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How to prove that the covariant derivative obeys the product rule
Covariant derivative and Leibniz ruleArbitrary tensor covariant derivativeProve that a derivative with respect to a covariant 4-vector is a contravariant vector operatorSpin connection and covariant derivativeCovariant Derivative of Basis Vector along another basis vector?Tensor Notation for Derivative and Covariant DerivativeOn covariant derivativeCovariant derivative vs partial derivative (in the context of Killing vectors)General covariance and Maxwell equationsQuestion about Ricci Rotation Coeficients
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$begingroup$
In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
beginequation
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_mualphaA^alpha
endequation
where $Gamma^nu_mualpha$ are Christoffel symbols.
My question: how do we prove it verifies the Leibniz product rule?
general-relativity differential-geometry tensor-calculus differentiation
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
beginequation
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_mualphaA^alpha
endequation
where $Gamma^nu_mualpha$ are Christoffel symbols.
My question: how do we prove it verifies the Leibniz product rule?
general-relativity differential-geometry tensor-calculus differentiation
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
$endgroup$
– DanielC
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
$endgroup$
– G. Smith
8 hours ago
4
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic♦
8 hours ago
2
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
$endgroup$
– octonion
8 hours ago
1
$begingroup$
You can’t start with only a formula for differentiating a contravariant vector.
$endgroup$
– G. Smith
7 hours ago
|
show 3 more comments
$begingroup$
In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
beginequation
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_mualphaA^alpha
endequation
where $Gamma^nu_mualpha$ are Christoffel symbols.
My question: how do we prove it verifies the Leibniz product rule?
general-relativity differential-geometry tensor-calculus differentiation
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
beginequation
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_mualphaA^alpha
endequation
where $Gamma^nu_mualpha$ are Christoffel symbols.
My question: how do we prove it verifies the Leibniz product rule?
general-relativity differential-geometry tensor-calculus differentiation
general-relativity differential-geometry tensor-calculus differentiation
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
edited 6 hours ago
Ben Crowell
58k6 gold badges173 silver badges334 bronze badges
58k6 gold badges173 silver badges334 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
asked 8 hours ago
Mohamed ELarbi GadjaMohamed ELarbi Gadja
62 bronze badges
62 bronze badges
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
$endgroup$
– DanielC
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
$endgroup$
– G. Smith
8 hours ago
4
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic♦
8 hours ago
2
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
$endgroup$
– octonion
8 hours ago
1
$begingroup$
You can’t start with only a formula for differentiating a contravariant vector.
$endgroup$
– G. Smith
7 hours ago
|
show 3 more comments
$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
$endgroup$
– DanielC
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
$endgroup$
– G. Smith
8 hours ago
4
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic♦
8 hours ago
2
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
$endgroup$
– octonion
8 hours ago
1
$begingroup$
You can’t start with only a formula for differentiating a contravariant vector.
$endgroup$
– G. Smith
7 hours ago
$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
$endgroup$
– DanielC
8 hours ago
$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
$endgroup$
– DanielC
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
$endgroup$
– G. Smith
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
$endgroup$
– G. Smith
8 hours ago
4
4
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic♦
8 hours ago
$begingroup$
Would Mathematics be a better home for this question?
$endgroup$
– Qmechanic♦
8 hours ago
2
2
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
$endgroup$
– octonion
8 hours ago
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
$endgroup$
– octonion
8 hours ago
1
1
$begingroup$
You can’t start with only a formula for differentiating a contravariant vector.
$endgroup$
– G. Smith
7 hours ago
$begingroup$
You can’t start with only a formula for differentiating a contravariant vector.
$endgroup$
– G. Smith
7 hours ago
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
On doubly contravariant vectors, for example, we have
$$
nabla_mu T^alphabeta= partial_mu T^alphabeta+ Gamma^alpha_lambda muT^lambdabeta+ Gamma^beta_lambdamu T^alphalambda
$$
and if you take $T^alphabeta=A^alpha B^beta$, you will see how Leibnitz works.
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
You need to define what the Leibniz product rule means for tensors of rank higher than 0.
One has:
$$nabla_V (bf Totimesbf U) := nabla_V bf T otimes bf U +bf Totimesnabla_V bf U $$
for $V$ a vector field, $nabla$ a connection, $bf T,U$ tensor fields.
Choose $V$ the coordinate base vector field, $bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $fracdmathbf vdt$ of some vector field $mathbf v$ tangent to some manifold $M $ in some $mathbb R^d$.
So for example $nabla_mathbf T mathbf v =frac nabla mathbf vdt = fracdmathbf vdt $ minus "the part normal to the manifold". So from this it should be obvious that $fracd(fmathbf v)dt $ satisifes the Leibniz rule
$frac dfdtmathbf v +ffracdmathbf vdt $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
$endgroup$
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
$begingroup$
On doubly contravariant vectors, for example, we have
$$
nabla_mu T^alphabeta= partial_mu T^alphabeta+ Gamma^alpha_lambda muT^lambdabeta+ Gamma^beta_lambdamu T^alphalambda
$$
and if you take $T^alphabeta=A^alpha B^beta$, you will see how Leibnitz works.
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
On doubly contravariant vectors, for example, we have
$$
nabla_mu T^alphabeta= partial_mu T^alphabeta+ Gamma^alpha_lambda muT^lambdabeta+ Gamma^beta_lambdamu T^alphalambda
$$
and if you take $T^alphabeta=A^alpha B^beta$, you will see how Leibnitz works.
