Further deriving the weak field limit in general relativity
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Consider the motion of a slow moving particle with worldline $x^a = x^a(t)$. We have
$$frac{dx^0}{dt}=1, frac{dx^1}{dt}=u_1,frac{dx^2}{dt}=u_2,frac{dx^3}{dt}=u_3$$
where $(u_1,u_2,u_3)$ is the velocity in the inertial coordinates on Minkowski space. In particular we have $gamma(u) sim 1$ and so we can identify the coordinate time $t$ with the proper time $tau$ along the worldline, and so approximate the four-velocty by
$$(V^a)=(1,u_1,u_2,u_3)$$
The geodesic equation is
$$frac{d^2x^a}{dtau^2}+Gamma_{bc}^afrac{dx^b}{dtau}frac{dx^c}{dtau}=0$$
Because we ignore products of the spatial components of the four-velocity with the Christoffel symbols, this is approximated by
$$frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$$
We also ignore terms involving time $(x^0)$ derivatives of the metric components.
1) Where has the $gamma(u) sim 1$ come from?
2) a) are we ignoring products due to being slow moving?
b) I still cant see how the last equation has been derived
3) We do we only ignore components of the time derivative? Why not the others?
differential-geometry mathematical-physics tensors general-relativity
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
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add a comment |
$begingroup$
Consider the motion of a slow moving particle with worldline $x^a = x^a(t)$. We have
$$frac{dx^0}{dt}=1, frac{dx^1}{dt}=u_1,frac{dx^2}{dt}=u_2,frac{dx^3}{dt}=u_3$$
where $(u_1,u_2,u_3)$ is the velocity in the inertial coordinates on Minkowski space. In particular we have $gamma(u) sim 1$ and so we can identify the coordinate time $t$ with the proper time $tau$ along the worldline, and so approximate the four-velocty by
$$(V^a)=(1,u_1,u_2,u_3)$$
The geodesic equation is
$$frac{d^2x^a}{dtau^2}+Gamma_{bc}^afrac{dx^b}{dtau}frac{dx^c}{dtau}=0$$
Because we ignore products of the spatial components of the four-velocity with the Christoffel symbols, this is approximated by
$$frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$$
We also ignore terms involving time $(x^0)$ derivatives of the metric components.
1) Where has the $gamma(u) sim 1$ come from?
2) a) are we ignoring products due to being slow moving?
b) I still cant see how the last equation has been derived
3) We do we only ignore components of the time derivative? Why not the others?
differential-geometry mathematical-physics tensors general-relativity
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
$endgroup$
add a comment |
$begingroup$
Consider the motion of a slow moving particle with worldline $x^a = x^a(t)$. We have
$$frac{dx^0}{dt}=1, frac{dx^1}{dt}=u_1,frac{dx^2}{dt}=u_2,frac{dx^3}{dt}=u_3$$
where $(u_1,u_2,u_3)$ is the velocity in the inertial coordinates on Minkowski space. In particular we have $gamma(u) sim 1$ and so we can identify the coordinate time $t$ with the proper time $tau$ along the worldline, and so approximate the four-velocty by
$$(V^a)=(1,u_1,u_2,u_3)$$
The geodesic equation is
$$frac{d^2x^a}{dtau^2}+Gamma_{bc}^afrac{dx^b}{dtau}frac{dx^c}{dtau}=0$$
Because we ignore products of the spatial components of the four-velocity with the Christoffel symbols, this is approximated by
$$frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$$
We also ignore terms involving time $(x^0)$ derivatives of the metric components.
1) Where has the $gamma(u) sim 1$ come from?
2) a) are we ignoring products due to being slow moving?
b) I still cant see how the last equation has been derived
3) We do we only ignore components of the time derivative? Why not the others?
differential-geometry mathematical-physics tensors general-relativity
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
$endgroup$
Consider the motion of a slow moving particle with worldline $x^a = x^a(t)$. We have
$$frac{dx^0}{dt}=1, frac{dx^1}{dt}=u_1,frac{dx^2}{dt}=u_2,frac{dx^3}{dt}=u_3$$
where $(u_1,u_2,u_3)$ is the velocity in the inertial coordinates on Minkowski space. In particular we have $gamma(u) sim 1$ and so we can identify the coordinate time $t$ with the proper time $tau$ along the worldline, and so approximate the four-velocty by
$$(V^a)=(1,u_1,u_2,u_3)$$
The geodesic equation is
$$frac{d^2x^a}{dtau^2}+Gamma_{bc}^afrac{dx^b}{dtau}frac{dx^c}{dtau}=0$$
Because we ignore products of the spatial components of the four-velocity with the Christoffel symbols, this is approximated by
$$frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$$
We also ignore terms involving time $(x^0)$ derivatives of the metric components.
