laws of physics take the standard form, and in GR the laws of physics are expressed in terms of tensor quantities in order to ensure that they take the same form in all coordinate systems"

Can't you say that an inertial frame is one in which SR applies locally?

This is partly just a matter of taste, but I'm in sympathy with DaleSpam. To me, the whole philosophy of GR is to explicitly stop trying to make these distinctions between valid and invalid frames of reference. That's the whole point of general covariance. Einstein has a very readable discussion of this in "The Foundation of the General Theory of Relativity" http://hem.bredband.net/b153434/Works/Einstein.htm . (Note that although the guy has his own copyright claim on the page, it's actually just a verbatim copy of the public-domain translation by Perrett and Jeffery.)

The way I read Einstein's explanation, it's basically a statement that the whole concept of a frame of reference is kind of meaningless in GR. By the time he gets done introducing general covariance, he's weakened the word "frame" so much that it no longer means what it originally meant. In SR "frame" carries the connotation of a lattice of clocks and rulers that fills space in a Euclidean way. In GR it just means a system of arbitrary labels for events.

 

[Altabeh]

Could the difference between George's and Altabeh's definitions (the acceleration def, not the length def) is that for George, a geodesic is an affinely parameterized geodesic, while for Altabeh, a geodesic can be a non-affinely parameterized curve traversing the same path?

This is not exactly my own discussion here, but you got it sort of right! If you had said a geodesic must be a non-affinely parameterized, I would have answered in the negative. You know, in GR the geodesic equation does not call necessarily for an affinely parameterized geodesic. If we set the tangent vector to a curve as u^{\mu}=dx^{\mu}/ds for some arbitrary parameter of curve, s, then the geodesic equation has the form

u^{\mu}u^{\nu}_{;\mu} = f(s)u^{\nu},

for some scalar function f(s). In general, the right hand side of this equation does not vanish unless we take s to be an affine parameter. But you can see that here the four-velocity u^{\nu} plays a role in determining the geodesics, but geodesics can be independent of them while they strongly depend on, first, metric (so Christoffel symbols) and then accelerations du^{\nu}/ds. This can be readily understood from Goerge's example (\alpha(t)=(t^2,0,0,0)) why here I don't rely on the four-velocity along a geodesic: As t is not an affine parameter and in a Minkowski spacetime the Christoffel symbols vanish, so the terms containing four-velocities disappear in the geodesic equation which reduces now to:

u^{\mu}u^{\nu}_{,\mu} = f(s)u^{\nu},

from which we have only the following non-vanishing equation

u^{0}u^{0}_{,0} = f(s)u^{0}.

So u^{0}_{,0}=f(s). Finally

f(s)=2.

As you can see, after considering the metric itself, it's the components of four-acceleration (which I'm fine with "acceleration" alone) along the curve determine whether \alpha is geodesic or not!

AB

 

[torquil]

I haven't read this mathematical discussion in detail.

For me, it seems reasonable that all physics should be completely independent of the choice of parametrization of a worldline path. So any observable property should only depend on the image of the path. Thus I hope that the mathematical formalism was independent of this.

Torquil

 

[madness]

This is partly just a matter of taste, but I'm in sympathy with DaleSpam. To me, the whole philosophy of GR is to explicitly stop trying to make these distinctions between valid and invalid frames of reference. That's the whole point of general covariance. Einstein has a very readable discussion of this in "The Foundation of the General Theory of Relativity" http://hem.bredband.net/b153434/Works/Einstein.htm . (Note that although the guy has his own copyright claim on the page, it's actually just a verbatim copy of the public-domain translation by Perrett and Jeffery.)

The way I read Einstein's explanation, it's basically a statement that the whole concept of a frame of reference is kind of meaningless in GR. By the time he gets done introducing general covariance, he's weakened the word "frame" so much that it no longer means what it originally meant. In SR "frame" carries the connotation of a lattice of clocks and rulers that fills space in a Euclidean way. In GR it just means a system of arbitrary labels for events.

I like that point of view too but unfortunately you can still tell whether or not you are in an inertial frame, which seems to detract from the absolute relativity of motion. Every treatment of GR I have read so far makes extensive use of inertial frames and the fact that you can always make a change of coordinates into an inertial frame.

 

[Dale]

I like that point of view too but unfortunately you can still tell whether or not you are in an inertial frame

How?

 

[madness]

By checking that SR applies locally. It's not much different to checking that N2 applies in Newtonian physics.

 

[Dale]

SR always applies locally, and the GR version of Newton's laws always applies.

 

[madness]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

 

[bcrowell]

I like that point of view too but unfortunately you can still tell whether or not you are in an inertial frame

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

Your first statement sounds like you're talking about using observations to tell whether you're in an inertial frame or not, where "in an inertial frame" presumably refers to the frame defined by the floor in your lab. But your second statement doesn't refer to observables, it refers to a choice of coordinates.

