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1

$begingroup$

Let $U^mu$ be a vector in 4-dimensional Minkowski space with norm $-1$ and $K^mu = V(x)U^mu$ a vector proportional to it. We can write $V(x) = sqrt-K_nu K^nu$.

(This setup comes from physics where $U^mu$ is a four-velocity, $K^mu$ is a normalized time-like Killing vector for an observer at infinity and $V(x)$ is called the redshift factor.)

Then, define $a^mu$ (the four-acceleration) by $U^sigma nabla_sigma U^mu$, where $nabla$ is the Christoffel connection.

According to Sean Carroll, Spacetime and Geometry, p. 247, $a_mu = nabla_muln V$. Why?

Attempt:

beginalignnabla_muln V &= frac 12V^2nabla_muleft(-K_nu K^nuright)\
&=-frac 12V^2left((nabla_mu K_nu)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left((nabla_mu g_rhonuK^rho)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left(K_rhonabla_mu K^rho + K_nunabla_mu K^nuright)\
&=-frac K_nunabla_mu K^nuV^2\
&= -frac 1VU_nunabla_muleft(VU^nuright)\
&= -U_nunabla_mu U^nu - frac 1 VU_nu U^nunabla_mu Vendalign

differential-geometry semi-riemannian-geometry

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asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

$endgroup$

add a comment | 

1

$begingroup$

Let $U^mu$ be a vector in 4-dimensional Minkowski space with norm $-1$ and $K^mu = V(x)U^mu$ a vector proportional to it. We can write $V(x) = sqrt-K_nu K^nu$.

(This setup comes from physics where $U^mu$ is a four-velocity, $K^mu$ is a normalized time-like Killing vector for an observer at infinity and $V(x)$ is called the redshift factor.)

Then, define $a^mu$ (the four-acceleration) by $U^sigma nabla_sigma U^mu$, where $nabla$ is the Christoffel connection.

According to Sean Carroll, Spacetime and Geometry, p. 247, $a_mu = nabla_muln V$. Why?

Attempt:

beginalignnabla_muln V &= frac 12V^2nabla_muleft(-K_nu K^nuright)\
&=-frac 12V^2left((nabla_mu K_nu)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left((nabla_mu g_rhonuK^rho)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left(K_rhonabla_mu K^rho + K_nunabla_mu K^nuright)\
&=-frac K_nunabla_mu K^nuV^2\
&= -frac 1VU_nunabla_muleft(VU^nuright)\
&= -U_nunabla_mu U^nu - frac 1 VU_nu U^nunabla_mu Vendalign

differential-geometry semi-riemannian-geometry

share|cite|improve this question

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

$endgroup$

add a comment | 

$begingroup$

Let $U^mu$ be a vector in 4-dimensional Minkowski space with norm $-1$ and $K^mu = V(x)U^mu$ a vector proportional to it. We can write $V(x) = sqrt-K_nu K^nu$.

(This setup comes from physics where $U^mu$ is a four-velocity, $K^mu$ is a normalized time-like Killing vector for an observer at infinity and $V(x)$ is called the redshift factor.)

Then, define $a^mu$ (the four-acceleration) by $U^sigma nabla_sigma U^mu$, where $nabla$ is the Christoffel connection.

According to Sean Carroll, Spacetime and Geometry, p. 247, $a_mu = nabla_muln V$. Why?

Attempt:

beginalignnabla_muln V &= frac 12V^2nabla_muleft(-K_nu K^nuright)\
&=-frac 12V^2left((nabla_mu K_nu)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left((nabla_mu g_rhonuK^rho)K^nu + K_nunabla_mu K^nuright)\
&=-frac 12V^2left(K_rhonabla_mu K^rho + K_nunabla_mu K^nuright)\
&=-frac K_nunabla_mu K^nuV^2\
&= -frac 1VU_nunabla_muleft(VU^nuright)\
&= -U_nunabla_mu U^nu - frac 1 VU_nu U^nunabla_mu Vendalign

differential-geometry semi-riemannian-geometry

share|cite|improve this question

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

$endgroup$

share|cite|improve this question

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

share|cite|improve this question

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231

asked Mar 21 at 13:36

RodrigoRodrigo

2,9751231