Difference Between Locally Flat & Locally Inertial?

[GR191511]

Are they equivalent to each other?Thank you.

 

[Dale]

No, they are not. Locally flat refers to the spacetime and locally inertial refers to a reference frame or an object.

Locally flat just means that tidal forces go to 0 faster than first order. It is always true for every spacetime regardless of the curvature.

Locally inertial means that we are using local coordinates in which the metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$. This is not always true for every coordinate system, even if the spacetime is flat.

 

[GR191511]

No, they are not. Locally flat refers to the spacetime and locally inertial refers to a reference frame or an object.

Locally flat just means that tidal forces go to 0 faster than first order. It is always true for every spacetime regardless of the curvature.

Locally inertial means that we are using local coordinates in which the metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$. This is not always true for every coordinate system, even if the spacetime is flat.

Thanks...In《A First Course in General Relativity》page 151:
"...So in a small region the manifold looks flat,and it is then natural to say that the derivative of a vector whose components are constant in this coordinate system is zero at that point.In particular,we say that the derivatives of the basis vectors of a locally inertial coordinate system are zero..."I don't understand

 

[Dale]

What specifically don't you understand? The more detail you can provide the more likely we are to be able to help.

 

[GR191511]

What specifically don't you understand? The more detail you can provide the more likely we are to be able to help.

The paragraph above seems to say that locally flat is equivalent to locally inertial.

 

[vanhees71]

We have a pseudo-Riemannian manifold, i.e., a manifold with a pseuo-Euclidean (more specifically a Lorentzian) metric. It is locally flat, because you have full freedom to choose coordinates. From this you can show that at any point $P$ of the manifold you can find a transformation such that the Taylor expansion of the pseudo-metric components around this point (with coordinates $q_0^{\mu}$) reads
$$g_{\mu \nu}(q) = \eta_{\mu \nu} + \frac{1}{2} \partial_{\rho} \partial_{\sigma} g_{\mu \nu}(q_0) (q^{\rho} -q_0^{\rho})(q^{\sigma}-q_0^{\sigma}) + \mathcal{O}(q-q_0)^3).$$
Here $\eta_{\mu \nu}$ are the pseudo-metric components in flat Minkowskispace, i.e., $\pm \mathrm{diag}(1,-1,-1,-1)$, with the overall sign depending on the convention used in your textbook. In these coordinates in a small neighborhood around the point $P$, at linear order, the pseudo-metric components look as in flat space. The corrections are of 2nd order in $(q-q_0)^{\rho}$. That's why one says the pseudo-Riemannian manifold is "locally flat".

The coordinates $q^{\rho}$ with these properties define also a local inertial frame, i.e., since all the 1st partial derivatives of the $g_{\mu \nu}$ vanish at the point $P$, also the Christoffel symbols vanish at this point, i.e., the coordinate lines are all geodesics through this point.

That doesn't imply that it is a globally flat affine manifold. For that the Riemannian curvature tensor must vanish everywhere.

 

[vanhees71]

The paragraph above seems to say that locally flat is equivalent to locally inertial.

That's true, because in a pseudo-Riemannian manifold by definition is torsion-free, i.e., the uniquely defined pseudo-metric-compatible affine connection is given by the usual Christoffel symbols.

 

[Dale]

The paragraph above seems to say that locally flat is equivalent to locally inertial.

How does the author define those two terms? Different authors may use the same term differently. And how do they define "this coordinate system"?

 

[vanhees71]

Good point. Which book are we talking about?

 

[Ibix]

Good point. Which book are we talking about?

Schutz.

@GR191511, textbooks usually get referred to by author since there's more variation in names than variations of "Something general relativity something". If you're only going to give one of title and author, make sure it's the author.

 

[vanhees71]

Then it's all in Chapter 6.2 of the 2nd edition.

 

[Nugatory]

Locally inertial means that we are using local coordinates in which the metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$. This is not always true for every coordinate system, even if the spacetime is flat.

How about “locally inertial means that we are using local coordinates in which the coordinate acceleration of a test mass is equal to its proper acceleration”? That allows for non-Cartesian coordinates (polar, for example) while still excluding rotating and accelerating frames.

 

[Sagittarius A-Star]

How about “locally inertial means that we are using local coordinates in which the coordinate acceleration of a test mass is equal to its proper acceleration”?

I would add: "... at the instance of coordinate time, when the proper accelerated test mass is momentarily at rest with reference to the local coordinates".

while still excluding rotating ... frames

... if the test mass does not move along the rotation axis, and including what I wrote above.

 

[PeterDonis]

The paragraph above seems to say that locally flat is equivalent to locally inertial.

No, it says that in a locally flat spacetime, it is always possible to find a locally inertial coordinate system at any point. That doesn't mean "locally flat" and "locally inertial" are the same.