As DaleSpam points out, any laboratory, and any set of experiments in that lab, can be described in locally Minkowski coordinates -- or in coordinates that are not Minkowski.

 

[madness]

Think about the famous thought experiment with a man in a lift. If he's floating, and objects float around him then he is in an inertial frame. If he is stuck to the floor and objects around him fall to the floor, then he is in a non-intertial frame. The difference is that one frame has a vanishing connection at that point and the other doesn't. In the latter case case you can transform to a frame where the connection does vanish, but for the man in the lift he is still in a non-inertial frame.

 

[Fredrik]

If he's floating, and objects float around him then he is in an inertial frame.

You need to be more careful with the terminology. A person or an object never just "is in an inertial frame". What you're trying to say is probably that both his velocity and acceleration are 0 in an inertial frame.

 

[madness]

"You need to be more careful with the terminology. A person or an object never just "is in an inertial frame". What you're trying to say is probably that both his velocity and acceleration are 0 in an inertial frame"

What I mean is that his frame of reference is inertial. If he is stuck to the floor of the lift and objects fall to the floor, it is possible to make a change of coordinates such that at his space-time point the connection vanishes, but the new coordinates are not his frame of reference. In other words, if we switch to a frame which is accelerating downwards with respect to him, then events at his space-time point appear to obey the laws of SR, but that for him he is still in a non-inertial frame.

 

[bcrowell]

In other words, if we switch to a frame which is accelerating downwards with respect to him, then events at his space-time point appear to obey the laws of SR, but that for him he is still in a non-inertial frame.

A lot of this is just a matter of taste. However, it's important to realize that the distinction you're drawing between inertial and noninertial frames is very different from, and much weaker than, the one made in Newtonian mechanics.

In Newtonian mechanics, you can (1) determine whether a certain object's motion is inertial. If it is, you can (2) build a global set of coordinates tied to that object. If it's not, then (3) the laws of physics are false if expressed in coordinates tied to that object.

In general relativity, you can still do (1), but you can't do (2), and (3) is no longer an issue. In Newtonian mechanics, 2 and 3 are the reason you care so much about 1. In general relativity, neither one constitutes a good reason to care about 1.

The way Einstein liked to present it was that because 2 and 3 are no longer considerations, 1 loses its interest entirely. You don't have to agree with his emphasis, but he clearly has a strong point.

 

[madness]

I don't see how it's a matter of choice and not a matter of fact. If there were no such thing as an inertial frame then the theory would be fully Machian. Then for example the twin paradox really would be a paradox in GR.

 

[bcrowell]

I don't see how it's a matter of choice and not a matter of fact. If there were no such thing as an inertial frame then the theory would be fully Machian. Then for example the twin paradox really would be a paradox in GR.

All you really need in order to resolve the twin paradox is the idea of inertial motion, not the idea of an inertial frame. The traveling twin can tell that his motion is noninertial, because he feels the floor pressing against his feet. If you want to start talking about frames, then that's a matter of taste, but the frames will lack properties 2 and 3 from #37, which to my taste makes them fairly uninteresting.

 

[Dale]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

I can see that definition, it is not common but it makes sense. Usually when most people talk about SR they just mean flat spacetime; at least that is what I meant by SR always applies locally. But I do understand your point.

My point is just that the GR versions of Newton's laws always apply as do the GR versions of Maxwell's equations, which IMO would be the more direct analogue of the original Newtonian meaning of "inertial frame". This is why I am of the opinion that the concept of inertial frame is not very significant in GR. Using your definition of inertial frame you can certainly say that they do exist in GR, but since GR can handle frames with non-vanishing connections just fine such frames are not terribly important. Certainly not the central role that they play in Newtonian mechanics.

 

[atyy]

SR only applies when the connection vanishes at that point in spacetime. This is achieved by making a change of coordinates, ie changing to an inertial frame.

I can see that definition, it is not common but it makes sense. Usually when most people talk about SR they just mean flat spacetime; at least that is what I meant by SR always applies locally. But I do understand your point.

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame". But the derivatives of the Christoffel symbols, ie. second derivatives of the metric, cannot be set to zero, and so the "local inertial frame" that exists anywhere in curved spacetime only pertains to measurements which do not measure curvature.

 

[pervect]

In coordinate independent language, one can simply talk about an observer following a geodesic. There's no need to talk about the specific coordinates this observer uses to define the notion of a "free-falling" or "geodesic" observer.

Note that there may be several coordinate systems through any given point that make the connection vanish. Even if you specify the basis vectors at that point, there are more than one set of coordinates that make the connection vanish.

For instance, the following quote from MTW's "Gravitation" discussing Riemann normal coordinates.

