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The equation of motion for a scalar field in curved spacetime in terms of the covariant derivative


Scalar field lagrangian in curved spacetimeScalar field lagrangian in curved spacetimeMaxwell's equation in curved spacetime - how come? And experimental evidence?What are the equations of motion for the scalar field in the tetrad formalism?A question about covariant derivative on spinor field (Vierbein formalism)Scalar field in curved spacesEoM for scalar field in Brans-Dicke TheoryQuestion on energy conservation from the stress tensor of a classical scalar field






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1












$begingroup$


The equation of motion for a scalar field in curved spacetime $$fracpartialmathcalLpartialphi=frac1sqrt-gpartial_muleft[sqrt-gfracpartialmathcalLpartialleft(partial_muphiright)right]tag1$$ can be written in terms of the covariant derivative as $$fracpartialmathcalLpartialphi=nabla_muleft[fracpartialmathcalLpartialleft(partial_muphiright)right].tag2$$



Here $mathcalL$ is a Lagrangian scalar function.
How is Eq.$(2)$ obtained from Eq.$(1)$? The action of a covariant on a vector field $A_mu$ is given by $$nabla_mu A_nu=partial_mu A_nu-Gamma_munu^rhoA_rho.$$










share|cite|improve this question











$endgroup$













  • $begingroup$
    source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
    $endgroup$
    – AccidentalFourierTransform
    7 hours ago










  • $begingroup$
    @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
    $endgroup$
    – mithusengupta123
    7 hours ago


















1












$begingroup$


The equation of motion for a scalar field in curved spacetime $$fracpartialmathcalLpartialphi=frac1sqrt-gpartial_muleft[sqrt-gfracpartialmathcalLpartialleft(partial_muphiright)right]tag1$$ can be written in terms of the covariant derivative as $$fracpartialmathcalLpartialphi=nabla_muleft[fracpartialmathcalLpartialleft(partial_muphiright)right].tag2$$



Here $mathcalL$ is a Lagrangian scalar function.
How is Eq.$(2)$ obtained from Eq.$(1)$? The action of a covariant on a vector field $A_mu$ is given by $$nabla_mu A_nu=partial_mu A_nu-Gamma_munu^rhoA_rho.$$










share|cite|improve this question











$endgroup$













  • $begingroup$
    source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
    $endgroup$
    – AccidentalFourierTransform
    7 hours ago










  • $begingroup$
    @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
    $endgroup$
    – mithusengupta123
    7 hours ago














1












1








1





$begingroup$


The equation of motion for a scalar field in curved spacetime $$fracpartialmathcalLpartialphi=frac1sqrt-gpartial_muleft[sqrt-gfracpartialmathcalLpartialleft(partial_muphiright)right]tag1$$ can be written in terms of the covariant derivative as $$fracpartialmathcalLpartialphi=nabla_muleft[fracpartialmathcalLpartialleft(partial_muphiright)right].tag2$$



Here $mathcalL$ is a Lagrangian scalar function.
How is Eq.$(2)$ obtained from Eq.$(1)$? The action of a covariant on a vector field $A_mu$ is given by $$nabla_mu A_nu=partial_mu A_nu-Gamma_munu^rhoA_rho.$$










share|cite|improve this question











$endgroup$




The equation of motion for a scalar field in curved spacetime $$fracpartialmathcalLpartialphi=frac1sqrt-gpartial_muleft[sqrt-gfracpartialmathcalLpartialleft(partial_muphiright)right]tag1$$ can be written in terms of the covariant derivative as $$fracpartialmathcalLpartialphi=nabla_muleft[fracpartialmathcalLpartialleft(partial_muphiright)right].tag2$$



Here $mathcalL$ is a Lagrangian scalar function.
How is Eq.$(2)$ obtained from Eq.$(1)$? The action of a covariant on a vector field $A_mu$ is given by $$nabla_mu A_nu=partial_mu A_nu-Gamma_munu^rhoA_rho.$$







general-relativity lagrangian-formalism differential-geometry field-theory differentiation






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edited 5 hours ago









Qmechanic

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asked 8 hours ago









mithusengupta123mithusengupta123

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  • $begingroup$
    source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
    $endgroup$
    – AccidentalFourierTransform
    7 hours ago










  • $begingroup$
    @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
    $endgroup$
    – mithusengupta123
    7 hours ago

















  • $begingroup$
    source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
    $endgroup$
    – AccidentalFourierTransform
    7 hours ago










  • $begingroup$
    @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
    $endgroup$
    – mithusengupta123
    7 hours ago
















$begingroup$
source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
$endgroup$
– AccidentalFourierTransform
7 hours ago




$begingroup$
source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks.
$endgroup$
– AccidentalFourierTransform
7 hours ago












$begingroup$
@AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
$endgroup$
– mithusengupta123
7 hours ago





$begingroup$
@AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1.
$endgroup$
– mithusengupta123
7 hours ago











1 Answer
1






active

oldest

votes


















2












$begingroup$

I'll give hints instead of a complete solution, in case this ends up being categorized as homework.



