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Why, when going from special to general relativity, do we just replace partial derivatives with covariant derivatives?
Difference between $partial$ and $nabla$ in general relativityMetric tensor in special and general relativityGeneral relativity from helicity 2 massless field theory by using Deser's argumentsProblem in General Relativity (metric tensor covariant derivative / indexes)Motivation for covariant derivative axioms in the context of General RelativityWhat is the motivation from Physics for the Levi-Civita connection on GR?Regarding $T^munu;_mu=0$ in general relativityOn covariant derivativeChristoffel symbol derivation in book by WaldWhen can we raise lower indices on “nontensors” as described in Dirac's book *General Theory of Relativity*?
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
I understand that covariant derivatives become partial derivatives in Minkowski space however is the reverse unique? Is there no other tensor operation which becomes a partial derivative / if so why do we not mention them?
general-relativity special-relativity differential-geometry tensor-calculus differentiation
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
$endgroup$
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
I understand that covariant derivatives become partial derivatives in Minkowski space however is the reverse unique? Is there no other tensor operation which becomes a partial derivative / if so why do we not mention them?
general-relativity special-relativity differential-geometry tensor-calculus differentiation
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
$endgroup$
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
I understand that covariant derivatives become partial derivatives in Minkowski space however is the reverse unique? Is there no other tensor operation which becomes a partial derivative / if so why do we not mention them?
general-relativity special-relativity differential-geometry tensor-calculus differentiation
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
$endgroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
I understand that covariant derivatives become partial derivatives in Minkowski space however is the reverse unique? Is there no other tensor operation which becomes a partial derivative / if so why do we not mention them?
general-relativity special-relativity differential-geometry tensor-calculus differentiation
general-relativity special-relativity differential-geometry tensor-calculus differentiation
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
edited Mar 30 at 15:08
Ben Crowell
55.4k6165318
55.4k6165318
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
asked Mar 30 at 11:36
Toby PeterkenToby Peterken
459216
459216
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Transforming partial derivatives to covariant derivatives when going from Minkowski to a general spacetime is just a rule of thumb, and should not be applied carelessly.
For example, when studying electromagnetism in the Lorenz gauge $(nabla_mu A^mu =0)$, working from first principles, one can show that the inhomogeneous wave equation reads:
$$nabla_nu nabla^nu A^mu - R^mu_,,nu A^nu = -j^mu$$
whereas in Minkowski the same equation reads:
$$partial_nu partial^nu A^mu = -j^mu$$
If we used $partialrightarrownabla$, we would not find the contribution of the curvature term. Although in general the $partialrightarrownabla$ might work, to be safe you should try to derive physical rules using a covariant approach (e.g. from an action principle).
edited Mar 30 at 12:35
DanielC
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
$endgroup$
add a comment |
$begingroup$
You are right that it is not unique. The rule you mention is called minimal coupling. It is similar to electromagnetism when we replace $p_mu$ by $p_mu - eA_mu$ in our first-order equations. This is the simplest approach one could take, in which you just add a term describing, e.g. electromagnetism, to the action, and then it just couples to gravity through the metric in the volume element.
There are other ways of doing so by contracting the Ricci tensor with the field strength tensor, for instance, but these are non-minimal. We make choices like these all the time, even in choosing the form of the connection in the covariant derivative. So the answer in the end is that this minimal approach agrees with experiment to their current accuracies, so why complicate things?
answered Mar 30 at 12:27
gmaroccogmarocco
1466
$endgroup$
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
Maybe but IMHO it's a wrong idea. Covariant derivatives are needed in SR too, if you wish to use arbitrary coordinates. Which is completely allowed even though generally inconvenient. But there are exceptions - see e.g. Rindler's coordinates.
Of course in a curved spacetime you're obliged to use coordinates where metric takes a complicated form, simply because a coordinate system which diagonalizes the metric tensor to constant components in a finite region doesn't exist. Then covariant derivative is an imperative tool.
But there's no warranty that it's a sufficient method to obtain the right physical laws in GR. @DanielC already gave a classical example.
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Transforming partial derivatives to covariant derivatives when going from Minkowski to a general spacetime is just a rule of thumb, and should not be applied carelessly.
