The weak field limit metric setup
$begingroup$
Assume that
$$g_{ab}=m_{ab}+h_{ab}$$
where $m_{ab}=text{diag}(1,-1,-1,-1)$ is the Minkowski space metric in an inertial coordinate system $x^a$, and $h_{ab$} is small and slowly varying.
$1)$ I thought that $text{diag}(1,-1,-1,-1)$ $textbf{was}$ $g_{ab}$???
Now the convariant metric is
$$g^{ab}= m^{ab}-m^{ac}m^{bd}h_{cd}$$
$2)$ I cannot see how the RHS of this has been derived, especially with respect to the minus sign.
differential-geometry mathematical-physics tensors general-relativity
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
$endgroup$
add a comment |
$begingroup$
Assume that
$$g_{ab}=m_{ab}+h_{ab}$$
where $m_{ab}=text{diag}(1,-1,-1,-1)$ is the Minkowski space metric in an inertial coordinate system $x^a$, and $h_{ab$} is small and slowly varying.
$1)$ I thought that $text{diag}(1,-1,-1,-1)$ $textbf{was}$ $g_{ab}$???
Now the convariant metric is
$$g^{ab}= m^{ab}-m^{ac}m^{bd}h_{cd}$$
$2)$ I cannot see how the RHS of this has been derived, especially with respect to the minus sign.
differential-geometry mathematical-physics tensors general-relativity
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
$endgroup$
add a comment |
$begingroup$
Assume that
$$g_{ab}=m_{ab}+h_{ab}$$
where $m_{ab}=text{diag}(1,-1,-1,-1)$ is the Minkowski space metric in an inertial coordinate system $x^a$, and $h_{ab$} is small and slowly varying.
$1)$ I thought that $text{diag}(1,-1,-1,-1)$ $textbf{was}$ $g_{ab}$???
Now the convariant metric is
$$g^{ab}= m^{ab}-m^{ac}m^{bd}h_{cd}$$
$2)$ I cannot see how the RHS of this has been derived, especially with respect to the minus sign.
differential-geometry mathematical-physics tensors general-relativity
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
$endgroup$
Assume that
$$g_{ab}=m_{ab}+h_{ab}$$
where $m_{ab}=text{diag}(1,-1,-1,-1)$ is the Minkowski space metric in an inertial coordinate system $x^a$, and $h_{ab$} is small and slowly varying.
$1)$ I thought that $text{diag}(1,-1,-1,-1)$ $textbf{was}$ $g_{ab}$???
Now the convariant metric is
$$g^{ab}= m^{ab}-m^{ac}m^{bd}h_{cd}$$
$2)$ I cannot see how the RHS of this has been derived, especially with respect to the minus sign.
differential-geometry mathematical-physics tensors general-relativity
differential-geometry mathematical-physics tensors general-relativity
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
edited Dec 22 '18 at 19:48
Permian
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
asked Dec 22 '18 at 17:43
PermianPermian
2,2681135
2,2681135
add a comment |
add a comment |
1 Answer
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1) That is only when you don't have to worry about gravity. Weak gravitational fields. That is the usual case. When you only need special relativity rather than general relativity.
2) Take the inverse of both sides of the equation $g_{ab}=m_{ab}+h_{ab}$. Rename the $a$ to $e$ here so can check whether $g_{ea} g^{ab} = delta_e^b$. That is to expand and simplify the following:
$$
(m_{ea}+h_{ea}) times (m^{ab}-m^{ac} m^{bd} h_{cd})\
$$
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
$endgroup$
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) That is only when you don't have to worry about gravity. Weak gravitational fields. That is the usual case. When you only need special relativity rather than general relativity.
2) Take the inverse of both sides of the equation $g_{ab}=m_{ab}+h_{ab}$. Rename the $a$ to $e$ here so can check whether $g_{ea} g^{ab} = delta_e^b$. That is to expand and simplify the following:
$$
(m_{ea}+h_{ea}) times (m^{ab}-m^{ac} m^{bd} h_{cd})\
$$
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
$endgroup$
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
add a comment |
$begingroup$
1) That is only when you don't have to worry about gravity. Weak gravitational fields. That is the usual case. When you only need special relativity rather than general relativity.
2) Take the inverse of both sides of the equation $g_{ab}=m_{ab}+h_{ab}$. Rename the $a$ to $e$ here so can check whether $g_{ea} g^{ab} = delta_e^b$. That is to expand and simplify the following:
$$
(m_{ea}+h_{ea}) times (m^{ab}-m^{ac} m^{bd} h_{cd})\
$$
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
$endgroup$
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
add a comment |
$begingroup$
1) That is only when you don't have to worry about gravity. Weak gravitational fields. That is the usual case. When you only need special relativity rather than general relativity.
2) Take the inverse of both sides of the equation $g_{ab}=m_{ab}+h_{ab}$. Rename the $a$ to $e$ here so can check whether $g_{ea} g^{ab} = delta_e^b$. That is to expand and simplify the following:
$$
(m_{ea}+h_{ea}) times (m^{ab}-m^{ac} m^{bd} h_{cd})\
$$
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
$endgroup$
1) That is only when you don't have to worry about gravity. Weak gravitational fields. That is the usual case. When you only need special relativity rather than general relativity.
2) Take the inverse of both sides of the equation $g_{ab}=m_{ab}+h_{ab}$. Rename the $a$ to $e$ here so can check whether $g_{ea} g^{ab} = delta_e^b$. That is to expand and simplify the following:
$$
(m_{ea}+h_{ea}) times (m^{ab}-m^{ac} m^{bd} h_{cd})\
$$
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
answered Dec 22 '18 at 20:01
AHusainAHusain
2,8282916
2,8282916
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
add a comment |
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
$begingroup$
How could we see this inverse a priori, rather than just checking (once it has been given) that it is the inverse?
$endgroup$
– Permian
Dec 23 '18 at 11:16
add a comment |
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