Mixing
Doubt about the vacua equations of General Relativity
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I'm facing a quite annoying conceptual problem concerning the Einstein Field Equations (EFE) in so called "vacuum". This problem is both physical and mathematical.
So, in a elementary point of view, when a manifold is flat we have:
$$ R^delta_mu gamma nu = 0tag1$$
i.e. the Riemannian curvature is zero.
Consider now the EFE in the form:
$$ R_mu nu = 0.tag2$$
This equation have both strong mathematical and physical meaning:
About the mathematical meaning this is a condition called Ricci flat and about the physical meaning this is the EFE in vaccum.
So, firstly, Ricci flat condition implies Riemann flat condition?
Secondly, I have a serious doubts about the concept of vacuum. What is vaccum in GR? I mean, Kerr metric is a vacuum solution but the Manifold is curved, so the absence of matter shouldn't implies a flat space?
general-relativity differential-geometry metric-tensor tensor-calculus curvature
edited 3 hours ago
Qmechanic♦
99k121781095
asked 4 hours ago
Jack Clerk
36619
add a comment |Â
up vote
1
down vote
favorite
I'm facing a quite annoying conceptual problem concerning the Einstein Field Equations (EFE) in so called "vacuum". This problem is both physical and mathematical.
So, in a elementary point of view, when a manifold is flat we have:
$$ R^delta_mu gamma nu = 0tag1$$
i.e. the Riemannian curvature is zero.
Consider now the EFE in the form:
$$ R_mu nu = 0.tag2$$
This equation have both strong mathematical and physical meaning:
About the mathematical meaning this is a condition called Ricci flat and about the physical meaning this is the EFE in vaccum.
So, firstly, Ricci flat condition implies Riemann flat condition?
Secondly, I have a serious doubts about the concept of vacuum. What is vaccum in GR? I mean, Kerr metric is a vacuum solution but the Manifold is curved, so the absence of matter shouldn't implies a flat space?
general-relativity differential-geometry metric-tensor tensor-calculus curvature
edited 3 hours ago
Qmechanic♦
99k121781095
asked 4 hours ago
Jack Clerk
36619
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
1
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm facing a quite annoying conceptual problem concerning the Einstein Field Equations (EFE) in so called "vacuum". This problem is both physical and mathematical.
So, in a elementary point of view, when a manifold is flat we have:
$$ R^delta_mu gamma nu = 0tag1$$
i.e. the Riemannian curvature is zero.
Consider now the EFE in the form:
$$ R_mu nu = 0.tag2$$
This equation have both strong mathematical and physical meaning:
About the mathematical meaning this is a condition called Ricci flat and about the physical meaning this is the EFE in vaccum.
So, firstly, Ricci flat condition implies Riemann flat condition?
Secondly, I have a serious doubts about the concept of vacuum. What is vaccum in GR? I mean, Kerr metric is a vacuum solution but the Manifold is curved, so the absence of matter shouldn't implies a flat space?
general-relativity differential-geometry metric-tensor tensor-calculus curvature
edited 3 hours ago
Qmechanic♦
99k121781095
asked 4 hours ago
Jack Clerk
36619
I'm facing a quite annoying conceptual problem concerning the Einstein Field Equations (EFE) in so called "vacuum". This problem is both physical and mathematical.
So, in a elementary point of view, when a manifold is flat we have:
$$ R^delta_mu gamma nu = 0tag1$$
i.e. the Riemannian curvature is zero.
Consider now the EFE in the form:
$$ R_mu nu = 0.tag2$$
This equation have both strong mathematical and physical meaning:
About the mathematical meaning this is a condition called Ricci flat and about the physical meaning this is the EFE in vaccum.
So, firstly, Ricci flat condition implies Riemann flat condition?
