Einstein says objects do not fall to the Earth?

[A.T.]

Perhaps you would say, "instead of using an inertial frame for this physical law, I will define a more complicated physical law for a frame fixed to the surface of the earth"?

For example.

How would you do that? I.e. how would you derive an equation for the corresponding physical law of inertia in this frame?

Why derive? Laws can be simply postulated based on observation.

Do you believe that fictitious forces should be considered as having physical existence?

That's a vague philosophical question of little interest to me.

 

[stevendaryl]

OK. I'm curious about what the alternative might be. How would you express the physical law below for any arbitrary frame? Or does each frame have it's own different physical laws? The latter concept is not an attractive approach to defining physical law.

Whenever you mistreat a non-inertial frame as inertial and apply the above physical law, fields appear that are not real, that is they have no existence independently from that frame.

There is a covariant way of expressing \vec{F} = \frac{d\vec{p}}{dt}. First, introduce a scalar parameter, s, and a trivial equation of motion for t itself:

Let t(s) be any linear function satisfying

\frac{d^2 t}{ds^2} = 0

Next, we make the equations 4-D instead of 3-D by making t into a coordinate:

x^0 = t, x^1 = x, x^2 = y, x^3 = z
F^0 = 0, F^1 = F_x, F^2 = F_y, F^3 = F_z

Now define the "velocity" with respect to the parameter s as follows:

U^\mu = \frac{d x^\mu}{ds}

In terms of U^\mu, we have:

m \frac{d}{ds} U^\mu = F^\mu

Then, to make it covariant, we just have to replace \frac{d}{ds} by the covariant notion of the directional derivative with respect to the parameterized path x^\mu(s):

m D_s U^\mu = F^\mu

or in terms of connection coefficients:

m ( \frac{d}{ds} U^\mu + \sum_{\nu \lambda} \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda) = F^\mu

This equation has the same form in any coordinate system (although the values of the coefficients \Gamma^\mu_{\nu \lambda} are different in different coordinate systems.

 

[CKH]

There is a covariant way of expressing \vec{F} = \frac{d\vec{p}}{dt}. First, introduce a scalar parameter, s, and a trivial equation of motion for t itself:

This equation has the same form in any coordinate system (although the values of the coefficients \Gamma^\mu_{\nu \lambda} are different in different coordinate systems.

Thanks. I wish I understood your equations better (the notations are quite unfamiliar), but perhaps I get the point. The equation while not written for a particular coordinate system has embedded in it a parameterized coordinate system. In order to use the equation for a coordinate system you plug in some coefficients? It does not seem that any finite set of scalar "coefficients" would be enough. So I have some more questions:

Don't you need a complete mapping of spacetime to some other spacetime?

How would that mapping be described? In other words, given an arbitrary coordinate system, how do you find these "coefficients"?

What are these coefficients in a inertial frame?

Is the force in that last equation (they right side) a real force or does it include fictitious force?

 

[CKH]

That's a vague philosophical question of little interest to me.

You are mistaken that it is a philosophical question. It is a physical question. Fictitious force fields are called "fictitious" for a real physical reason. They exist only in the frame in which they are measured. They are entirely accounted for by the frame's proper acceleration and have no independent cause nor independent existence. That is, fictitious forces have no frame-independent reality, any more than your shadow has a separate reality from yourself . Your shadow is entirely caused by your own position and motion relative to the sun, it has no independent existence. That fact (for both your shadow and the fictitious forces that accompany your non-inertial motion) is easily verified by experiment. We are easily disabused of the notion that our shadow exists without us, we should be as easily disabused of the notion that fictitious forces exist around us in our environment.

 

[stevendaryl]

Thanks. I wish I understood your equations better (the notations are quite unfamiliar), but perhaps I get the point. The equation while not written for a particular coordinate system has embedded in it a parameterized coordinate system.

I don't think that's quite the way I would put it. The point is that the equations only vary from one coordinate system to another through the "connection coefficients" \Gamma^\mu_{\nu \lambda}. Those coefficients have different values in different coordinate systems. Essentially, these coefficients record the "fictitious forces" associated with the coordinate system.