1) Where has the $gamma(u) sim 1$ come from?
2) a) are we ignoring products due to being slow moving?
b) I still cant see how the last equation has been derived
3) We do we only ignore components of the time derivative? Why not the others?
differential-geometry mathematical-physics tensors general-relativity
differential-geometry mathematical-physics tensors general-relativity
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
asked Dec 23 '18 at 11:28
PermianPermian
2,2781135
2,2781135
add a comment |
add a comment |
1 Answer
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A short answner:
For your first question, here a hint: consider that $gamma = frac{1}{sqrt{1 - beta^2}}$
where $beta = v/c$, approximate $gamma$ when $ v << c$ (Slow moving particle).
for the second and third question's:
yes, $frac{dx^i}{dtau}frac{dx^j}{dtau}=gamma^2 u^i u^j$,
$gamma^2 approx 1$, then $frac{dx^i}{dtau}frac{dx^j}{dtau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $Gamma_{00}^afrac{dx^0}{dtau}frac{dx^0}{dtau}$, but $frac{dx^0}{dt}=1$, then $frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.
answered Dec 24 '18 at 16:02
ZoberZober
715
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1 Answer
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1 Answer
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$begingroup$
A short answner:
For your first question, here a hint: consider that $gamma = frac{1}{sqrt{1 - beta^2}}$
where $beta = v/c$, approximate $gamma$ when $ v << c$ (Slow moving particle).
for the second and third question's:
yes, $frac{dx^i}{dtau}frac{dx^j}{dtau}=gamma^2 u^i u^j$,
$gamma^2 approx 1$, then $frac{dx^i}{dtau}frac{dx^j}{dtau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $Gamma_{00}^afrac{dx^0}{dtau}frac{dx^0}{dtau}$, but $frac{dx^0}{dt}=1$, then $frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.
answered Dec 24 '18 at 16:02
ZoberZober
715
$endgroup$
add a comment |
$begingroup$
A short answner:
For your first question, here a hint: consider that $gamma = frac{1}{sqrt{1 - beta^2}}$
where $beta = v/c$, approximate $gamma$ when $ v << c$ (Slow moving particle).
for the second and third question's:
yes, $frac{dx^i}{dtau}frac{dx^j}{dtau}=gamma^2 u^i u^j$,
$gamma^2 approx 1$, then $frac{dx^i}{dtau}frac{dx^j}{dtau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $Gamma_{00}^afrac{dx^0}{dtau}frac{dx^0}{dtau}$, but $frac{dx^0}{dt}=1$, then $frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.
answered Dec 24 '18 at 16:02
ZoberZober
715
$endgroup$
add a comment |
$begingroup$
A short answner:
For your first question, here a hint: consider that $gamma = frac{1}{sqrt{1 - beta^2}}$
where $beta = v/c$, approximate $gamma$ when $ v << c$ (Slow moving particle).
for the second and third question's:
yes, $frac{dx^i}{dtau}frac{dx^j}{dtau}=gamma^2 u^i u^j$,
$gamma^2 approx 1$, then $frac{dx^i}{dtau}frac{dx^j}{dtau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $Gamma_{00}^afrac{dx^0}{dtau}frac{dx^0}{dtau}$, but $frac{dx^0}{dt}=1$, then $frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.
answered Dec 24 '18 at 16:02
ZoberZober
715
$endgroup$
A short answner:
For your first question, here a hint: consider that $gamma = frac{1}{sqrt{1 - beta^2}}$
where $beta = v/c$, approximate $gamma$ when $ v << c$ (Slow moving particle).
for the second and third question's:
yes, $frac{dx^i}{dtau}frac{dx^j}{dtau}=gamma^2 u^i u^j$,
$gamma^2 approx 1$, then $frac{dx^i}{dtau}frac{dx^j}{dtau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $Gamma_{00}^afrac{dx^0}{dtau}frac{dx^0}{dtau}$, but $frac{dx^0}{dt}=1$, then $frac{d^2x^a}{dtau^2}+Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.
answered Dec 24 '18 at 16:02
ZoberZober
715
edited Dec 24 '18 at 17:57
answered Dec 24 '18 at 16:02
ZoberZober
715
answered Dec 24 '18 at 16:02
ZoberZober
715
answered Dec 24 '18 at 16:02
ZoberZober
715
715
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