Understanding the General Relativity view of gravity on Earth - Comments

[cianfa72]

Is it possible to have a region of curved spacetime with geodesic deviation and a chart where all inertial objects have worldlines that are straight lines in the chart? I can’t think of an example, but certainly admit that there may be one that I haven’t considered.

For sure you can map a region of curved spacetime with a chart in which a full congruence of timelike geodesics are "at rest" in it (i.e. they have constant spatial coordinates in that chart).

 

[Dale]

a full congruence of timelike geodesics are "at rest"

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

 

[cianfa72]

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

Ah ok, I see. I don't know if it is actually possible.

About the first statement in #104 do you actually refer to zero coordinate acceleration as vanishing 2nd coordinate derivative w.r.t. the timelike coordinate and/or other coordinates in the given coordinate chart ?

 

[Dale]

About the first statement in #104 do you actually refer to zero coordinate acceleration as vanishing 2nd coordinate derivative w.r.t. the timelike coordinate

Yes, this was one of the other “important concepts” bullet points.

Note, I didn’t intend for those to be rigorous definitions. I was covering too many to be rigorous. I just wanted to get the “kernel” of the idea to an audience with some calculus and physics background (e.g. two semesters undergraduate-level physics and calculus) but no GR background.

 

[cianfa72]

Yes, but what about all geodesics? Or even just all timelike geodesics? I don’t think that is possible, but perhaps I am wrong.

The point discussed in this thread was that in flat spacetime (i.e. Minkowski) there is only a family/congruence of timelike geodesics filling the entire manifold "at rest" in a given specific coordinate chart. If we further add the condition of zero shear/expansion and zero vorticity for those timelike geodesics we get the definition of coordinate chart as global inertial.

Now I believe the point to be investigated is whether those additional conditions are actually equivalent to require that all timelike geodesics are actually straight in that chart (even if not all "at rest" in it).

 

[cianfa72]

Any thought w.r.t. the previous post ?

 

[Dale]

I was considering not just accelerometers at rest. I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation. I cannot prove it, but the linked post is specifically dealing with a congruence of “at rest” objects, so the requirement there is not applicable.

 

[cianfa72]

I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation.

Reading again this post, I had the following doubt: consider an inertial object at rest in a given frame (coordinate chart). The above definition of inertial coordinate chart requires that for this 'at rest' inertial object the coordinate acceleration must vanish. In this special case the spatial coordinate acceleration (i.e. the derivative of spatial coordinates w.r.t. coordinate time) is of course null by definition of 'at rest in it'.

So, for this case, which is the meaning of zero coordinate acceleration ?

 

[Dale]

So, for this case, which is the meaning of zero coordinate acceleration ?

Zero coordinate acceleration means that the coordinate system consists of one timelike coordinate ##t## and three spacelike coordinates ##\vec x## and that $$\frac{d^2\vec x}{dt^2}=0$$

 

[DrGreg]

$$\frac{d\vec x}{dt}=0$$

I'm sure that's a typo and you meant to say $$\frac{d^2\vec x}{dt^2}=0$$

 

[Dale]

I'm sure that's a typo and you meant to say $$\frac{d^2\vec x}{dt^2}=0$$

Oops, yes, I corrected it

 

[PeterDonis]

I think that “a coordinate system where inertial objects have no coordinate acceleration” doesn’t need the addition of a statement about geodesic deviation.

I think it does, since the standard FRW coordinates used in cosmology are coordinates in which inertial objects--comoving objects--have zero coordinate acceleration, but there is nonzero geodesic deviation.

The usual definition of an inertial chart in SR does include an (often unstated) assumption of zero geodesic deviation, since of course such a chart is intended for use in flat Minkowski spacetime.

 

[cianfa72]

Zero coordinate acceleration means that the coordinate system consists of one timelike coordinate ##t## and three spacelike coordinates ##\vec x## and that $$\frac{d^2\vec x}{dt^2}=0$$

So that actually means we cannot decide whether a coordinate chart is inertial or not by looking just at the coordinate acceleration of inertial objects at rest in it -- indeed by definition inertial objects at rest in any (1 timelike + 3 spacelike) chart have zero coordinate acceleration in it.