In order to use the equation for a coordinate system you plug in some coefficients? It does not seem that any finite set of scalar "coefficients" would be enough.

For an specific coordinate system, there are only finitely many connection coefficients, of the form \Gamma^\mu_{\nu \lambda}. There are 64 coefficients, in general, although in many cases, you find many of those are zeros and others are repeats. Except that the coefficients can vary from location to location, so in that sense, there are infinitely many parameters.

So I have some more questions:

Don't you need a complete mapping of spacetime to some other spacetime?

No, you don't need anything more than the connection coefficients associated with a specific coordinate system. This is equivalent to knowing how vectors transport from place to place (that is, knowing which vectors over there are parallel to which vectors over here.)

How would that mapping be described? In other words, given an arbitrary coordinate system, how do you find these "coefficients"?

The connection coefficients can be found through parallel transport, which can be determined empirically (at least, if you make a few assumptions about what is "force-free" motion).

What are these coefficients in a inertial frame?

The special fact about Cartesian coordinates in an inertial frame is that the connection coefficients are all zero. It's not enough that the frame be inertial--spherical coordinates have nonzero connection coefficients even in an inertial frame.

Is the force in that last equation (they right side) a real force or does it include fictitious force?

No, the force on the right is "real" forces, and all the "fictitious" forces are captured by the term

m \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda

Fictitious forces are all velocity-dependent, even though that isn't always apparent.

 

[Dale]

Don't you need a complete mapping of spacetime to some other spacetime?

You need a mapping from an open subset of spacetime to an open subset of R4. This mapping is called a coordinate chart (hopefully that term sounds familiar to you now).

Every spacetime has a metric. This is the quantity which represents distances and durations in the spacetime. To use a coordinate system you have to know how to express the metric in terms of the coordinates. That is the "coefficients" you mention earlier.

All of this is covered in chapters 1 and 2 of Sean Carrolls lecture notes on general relativity. Your questions and comments keep on running headlong into this material, so I would again encourage you to study it.

EDIT: note that stevendaryl spoke of the Christoffel symbols as your "coefficients" and I spoke of the metric tensor in terms of the coordinates. You can obtain one from the other, so they are closely related.

 

[A.T.]

You are mistaken that it is a philosophical question. It is a physical question.

If it doesn't affect the quantitative outcome of predictions, then it isn't a physics question.

Fictitious force fields are called "fictitious" for a real physical reason.

To physics it's irrelevant how you call them. It doesn't affect the quantitative outcome of predictions.

 

[stevendaryl]

If it doesn't affect the quantitative outcome of predictions, then it isn't a physics question.

To physics it's irrelevant how you call them. It doesn't affect the quantitative outcome of predictions.

To me, there certainly is a physical distinction between "fictitious" and non-fictitious forces. For one big difference, the latter obey Newton's 3rd law of motion (or a generalization) and the former do not. For another, fictitious forces can be made to vanish through a coordinate change. For another, fictitious forces always have a similar form--they are quadratic in the generalized 4-velocity, and are proportional to mass. All of these differences are physically and quantitatively important. I would not lump this in with a "philosophical" distinction that makes no difference to the physics.

 

[A.T.]

To me, there certainly is a physical distinction between "fictitious" and non-fictitious forces.

CKH didn't ask me about that distinction, but about "physical existence". Musing about what "really exists" and what doesn't is what I consider philosophy.

 

[stevendaryl]

CKH didn't ask me about that distinction, but about "physical existence". Musing about what "really exists" and what doesn't is what I consider philosophy.

Well, in my opinion, the best way to respond to philosophical questions is to steer them toward conceptually related questions that are more physics-oriented, such as: In what concrete ways are fictitious forces different from nonfictitious forces.

 

[A.T.]

Well, in my opinion, the best way to respond to philosophical questions is to steer them toward conceptually related questions that are more physics-oriented, such as: In what concrete ways are fictitious forces different from nonfictitious forces.

That's a good approach. Thanks.