Secondly, I have a serious doubts about the concept of vacuum. What is vaccum in GR? I mean, Kerr metric is a vacuum solution but the Manifold is curved, so the absence of matter shouldn't implies a flat space?
general-relativity differential-geometry metric-tensor tensor-calculus curvature
general-relativity differential-geometry metric-tensor tensor-calculus curvature
edited 3 hours ago
Qmechanic♦
99k121781095
asked 4 hours ago
Jack Clerk
36619
edited 3 hours ago
Qmechanic♦
99k121781095
asked 4 hours ago
Jack Clerk
36619
edited 3 hours ago
Qmechanic♦
99k121781095
edited 3 hours ago
Qmechanic♦
99k121781095
edited 3 hours ago
Qmechanic♦
99k121781095
99k121781095
asked 4 hours ago
Jack Clerk
36619
asked 4 hours ago
Jack Clerk
36619
asked 4 hours ago
Jack Clerk
36619
36619
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
1
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago
add a comment |Â
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
1
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
1
1
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $R_mu nu$ in that region.
The way you solve the Kerr solution is by saying "what if I have $R_mu nu = 0$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_mu nu rho sigma = 0$ your space must be Minkowski everwhere. However, you can have $R_mu nu rho sigma neq 0$ somewhere and still have $R_mu nu = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_mu nu = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.
answered 4 hours ago
user1379857
1,842721
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
add a comment |Â
up vote
1
down vote
The vanishing of the curvature tensor $R_abcd=0$ indeed implies that the Ricci tensor also vanishes, $R_ab=0$, but not the other way around.
It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$ (assuming it is non-zero elsewhere causing curvature in the first place).
answered 4 hours ago
Garf
1,095217
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
add a comment |Â
up vote
1
down vote
The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.
It is a common misconception to think that zero $R_ab$ implies no curvature, or that zero $T_ab$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $R_ab$, you can think of $R_ab$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.
Next, to get a feel for the role of $T_ab$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $T_ab$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.
answered 3 hours ago
Andrew Steane
5346
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $R_mu nu$ in that region.
The way you solve the Kerr solution is by saying "what if I have $R_mu nu = 0$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_mu nu rho sigma = 0$ your space must be Minkowski everwhere. However, you can have $R_mu nu rho sigma neq 0$ somewhere and still have $R_mu nu = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_mu nu = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.
answered 4 hours ago
user1379857
1,842721
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $R_mu nu$ in that region.
The way you solve the Kerr solution is by saying "what if I have $R_mu nu = 0$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_mu nu rho sigma = 0$ your space must be Minkowski everwhere. However, you can have $R_mu nu rho sigma neq 0$ somewhere and still have $R_mu nu = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_mu nu = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.
answered 4 hours ago
user1379857
1,842721
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $R_mu nu$ in that region.
The way you solve the Kerr solution is by saying "what if I have $R_mu nu = 0$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_mu nu rho sigma = 0$ your space must be Minkowski everwhere. However, you can have $R_mu nu rho sigma neq 0$ somewhere and still have $R_mu nu = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_mu nu = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.
answered 4 hours ago
user1379857
1,842721
The Kerr solution is a vacuum solution everywhere BUT the singularity! If you were to "smooth out" the singularity, rest assured there would be a very big $R_mu nu$ in that region.
The way you solve the Kerr solution is by saying "what if I have $R_mu nu = 0$ everwhere in a rotationally symmetric space time with the boundary conditions that my metric becomes Minkowski at infinity?"
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_mu nu rho sigma = 0$ your space must be Minkowski everwhere. However, you can have $R_mu nu rho sigma neq 0$ somewhere and still have $R_mu nu = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_mu nu = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat.
answered 4 hours ago
user1379857
1,842721
edited 3 hours ago
answered 4 hours ago
user1379857
1,842721
answered 4 hours ago
user1379857
1,842721
answered 4 hours ago
user1379857
1,842721
1,842721
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
add a comment |Â
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
1
1
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
I like this answer, and I always appreciated the role of the Weyl Tensor: a nice way of viewing the difference between flatness of the Ricci tensor vs flatness of the Riemann tensor, since the Riemann tensor is zero precisely when the Weyl tensor is zero. en.wikipedia.org/wiki/Weyl_tensor
– N. Steinle
3 hours ago
add a comment |Â
up vote
1
down vote
The vanishing of the curvature tensor $R_abcd=0$ indeed implies that the Ricci tensor also vanishes, $R_ab=0$, but not the other way around.
It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$ (assuming it is non-zero elsewhere causing curvature in the first place).
answered 4 hours ago
Garf
1,095217
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
add a comment |Â
up vote
1
down vote
The vanishing of the curvature tensor $R_abcd=0$ indeed implies that the Ricci tensor also vanishes, $R_ab=0$, but not the other way around.
