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non-zero divergence of stress energy tensor
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are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
if the divergence of the energy momentum tensor is non-zero, what does that tell us about the spacetime aside from the fact that energy momentum is not conserved?
general-relativity stress-energy-momentum-tensor
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Ben Crowell
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jboy
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up vote
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are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
if the divergence of the energy momentum tensor is non-zero, what does that tell us about the spacetime aside from the fact that energy momentum is not conserved?
general-relativity stress-energy-momentum-tensor
edited 1 hour ago
Ben Crowell
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asked 1 hour ago
jboy
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@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago
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up vote
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up vote
1
down vote
favorite
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
if the divergence of the energy momentum tensor is non-zero, what does that tell us about the spacetime aside from the fact that energy momentum is not conserved?
general-relativity stress-energy-momentum-tensor
edited 1 hour ago
Ben Crowell
44.5k3146271
asked 1 hour ago
jboy
61
New contributor
jboy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
if the divergence of the energy momentum tensor is non-zero, what does that tell us about the spacetime aside from the fact that energy momentum is not conserved?
general-relativity stress-energy-momentum-tensor
general-relativity stress-energy-momentum-tensor
edited 1 hour ago
Ben Crowell
44.5k3146271
asked 1 hour ago
jboy
61
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jboy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Ben Crowell
44.5k3146271
asked 1 hour ago
jboy
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edited 1 hour ago
Ben Crowell
44.5k3146271
edited 1 hour ago
Ben Crowell
44.5k3146271
edited 1 hour ago
Ben Crowell
44.5k3146271
44.5k3146271
asked 1 hour ago
jboy
61
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asked 1 hour ago
jboy
61
asked 1 hour ago
jboy
61
61
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jboy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
jboy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jboy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago
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@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago
@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago
@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago
add a comment |Â
2 Answers
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General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.
answered 1 hour ago
Ben Crowell
44.5k3146271
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
add a comment |Â
up vote
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Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as
$x'^mu = x^mu-epsilon^mu$
If we require the field $phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4times4$ matrix, we get the Stress-Energy Tensor $T^munu$
$T^munu = fracpartialmathcalLpartial(partial_mu phi)partial_nu phi-delta^munumathcalL$
It can be easly shown that, since the $T^munu$ is made up of conserved currents, it has zero divergence:
$partial_nu T^munu = 0$
answered 7 mins ago
Matteo Campagnoli
814
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Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.
answered 1 hour ago
Ben Crowell
44.5k3146271
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
add a comment |Â
up vote
5
down vote
General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.
answered 1 hour ago
Ben Crowell
44.5k3146271
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.
answered 1 hour ago
Ben Crowell
44.5k3146271
General relativity does not allow a stress-energy tensor with a nonvanishing divergence. The Einstein field equations set the Einstein curvature tensor proportional to the stress-energy tensor. The Einstein curvature tensor is constructed in such a way that it is automatically divergence-free.
are there spacetime metrics where $nabla_nuT^munu ne 0$ ?
No. If you write down a metric, the way you find the corresponding stress-energy tensor is to use the Einstein field equations, but the field equations will always give a divergence-free stress-energy tensor.
answered 1 hour ago
Ben Crowell
44.5k3146271
answered 1 hour ago
Ben Crowell
44.5k3146271
answered 1 hour ago
Ben Crowell
44.5k3146271
answered 1 hour ago
Ben Crowell
44.5k3146271
44.5k3146271
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
add a comment |Â
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
The only comment I would add is the fact that there are space-times that "almost" have a violation of stress-energy that is created by assuming a situation that cannot be in equilibrium (such as two static black holes next to each other). These kinds of assumptions result in the various "strut" and "string" singularities in solutions to Einstein equations.
– Void
1 hour ago
add a comment |Â
up vote
1
down vote
Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as
$x'^mu = x^mu-epsilon^mu$
If we require the field $phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4times4$ matrix, we get the Stress-Energy Tensor $T^munu$
$T^munu = fracpartialmathcalLpartial(partial_mu phi)partial_nu phi-delta^munumathcalL$
It can be easly shown that, since the $T^munu$ is made up of conserved currents, it has zero divergence:
$partial_nu T^munu = 0$
answered 7 mins ago
Matteo Campagnoli
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as
$x'^mu = x^mu-epsilon^mu$
If we require the field $phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4times4$ matrix, we get the Stress-Energy Tensor $T^munu$
$T^munu = fracpartialmathcalLpartial(partial_mu phi)partial_nu phi-delta^munumathcalL$
It can be easly shown that, since the $T^munu$ is made up of conserved currents, it has zero divergence:
$partial_nu T^munu = 0$
answered 7 mins ago
Matteo Campagnoli
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as
$x'^mu = x^mu-epsilon^mu$
If we require the field $phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4times4$ matrix, we get the Stress-Energy Tensor $T^munu$
$T^munu = fracpartialmathcalLpartial(partial_mu phi)partial_nu phi-delta^munumathcalL$
It can be easly shown that, since the $T^munu$ is made up of conserved currents, it has zero divergence:
$partial_nu T^munu = 0$
answered 7 mins ago
Matteo Campagnoli
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Stress-Energy Tensor is defined, at least in QFT, as the Noehter current associated with infinitesimal space-time translations. Let me explain, consider an infinitesimal space-time translations such as
$x'^mu = x^mu-epsilon^mu$
If we require the field $phi$ to be invariant under the transformation we've just written down, we can then apply Noether's Theorem. Thus, we say there are four conserved currents, one for each space-time direction. Putting this four currents all together in a big $4times4$ matrix, we get the Stress-Energy Tensor $T^munu$
$T^munu = fracpartialmathcalLpartial(partial_mu phi)partial_nu phi-delta^munumathcalL$
It can be easly shown that, since the $T^munu$ is made up of conserved currents, it has zero divergence:
$partial_nu T^munu = 0$
answered 7 mins ago
Matteo Campagnoli
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 mins ago
Matteo Campagnoli
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 mins ago
Matteo Campagnoli
814
answered 7 mins ago
Matteo Campagnoli
814
814
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matteo Campagnoli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
jboy is a new contributor. Be nice, and check out our Code of Conduct.
jboy is a new contributor. Be nice, and check out our Code of Conduct.
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@BenCrowell Yes. Still the fact that you have to use the field equations in your answer worries me. Maybe you should add that these are classical statements. (But you are right, I misread the question.)
– marmot
1 hour ago