- #1

- 72

- 0

In a local inertial frame with [itex]g_{ij}=\eta_{ij}[/itex] and [itex]\Gamma^i_{jk}=0[/itex].

why in such a frame, curvature tensor isn't zero?

curvature tensor is made of metric,first and second derivative of metric.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter sadegh4137
- Start date

- #1

- 72

- 0

In a local inertial frame with [itex]g_{ij}=\eta_{ij}[/itex] and [itex]\Gamma^i_{jk}=0[/itex].

why in such a frame, curvature tensor isn't zero?

curvature tensor is made of metric,first and second derivative of metric.

- #2

WannabeNewton

Science Advisor

- 5,800

- 537

- #3

- 72

- 0

so why first derivative become zero?

- #4

WannabeNewton

Science Advisor

- 5,800

- 537

- #5

- 72

- 0

I know what's derivative but first derivative become zero!!! and second isn't!!

is it regular?

according your reason, first derivative isn't zero!!

I can't understand you!!!

- #6

WannabeNewton

Science Advisor

- 5,800

- 537

No, Riemann normal coordinates tell us that the metric becomes Minkowski at a given point **and separately** that the Christoffel symbols vanish identically at said point. The vanishing of the Christoffel symbols at that point is **not** a direct consequence of the metric being Minkowski at that point *just through differentiation of the Minkowski metric*; this would make zero sense mathematically.

Last edited:

- #7

phyzguy

Science Advisor

- 4,817

- 1,774

I know what's derivative but first derivative become zero!!! and second isn't!!

is it regular?

according your reason, first derivative isn't zero!!

I can't understand you!!!

Think of the function y = x^2.

The function and it's first derivative are both equal to 0 at x=0, but the second derivative (it's curvature) is not equal to zero at x=0. In the same way, if you set up Riemann normal coordinates at a point, the metric is Minkowski, and the Christoffel symbols(first derivative) are zero, but the curvature tensor (second derivative) is not zero.

- #8

Nugatory

Mentor

- 13,389

- 6,367

I know what's derivative but first derivative become zero!!! and second isn't!!

is it regular?

according your reason, first derivative isn't zero!!

I can't understand you!!!

First derivative equal to zero, means that it flat at that point.

Second derivative not equal to zero, means that the first derivative will change (no longer be zero) when you move off that point.

- #9

- 8,776

- 2,806

I know what's derivative but first derivative become zero!!! and second isn't!!

is it regular?

according your reason, first derivative isn't zero!!

I can't understand you!!!

Let's take a particular example: The surface of a sphere of radius 1 meter can be described by coordinates [itex]\theta[/itex] and [itex]\phi[/itex]. (You can think of [itex]\theta[/itex] as latitude and [itex]\phi[/itex] as longitude, although the mathematical convention is to have [itex]\theta[/itex] run from 0 to [itex]\pi[/itex], rather than from -90 to +90, and [itex]\phi[/itex] runs from 0 to 2[itex]\pi[/itex], rather than from -180 to +180)

The components of the metric tensor in this coordinate system are:

[itex]g_{\theta \theta} = 1[/itex]

[itex]g_{\phi \phi} = sin^2(\theta)[/itex]

Take a first derivative to get:

[itex]\dfrac{\partial}{\partial \theta} g_{\phi \phi} = 2 sin(\theta) cos(\theta)[/itex]

Take a second derivative to get:

[itex]\dfrac{\partial^2}{\partial \theta^2} g_{\phi \phi} = 2 (cos^2(\theta) - sin^2(\theta))[/itex]

At [itex]\theta = \dfrac{\pi}{2}[/itex], we have

[itex]g_{\theta \theta} = 1[/itex]

[itex]g_{\phi \phi} = 1[/itex]

[itex]\dfrac{\partial}{\partial \gamma} g_{\alpha\beta} = 0[/itex]

where [itex]\alpha, \beta, \gamma[/itex] are either [itex]\theta[/itex] or [itex]\phi[/itex]

So, the metric components and their first derivatives look just like flat space. But

the second derivative is nonzero, which means that the Riemann curvature tensor can be nonzero.

- #10

WannabeNewton

Science Advisor

- 5,800

- 537

Also, you may already know this, but keep in mind that a frame is not a coordinate system and that a coordinate system is not a frame. What you are describing are Riemann normal coordinates, which are unfortunately called "locally inertial frames" in some texts; other texts more appropriately call them "locally inertial coordinates". What one does is use a frame to *define* a coordinate system, which is how Riemann normal coordinates are constructed (on the physics side anyways).

