Mixing
The degrees of freedom in 3+1 General Relativity
$begingroup$
I would like to understand what are the degrees of freedom in GR. I have read a few previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.
Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...
1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{mu nu}$ solve the EFE, then $g_{mu nu} rightarrow g'_{mu nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?
2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations.
But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if %G_{mu nu} = 0%, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $nabla_t G^{00} + nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $nabla_t G^{00} = 0$, but we can't get any further?
3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does??
4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $nabla_mu T^{mu nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?
Thank you if you managed to read all this and thanks in advance for your responses.
general-relativity gauge-theory
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
$endgroup$
add a comment |
$begingroup$
I would like to understand what are the degrees of freedom in GR. I have read a few previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.
Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...
1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{mu nu}$ solve the EFE, then $g_{mu nu} rightarrow g'_{mu nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?
2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations.
But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if %G_{mu nu} = 0%, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $nabla_t G^{00} + nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $nabla_t G^{00} = 0$, but we can't get any further?
3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does??
4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $nabla_mu T^{mu nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?
Thank you if you managed to read all this and thanks in advance for your responses.
general-relativity gauge-theory
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
$endgroup$
$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34
add a comment |
$begingroup$
I would like to understand what are the degrees of freedom in GR. I have read a few previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.
Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...
1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{mu nu}$ solve the EFE, then $g_{mu nu} rightarrow g'_{mu nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?
2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations.
But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if %G_{mu nu} = 0%, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $nabla_t G^{00} + nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $nabla_t G^{00} = 0$, but we can't get any further?
3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does??
4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $nabla_mu T^{mu nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?
Thank you if you managed to read all this and thanks in advance for your responses.
general-relativity gauge-theory
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
$endgroup$
I would like to understand what are the degrees of freedom in GR. I have read a few previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.
Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...
1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{mu nu}$ solve the EFE, then $g_{mu nu} rightarrow g'_{mu nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?
2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations.
But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if %G_{mu nu} = 0%, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $nabla_t G^{00} + nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $nabla_t G^{00} = 0$, but we can't get any further?
3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does??
4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $nabla_mu T^{mu nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?
Thank you if you managed to read all this and thanks in advance for your responses.
general-relativity gauge-theory
general-relativity gauge-theory
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
asked Oct 17 '18 at 13:10
Patrick.BPatrick.B
11
11
$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34
add a comment |
$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34
$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- Yes, you can play the same game with wave equations: just change coordinates, so you have multiple coordinate descriptions of the same physical situation. This is not unique to relativity: you might like John Norton's article "The Physical Content of General Covariance". As for the metric being "undetermined", well the metric tensor is not, but the coordinate description of it is. This is a central consideration in numerical relativity, where a coordinate gauge is chosen, to avoid undeterminism in the sense you use it here (coordinate components).
2 and 3. As Wikipedia states, the 4 Bianchi identities and 4 coordinate degrees of freedom do not add as $4+4=8$, but are the same dof's:
The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
- When you choose a stress energy tensor (and hence solve for the metric), this constrains the Einstein tensor, it does not add more freedoms. In vacuum with no cosmological constant, the Einstein tensor is zero.
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
- Yes, you can play the same game with wave equations: just change coordinates, so you have multiple coordinate descriptions of the same physical situation. This is not unique to relativity: you might like John Norton's article "The Physical Content of General Covariance". As for the metric being "undetermined", well the metric tensor is not, but the coordinate description of it is. This is a central consideration in numerical relativity, where a coordinate gauge is chosen, to avoid undeterminism in the sense you use it here (coordinate components).
2 and 3. As Wikipedia states, the 4 Bianchi identities and 4 coordinate degrees of freedom do not add as $4+4=8$, but are the same dof's:
The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
- When you choose a stress energy tensor (and hence solve for the metric), this constrains the Einstein tensor, it does not add more freedoms. In vacuum with no cosmological constant, the Einstein tensor is zero.
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
$begingroup$
- Yes, you can play the same game with wave equations: just change coordinates, so you have multiple coordinate descriptions of the same physical situation. This is not unique to relativity: you might like John Norton's article "The Physical Content of General Covariance". As for the metric being "undetermined", well the metric tensor is not, but the coordinate description of it is. This is a central consideration in numerical relativity, where a coordinate gauge is chosen, to avoid undeterminism in the sense you use it here (coordinate components).
2 and 3. As Wikipedia states, the 4 Bianchi identities and 4 coordinate degrees of freedom do not add as $4+4=8$, but are the same dof's:
The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
- When you choose a stress energy tensor (and hence solve for the metric), this constrains the Einstein tensor, it does not add more freedoms. In vacuum with no cosmological constant, the Einstein tensor is zero.
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
$begingroup$
- Yes, you can play the same game with wave equations: just change coordinates, so you have multiple coordinate descriptions of the same physical situation. This is not unique to relativity: you might like John Norton's article "The Physical Content of General Covariance". As for the metric being "undetermined", well the metric tensor is not, but the coordinate description of it is. This is a central consideration in numerical relativity, where a coordinate gauge is chosen, to avoid undeterminism in the sense you use it here (coordinate components).
2 and 3. As Wikipedia states, the 4 Bianchi identities and 4 coordinate degrees of freedom do not add as $4+4=8$, but are the same dof's:
The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
- When you choose a stress energy tensor (and hence solve for the metric), this constrains the Einstein tensor, it does not add more freedoms. In vacuum with no cosmological constant, the Einstein tensor is zero.
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
- Yes, you can play the same game with wave equations: just change coordinates, so you have multiple coordinate descriptions of the same physical situation. This is not unique to relativity: you might like John Norton's article "The Physical Content of General Covariance". As for the metric being "undetermined", well the metric tensor is not, but the coordinate description of it is. This is a central consideration in numerical relativity, where a coordinate gauge is chosen, to avoid undeterminism in the sense you use it here (coordinate components).
2 and 3. As Wikipedia states, the 4 Bianchi identities and 4 coordinate degrees of freedom do not add as $4+4=8$, but are the same dof's:
The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
- When you choose a stress energy tensor (and hence solve for the metric), this constrains the Einstein tensor, it does not add more freedoms. In vacuum with no cosmological constant, the Einstein tensor is zero.
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 10:32
Colin MacLaurinColin MacLaurin
1065
1065
add a comment |
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$begingroup$
This would fit much better on physicsSE.
$endgroup$
– Luke
Oct 17 '18 at 13:19
$begingroup$
partially. To some extend, my question(s) are also revolving about solutions of PDE's and and their well-posedness to some extend. I will ask the question over to the physics guys tough as well.
$endgroup$
– Patrick.B
Oct 18 '18 at 13:34