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How to prove that the covariant derivative obeys the product rule

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2

$begingroup$

In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
begin{equation}
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_{mualpha}A^alpha
end{equation}
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

share|cite|improve this question

edited 8 hours ago

Ben Crowell

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asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

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New contributor

Mohamed ELarbi Gadja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
  $endgroup$
  – G. Smith
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Would Mathematics be a better home for this question?
  $endgroup$
  – Qmechanic♦
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can’t start with only a formula for differentiating a contravariant vector.
  $endgroup$
  – G. Smith
  9 hours ago
  

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
begin{equation}
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_{mualpha}A^alpha
end{equation}
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

share|cite|improve this question

edited 8 hours ago

Ben Crowell

58k66 gold badges173173 silver badges334334 bronze badges

asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what?
  $endgroup$
  – G. Smith
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Would Mathematics be a better home for this question?
  $endgroup$
  – Qmechanic♦
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can’t start with only a formula for differentiating a contravariant vector.
  $endgroup$
  – G. Smith
  9 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

In General Relativity the covariant derivative of contravariant vectors $A^mu$ is:
begin{equation}
nabla_mu A^nu=partial_mu A^nu+Gamma^nu_{mualpha}A^alpha
end{equation}
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

share|cite|improve this question

edited 8 hours ago

Ben Crowell

58k66 gold badges173173 silver badges334334 bronze badges

asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 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.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Ben Crowell

58k66 gold badges173173 silver badges334334 bronze badges

asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 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.

share|cite|improve this question

edited 8 hours ago

Ben Crowell

58k66 gold badges173173 silver badges334334 bronze badges

asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 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 8 hours ago

Ben Crowell

58k66 gold badges173173 silver badges334334 bronze badges

edited 8 hours ago

Ben Crowell

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asked 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 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 10 hours ago

Mohamed ELarbi GadjaMohamed ELarbi Gadja

1122 bronze badges

3 Answers
3

active

oldest

votes

5

$begingroup$

On doubly contravariant vectors, for example, we have
$$
nabla_mu T^{alphabeta}= partial_mu T^{alphabeta}+ Gamma^alpha_{lambda mu}T^{lambdabeta}+ Gamma^beta_{lambdamu} T^{alphalambda}
$$
and if you take $T^{alphabeta}=A^alpha B^beta$, you will see how Leibnitz works.

share|cite|improve this answer

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

You need to define what the Leibniz product rule means for tensors of rank higher than 0.

One has:

$$nabla_V ({bf T}otimes{bf U}) := nabla_V {bf T} otimes {bf U} +{bf T}otimesnabla_V {bf U} $$

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.

share|cite|improve this answer

edited 9 hours ago

AccidentalFourierTransform

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answered 10 hours ago

DanielCDanielC

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$endgroup$

add a comment | 

0

$begingroup$

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 $frac{dmathbf v}{dt}$ 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 v}{dt} = frac{dmathbf v}{dt} $ minus "the part normal to the manifold". So from this it should be obvious that $frac{d(fmathbf v)}{dt} $ satisifes the Leibniz rule

$frac {df}{dt}mathbf v +ffrac{dmathbf v}{dt} $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

On doubly contravariant vectors, for example, we have
$$
nabla_mu T^{alphabeta}= partial_mu T^{alphabeta}+ Gamma^alpha_{lambda mu}T^{lambdabeta}+ Gamma^beta_{lambdamu} T^{alphalambda}
$$
and if you take $T^{alphabeta}=A^alpha B^beta$, you will see how Leibnitz works.

share|cite|improve this answer

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

On doubly contravariant vectors, for example, we have
$$
nabla_mu T^{alphabeta}= partial_mu T^{alphabeta}+ Gamma^alpha_{lambda mu}T^{lambdabeta}+ Gamma^beta_{lambdamu} T^{alphalambda}
$$
and if you take $T^{alphabeta}=A^alpha B^beta$, you will see how Leibnitz works.

share|cite|improve this answer

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

$endgroup$

add a comment | 

$begingroup$

On doubly contravariant vectors, for example, we have
$$
nabla_mu T^{alphabeta}= partial_mu T^{alphabeta}+ Gamma^alpha_{lambda mu}T^{lambdabeta}+ Gamma^beta_{lambdamu} T^{alphalambda}
$$
and if you take $T^{alphabeta}=A^alpha B^beta$, you will see how Leibnitz works.

share|cite|improve this answer

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

$endgroup$

share|cite|improve this answer

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

edited 3 hours ago

DanielC

1,93511 gold badge99 silver badges2020 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

answered 8 hours ago

mike stonemike stone

8,93211 gold badge1313 silver badges2828 bronze badges

3

$begingroup$

You need to define what the Leibniz product rule means for tensors of rank higher than 0.

One has:

$$nabla_V ({bf T}otimes{bf U}) := nabla_V {bf T} otimes {bf U} +{bf T}otimesnabla_V {bf U} $$

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.

share|cite|improve this answer

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

answered 10 hours ago

DanielCDanielC

1,93511 gold badge99 silver badges2020 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

You need to define what the Leibniz product rule means for tensors of rank higher than 0.

One has:

$$nabla_V ({bf T}otimes{bf U}) := nabla_V {bf T} otimes {bf U} +{bf T}otimesnabla_V {bf U} $$

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.

share|cite|improve this answer

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

answered 10 hours ago

DanielCDanielC

1,93511 gold badge99 silver badges2020 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:

$$nabla_V ({bf T}otimes{bf U}) := nabla_V {bf T} otimes {bf U} +{bf T}otimesnabla_V {bf U} $$

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.

share|cite|improve this answer

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

answered 10 hours ago

DanielCDanielC

1,93511 gold badge99 silver badges2020 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

answered 10 hours ago

DanielCDanielC

1,93511 gold badge99 silver badges2020 bronze badges

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

edited 9 hours ago

AccidentalFourierTransform

26.2k1414 gold badges7575 silver badges133133 bronze badges

answered 10 hours ago

DanielCDanielC

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answered 10 hours ago

DanielCDanielC

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0

$begingroup$

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 $frac{dmathbf v}{dt}$ 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 v}{dt} = frac{dmathbf v}{dt} $ minus "the part normal to the manifold". So from this it should be obvious that $frac{d(fmathbf v)}{dt} $ satisifes the Leibniz rule

$frac {df}{dt}mathbf v +ffrac{dmathbf v}{dt} $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

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 $frac{dmathbf v}{dt}$ 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 v}{dt} = frac{dmathbf v}{dt} $ minus "the part normal to the manifold". So from this it should be obvious that $frac{d(fmathbf v)}{dt} $ satisifes the Leibniz rule

$frac {df}{dt}mathbf v +ffrac{dmathbf v}{dt} $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 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 $frac{dmathbf v}{dt}$ 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 v}{dt} = frac{dmathbf v}{dt} $ minus "the part normal to the manifold". So from this it should be obvious that $frac{d(fmathbf v)}{dt} $ satisifes the Leibniz rule

$frac {df}{dt}mathbf v +ffrac{dmathbf v}{dt} $ minus "the part normal to the manifold" $ = nabla_mathbf T (fmathbf v) = mathbf T(f)mathbf v + f nabla_mathbf T mathbf v $

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges

answered 6 hours ago

ctsmdctsmd

3811 silver badge88 bronze badges