Last edited:

- #11

- 72

- 0

there are another examples that first and second derivative equal zero at point

for example,y=(x-1)^3 +1 at x=1, this function is simple example!

it isn't a reason that if some functions like y=x^2 have this properties, another has also!!!

this is not a prove!!

I think your prove is based according properties of metric tensor!!

I don't know that!! but I think like this.

wannabenewton

i don't know exactly what's difference between "coordinate system" and "frame"!!

is it possible to explain it for me?

thanks.

- #12

WannabeNewton

Science Advisor

- 5,800

- 537

- #13

- 72

- 0

what's Riemann normal coordinates?

- #14

phyzguy

Science Advisor

- 4,817

- 1,774

i can't understand your reasons!! 'stevendyal' and 'phyzguy'

there are another examples that first and second derivative equal zero at point

for example,y=(x-1)^3 +1 at x=1, this function is simple example!

it isn't a reason that if some functions like y=x^2 have this properties, another has also!!!

this is not a prove!!

I think your prove is based according properties of metric tensor!!

I don't know that!! but I think like this.

What exactly is your question? I thought from your first post that the question was, "In a local inertial frame with a Minkowski metric and Christoffel symbols equal to zero at a point, why isn't the curvature tensor zero?" So I provided an example function (y = x^2) whose value and first derivative are zero at a point, but whose second derivative at that same point is non-zero. I thought this might help you understand, but apparently it didn't.

Please tell us exactly what your question is.

- #15

WannabeNewton

Science Advisor

- 5,800

- 537

Looks like there's a bit of background you have to cover. "Riemann normal coordinates" is the mathematical name for the "locally inertial coordinates" you mentioned in your original post. Can you tell me what textbook you're using for GR? And do you know how "locally inertial coordinates" are actually constructed mathematically? Knowing this will clear up your confusions.what's Riemann normal coordinates?

- #16

- 8,776

- 2,806

i can't understand your reasons!! 'stevendyal' and 'phyzguy'

there are another examples that first and second derivative equal zero at point

for example,y=(x-1)^3 +1 at x=1, this function is simple example!

If both the first and the second derivatives of the metric tensor are zero, then the curvature tensor is zero. At least at that single point. A curved space can have a curvature tensor that has different values at different points.

- #17

- 72

- 0

yes i know how to construct it.

i study padmanabhan and caroll

my question is how to prove second derivative of metric isn't zero, while first is zero.

hoe we can prove it?

yes, there are some examples that first is zero but second isn't, we speak about metric function!!!!

- #18

- 8,776

- 2,806

what's Riemann normal coordinates?

Let me illustrate for the simplest case, of 2D space (no time dimension). Suppose you have some local coordinate system [itex]x, y[/itex], and you have two points

[itex]P_1[/itex] with coordinates [itex](x,y)[/itex]

[itex]P_2[/itex] with coordinates [itex](x + \delta x, y + \delta y)[/itex]

The distance between these points is given approximately (for small [itex]\delta x[/itex] and [itex]\delta y[/itex]) by:

[itex]\delta s^2 = g_{xx} \delta x^2 + 2 g_{xy} \delta x \delta y + g_{yy} \delta y^2[/itex]

where [itex]g_{xx}, g_{xy}, g_{yy}[/itex] are three functions of position.

The coordinates are Riemann normal coordinates for the point [itex]P_1[/itex] provided that:

[itex]g_{xx} = g_{yy} = 1[/itex] at [itex]P_1[/itex]

[itex]g_{xy} = 0[/itex] at [itex]P_1[/itex]

[itex]\frac{\partial}{\partial x} g_{xx} = \frac{\partial}{\partial y} g_{xx} = 0[/itex] at [itex]P_1[/itex]

[itex]\frac{\partial}{\partial x} g_{xy} = \frac{\partial}{\partial y} g_{xy} = 0[/itex] at [itex]P_1[/itex]

[itex]\frac{\partial}{\partial x} g_{yy} = \frac{\partial}{\partial y} g_{yy} = 0[/itex] at [itex]P_1[/itex]

Riemann normal coordinates are the closest you can get to Cartesian coordinates.

- #19

- 8,776

- 2,806

yes i know how to construct it.

i study padmanabhan and caroll

my question is how to prove second derivative of metric isn't zero, while first is zero.

hoe we can prove it?

yes, there are some examples that first is zero but second isn't, we speak about metric function!!!!