 

[Nugatory]

I would add: "... at the instance of coordinate time, when the proper accelerated test mass is momentarily at rest with reference to the local coordinates".

I think that requirement is not necessary (non-zero velocity with zero coordinate and proper acceleration is generally consistent with inertial motion) except to close loopholes such as

... if the test mass does not move along the rotation axis

and it's a very big hammer for that purpose. Surely there's a way of capturing the notion of "no fictitious forces needed" exactly?

 

[Dale]

How about “locally inertial means that we are using local coordinates in which the coordinate acceleration of a test mass is equal to its proper acceleration”? That allows for non-Cartesian coordinates (polar, for example) while still excluding rotating and accelerating frames.

In polar coordinates, doesn’t a tangentially moving inertial object have coordinate acceleration for both $r$ and $\theta$

 

[Sagittarius A-Star]

I think that requirement is not necessary

Isn't it is possible to define an accelerated reference frame, in which the coordinate-acceleration of a moving and accelerated particle is equal to the particle's proper acceleration?

 

[Ibix]

Isn't it is possible to define an accelerated reference frame, in which the coordinate-acceleration of a moving and accelerated particle is equal to the particle's proper acceleration?

I think so, yes. For example, consider motion in the y or z direction in a Rindler frame. However, if we add that the coordinate and proper accelerations must be equal in any direction, I don't think so.

 

[Ibix]

To expand slightly - if an object whose four velocity is $v^\mu$ has zero proper acceleration then $$\begin{eqnarray*}
0&=&v^\nu\nabla_\nu v^\mu\\
&=&v^\nu\left(\partial_\nu v^\mu+\Gamma_{\nu\lambda}^\mu v^\lambda\right)
\end{eqnarray*}$$Requiring that the object has zero coordinate acceleration ($\partial_\nu v^\mu=0$) would seem to boil down to $0=v^\nu\Gamma_{\nu\lambda}^\mu v^\lambda$. If we're allowed to pick our $v^\mu$, as I did in my Rindler example above, then it is possible to find four velocities that have zero coordinate acceleration and zero proper acceleration even in non-inertial frames. If we want an arbitrary $v^\mu$ to have both zero, though, we require $\Gamma_{\nu\lambda}^\mu=0$.

 

[Sagittarius A-Star]

If we're allowed to pick our $v^\mu$, as I did in my Rindler example above, then it is possible to find four velocities that have zero coordinate acceleration and zero proper acceleration even in non-inertial frames.

Consider the $x$ direction of a Rindler frame and following scenarios:

1. Particle is at rest: proper acceleration is unequal zero, coordinate acceleration is zero.
2. Particle is in free-fall: proper acceleration is zero, coordinate acceleration is unequal zero.

Isn't there a scenario "in between", were both are equal?

 

[vanhees71]

No, it says that in a locally flat spacetime, it is always possible to find a locally inertial coordinate system at any point. That doesn't mean "locally flat" and "locally inertial" are the same.

True, but in GR they are, because you assume that spacetime is a pseudo-Riemannian manifold (i.e., with no torsion and the affine connection thus the unique metric-compatible one given by the usual Christoffel symbols). Weinberg derives this from the assumption that there's always a local inertial frame in any point of spacetime.

 

[Ibix]

Isn't there a scenario "in between", were both are equal?

Casually, I would say equal and opposite. Rindler is well matched to an accelerating lift scenario, and a free faller would see the lift floor coming up with the same magnitude of acceleration an observer on the floor would see the free faller coming down with. The "midpoint" would be a downward accelerating observer seeing the floor accelerating upwards with the same coordinate acceleration magnitude as their own proper acceleration. I suspect that's only instantaneously correct, though - sooner or later the observers must pass through each other's Rindler horizons, and then their coordinate acceleration becomes undefined.

Formally, the four acceleration is a four vector and the coordinate acceleration isn't (unless all the $\Gamma$ are zero), so I think there's a bit of work to be done to agree what it means to compare vectorial apples to non-vectorial oranges. I suspect what I did in the previous paragraph was compare components in some basis that I didn't really formally define. I don't think that's wrong, exactly, but it definitely needs some tightening up.

 

[Sagittarius A-Star]

Casually, I would say equal and opposite.

You are right. I forgot the sign. Then consider in the Rindler frame a rocket, moving in $x$ direction (="upwards") with "almost $c$" and a proper acceleration in the opposite direction (downwards). Time dilation reduces the ratio between coordinate acceleration and proper acceleration.

Is there a scenario, were both are equal?