I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
$$
V^mu := fracpartial cal Lpartial(partial_muphi)
tag1
$$

is a vector (I mean, the components of a vector), so the question is how to derive
$$
frac1sqrtpartial_muleft(sqrt V^muright)
=nabla_mu V^mu
tag2
$$

for a vector $V$. Use the definition of $nabla_mu$ to see that (2) can also be written
$$
frac1sqrt V^mupartial_musqrt
= Gamma^alpha_alphamuV^mu.
tag3
$$

The left-hand side of (3) involves the quantity
$$
frac1sqrtpartial_musqrt
=
frac12partial_mulog|g|.
tag4
$$

We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $lambda_n$. Rewrite the right-hand side of (4) in terms of $lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^-1$ and $partial_mu g$. Written in terms of indices, this becomes
$$
frac1sqrtpartial_musqrt
=frac12g^alphabetapartial_mu g_alphabeta.
tag5
$$

We can use this to rewrite the left-hand side of (3). Now write the $Gamma^alpha_alphamu$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.



Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I'll give hints instead of a complete solution, in case this ends up being categorized as homework.



    I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
    $$
    V^mu := fracpartial cal Lpartial(partial_muphi)
    tag1
    $$

    is a vector (I mean, the components of a vector), so the question is how to derive
    $$
    frac1sqrtpartial_muleft(sqrt V^muright)
    =nabla_mu V^mu
    tag2
    $$

    for a vector $V$. Use the definition of $nabla_mu$ to see that (2) can also be written
    $$
    frac1sqrt V^mupartial_musqrt
    = Gamma^alpha_alphamuV^mu.
    tag3
    $$

    The left-hand side of (3) involves the quantity
    $$
    frac1sqrtpartial_musqrt
    =
    frac12partial_mulog|g|.
    tag4
    $$

    We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $lambda_n$. Rewrite the right-hand side of (4) in terms of $lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^-1$ and $partial_mu g$. Written in terms of indices, this becomes
    $$
    frac1sqrtpartial_musqrt
    =frac12g^alphabetapartial_mu g_alphabeta.
    tag5
    $$

    We can use this to rewrite the left-hand side of (3). Now write the $Gamma^alpha_alphamu$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.



    Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.






    share|cite|improve this answer









    $endgroup$



















      2












      $begingroup$

      I'll give hints instead of a complete solution, in case this ends up being categorized as homework.



      I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
      $$
      V^mu := fracpartial cal Lpartial(partial_muphi)
      tag1
      $$

      is a vector (I mean, the components of a vector), so the question is how to derive
      $$
      frac1sqrtpartial_muleft(sqrt V^muright)
      =nabla_mu V^mu
      tag2
      $$

      for a vector $V$. Use the definition of $nabla_mu$ to see that (2) can also be written
      $$
      frac1sqrt V^mupartial_musqrt
      = Gamma^alpha_alphamuV^mu.
      tag3
      $$

      The left-hand side of (3) involves the quantity
      $$
      frac1sqrtpartial_musqrt
      =
      frac12partial_mulog|g|.
      tag4
      $$

      We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $lambda_n$. Rewrite the right-hand side of (4) in terms of $lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^-1$ and $partial_mu g$. Written in terms of indices, this becomes
      $$
      frac1sqrtpartial_musqrt
      =frac12g^alphabetapartial_mu g_alphabeta.
      tag5
      $$

      We can use this to rewrite the left-hand side of (3). Now write the $Gamma^alpha_alphamu$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.



      Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.






      share|cite|improve this answer









      $endgroup$

















        2












        2








        2





        $begingroup$

        I'll give hints instead of a complete solution, in case this ends up being categorized as homework.



        I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
        $$
        V^mu := fracpartial cal Lpartial(partial_muphi)
        tag1
        $$

        is a vector (I mean, the components of a vector), so the question is how to derive
        $$
        frac1sqrtpartial_muleft(sqrt V^muright)
        =nabla_mu V^mu
        tag2
        $$

        for a vector $V$. Use the definition of $nabla_mu$ to see that (2) can also be written
        $$
        frac1sqrt V^mupartial_musqrt
        = Gamma^alpha_alphamuV^mu.
        tag3
        $$

        The left-hand side of (3) involves the quantity
        $$
        frac1sqrtpartial_musqrt
        =
        frac12partial_mulog|g|.
        tag4
        $$

        We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $lambda_n$. Rewrite the right-hand side of (4) in terms of $lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^-1$ and $partial_mu g$. Written in terms of indices, this becomes
        $$
        frac1sqrtpartial_musqrt
        =frac12g^alphabetapartial_mu g_alphabeta.
        tag5
        $$

        We can use this to rewrite the left-hand side of (3). Now write the $Gamma^alpha_alphamu$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.



        Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.






        share|cite|improve this answer









        $endgroup$



        I'll give hints instead of a complete solution, in case this ends up being categorized as homework.



        I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity
        $$
        V^mu := fracpartial cal Lpartial(partial_muphi)
        tag1
        $$

        is a vector (I mean, the components of a vector), so the question is how to derive
        $$
        frac1sqrtpartial_muleft(sqrt V^muright)
        =nabla_mu V^mu
        tag2
        $$

        for a vector $V$. Use the definition of $nabla_mu$ to see that (2) can also be written
        $$
        frac1sqrt V^mupartial_musqrt
        = Gamma^alpha_alphamuV^mu.
        tag3
        $$

        The left-hand side of (3) involves the quantity
        $$
        frac1sqrtpartial_musqrt
        =
        frac12partial_mulog|g|.
        tag4
        $$

        We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $lambda_n$. Rewrite the right-hand side of (4) in terms of $lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^-1$ and $partial_mu g$. Written in terms of indices, this becomes
        $$
        frac1sqrtpartial_musqrt
        =frac12g^alphabetapartial_mu g_alphabeta.
        tag5
        $$

        We can use this to rewrite the left-hand side of (3). Now write the $Gamma^alpha_alphamu$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.



        Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










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