Boundedness of general relativity Hamiltonian

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When one consider a lagrangian and construct hamiltonian, we expect to be bounded below.
While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.



  1. How this can be shown?


  2. Is it related to the positive energy theorem?




general-relativity energy hamiltonian-formalism




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migrated from math.stackexchange.com Jul 20 at 11:05


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  • What Lagranian and Hamiltonian are you considering?
    – md2perpe
    Jul 19 at 10:19










  • The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
    – anubis
    Jul 19 at 11:14














up vote
2
down vote

favorite












When one consider a lagrangian and construct hamiltonian, we expect to be bounded below.
While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.



  1. How this can be shown?


  2. Is it related to the positive energy theorem?




general-relativity energy hamiltonian-formalism




share|cite|improve this question













migrated from math.stackexchange.com Jul 20 at 11:05


This question came from our site for people studying math at any level and professionals in related fields.














  • What Lagranian and Hamiltonian are you considering?
    – md2perpe
    Jul 19 at 10:19










  • The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
    – anubis
    Jul 19 at 11:14












up vote
2
down vote

favorite









up vote
2
down vote

favorite











When one consider a lagrangian and construct hamiltonian, we expect to be bounded below.
While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.



  1. How this can be shown?


  2. Is it related to the positive energy theorem?




general-relativity energy hamiltonian-formalism




share|cite|improve this question













When one consider a lagrangian and construct hamiltonian, we expect to be bounded below.
While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.



  1. How this can be shown?


  2. Is it related to the positive energy theorem?





general-relativity energy hamiltonian-formalism





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 12:07









Qmechanic♦

95.8k121621004




95.8k121621004









asked Jul 19 at 9:43









anubis

756




756




migrated from math.stackexchange.com Jul 20 at 11:05


This question came from our site for people studying math at any level and professionals in related fields.






migrated from math.stackexchange.com Jul 20 at 11:05


This question came from our site for people studying math at any level and professionals in related fields.













  • What Lagranian and Hamiltonian are you considering?
    – md2perpe
    Jul 19 at 10:19










  • The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
    – anubis
    Jul 19 at 11:14
















  • What Lagranian and Hamiltonian are you considering?
    – md2perpe
    Jul 19 at 10:19










  • The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
    – anubis
    Jul 19 at 11:14















What Lagranian and Hamiltonian are you considering?
– md2perpe
Jul 19 at 10:19




What Lagranian and Hamiltonian are you considering?
– md2perpe
Jul 19 at 10:19












The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
– anubis
Jul 19 at 11:14




The Lagrangian of general relativity $L=sqrt-gR$ and the ADM hamiltonian
– anubis
Jul 19 at 11:14










1 Answer
1






active

oldest

votes

















up vote
2
down vote













This is exactly the content of the positive energy theorem:



Let $(Sigma, h_ab, K_ab)$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_ADM geq sqrtP_i P_i$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.



The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".






share|cite|improve this answer





















  • Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
    – anubis
    Jul 20 at 15:39










  • If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
    – John Donne
    Jul 20 at 21:03










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













This is exactly the content of the positive energy theorem:



Let $(Sigma, h_ab, K_ab)$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_ADM geq sqrtP_i P_i$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.



The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".






share|cite|improve this answer





















  • Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
    – anubis
    Jul 20 at 15:39










  • If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
    – John Donne
    Jul 20 at 21:03














up vote
2
down vote













This is exactly the content of the positive energy theorem:



Let $(Sigma, h_ab, K_ab)$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_ADM geq sqrtP_i P_i$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.



The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".






share|cite|improve this answer





















  • Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
    – anubis
    Jul 20 at 15:39










  • If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
    – John Donne
    Jul 20 at 21:03












up vote
2
down vote










up vote
2
down vote









This is exactly the content of the positive energy theorem:



Let $(Sigma, h_ab, K_ab)$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_ADM geq sqrtP_i P_i$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.



The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".






share|cite|improve this answer













This is exactly the content of the positive energy theorem:



Let $(Sigma, h_ab, K_ab)$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_ADM geq sqrtP_i P_i$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.



The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 20 at 11:12









John Donne

2,4531724




2,4531724











  • Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
    – anubis
    Jul 20 at 15:39










  • If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
    – John Donne
    Jul 20 at 21:03
















  • Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
    – anubis
    Jul 20 at 15:39










  • If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
    – John Donne
    Jul 20 at 21:03















Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
– anubis
Jul 20 at 15:39




Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof
– anubis
Jul 20 at 15:39












If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
– John Donne
Jul 20 at 21:03




If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild)
– John Donne
Jul 20 at 21:03












 

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