Is this the only coordinate system at P0 that is locally inertial at P0 ... and is tied to the basis vectors there? No. But all such coordinate systems, (called 'normal coordinates') will be the same to second order.

In the elipses of the above quote (...) MTW explains that a coordinate system that is locally inertial at a point has a vanishing connection.

 

[Altabeh]

Note that there may be several coordinate systems through any given point that make the connection vanish. Even if you specify the basis vectors at that point, there are more than one set of coordinates that make the connection vanish.

Would you mind giving us an example of what you just claimed above?! It sounds completely wrong to me as if you are ready, I'm going to explain it matheamtically!

AB

 

[Altabeh]

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame". But the derivatives of the Christoffel symbols, ie. second derivatives of the metric, cannot be set to zero, and so the "local inertial frame" that exists anywhere in curved spacetime only pertains to measurements which do not measure curvature.

Completely true! You don't need to even think of "local flatness" in terms of Christoffel symbols when in GR. One just needs to set the first derivatives of metric tensor equal to zero which fits within the definition of "geodesic coordinates" which is not actually an actual coordinate system (read the point 1 below)! But remember that there is a strongly local coordinates in the sense that just at a point you can make the values of metric equal those of Minkowski metric. Here two points must be made:

1- We don't have a "coordinate transformation" x^{\mu}\rightarrow {\bar{x}}^{\mu} by which at a point P, one has g^{\mu\nu}(P)={\eta}^{\mu\nu}(P) iff g^{\mu\nu}(P) is symmetric. If it is diagonal, the number of degrees of freedom of its matrix would be the same as the number of equations of its transformation so there is just one set of coordinates being in charge of the transformation g^{\mu\nu}(P)\rightarrow {\eta}^{\mu\nu}(P) and of course for this case one can show a coordinate system.

2- Geodesic coordinates just account for the first derivatives of metric being all equal to zero, so this way of leading to the local flatness at some point is another alternative.

3- Following 1, in the neighborhood of P the spacetime is "nearly" flat within a range that the equivalence principle issues. This is because we don't know what is meant by "neighborhood" in GR and this is only evaluated\estimated by EP. This might not be applicable for the case 2.

AB

 

[Dale]

Isn't this the same difference? Roughly identify "connection" with "Christoffel symbols". The Christoffel symbols are first derivatives of the metric and can be made to vanish anywhere in curved spacetime, which is why there is always a "local inertial frame".

Yes, they can be made to vanish anywhere through suitable choice of coordinate system, but there do exist coordinate systems where they do not vanish at any given point. Madness is defining such coordinate systems as locally non-inertial, which is not unreasonable IMO.

 

[madness]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.
I studied Ray d'Inverno's book before, which is where I got the idea about vanishing affine connections and locally inertial frames. He shows that at any point on a manifold, it is possible to make a change of coordinates such that the connection vanishes at that point. This can be generalised to a curve but not usually to the whole manifold. I interpreted this as transforming to a locally inertial frame.
Further, d'Inverno seemed to present the affine connection in the geodesic equation as describing non-inertial forces or gravitational forces, which lends to the idea that a vanishing affine connection means a locally inertial frame.

 

[Fredrik]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.

They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details).

 

[madness]

"They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details)."

Well for any LIF, there are infinitely many other LIF's at that point related by a Lorentz transform. I'm not sure if that's why you mean by unique.

 

[Altabeh]

"They exist, but there's no standardized terminology, and apparently (I didn't know this a few days ago), they aren't unique, and can also be very difficult to write down explicitly, which makes them less useful than I expected. (I hope I got that right. I still don't understand the details)."

Well for any LIF, there are infinitely many other LIF's at that point related by a Lorentz transform. I'm not sure if that's why you mean by unique.

Madness, you can find one and only one "coordinates transformation" (be careful with this; read my last post here) that makes your connection (first derivatives of metric tensor) vanish at some point. We never ever can use Lorentz transformations at that point to make lots of frames since in general those coordinates transformations which would meet our demands here, are not explicitly written or shown in simple expressions.

AB

 

[bcrowell]

My lecture notes rely extensively on the concept of a "locally inertial frame", so much so that it is abbreviated to LIF. I would even go so far as to say the course is structured around this concept. This is why it's confusing for me to hear that the concept doesn't exist or isn't useful in GR.

The "L" in LIF is one reason that LIFs are much less central than in Newtonian mechanics. The other reason they're less important is that the laws of GR (unlike Newton's laws) are still valid in coordinates that don't describe an LIF.

If you look at a presentation of relativity like Einstein's "The Foundation of the General Theory of Relativity," ( http://hem.bredband.net/b153434/Works/Einstein.htm ) that doesn't mention the concept of inertial frames (or mentions them only in order to pooh-pooh the concept), what's happening is that he makes the equivalence principle more central conceptually. One way of stating the equivalence principle is that LIFs exist.