Mixing
Stress Energy Tensor in language of differential forms
Clash Royale CLAN TAG#URR8PPP
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The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
differential-geometry metric-tensor tensor-calculus stress-energy-momentum-tensor continuum-mechanics
edited Aug 29 at 2:27
Mike
11.7k13753
asked Aug 28 at 21:39
puzzleshark
333
add a comment |Â
up vote
3
down vote
favorite
The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
differential-geometry metric-tensor tensor-calculus stress-energy-momentum-tensor continuum-mechanics
edited Aug 29 at 2:27
Mike
11.7k13753
asked Aug 28 at 21:39
puzzleshark
333
1
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
differential-geometry metric-tensor tensor-calculus stress-energy-momentum-tensor continuum-mechanics
edited Aug 29 at 2:27
Mike
11.7k13753
asked Aug 28 at 21:39
puzzleshark
333
The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
differential-geometry metric-tensor tensor-calculus stress-energy-momentum-tensor continuum-mechanics
edited Aug 29 at 2:27
Mike
11.7k13753
asked Aug 28 at 21:39
puzzleshark
333
edited Aug 29 at 2:27
Mike
11.7k13753
edited Aug 29 at 2:27
Mike
11.7k13753
edited Aug 29 at 2:27
Mike
11.7k13753
11.7k13753
asked Aug 28 at 21:39
puzzleshark
333
asked Aug 28 at 21:39
puzzleshark
333
asked Aug 28 at 21:39
puzzleshark
333
333
1
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48
add a comment |Â
1
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48
1
1
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48
add a comment |Â
3 Answers
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accepted
If you invite "vector bundle-valued differential forms" then you can define $$ T^mu=T^mu_ nudx^nu $$ as a "vector field-valued 1-form", then you have $$ d^nablastar T^musimnabla_nu T^munu. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^mu=fracdelta S_mdelta A_mu, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
answered Aug 29 at 9:56
Uldreth
3,925417
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
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Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_mu nu := -2 fracdelta mathcalLdelta g^munu + g_mu nu mathcalL,$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
answered Aug 29 at 1:53
tparker
20.8k42112
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
add a comment |Â
up vote
2
down vote
With the help of a Killing vector field $xi$ one can define the current 3-form
$$J_xi = star iota_xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$operatorname d J_xi = 0.$$
Note that the metric is hidden inside both $T$ and $xi$.
answered Aug 28 at 23:56
Phoenix87
7,57911026
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If you invite "vector bundle-valued differential forms" then you can define $$ T^mu=T^mu_ nudx^nu $$ as a "vector field-valued 1-form", then you have $$ d^nablastar T^musimnabla_nu T^munu. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^mu=fracdelta S_mdelta A_mu, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
answered Aug 29 at 9:56
Uldreth
3,925417
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
 |Â
show 4 more comments
up vote
3
down vote
accepted
If you invite "vector bundle-valued differential forms" then you can define $$ T^mu=T^mu_ nudx^nu $$ as a "vector field-valued 1-form", then you have $$ d^nablastar T^musimnabla_nu T^munu. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^mu=fracdelta S_mdelta A_mu, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
answered Aug 29 at 9:56
Uldreth
3,925417
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
 |Â
show 4 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If you invite "vector bundle-valued differential forms" then you can define $$ T^mu=T^mu_ nudx^nu $$ as a "vector field-valued 1-form", then you have $$ d^nablastar T^musimnabla_nu T^munu. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^mu=fracdelta S_mdelta A_mu, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
answered Aug 29 at 9:56
Uldreth
3,925417
If you invite "vector bundle-valued differential forms" then you can define $$ T^mu=T^mu_ nudx^nu $$ as a "vector field-valued 1-form", then you have $$ d^nablastar T^musimnabla_nu T^munu. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^mu=fracdelta S_mdelta A_mu, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the current 3-form $J$, you need to take the Hodge dual $J=star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
answered Aug 29 at 9:56
Uldreth
3,925417
answered Aug 29 at 9:56
Uldreth
3,925417
answered Aug 29 at 9:56
Uldreth
3,925417
answered Aug 29 at 9:56
Uldreth
3,925417
3,925417
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
 |Â
show 4 more comments
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
Isn't the first half of your answer kind of trivial, because you can think of any type-$(p,q+1)$ tensor at all with at least one covariant index as a "type-$(p,q)$-valued one-form" by "picking off" one covariant index?
– tparker
Aug 31 at 4:24
1
1
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
Also, I disagree with the second half of your answer; I think that the three-form $J$ is more "natural" for expressing the current density than the vector $j$. The current density is naturally defined as the flux of electric charge through a spatial surface over time, i.e. you integrate it over a timelike hypersurface to get a total charge. This is natural for a three-form, not a vector. Moreover, the coupling term $A_mu j^mu$ in the Lagrangian can be expressed in terms of $J$ as $star(A wedge J)$. So the continuity equation really is metric-independent.
– tparker
Aug 31 at 4:33
1
1
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
On second thought, you don't even need the Hodge star, because within the framework of differential forms the Lagrangian density is naturally thought of as a $D$-form rather than a scalar.
– tparker
Aug 31 at 4:58
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To answer the first of your inquiries, while you can do this with any tensor field you'd like, and here it is kind of cheating, there are many cases when it is very natural to think of some tensor fields as differential forms with values in a vector bundle. For example, viewing the curvature tensor as $Omega^mu_ nu=frac12R^mu_ nurhosigmadx^rhowedge dx^sigma$ is fairly natural, and considerable simplifications can be gained this way. (contd.)