 

[cianfa72]

I think it does, since the standard FRW coordinates used in cosmology are coordinates in which inertial objects--comoving objects--have zero coordinate acceleration, but there is nonzero geodesic deviation

Yes but comoving objects are actually at rest in standard FRW coordinates. If we include the entire class of inertially moving objects (not just those at rest) then is the requirement about nonzero geodesic deviation still needed to declare inertial a coordinate chart ?

 

[Dale]

So that actually means we cannot decide whether a coordinate chart is inertial or not by looking just at the coordinate acceleration of inertial objects at rest in it

No, that is incorrect.

indeed by definition inertial objects at rest in any (1 timelike + 3 spacelike) chart have zero coordinate acceleration in it.

Yes, but it is not possible to find inertial objects at rest is all charts. For example, in a rotating coordinate system an object at rest is not inertial and inertial objects have non-zero coordinate acceleration. So the proposed definition correctly identifies the rotating chart as non-inertial

 

[Dale]

I think it does, since the standard FRW coordinates used in cosmology are coordinates in which inertial objects--comoving objects--have zero coordinate acceleration, but there is nonzero geodesic deviation.

Yes, but only the comoving objects are both inertial and have zero coordinate acceleration. Inertial objects that are not comoving have non-zero coordinate acceleration in standard FLRW coordinates. So in the FLRW spacetime there are inertial objects that have coordinate acceleration, and thus the chart is non-inertial.

A better example, which I am torn about, is Anderson coordinates with an anisotropic one-way speed of light. These coordinates use a synchronization convention where the one way speed of light is ##c/(1+\kappa)## in one direction and ##c/(1-\kappa)## in the other. These coordinates have all the Christoffel symbols are zero, so all inertial objects have no coordinate acceleration. There is also zero geodesic deviation so adding the geodesic deviation restriction does nothing. But they are not Einstein synchronized. So while they meet Newton’s definition of inertial, they don’t meet Einstein’s definition of inertial. I decided to classify those as inertial just for simplicity.

 

[cianfa72]

Yes, but it is not possible to find inertial objects at rest is all charts.

Yes, that's true.

For example, in a rotating coordinate system an object at rest is not inertial and inertial objects have non-zero coordinate acceleration. So the proposed definition correctly identifies the rotating chart as non-inertial

Yes, however my point was that by restricting to look only at inertial objects at rest in a given chart, we are not allowed to conclude whether the given chart is inertial or not.

For instance in FWR spacetime in standard FWR coordinates, if we look only at inertial objects at rest in it (i.e. comoving objects) they have zero coordinate acceleration, yet the standard FWR coordinate chart is not inertial.

 

[Dale]

by restricting to look only at inertial objects at rest in a given chart, we cannot conclude whether the given chart is inertial or not.

Which is why I specifically and explicitly said:

I was considering not just accelerometers at rest.

I recognize that issue and thus explicitly reject that restriction.

For a chart to be inertial all inertial objects, not just some, must have 0 coordinate acceleration.

 

[cianfa72]

For a chart to be inertial all inertial objects, not just some, must have 0 coordinate acceleration.

Ok, yes. The point worth investigating, as you pointed out earlier in this thread, is whether the restriction on zero geodesic deviation is really required or if just your quoted claim --by itself-- defines a chart as inertial.

 

[Dale]

Ok, yes. The point to be investigated, as you pointed out before in this thread, is whether the restriction about zero geodesic deviation is really necessary or if just your quoted claim by itself defines a chart as inertial.

Yes, I don’t think that the zero geodesic deviation is helpful, but I cannot prove it. The one “edge case” that I know about is the anisotropic speed of light coordinates. The geodesic deviation doesn’t catch that case.

 

[PeterDonis]

only the comoving objects are both inertial and have zero coordinate acceleration. Inertial objects that are not comoving have non-zero coordinate acceleration in standard FLRW coordinates. So in the FLRW spacetime there are inertial objects that have coordinate acceleration, and thus the chart is non-inertial.

Ah, I see. I had misunderstood the proposed definition.