 

[Dale]

Musing about what "really exists" and what doesn't is what I consider philosophy

Specifically the subdiscipline of metaphysics called ontology.

 

[CKH]

Those coefficients have different values in different coordinate systems. Essentially, these coefficients record the "fictitious forces" associated with the coordinate system.

The connection coefficients can be found through parallel transport, which can be determined empirically (at least, if you make a few assumptions about what is "force-free" motion).

The special fact about Cartesian coordinates in an inertial frame is that the connection coefficients are all zero. It's not enough that the frame be inertial--spherical coordinates have nonzero connection coefficients even in an inertial frame.

No, the force on the right is "real" forces, and all the "fictitious" forces are captured by the term

m \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda

Fictitious forces are all velocity-dependent, even though that isn't always apparent.

Thanks, nice explanation. You have removed the mystery of "coordinate independence" in this equation. As I understand it, this equations applies at an event, a single point in spacetime or infinitesimal patch of spacetime, if you prefer.

Where is the "physical law" in this covariant equation. It seems to be the same physical law as:

One difference is that the covariant version adds a coordinate transformation. That inclusion is not necessary to express the underlying physical law. You can state the physical law without a transformation and then just apply the transformation when you need to. The transformation is an "add on" for mathematical purposes I imagine.

Another difference is that this covariant version applies only to a point while the simple version applies to all inertial frames (throughout time and space).

Using this covariant version by plugging in the coefficients for a given coordinate system, can we claim a "different physical law" for that coordinate system?

Did we choose an inertial frame in which to express the coefficients just for convenience?

Both versions of the physical law are based on the concept of an inertial frame. The inertial frame is essential for stating the physics of motion. SR and GR are constructed upon this concept. The use of an inertial frame is not simply a preference for expressing physical law in a simpler formula, rather it is the foundation of the physical law.

Newton figured it out. He knew he should not include environmental conditions in his equations for motion. He ignored gravity, air resistance, spin of the Earth and so on, in order to discover the underlying physical laws of motion. That was a major step forward for physics and had nothing to do with a "preference" for an inertial frame on Newton's part, it is rather the discovery of the inertial frame (although it seems so obvious to us now) that made the laws of motion possible.

 

[Dale]

Another difference is that this covariant version applies only to a point while the simple version applies to all inertial frames (throughout time and space).

No, the covariant version is more general than the simple version. There is no case where the simple version applies that the covariant version does not also apply. In addition, there are cases where the simple version does not apply, and in those cases the covariant version does (e.g. non-inertial frames).

In other words, the simple version requires the use of an inertial frame. It is not an option if you want to use the simple version. However, there is a generalization of the simple version which does not require the use of an inertial frame. If you want to use the general version then the use of an inertial frame is optional.

If you use the general version in an inertial frame then it will automatically reduce to the simple version.

Using this covariant version by plugging in the coefficients for a given coordinate system, can we claim a "different physical law" for that coordinate system?

I wouldn't claim it as a "different physical law". I would claim it as a generalization of the simplified physical law. In cases where the simplified law applies, the generalized law matches it. In addition, the generalized law applies to other cases where the simplified one does not.

 

[CKH]

If you want to use the general version then the use of an inertial frame is optional.

But isn't an inertial frame still required since the coefficients in the transformation term are defined relative to an inertial frame (with Cartesian coordinates)? The transformation term disappears in an inertial frame, thus the transformation term itself is relative to an inertial frame.

The connection coefficients can be found through parallel transport, which can be determined empirically (at least, if you make a few assumptions about what is "force-free" motion).

Where "force-free" motion == local inertial motion or no proper acceleration?

The special fact about Cartesian coordinates in an inertial frame is that the connection coefficients are all zero.

 

[Dale]

But isn't an inertial frame still required since the coefficients in the transformation term are defined relative to an inertial frame (with Cartesian coordinates)? The transformation term disappears in an inertial frame, thus the transformation term itself is relative to an inertial frame.