Take the second derivative, and see if it's nonzero. That's how you prove that the second derivative is nonzero. I'm not sure I understand what you are asking for. Sometimes the first derivative is zero, but not the second. Sometimes both the first and second derivatives are zero.

The point about the curvature tensor is that if it is nonzero in one coordinate system, then it will be nonzero in every coordinate system. If it is zero in one coordinate system, then it will be zero in every coordinate system. So the curvature tensor is independent of your coordinate system in a way that the Christoffel coefficients [itex]\Gamma^i_{jk}[/itex] are not.

- #20

WannabeNewton

Science Advisor

- 5,800

- 537

Steven has already explained it brilliantly. I really have nothing else to add with regards to that issue.

As far as coordinates and frames go: given an ##n##-manifold ##M##, a frame at an event ##p \in M## is just an orthonormal basis ##\{e_{\mu}\}## for ##T_p M##; a coordinate system is a pair ##(U,x^{\mu})## where ##U \subseteq M## is open and ##x^{\mu}: M \rightarrow \mathbb{R}## are a set of ##n## coordinate functions. In the context of GR, a frame corresponds to the instantaneous rest frame of an ideal observer with ##(e_0) = u##, where ##u## is the 4-velocity of said observer, and with ##(e_i), i = 1,2,3## representing the spatial axes of the instantaneous rest frame. These instantaneous rest frames are also called local inertial frames or local Lorentz frames. One can then*use* these frames to construct coordinates; a local inertial frame at an event ##p## can be used to construct a set of locally inertial coordinates ##x^{\mu}## in a neighborhood of ##p## using the exponential map (see p.112 of Carroll) in which ##g_{\mu\nu}(p) = \eta_{\mu\nu}## and separately ##\Gamma^{\gamma}_{\mu\nu}(p) = 0##.

As far as coordinates and frames go: given an ##n##-manifold ##M##, a frame at an event ##p \in M## is just an orthonormal basis ##\{e_{\mu}\}## for ##T_p M##; a coordinate system is a pair ##(U,x^{\mu})## where ##U \subseteq M## is open and ##x^{\mu}: M \rightarrow \mathbb{R}## are a set of ##n## coordinate functions. In the context of GR, a frame corresponds to the instantaneous rest frame of an ideal observer with ##(e_0) = u##, where ##u## is the 4-velocity of said observer, and with ##(e_i), i = 1,2,3## representing the spatial axes of the instantaneous rest frame. These instantaneous rest frames are also called local inertial frames or local Lorentz frames. One can then

Last edited:

- #21

- 3,507

- 26

In a local inertial frame with [itex]g_{ij}=\eta_{ij}[/itex] and [itex]\Gamma^i_{jk}=0[/itex].

why in such a frame, curvature tensor isn't zero?

curvature tensor is made of metric,first and second derivative of metric.

Maybe the OP is asking about the properties of the metric tensor(a symmetric bilinear form) that lead to having putative vanishing first derivatives in some coordinates and coordinate-independent non-vanishing second derivatives for general manifolds (vanishing too for the special flat case) for

I could be wrong but I believe that both being a symmetric and nondegenerate(having nonzero determinant for its associated matrix in matrix language) bilinear form is important since in general the second partial derivatives only depend on the differential structure but are only defined for critical points in the absence of a Riemannian connection induced by a metric tensor. But we want curvature (the Hessian, defined as ∇df with ∇ being the Levi-Civita connection) to be defined for

- #22

- 72

- 0

bilinear form is important since in general the second partial derivatives only depend on the differential structure but are only defined for critical points in the absence of a Riemannian connection induced by a metric tensor. But we want curvature (the Hessian said:any[/I] point in the manifold.

i couldn't understand you!

can you explain it for me more please.

thanks

- #23

- 8,776

- 2,806

i couldn't understand you!

can you explain it for me more please.

thanks

I think you need to formulate more specific questions, because it's not clear what it is that you don't understand.

Do you understand that the components of the metric tensor [itex]g_{\mu \nu}[/itex] change when you change to a different coordinate system?

Do you understand that the connection coefficients [itex]\Gamma^\mu_{\nu \lambda}[/itex] (which are constructed from derivatives of the metric tensor components) change when you change to a different coordinate system? For one coordinate system, it might be zero at a point, but for another coordinate system, it might be nonzero at that point.

Do you understand that the Riemann tensor, constructed from the first and second derivatives of the metric tensor components, has the property that if it is nonzero in one coordinate system, then it is nonzero in every coordinate system?

Share: