Non-static spherical symmetry spacetime
$begingroup$
The Schwarzschild solution is a static spherically symmetric metric. But I wanted to know that how would the space-time interval look in a Non-Static case. I tried to work it out and got
$$ds²= Bdt² - Cdtdr - Adr² - r²dΩ²$$
Where $A,B,C$ are functions if $r$ and $t$. Being an amateur I am not sure whether this is correct.
general-relativity spacetime differential-geometry symmetry metric-tensor
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
$endgroup$
add a comment |
$begingroup$
The Schwarzschild solution is a static spherically symmetric metric. But I wanted to know that how would the space-time interval look in a Non-Static case. I tried to work it out and got
$$ds²= Bdt² - Cdtdr - Adr² - r²dΩ²$$
Where $A,B,C$ are functions if $r$ and $t$. Being an amateur I am not sure whether this is correct.
general-relativity spacetime differential-geometry symmetry metric-tensor
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
$endgroup$
add a comment |
$begingroup$
The Schwarzschild solution is a static spherically symmetric metric. But I wanted to know that how would the space-time interval look in a Non-Static case. I tried to work it out and got
$$ds²= Bdt² - Cdtdr - Adr² - r²dΩ²$$
Where $A,B,C$ are functions if $r$ and $t$. Being an amateur I am not sure whether this is correct.
general-relativity spacetime differential-geometry symmetry metric-tensor
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
$endgroup$
The Schwarzschild solution is a static spherically symmetric metric. But I wanted to know that how would the space-time interval look in a Non-Static case. I tried to work it out and got
$$ds²= Bdt² - Cdtdr - Adr² - r²dΩ²$$
Where $A,B,C$ are functions if $r$ and $t$. Being an amateur I am not sure whether this is correct.
general-relativity spacetime differential-geometry symmetry metric-tensor
general-relativity spacetime differential-geometry symmetry metric-tensor
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
edited Jan 13 at 14:13
Qmechanic♦
103k121871184
103k121871184
asked Jan 13 at 11:55
user215742user215742
505
asked Jan 13 at 11:55
user215742user215742
505
asked Jan 13 at 11:55
user215742user215742
505
505
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
While Qmechanic's answer is not incorrect, I think it is misleading and misses the point of the question.
First, there is the well-known "Vaidya metric", which is a simple generalization of Schwarzschild to achieve a non-static spherically symmetric spacetime. Of course, in agreement with Birkhoff's theorem, this is also a non-empty (non-vacuum) spacetime — but the OP never restricted to vacuum solutions.
Second, the question strikes me as being more immediately about the general form of a spherically symmetric metric. There's a nice treatment given by Schutz in his chapter 10.1, which is also mostly reproduced here, as well as chapter 23.2 and box 23.3 of MTW. The answer is that yes, a spherically symmetric metric can generally be put in the form given by the OP's equation. But it's also worth pointing out that the $dt, dr$ term can be eliminated by redefining $t$ by a linear combination of $t$ and $r$ involving $B$ and $C$ (as MTW explain).
answered Jan 13 at 14:55
MikeMike
12k13855
$endgroup$
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
|
show 2 more comments
$begingroup$
It is actually a consequence of Birkhoff's theorem that a spherically symmetric vacuum solution to EFE is necessarily static. See also this related Phys.SE post.
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
$endgroup$
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
While Qmechanic's answer is not incorrect, I think it is misleading and misses the point of the question.
First, there is the well-known "Vaidya metric", which is a simple generalization of Schwarzschild to achieve a non-static spherically symmetric spacetime. Of course, in agreement with Birkhoff's theorem, this is also a non-empty (non-vacuum) spacetime — but the OP never restricted to vacuum solutions.
Second, the question strikes me as being more immediately about the general form of a spherically symmetric metric. There's a nice treatment given by Schutz in his chapter 10.1, which is also mostly reproduced here, as well as chapter 23.2 and box 23.3 of MTW. The answer is that yes, a spherically symmetric metric can generally be put in the form given by the OP's equation. But it's also worth pointing out that the $dt, dr$ term can be eliminated by redefining $t$ by a linear combination of $t$ and $r$ involving $B$ and $C$ (as MTW explain).
answered Jan 13 at 14:55
MikeMike
12k13855
$endgroup$
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
|
show 2 more comments
$begingroup$
While Qmechanic's answer is not incorrect, I think it is misleading and misses the point of the question.
First, there is the well-known "Vaidya metric", which is a simple generalization of Schwarzschild to achieve a non-static spherically symmetric spacetime. Of course, in agreement with Birkhoff's theorem, this is also a non-empty (non-vacuum) spacetime — but the OP never restricted to vacuum solutions.
Second, the question strikes me as being more immediately about the general form of a spherically symmetric metric. There's a nice treatment given by Schutz in his chapter 10.1, which is also mostly reproduced here, as well as chapter 23.2 and box 23.3 of MTW. The answer is that yes, a spherically symmetric metric can generally be put in the form given by the OP's equation. But it's also worth pointing out that the $dt, dr$ term can be eliminated by redefining $t$ by a linear combination of $t$ and $r$ involving $B$ and $C$ (as MTW explain).
answered Jan 13 at 14:55
MikeMike
12k13855
$endgroup$
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
|
show 2 more comments
$begingroup$
While Qmechanic's answer is not incorrect, I think it is misleading and misses the point of the question.