No. There is no need for a transformation at all. If I give you some coordinates, say $(k,l,m,n)$ and a metric $ds^2 = n^2~dk^2+ dl^2 + dm^2 - k^2~dn^2$ then you can do all of the physics directly in that frame without ever calculating or worrying about the transformation between this frame and an inertial frame.

If you do not know the metric in a given frame then you can either calculate it by transforming from an inertial frame, or you can simply measure it experimentally. Usually the former is easier to do, but that does not mean that it is required.

 

[stevendaryl]

Thanks, nice explanation. You have removed the mystery of "coordinate independence" in this equation. As I understand it, this equations applies at an event, a single point in spacetime or infinitesimal patch of spacetime, if you prefer.

Where is the "physical law" in this covariant equation. It seems to be the same physical law as:

Yes, it's the same law, just the form is generalized so that it works equally well in any coordinate system.

One difference is that the covariant version adds a coordinate transformation.

No, there's no coordinate transformation involved. The meaning of the connection coefficients \Gamma^\mu_{\nu \lambda} is in terms of parallel transport. There is an observer that is traveling with a 4-velocity U^\mu. There is a vector A^\nu and the observer carries that vector without changing its direction or magnitude, then the components of that vector will change at a rate given by:

\frac{d A^\nu}{ds} + \Gamma^\nu_{\mu \lambda} A^\mu U^\lambda = 0

Why are the components changing, if you are keeping the vector A^\mu constant in magnitude and direction? Think about a 2D plane (the x-axis oriented left-right and the y-axis oriented up-down, and you are carrying a vector of length 1 that is pointing in the y-direction. You move this vector from a point on the x-axis to a point on the y-axis. This is shown in the figure below

But when expressed in terms of polar coordinates, the components change as you move the vector around. In polar coordinates, the basis vectors are e_\theta, which points in the direction of increasing \theta (where \theta = 0 along the x-axis and \theta = \frac{\pi}{2} on the y-axis, and e_r, which points in the direction of increasing r, the distance from the center. On the x-axis, the vector is pointing in the direction of e_\theta, so its components would be A^r = 0, A^\theta > 0. On the y-axis, the same vector is pointing in the direction of e_r, so its components would be A^r > 0, A^\theta = 0. So as you move the vector along the circular path shown, \frac{dA^r}{ds} > 0 and \frac{dA^\theta}{ds} < 0.

Another difference is that this covariant version applies only to a point while the simple version applies to all inertial frames (throughout time and space).

That's true if space is "flat", meaning that it is describable using Cartesian coordinates that apply everywhere. If space is not flat, then you can't do any better than have connection coefficients that apply at a single point.

Using this covariant version by plugging in the coefficients for a given coordinate system, can we claim a "different physical law" for that coordinate system?

No, it's just that the details for how things are described are different in different coordinate systems, just like things are described differently using polar or Cartesian coordinates.

Both versions of the physical law are based on the concept of an inertial frame.

I don't think that's quite correct. What it's based on is the notion of "parallel transport", which is the notion that one vector at one point is parallel (the same magnitude and direction) to another vector at another point. What's nice about inertial Cartesian coordinates is that it's really simple to know when one vector is parallel to another vector: you just see if the components are equal.

The inertial frame is essential for stating the physics of motion. SR and GR are constructed upon this concept. The use of an inertial frame is not simply a preference for expressing physical law in a simpler formula, rather it is the foundation of the physical law.

I don't quite agree. Newton wasn't up to speed on differential geometry, because he'd just invented calculus, and differential geometry would still be 200 years later. But really what's more fundamental in physics is the notion of parallel transport. Inertial motion can be defined in terms of parallel transport, or the other way around. They are more or less equivalent concepts. But what you don't need is the notion of an inertial frame. In curved space (or curved spacetime), there are no inertial frames, although there is still inertial motion.