– Uldreth
Aug 31 at 9:33
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
@tparker To address the second of your inquiries, I guess that depends on your point of view. From a formal/field theoretic point of view, a current is a response to some kind of symmetry. In the case of EM, the charge current is the response to gauge transformations, whilst the SEM tensor is a kind of "diffeomorphism current". The space here is too little for a detailed discussion, however the point is that if you have a Lagrangian that is completely independent of any metrics, and it has a symmetry, the resulting current is metric-independent and so is its conservation law, but (contd.)
– Uldreth
Aug 31 at 9:37
 |Â
show 4 more comments
up vote
3
down vote
Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_mu nu := -2 fracdelta mathcalLdelta g^munu + g_mu nu mathcalL,$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
answered Aug 29 at 1:53
tparker
20.8k42112
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
add a comment |Â
up vote
3
down vote
Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_mu nu := -2 fracdelta mathcalLdelta g^munu + g_mu nu mathcalL,$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
answered Aug 29 at 1:53
tparker
20.8k42112
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_mu nu := -2 fracdelta mathcalLdelta g^munu + g_mu nu mathcalL,$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
answered Aug 29 at 1:53
tparker
20.8k42112
Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_mu nu := -2 fracdelta mathcalLdelta g^munu + g_mu nu mathcalL,$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
answered Aug 29 at 1:53
tparker
20.8k42112
answered Aug 29 at 1:53
tparker
20.8k42112
answered Aug 29 at 1:53
tparker
20.8k42112
answered Aug 29 at 1:53
tparker
20.8k42112
20.8k42112
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
add a comment |Â
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
Are you sure your second term should be there? The standard definition of $T_munu$ is just the derivative of the (matter) Lagrangian with respect to the metric, without the diagonal term. Maybe it is a matter of conventions, and you are using one that I am unfamiliar with.
– AccidentalFourierTransform
Aug 29 at 2:02
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
@AccidentalFourierTransform You can combine the two terms into a single derivative by putting a $sqrtg$ inside the derivative: $-2/sqrtg partial(sqrtg mathcalL) / partial g^munu$. Doing the product rule eliminates the $sqrtg$'s but leads to the two terms in my answer.
– tparker
Aug 29 at 2:04
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Oh, I see. It basically boils down to the redefinition $mathscr L=sqrtgmathcal L$.
– AccidentalFourierTransform
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
Yes, exactly...
– tparker
Aug 29 at 2:07
add a comment |Â
up vote
2
down vote
With the help of a Killing vector field $xi$ one can define the current 3-form
$$J_xi = star iota_xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$operatorname d J_xi = 0.$$
Note that the metric is hidden inside both $T$ and $xi$.
answered Aug 28 at 23:56
Phoenix87
7,57911026
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
add a comment |Â
up vote
2
down vote
With the help of a Killing vector field $xi$ one can define the current 3-form
$$J_xi = star iota_xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$operatorname d J_xi = 0.$$
Note that the metric is hidden inside both $T$ and $xi$.
answered Aug 28 at 23:56
Phoenix87
7,57911026
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
add a comment |Â
up vote
2
down vote
up vote
2
down vote
With the help of a Killing vector field $xi$ one can define the current 3-form
$$J_xi = star iota_xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$operatorname d J_xi = 0.$$
Note that the metric is hidden inside both $T$ and $xi$.
answered Aug 28 at 23:56
Phoenix87
7,57911026
With the help of a Killing vector field $xi$ one can define the current 3-form
$$J_xi = star iota_xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$operatorname d J_xi = 0.$$
Note that the metric is hidden inside both $T$ and $xi$.
answered Aug 28 at 23:56
Phoenix87
7,57911026
answered Aug 28 at 23:56
Phoenix87
7,57911026
answered Aug 28 at 23:56
Phoenix87
7,57911026
answered Aug 28 at 23:56
Phoenix87
7,57911026
7,57911026
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
add a comment |Â
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
Could you explain your notation $iota_xi$?
– tparker
Aug 29 at 1:44
1
1
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
@tparker Interior product.
– AccidentalFourierTransform
Aug 29 at 2:00
1
1
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
Just a minor point of order, but I've only ever seen the interior product defined on forms, and I've never seen $T$ in form-form (sorry) — rather, I'd say that this $iota_xi T$ quantity is more correctly referred to as contraction as physicists usually describe it: the one-form $xi^a T_ab$.
– Mike
Aug 29 at 2:16
2
2
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
Also, I'd sooner emphasize that the metric is (barely) hidden inside the Hodge dual.
– Mike
Aug 29 at 2:17
1
1
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
...and definitely in the very notion of a Killing field.
– Mike
Aug 29 at 2:26
add a comment |Â
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1
Exterior derivatives come for free: you don't need any extra structure to define them. Covariant derivatives, on the other hand, need a connection, which may or may not be metric compatible. If it is not, you don't even need a metric to define it. But if you do have a metric, then the metric-compatible connection is in fact unique, so it is in this sense special.
– AccidentalFourierTransform
Aug 29 at 2:05
@AccidentalFourierTransform Minor correction: The torsion-free metric-compatible connection is unique (Levi-Civita). I think you can have multiple torsive metric-compatible connections. But they are clearly less natural so your point stands.
– tparker
Aug 29 at 2:09
@tparker who cares for torsion anyway? :-P
– AccidentalFourierTransform
Aug 29 at 2:10
Regarding the need for a metric: I guess you might argue that the current vector is the basic object, and then you current three-form involves a Hodge star (i.e. the metric). Then again, you could consider the three-form to be the basic object and apply the Hodge dual the other way round...
– Toffomat
Aug 29 at 10:56
@tparker: Thanks for the correction. I deleted my comment.
– Ben Crowell
Aug 31 at 1:48