 

[cianfa72]

A better example, which I am torn about, is Anderson coordinates with an anisotropic one-way speed of light. These coordinates use a synchronization convention where the one way speed of light is ##c/(1+\kappa)## in one direction and ##c/(1-\kappa)## in the other.

I take it as in flat spacetime (i.e. Minkowski spacetime) Anderson coordinate system is actually a specific example of Reichenbach non-standard synchronization convention (i.e. two-way speed of light is the invariant constant ##c## even though one-way speed is not isotropic).

So while they meet Newton’s definition of inertial

i.e. any inertial object has zero coordinate acceleration in Anderson coordinates.

But they are not Einstein synchronized.

since Einstein's definition of inertial coordinate system requires furthermore constant, invariant and isotropic one-way speed of light.

 

[Dale]

Anderson coordinate system is actually a specific example of Reichenbach non-standard synchronization convention

Yes, there is a simple formula relating Anderson’s ##\kappa## to Reichenbach’s ##\epsilon##, but I don’t remember the formula off the top of my head. Einstein synchronization is recovered for ##\kappa=0## and ##\epsilon=0.5##

i.e. any inertial object has zero coordinate acceleration in Anderson coordinates.

since Einstein's definition of inertial coordinate system requires furthermore constant, invariant and isotropic one-way speed of light.

Yes, and yes.

 

[cianfa72]

Einstein synchronization is recovered for ##\kappa=0## and ##\epsilon=0.5##

So for Reichenbach ##\epsilon \neq 0.5## which is the form of the metric for the underlying Minkowski flat spacetime ?

I believe the form of metric must be such that the condition ##ds^2=0## has to cope with the anisotropy of the one-way speed of light.

 

[Dale]

So for Reichenbach ϵ≠0.5 which is the form of the metric for the underlying Minkowski flat spacetime ?

I don’t know Reichenbach’s form, but Anderson’s form is: $$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$

 

[cianfa72]

I don’t know Reichenbach’s form, but Anderson’s form is: $$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa) dx^2 + dy^2 + dz^2$$

ok, so from the equation ##ds^2 = 0## for a light beam in the ##x## direction we get ##dt^2 - 2 \kappa \ dt dx + (1-\kappa) dx^2= 0##

It is a quadratic equation in ##dt## so we get two different solutions. I believe those solutions allow for the two different speed of light in each of the two direction on the ##x## axis (anisotropic speed of light in the ##x## direction).

 

[DrGreg]

I don’t know Reichenbach’s form, but Anderson’s form is: $$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa) dx^2 + dy^2 + dz^2$$

I think that should be$$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$(from post #131 and knowing what the solutions must be).

 

[Dale]

I think that should be$$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$(from post #131 and knowing what the solutions must be).

Oops, yes you are right. I just checked R Anderson et al. Physics Reports 295 (1998) 93-180 and on p 111 it is indeed ##(1-\kappa^2)##, I have corrected the earlier post.

 

[cianfa72]

I think that should be$$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$(from post #131 and knowing what the solutions must be).

Do you mean starting from ##ds^2=0## and from the fact that we know in advance which are the values of one-way speed of light along the two directions on the ##x## axis ?

 

[DrGreg]

Do you mean starting from ##ds^2=0## and from the fact that we know in advance which are the values of speed of light along the two directions on the ##x## axis ?

Yes. From the definition of ##\kappa## in post #121, we know the solutions of ##ds = 0 = dy = dz## have got to be ##dt = \pm (1 + \kappa) dx## and ##dt = \mp (1 - \kappa) dx## (with ##c=1## of course).

(The ##\pm## is there because you also have to choose which direction is the positive ##x## direction.)

 

[Dale]

ok, so from the equation ##ds^2 = 0## for a light beam in the ##x## direction we get ##dt^2 - 2 \kappa \ dt dx + (1-\kappa) dx^2= 0##

It is a quadratic equation in ##dt## so we get two different solutions. I believe those solutions allow for the two different speed of light in each of the two direction on the ##x## axis (anisotropic speed of light in the ##x## direction).

Yes. You can even do a little better. You can divide everything by ##dt^2## and then ##dx/dt=v_x## and so forth. Then it is quadratic in velocity immediately.