 

[CKH]

The meaning of the connection coefficients \Gamma^\mu_{\nu \lambda} is in terms of parallel transport. There is an observer that is traveling with a 4-velocity U^\mu. There is a vector A^\nu and the observer carries that vector without changing its direction or magnitude, then the components of that vector will change at a rate given by:

\frac{d A^\nu}{ds} + \Gamma^\nu_{\mu \lambda} A^\mu U^\lambda = 0

OK. this is more than a bit over my head. I understand we are generalizing a law so that it may be stated in "any" coordinate system, but we are only doing it at one point. So perhaps we are only interested in the rates of change in coordinates at this point; maybe first and second derivatives?

Unfortunately I don't understand the notation so it's still a little foggy what's going on. What is it that designates the coordinate system? Can you tell me easily what each letter designates in the full equation or provide a reference?

This A^\nu vector is a unit 4-vector indicating a direction in which coordinates?

You said that the coefficients are all zero in any (Cartesian) inertial frame. That seems to imply that the "from" coordinates are Cartesian coordinates in an inertial frame. In that special case there are no fictitious forces to account for, so that entire term drops out.

There are a lot of coefficients (64). What are all these coefficients?

 

[vin300]

Why is a fictitious force always quadratic?

 

[stevendaryl]

OK. this is more than a bit over my head. I understand we are generalizing a law so that it may be stated in "any" coordinate system, but we are only doing it at one point. So perhaps we are only interested in the rates of change in coordinates at this point; maybe first and second derivatives?

Unfortunately I don't understand the notation so it's still a little foggy what's going on. What is it that designates the coordinate system? Can you tell me easily what each letter designates in the full equation or provide a reference?

As I said, U^\mu = \frac{d x^\mu}{ds}, so that's the "4-velocity". A^\mu is any other 4-vector.

This A^\nu vector is a unit 4-vector indicating a direction in which coordinates?

A^\mu is any 4-vector (not necessarily a unit vector). You pick whatever coordinate system you like, and then that coordinate system determines the components of A and U and determines the coefficients \Gamma^\mu_{\nu \lambda}.

You said that the coefficients are all zero in any (Cartesian) inertial frame. That seems to imply that the "from" coordinates are Cartesian coordinates in an inertial frame. In that special case there are no fictitious forces to account for, so that entire term drops out.

It's not a transformation, so there are no "from" coordinates. However, you can use transformations to figure out connection coefficients in one coordinate system from the connection coefficients in another coordinate system. The connection coefficients themselves are not transformations, though.

There are a lot of coefficients (64). What are all these coefficients?

They describe how the components of vectors change from place to place. Let me go through the example of 2D flat space in polar and cartesian coordinates. In the case of Cartesian coordinates x,y, if an object is moving around, then its velocity is given by the vector U = U^x e_x + U^y e_y where U^x = \frac{dx}{dt} and U^y = \frac{dy}{dt}, and where e_x is the basis vector in the x-direction and e_y is the basis vector in the y-direction. The nice thing about Cartesian coordinates is that the basis vectors are constant. e_x points in the same direction no matter what your location.

But now switch to polar coordinates r and \theta. Then if you have an object that is moving around, its velocity is given by U^r e_r + U^\theta e_\theta where U^r = \frac{dr}{dt} and U^\theta = \frac{d\theta}{dt}, and where e_r and e_\theta are basis vectors in the r and \theta directions, respectively. If you work out what the basis vectors e_r and e_r are in terms of a cartesian basis e_x and e_y, you will find:

e_r = cos(\theta) e_x + sin(\theta) e_y
e_\theta = - r sin(\theta) e_x + r cos(\theta) e_y

Note that e_r is not constant. When \theta = 0, it points in the x-direction. When \theta = \frac{\pi}{2}, it points in the y-direction. Similarly, e_\theta is not constant, either. We can characterize how e_r and e_\theta change with position by taking derivatives:

\partial_\theta e_r = - sin(\theta) e_x + cos(\theta) e_y = \frac{1}{r} e_\theta
\partial_\theta e_\theta = - r e_r
\partial_r e_\theta = \frac{1}{r} e_\theta

The general pattern is that \partial_\nu e_\lambda is equal to some combination of other basis vectors. The coefficient \Gamma^\mu_{\nu \lambda} is the number multiplying e_\mu. So for our example, we can read off:

\Gamma^\theta_{\theta r} = \frac{1}{r}
\Gamma^r_{\theta \theta} = -r
\Gamma^\theta_{r \theta} = \frac{1}{r}

 

[stevendaryl]

Why is a fictitious force always quadratic?