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
$endgroup$
add a comment |
$begingroup$
On doubly contravariant vectors, for example, we have
$$
nabla_mu T^alphabeta= partial_mu T^alphabeta+ Gamma^alpha_lambda muT^lambdabeta+ Gamma^beta_lambdamu T^alphalambda
$$
and if you take $T^alphabeta=A^alpha B^beta$, you will see how Leibnitz works.
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
$endgroup$
On doubly contravariant vectors, for example, we have
$$
nabla_mu T^alphabeta= partial_mu T^alphabeta+ Gamma^alpha_lambda muT^lambdabeta+ Gamma^beta_lambdamu T^alphalambda
$$
and if you take $T^alphabeta=A^alpha B^beta$, you will see how Leibnitz works.
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
edited 46 mins ago
DanielC
1,9251 gold badge9 silver badges20 bronze badges
1,9251 gold badge9 silver badges20 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
answered 6 hours ago
mike stonemike stone
8,9221 gold badge13 silver badges28 bronze badges
8,9221 gold badge13 silver badges28 bronze badges
add a comment |
add a comment |
$begingroup$
You need to define what the Leibniz product rule means for tensors of rank higher than 0.
One has:
$$nabla_V (bf Totimesbf U) := nabla_V bf T otimes bf U +bf Totimesnabla_V bf U $$
for $V$ a vector field, $nabla$ a connection, $bf T,U$ tensor fields.
Choose $V$ the coordinate base vector field, $bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
You need to define what the Leibniz product rule means for tensors of rank higher than 0.
One has:
$$nabla_V (bf Totimesbf U) := nabla_V bf T otimes bf U +bf Totimesnabla_V bf U $$
for $V$ a vector field, $nabla$ a connection, $bf T,U$ tensor fields.
Choose $V$ the coordinate base vector field, $bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
You need to define what the Leibniz product rule means for tensors of rank higher than 0.
One has:
$$nabla_V (bf Totimesbf U) := nabla_V bf T otimes bf U +bf Totimesnabla_V bf U $$
for $V$ a vector field, $nabla$ a connection, $bf T,U$ tensor fields.
Choose $V$ the coordinate base vector field, $bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
$endgroup$
You need to define what the Leibniz product rule means for tensors of rank higher than 0.
One has:
$$nabla_V (bf Totimesbf U) := nabla_V bf T otimes bf U +bf Totimesnabla_V bf U $$
for $V$ a vector field, $nabla$ a connection, $bf T,U$ tensor fields.
Choose $V$ the coordinate base vector field, $bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
edited 7 hours ago
AccidentalFourierTransform
26.2k14 gold badges75 silver badges133 bronze badges
26.2k14 gold badges75 silver badges133 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
answered 7 hours ago
DanielCDanielC
1,9251 gold badge9 silver badges20 bronze badges
1,9251 gold badge9 silver badges20 bronze badges
add a comment |
add a comment |
$begingroup$
The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $fracdmathbf vdt$ of some vector field $mathbf v$ tangent to some manifold $M $ in some $mathbb R^d$.
So for example $nabla_mathbf T mathbf v =frac nabla mathbf vdt = fracdmathbf vdt $ minus "the part normal to the manifold". So from this it should be obvious that $fracd(fmathbf v)dt $ satisifes the Leibniz rule
$frac dfdtmathbf v +ffracdmathbf vdt $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
$endgroup$
add a comment |
$begingroup$
The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $fracdmathbf vdt$ of some vector field $mathbf v$ tangent to some manifold $M $ in some $mathbb R^d$.
So for example $nabla_mathbf T mathbf v =frac nabla mathbf vdt = fracdmathbf vdt $ minus "the part normal to the manifold". So from this it should be obvious that $fracd(fmathbf v)dt $ satisifes the Leibniz rule
$frac dfdtmathbf v +ffracdmathbf vdt $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
$endgroup$
add a comment |
$begingroup$
The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $fracdmathbf vdt$ of some vector field $mathbf v$ tangent to some manifold $M $ in some $mathbb R^d$.
So for example $nabla_mathbf T mathbf v =frac nabla mathbf vdt = fracdmathbf vdt $ minus "the part normal to the manifold". So from this it should be obvious that $fracd(fmathbf v)dt $ satisifes the Leibniz rule
$frac dfdtmathbf v +ffracdmathbf vdt $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
$endgroup$
The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $fracdmathbf vdt$ of some vector field $mathbf v$ tangent to some manifold $M $ in some $mathbb R^d$.
So for example $nabla_mathbf T mathbf v =frac nabla mathbf vdt = fracdmathbf vdt $ minus "the part normal to the manifold". So from this it should be obvious that $fracd(fmathbf v)dt $ satisifes the Leibniz rule
$frac dfdtmathbf v +ffracdmathbf vdt $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
edited 4 hours ago
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
answered 4 hours ago
ctsmdctsmd
381 silver badge8 bronze badges
381 silver badge8 bronze badges
add a comment |
add a comment |
Mohamed ELarbi Gadja is a new contributor. Be nice, and check out our Code of Conduct.
Mohamed ELarbi Gadja is a new contributor. Be nice, and check out our Code of Conduct.
Mohamed ELarbi Gadja is a new contributor. Be nice, and check out our Code of Conduct.
Mohamed ELarbi Gadja is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I adjusted your notation from $D_mu$ to $nabla_mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model
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– DanielC
8 hours ago
$begingroup$
You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
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– G. Smith
8 hours ago
4
$begingroup$
Would Mathematics be a better home for this question?
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– Qmechanic♦
8 hours ago
2
$begingroup$
To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said.
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– octonion
8 hours ago
1
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You can’t start with only a formula for differentiating a contravariant vector.
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– G. Smith
7 hours ago