 

[Dale]

True, but in GR they are, because you assume that spacetime is a pseudo-Riemannian manifold

No. Because “flat” is a term that describes a spacetime and “inertial” is a term that describes a reference frame, worldline, or coordinate chart. You wouldn’t say “this is an inertial manifold” and you wouldn’t say “this is a flat reference frame”, so the terms are not interchangeable.

 

[vanhees71]

What I meant is that if you assume spacetime to be a (pseudo-)Riemannian manifold this implies that it's locally flat, i.e., that you can always choose coordinates, where $g_{\mu \nu}(q)=\eta_{\mu \nu} +\mathcal{O}(q^2)$, i.e., $\Gamma_{\mu \nu}^{\rho}=0$, and that defines by definition a locally inertial frame. That's the approach by the here discussed book by Schutz.

You can also argue more from the physics side and assume that at each spacetime point there's a locally inertial frame, from which you can deduce that spacetime should be a pseudo-Riemannian manifold and thus that it is locally flat at each point. That's Weinberg's approach in his book (1971).

 

[Dale]

What I meant is that if you assume spacetime to be a (pseudo-)Riemannian manifold this implies that it's locally flat,

Yes, that I agree with

 

[GR191511]

$$g_{\mu \nu}(q) = \eta_{\mu \nu} + \frac{1}{2} \partial_{\rho} \partial_{\sigma} g_{\mu \nu}(q_0) (q^{\rho} -q_0^{\rho})(q^{\sigma}-q_0^{\sigma}) + \mathcal{O}(q-q_0)^3).$$

Where is the zero order?$\eta_{\mu \nu}$?

 

[vanhees71]

It's explicitly on the right-hand side of the quoted equation.

 

[Nugatory]

In polar coordinates, doesn’t a tangentially moving inertial object have coordinate acceleration for both $r$ and $\theta$

The second derivatives of $r(t)$ and $\theta(t)$ are non-zero but it’s the position vector $\textbf{R}=r(t)\hat{r}$ that we care about, and its second derivative $\ddot{\textbf{R}}$ will be zero.

(Belated reply because I just noticed that this saved draft has been rotting unposted for the past few days)

 

[Dale]

The second derivatives of $r(t)$ and $\theta(t)$ are non-zero but it’s the position vector $\textbf{R}=r(t)\hat{r}$ that we care about, and its second derivative $\ddot{\textbf{R}}$ will be zero.

(Belated reply because I just noticed that this saved draft has been rotting unposted for the past few days)

Ah, I see. You are using “frame” to mean something different from “coordinate chart”. Possibly you mean something like a tetrad? So you could describe an inertial tetrad in terms of a polar coordinate basis.

I was using “frame” to mean “coordinate chart”

 

[cianfa72]

Ah, I see. You are using “frame” to mean something different from “coordinate chart”. Possibly you mean something like a tetrad? So you could describe an inertial tetrad in terms of a polar coordinate basis.

I was using “frame” to mean “coordinate chart”

Do you mean that a tetrad/verbatin can be said inertial when its unit timelike vector field is indeed inertial (i.e. its timelike integral curve results in a geodesic of underlying spacetime) ?

 

[Dale]

Do you mean that a tetrad/verbatin can be said inertial when its unit timelike vector field is indeed inertial (i.e. its timelike integral curve results in a geodesic of underlying spacetime) ?

Yes, with the additional restriction that the spacelike vector fields do not rotate along the timelike integral curves

 

[vanhees71]

To define a general rotation free (non-inertial) frame of reference along an arbitrary time-like curve you use Fermi-Walker transport of the three space-like vectors of a tetrad with the time-like vector being, of course, the tangent vector on this time-like curve.

If the time-like curve is a geodesic this defines a local inertial frame along the time-like curve, and the Fermi-Walker transport of the space-like vectors of the tetrad is identical with parallel transport.

 

[PAllen]

No. Because “flat” is a term that describes a spacetime and “inertial” is a term that describes a reference frame, worldline, or coordinate chart. You wouldn’t say “this is an inertial manifold” and you wouldn’t say “this is a flat reference frame”, so the terms are not interchangeable.

I agree and would add some further distinctions:

- inertial motion is a characteristic of world line (and of a body to the extent it can be approximated by a world line). In SR and GR, this corresponds to the world line being a timelike geodesic.

- an inertial frame or coordinate system is one whose (t,0,0,0) world line is inertial and the rest is built by a standard procedure from the origin world line.

- locally flat in GR is simply the property that in small regions and short times, SR is a good approximation to GR. Or, mathematically, that all curvature effects (e.g. triangle angular defect/surplus) go to zero as size goes to zero. Curvature at a point is realized by normalizing by the area. Without normalization, the effect goes to zero as the square of the size. This has nothing to do with the coordinates or motion, and is simply part ot the definition of a Riemannian or pseudo-Riemannian manifold.