For example, when studying electromagnetism in the Lorenz gauge $(nabla_mu A^mu =0)$, working from first principles, one can show that the inhomogeneous wave equation reads:
$$nabla_nu nabla^nu A^mu - R^mu_,,nu A^nu = -j^mu$$
whereas in Minkowski the same equation reads:
$$partial_nu partial^nu A^mu = -j^mu$$
If we used $partialrightarrownabla$, we would not find the contribution of the curvature term. Although in general the $partialrightarrownabla$ might work, to be safe you should try to derive physical rules using a covariant approach (e.g. from an action principle).
edited Mar 30 at 12:35
DanielC
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
$endgroup$
add a comment |
$begingroup$
Transforming partial derivatives to covariant derivatives when going from Minkowski to a general spacetime is just a rule of thumb, and should not be applied carelessly.
For example, when studying electromagnetism in the Lorenz gauge $(nabla_mu A^mu =0)$, working from first principles, one can show that the inhomogeneous wave equation reads:
$$nabla_nu nabla^nu A^mu - R^mu_,,nu A^nu = -j^mu$$
whereas in Minkowski the same equation reads:
$$partial_nu partial^nu A^mu = -j^mu$$
If we used $partialrightarrownabla$, we would not find the contribution of the curvature term. Although in general the $partialrightarrownabla$ might work, to be safe you should try to derive physical rules using a covariant approach (e.g. from an action principle).
edited Mar 30 at 12:35
DanielC
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
$endgroup$
add a comment |
$begingroup$
Transforming partial derivatives to covariant derivatives when going from Minkowski to a general spacetime is just a rule of thumb, and should not be applied carelessly.
For example, when studying electromagnetism in the Lorenz gauge $(nabla_mu A^mu =0)$, working from first principles, one can show that the inhomogeneous wave equation reads:
$$nabla_nu nabla^nu A^mu - R^mu_,,nu A^nu = -j^mu$$
whereas in Minkowski the same equation reads:
$$partial_nu partial^nu A^mu = -j^mu$$
If we used $partialrightarrownabla$, we would not find the contribution of the curvature term. Although in general the $partialrightarrownabla$ might work, to be safe you should try to derive physical rules using a covariant approach (e.g. from an action principle).
edited Mar 30 at 12:35
DanielC
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
$endgroup$
Transforming partial derivatives to covariant derivatives when going from Minkowski to a general spacetime is just a rule of thumb, and should not be applied carelessly.
For example, when studying electromagnetism in the Lorenz gauge $(nabla_mu A^mu =0)$, working from first principles, one can show that the inhomogeneous wave equation reads:
$$nabla_nu nabla^nu A^mu - R^mu_,,nu A^nu = -j^mu$$
whereas in Minkowski the same equation reads:
$$partial_nu partial^nu A^mu = -j^mu$$
If we used $partialrightarrownabla$, we would not find the contribution of the curvature term. Although in general the $partialrightarrownabla$ might work, to be safe you should try to derive physical rules using a covariant approach (e.g. from an action principle).
edited Mar 30 at 12:35
DanielC
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
edited Mar 30 at 12:35
DanielC
1,7301920
edited Mar 30 at 12:35
DanielC
1,7301920
edited Mar 30 at 12:35
DanielC
1,7301920
1,7301920
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
answered Mar 30 at 12:15
Filipe MiguelFilipe Miguel
392213
392213
add a comment |
add a comment |
$begingroup$
You are right that it is not unique. The rule you mention is called minimal coupling. It is similar to electromagnetism when we replace $p_mu$ by $p_mu - eA_mu$ in our first-order equations. This is the simplest approach one could take, in which you just add a term describing, e.g. electromagnetism, to the action, and then it just couples to gravity through the metric in the volume element.
There are other ways of doing so by contracting the Ricci tensor with the field strength tensor, for instance, but these are non-minimal. We make choices like these all the time, even in choosing the form of the connection in the covariant derivative. So the answer in the end is that this minimal approach agrees with experiment to their current accuracies, so why complicate things?
answered Mar 30 at 12:27
gmaroccogmarocco
1466
$endgroup$
add a comment |
$begingroup$
You are right that it is not unique. The rule you mention is called minimal coupling. It is similar to electromagnetism when we replace $p_mu$ by $p_mu - eA_mu$ in our first-order equations. This is the simplest approach one could take, in which you just add a term describing, e.g. electromagnetism, to the action, and then it just couples to gravity through the metric in the volume element.