Here's a heuristic explanation:

You have a 4-velocity U. It can be written as a linear combination of basis vectors:

U = U^0 e_0 + U^1 e_1 + U^2 e_2 + U^3 e_3 which we summarize as U^\mu e_\mu (where the repeated index \mu means that it's to be summed over all possible indices).

Now, let's compute \frac{d}{ds} U:

\frac{d}{ds} U = \frac{d}{ds} (U^\mu e_\mu) = \frac{dU^\mu}{ds} e_\mu + U^\mu \frac{de_\mu}{ds} (by the product rule).

Now, what is the meaning of \frac{de_\mu}{ds}? It's the rate of change of the basis vector e_\mu. Note: if you're not using inertial Cartesian coordinates, then your basis vectors are not covariantly constant. The basis vectors are actually functions of position x^\mu. So if your position is changing by:

\frac{dx^\mu}{ds} = U^\mu (the \mu component of the 4-velocity)

then e_\mu will change at the rate:

\frac{d e_\mu}{ds} = \frac{\partial e_\mu}{\partial x^\nu} \frac{dx^\nu}{ds} by the chain rule. But \frac{dx^\nu}{ds} = U^\nu. So we have:

\frac{d e_\mu}{ds} = \frac{\partial e_\mu}{\partial x^\nu} U^\nu

Substituting that into the expression for \frac{dU}{ds} gives:

\frac{dU}{ds} = \frac{dU^\mu}{ds} e_\mu + U^\mu U^\nu \frac{\partial e_\mu}{\partial x^\nu}

The second term is the fictitious force, and it involves two factors of U.

(Everywhere I use derivatives or partial derivatives, I should be using covariant versions.)

 

[pervect]

Let's take the simplest possible case. You've got a plane, a flat plane (no curvature). On the plane you have cartesian coordinates (x,y) and some more general coordinates. Rather than making them truly general, at this point we'll just use polar coordinates, r and $\theta$.

Suppose you have two points on the plane, P and a nearby point Z. Note we are imagining that P and Z are the same set of points - we've only changed the description of them, i.e. the coordinates.

Now we can create a vector that represents the generalized displacement from P to Z by subtracting the coordinates. We will have in one case $\delta_x$ and $\delta_y$ as components of the vector, in the other case we will have $\delta_r$ and $\delta_\theta$

There will be some linear transformation between the two descriptions of nearby points, if you remember the chain rule of calculus you might write for instance

\delta_x = \frac{dx}{dr} \delta_r + \frac{dx}{d\theta} \delta_\theta \quad \delta_y = \frac{dy}{dr} \delta_r + \frac{dy}{d\theta} \delta_\theta<br />The important thing is that there is a linear relationship between the two descriptions, the particular notation I used assumes that you have some function x(r, $\theta$) and another function y(r, $\theta$) in order to calculate the values of the derivatives which give the linear relationship.

The linear part is more in line of an assumption, by the way - we are assuming that we always stick close enough to P that only the linear terms matter, and that we can ignore any second order terms that might exist.

This part is a super-brief description of what we'd describe in abstract mathematical terms as the existence of a "vector space" near P. And we've also implicitly specified a specific set of what are called "basis vectors" near P, which are the coordinate basis. I didn't introduce the necessary notation, rather than talk around the issue let me just illustrate a modernish non-tensor form of the notation that may not actually match any particular textbook you might use. My textbooks all use tensor notation, because I don't have any textbooks to consult for non-tensor notation, I'm hoping that my possibly non-standard notation will make sense and get the point across.

$\delta_x$ and $\delta_y$ are just numbers. To have a true vector, we notationally write $\delta_x \vec{x} + \delta_y \vec{y}$. The things with the arrows over them are the actual vectors, what I called the basis vectors. The numbers that multiply them are really scalars that multiply the value of the basis vectors.

