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Does it necessarily follow that the integral curves of $k^a$ are null geodesics?

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5

1

$begingroup$

Let $f$ be a function on a spacetime $(M, g_ab)$ whose gradient, $k_a = nabla_a f$, ie everywhere null, i.e., $k_ak^a = 0$ throughout $M$. Does it necessarily follow that the integral curves of $k^a$ are null geodesics?

differential-geometry manifolds riemannian-geometry mathematical-physics general-relativity

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edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

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5

1

$begingroup$

Let $f$ be a function on a spacetime $(M, g_ab)$ whose gradient, $k_a = nabla_a f$, ie everywhere null, i.e., $k_ak^a = 0$ throughout $M$. Does it necessarily follow that the integral curves of $k^a$ are null geodesics?

differential-geometry manifolds riemannian-geometry mathematical-physics general-relativity

share|cite|improve this question

edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

271317

$endgroup$

add a comment | 

$begingroup$

Let $f$ be a function on a spacetime $(M, g_ab)$ whose gradient, $k_a = nabla_a f$, ie everywhere null, i.e., $k_ak^a = 0$ throughout $M$. Does it necessarily follow that the integral curves of $k^a$ are null geodesics?

differential-geometry manifolds riemannian-geometry mathematical-physics general-relativity

share|cite|improve this question

edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

271317

$endgroup$

share|cite|improve this question

edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

271317

share|cite|improve this question

edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

271317

edited yesterday

Andrews

1,2641420

edited yesterday

Andrews

1,2641420

asked Apr 4 '16 at 3:18

StudentStudent

271317

asked Apr 4 '16 at 3:18

StudentStudent

271317

1 Answer
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3

$begingroup$

The condition that $k$ is a gradient implies that it is curl-free; i.e. $nabla_i k_j = nabla_j k_i$. Let $gamma$ be an integral curve of $k^i$; i.e. $dot gamma^i(t) = k^i(gamma(t))$. Then replacing $dot gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$ddot gamma^i = dot gamma^j nabla_j dot gamma^i = k^j nabla_j k^i = k^j nabla^i k_j.$$

Now we're basically done - just recognize the RHS as $frac12 nabla^i (k^j k_j)$, at which point the fact that $k$ is null ($k^j k_j = 0$) tells us that $ddot gamma = 0$; i.e. $gamma$ is a geodesic.

share|cite|improve this answer

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

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3

$begingroup$

The condition that $k$ is a gradient implies that it is curl-free; i.e. $nabla_i k_j = nabla_j k_i$. Let $gamma$ be an integral curve of $k^i$; i.e. $dot gamma^i(t) = k^i(gamma(t))$. Then replacing $dot gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$ddot gamma^i = dot gamma^j nabla_j dot gamma^i = k^j nabla_j k^i = k^j nabla^i k_j.$$

Now we're basically done - just recognize the RHS as $frac12 nabla^i (k^j k_j)$, at which point the fact that $k$ is null ($k^j k_j = 0$) tells us that $ddot gamma = 0$; i.e. $gamma$ is a geodesic.

share|cite|improve this answer

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

$endgroup$

add a comment | 

3

$begingroup$

The condition that $k$ is a gradient implies that it is curl-free; i.e. $nabla_i k_j = nabla_j k_i$. Let $gamma$ be an integral curve of $k^i$; i.e. $dot gamma^i(t) = k^i(gamma(t))$. Then replacing $dot gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$ddot gamma^i = dot gamma^j nabla_j dot gamma^i = k^j nabla_j k^i = k^j nabla^i k_j.$$

Now we're basically done - just recognize the RHS as $frac12 nabla^i (k^j k_j)$, at which point the fact that $k$ is null ($k^j k_j = 0$) tells us that $ddot gamma = 0$; i.e. $gamma$ is a geodesic.

share|cite|improve this answer

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

$endgroup$

add a comment | 

$begingroup$

The condition that $k$ is a gradient implies that it is curl-free; i.e. $nabla_i k_j = nabla_j k_i$. Let $gamma$ be an integral curve of $k^i$; i.e. $dot gamma^i(t) = k^i(gamma(t))$. Then replacing $dot gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$ddot gamma^i = dot gamma^j nabla_j dot gamma^i = k^j nabla_j k^i = k^j nabla^i k_j.$$

Now we're basically done - just recognize the RHS as $frac12 nabla^i (k^j k_j)$, at which point the fact that $k$ is null ($k^j k_j = 0$) tells us that $ddot gamma = 0$; i.e. $gamma$ is a geodesic.

share|cite|improve this answer

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

$endgroup$

share|cite|improve this answer

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966

answered Apr 4 '16 at 12:13

Anthony CarapetisAnthony Carapetis

27.4k32966