First, there is the well-known "Vaidya metric", which is a simple generalization of Schwarzschild to achieve a non-static spherically symmetric spacetime. Of course, in agreement with Birkhoff's theorem, this is also a non-empty (non-vacuum) spacetime — but the OP never restricted to vacuum solutions.
Second, the question strikes me as being more immediately about the general form of a spherically symmetric metric. There's a nice treatment given by Schutz in his chapter 10.1, which is also mostly reproduced here, as well as chapter 23.2 and box 23.3 of MTW. The answer is that yes, a spherically symmetric metric can generally be put in the form given by the OP's equation. But it's also worth pointing out that the $dt, dr$ term can be eliminated by redefining $t$ by a linear combination of $t$ and $r$ involving $B$ and $C$ (as MTW explain).
answered Jan 13 at 14:55
MikeMike
12k13855
$endgroup$
While Qmechanic's answer is not incorrect, I think it is misleading and misses the point of the question.
First, there is the well-known "Vaidya metric", which is a simple generalization of Schwarzschild to achieve a non-static spherically symmetric spacetime. Of course, in agreement with Birkhoff's theorem, this is also a non-empty (non-vacuum) spacetime — but the OP never restricted to vacuum solutions.
Second, the question strikes me as being more immediately about the general form of a spherically symmetric metric. There's a nice treatment given by Schutz in his chapter 10.1, which is also mostly reproduced here, as well as chapter 23.2 and box 23.3 of MTW. The answer is that yes, a spherically symmetric metric can generally be put in the form given by the OP's equation. But it's also worth pointing out that the $dt, dr$ term can be eliminated by redefining $t$ by a linear combination of $t$ and $r$ involving $B$ and $C$ (as MTW explain).
answered Jan 13 at 14:55
MikeMike
12k13855
answered Jan 13 at 14:55
MikeMike
12k13855
answered Jan 13 at 14:55
MikeMike
12k13855
answered Jan 13 at 14:55
MikeMike
12k13855
12k13855
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
|
show 2 more comments
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
If I may ask, how is the Schwarzschild solution static inside the horizon? If I understand this correctly, static means that the metric coefficients don't depend on time, but the radial coordinate is time inside the horizon and the metric does depend on this coordinate inside.
$endgroup$
– safesphere
Jan 13 at 17:48
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
PS. I see that Wki states (under Examples) that only exterior Schwarzschild solution is static. However, this whole solution is derived based on the assumption of being static. Thus the interior part of the solution must not be valid. No? en.wikipedia.org/wiki/Static_spacetime
$endgroup$
– safesphere
Jan 13 at 17:55
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
You might want to ask these questions as a separate question for fuller responses, but I'll try to answer. First, "static" does not mean that "the metric coefficients don't depend on time" in any particular coordinate system; rather it means that there exist coordinates in which the metric coefficients don't depend on time. To approach it from another direction, suppose you had a metric with coefficients that don't depend on time. You know that a smooth change of coordinates doesn't change anything physically, but you can probably see how to add time dependence to those coordinates.
$endgroup$
– Mike
Jan 14 at 17:12
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Second, you are correct that the region inside the Schwarzschild horizon is not static — meaning that it's impossible to find coordinates where the metric coefficient is time-independent. In fact, in the Kerr metric there's even a region that's outside the horizon but is not static.
$endgroup$
– Mike
Jan 14 at 17:15
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
$begingroup$
Third, it is true that the Schwarzschild solution is usually derived using the assumption that it is static, but it is also true that the result you get with that approach is a metric that makes sense (is nonsingular) everywhere except $r=0$ and $r=2M$, and you can just check to make sure that even for $r leq 2M$ the metric does solve the vacuum Einstein equations. There's no reason it has to work that way; it just does. So the inside of the horizon is a valid solution.
$endgroup$
– Mike
Jan 14 at 17:19
|
show 2 more comments
$begingroup$
It is actually a consequence of Birkhoff's theorem that a spherically symmetric vacuum solution to EFE is necessarily static. See also this related Phys.SE post.
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
$endgroup$
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
add a comment |
$begingroup$
It is actually a consequence of Birkhoff's theorem that a spherically symmetric vacuum solution to EFE is necessarily static. See also this related Phys.SE post.
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
$endgroup$
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
add a comment |
$begingroup$
It is actually a consequence of Birkhoff's theorem that a spherically symmetric vacuum solution to EFE is necessarily static. See also this related Phys.SE post.
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
$endgroup$
It is actually a consequence of Birkhoff's theorem that a spherically symmetric vacuum solution to EFE is necessarily static. See also this related Phys.SE post.
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
answered Jan 13 at 12:09
Qmechanic♦Qmechanic
103k121871184
103k121871184
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
add a comment |
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
So a metric that is Non-Static can not be Symmetric?
$endgroup$
– user215742
Jan 13 at 12:13
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
$begingroup$
One can still pick a coordinate system, where the symmetries of the system is not manifest, if that's what you're asking.
$endgroup$
– Qmechanic♦
Jan 13 at 12:21
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