There's a different set of basis vectors for every coordinate system, we have one set for our cartesian coordinates, another set for our polar coordinates.

Now on to the connection, where things gets a bit more complicated.

Suppose we have a vector near P, we want to map it into a parallel vector near Q. Because we are on a plane, there is a natural notion of parallelism, so given P and Q there is exactly one vector at Q that is parallel to P and has the same length as the vector near P.

In cartesian coordinates this process is easy. The process of mapping a vector near P to a vector near Q while keeping them parallel and their length constant involves not allowing the components of the vector to change.

In polar coordinates, this simple prescription won't work. We are trying to provide a coordinate independent description of the physics, though, so we want a coordinate independent way of writing down the prescription for transporting vectors.

What we do know that if Q is near P, we know that the desired result, what I will call the "output vector" near Q, will be some bi-linear function of the input vector (near P) and the displacement from Q to P, which (because we did things on a flat plane to keep things simple) is another vector, which we will call D.

It takes 8 numbers to write the most general possible bi-linear relation between the output vector Q, and the input vectors P and D. It's a rank 3 tensor, if you are unfamiliar with tensors you can think of it as a sort-of 3 dimensional matrix. By bilinear, we mean that the relationship between Q and P holding D fixed is linear, and the relationship between Q and D holding P fixed is linear.

It takes 8 numbers because our simplified problem is 2 dimensional. If it were 4 dimensional, it'd take 4x4x4 = 64, as you've noted previously.

Those 8 numbers are the connection coefficients in two dimensions, which are notationally give the representation $\Gamma^{i}{}_{jk}$. While the most general possible set of connection coefficients have 8 elements, in order to preserve distances and angles as we have described, given the metric, there is only one set of connection coefficients that will work. Knowing the metric specifies the connection coefficients, though I haven't explained why. Honestly, I'd have to think quite a bit about "why", but I know that it does.

Onto the metric. Hopefully, this part is easy. The metric for the cartesian coordinates is just dx^2 + dy^2. The metric for the polar coordinates is dr^2 + r^2 d\theta^2. That's it for the metric! It seems short, but I'm not sure that more needs to be said, hopefully this was already familiar.

If you want the formula for how to compute the Christoffel symbols from the metric, wiki has a brief though perhaps hard to follow description at http://en.wikipedia.org/wiki/Christoffel_symbols that doesn't explain how it got the answer either :-).

 

[Dale]

Can you tell me easily what each letter designates in the full equation or provide a reference?

http://preposterousuniverse.com/grnotes/
This includes a very brief intro:
http://preposterousuniverse.com/grnotes/grtinypdf.pdf

 

[Dale]

we are only doing it at one point

I don't know where you got this idea. It is not correct. The law applies at all points on any world line, just as with the simplified version.

 

[stevendaryl]

I don't know where you got this idea. It is not correct. The law applies at all points on any world line, just as with the simplified version.

I think what he means is that the connection coefficients have different values at every point, in general.

 

[Dale]

I think what he means is that the connection coefficients have different values at every point, in general.

As does the force and the acceleration and even the mass, in general. It is simply wrong to conclude from that that the law itself only applies at a single point. All of the quantities in these equations can vary, the law is the relationship between them which holds everywhere.

 

[CKH]

Gosh, thanks for all the help guys. I need to digest all this for a while. Dale, thanks I really need that help with notation. (I have to refresh my calculus a bit as well since it's been a long time without much use.)

I think what he means is that the connection coefficients have different values at every point, in general.

That's right and perhaps in some cases you can make the connection coefficients functions of position and thus cover some part of spacetime as suggested by Dale and in stevendaryl's example where the variable 'r' is used to express some coefficients.

BTW I have trouble quoting equations in parts of your messages. Perhaps there are some global tags in the messages that need to be copied?

 

[CKH]

Let's take the simplest possible case. . .

Thanks. That's helpful as a bridge toward understanding what's happening in the more mysterious compact notation.