It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$ (assuming it is non-zero elsewhere causing curvature in the first place).
answered 4 hours ago
Garf
1,095217
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The vanishing of the curvature tensor $R_abcd=0$ indeed implies that the Ricci tensor also vanishes, $R_ab=0$, but not the other way around.
It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$ (assuming it is non-zero elsewhere causing curvature in the first place).
answered 4 hours ago
Garf
1,095217
The vanishing of the curvature tensor $R_abcd=0$ indeed implies that the Ricci tensor also vanishes, $R_ab=0$, but not the other way around.
It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$ (assuming it is non-zero elsewhere causing curvature in the first place).
answered 4 hours ago
Garf
1,095217
edited 3 hours ago
answered 4 hours ago
Garf
1,095217
answered 4 hours ago
Garf
1,095217
answered 4 hours ago
Garf
1,095217
1,095217
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
add a comment |Â
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
1
1
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
"It is possible to have curved spacetime in a region where the stress-energy tensor $T_munu=0$", ok, why?
– Jack Clerk
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
I've just edited my answer - as long as $T_munu$ is non-zero somewhere, this will give rise to curvature everywhere in spacetime. If however we're only interested in what's happening in the vacuum, we can just say we wish to solve $R_munu=0$.
– Garf
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
When the energy-momentum tensor is zero, this means that there's no matter or you are faraway from the souce of curvature?
– Jack Clerk
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
No I don't think this is correct - for example a very basic form of $T_munu$ is for a non-relativistic fluid $$T_munu=rho u_mu u_nu$$ and if we say $rho$ is non-zero in a finite region (say a sphere) and is $0$ elsewhere, then $T_munu$ behaves the same - however, spacetime is still curved in the region where $rho=0$.
– Garf
3 hours ago
add a comment |Â
up vote
1
down vote
The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.
It is a common misconception to think that zero $R_ab$ implies no curvature, or that zero $T_ab$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $R_ab$, you can think of $R_ab$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.
Next, to get a feel for the role of $T_ab$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $T_ab$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.
answered 3 hours ago
Andrew Steane
5346
add a comment |Â
up vote
1
down vote
The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.
It is a common misconception to think that zero $R_ab$ implies no curvature, or that zero $T_ab$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $R_ab$, you can think of $R_ab$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.
Next, to get a feel for the role of $T_ab$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $T_ab$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.
answered 3 hours ago
Andrew Steane
5346
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.
It is a common misconception to think that zero $R_ab$ implies no curvature, or that zero $T_ab$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $R_ab$, you can think of $R_ab$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.
Next, to get a feel for the role of $T_ab$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $T_ab$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.
answered 3 hours ago
Andrew Steane
5346
The answers already given are correct, but I will elaborate a little in hopes of giving some physical insight.
It is a common misconception to think that zero $R_ab$ implies no curvature, or that zero $T_ab$ implies no curvature and hence no tidal effects (geodesic deviation and the like). To avoid the misconception about $R_ab$, you can think of $R_ab$ as a statement about direction-averaged curvature, that is, curvature after averaging over direction. Then what we are saying for vacuum is not there is no curvature, but rather if there is some positive curvature in one direction, then there must be negative curvature in another direction.
Next, to get a feel for the role of $T_ab$, compare it to the simpler case of charge and current as a source of electromagnetic fields. If there is a region of space with no charge and current, it does not necessarily mean the electromagnetic field tensor has to be zero there. Rather, its divergence is zero. Similarly, if $T_ab$ is zero, that in itself does not say whether the Riemann curvature is zero, but it does restrict the way the Riemann tensor can vary from one place to another.
answered 3 hours ago
Andrew Steane
5346
answered 3 hours ago
Andrew Steane
5346
answered 3 hours ago
Andrew Steane
5346
answered 3 hours ago
Andrew Steane
5346
5346
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f439749%2fdoubt-about-the-vacua-equations-of-general-relativity%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
That's not Einsteins Field Equations. That's just an arbitrary condition you've just applied.
– Mozibur Ullah
3 hours ago
1
@MoziburUllah In vacuum the Einstein field equations really do simplify to $R_munu = 0$.
– Mike
3 hours ago