There are other ways of doing so by contracting the Ricci tensor with the field strength tensor, for instance, but these are non-minimal. We make choices like these all the time, even in choosing the form of the connection in the covariant derivative. So the answer in the end is that this minimal approach agrees with experiment to their current accuracies, so why complicate things?
answered Mar 30 at 12:27
gmaroccogmarocco
1466
$endgroup$
add a comment |
$begingroup$
You are right that it is not unique. The rule you mention is called minimal coupling. It is similar to electromagnetism when we replace $p_mu$ by $p_mu - eA_mu$ in our first-order equations. This is the simplest approach one could take, in which you just add a term describing, e.g. electromagnetism, to the action, and then it just couples to gravity through the metric in the volume element.
There are other ways of doing so by contracting the Ricci tensor with the field strength tensor, for instance, but these are non-minimal. We make choices like these all the time, even in choosing the form of the connection in the covariant derivative. So the answer in the end is that this minimal approach agrees with experiment to their current accuracies, so why complicate things?
answered Mar 30 at 12:27
gmaroccogmarocco
1466
$endgroup$
You are right that it is not unique. The rule you mention is called minimal coupling. It is similar to electromagnetism when we replace $p_mu$ by $p_mu - eA_mu$ in our first-order equations. This is the simplest approach one could take, in which you just add a term describing, e.g. electromagnetism, to the action, and then it just couples to gravity through the metric in the volume element.
There are other ways of doing so by contracting the Ricci tensor with the field strength tensor, for instance, but these are non-minimal. We make choices like these all the time, even in choosing the form of the connection in the covariant derivative. So the answer in the end is that this minimal approach agrees with experiment to their current accuracies, so why complicate things?
answered Mar 30 at 12:27
gmaroccogmarocco
1466
answered Mar 30 at 12:27
gmaroccogmarocco
1466
answered Mar 30 at 12:27
gmaroccogmarocco
1466
answered Mar 30 at 12:27
gmaroccogmarocco
1466
1466
add a comment |
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
Maybe but IMHO it's a wrong idea. Covariant derivatives are needed in SR too, if you wish to use arbitrary coordinates. Which is completely allowed even though generally inconvenient. But there are exceptions - see e.g. Rindler's coordinates.
Of course in a curved spacetime you're obliged to use coordinates where metric takes a complicated form, simply because a coordinate system which diagonalizes the metric tensor to constant components in a finite region doesn't exist. Then covariant derivative is an imperative tool.
But there's no warranty that it's a sufficient method to obtain the right physical laws in GR. @DanielC already gave a classical example.
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
$endgroup$
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
Maybe but IMHO it's a wrong idea. Covariant derivatives are needed in SR too, if you wish to use arbitrary coordinates. Which is completely allowed even though generally inconvenient. But there are exceptions - see e.g. Rindler's coordinates.
Of course in a curved spacetime you're obliged to use coordinates where metric takes a complicated form, simply because a coordinate system which diagonalizes the metric tensor to constant components in a finite region doesn't exist. Then covariant derivative is an imperative tool.
But there's no warranty that it's a sufficient method to obtain the right physical laws in GR. @DanielC already gave a classical example.
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
$endgroup$
add a comment |
$begingroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
Maybe but IMHO it's a wrong idea. Covariant derivatives are needed in SR too, if you wish to use arbitrary coordinates. Which is completely allowed even though generally inconvenient. But there are exceptions - see e.g. Rindler's coordinates.
Of course in a curved spacetime you're obliged to use coordinates where metric takes a complicated form, simply because a coordinate system which diagonalizes the metric tensor to constant components in a finite region doesn't exist. Then covariant derivative is an imperative tool.
But there's no warranty that it's a sufficient method to obtain the right physical laws in GR. @DanielC already gave a classical example.
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
$endgroup$
I've come across several references to the idea that to upgrade a law of physics to general relativity all you have to do is replace any partial derivatives with covariant derivatives.
Maybe but IMHO it's a wrong idea. Covariant derivatives are needed in SR too, if you wish to use arbitrary coordinates. Which is completely allowed even though generally inconvenient. But there are exceptions - see e.g. Rindler's coordinates.
Of course in a curved spacetime you're obliged to use coordinates where metric takes a complicated form, simply because a coordinate system which diagonalizes the metric tensor to constant components in a finite region doesn't exist. Then covariant derivative is an imperative tool.
But there's no warranty that it's a sufficient method to obtain the right physical laws in GR. @DanielC already gave a classical example.
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
answered Mar 30 at 20:44
Elio FabriElio Fabri
3,9861214
3,9861214
add a comment |
add a comment |
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