Conserved charge

  • Thread starter kakarukeys
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  • #1
kakarukeys
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Is there an inconsistency between the definition of conserved charge and conserved current in Hamiltonian and Lagrangian formulation?

For example, [tex]H = \int T^{00} d^3x[/tex] is a conserved charge,
[tex]\frac{dH}{dt} = \{H, H\} = 0[/tex]

But we have [tex]\partial_\mu T^{\mu\nu} = 0[/tex] implies
[tex]\int (\partial_\mu T^{\mu 0}) d^3x = \int (\partial_0 T^{00} + \partial_i T^{i0}) d^3x = 0[/tex] so it seems
[tex]\frac{d}{dt}\int T^{00}d^3x = - \int \partial_i T^{i0} d^3x \neq 0[/tex]

I'm very puzzled.
 
Last edited:

Answers and Replies

  • #2
Crosson
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Why is the last integral non-zero? If the charge is constant, shouldn't the current be zero?
 
  • #3
kakarukeys
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So there's a contradiction. In general if there is a boundary, the last integral is not zero.
 
  • #4
dextercioby
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Yes, of course. It's the case on curved manifolds which occur in GR, for example. The boundary terms are very important. However, as it's usually presented in field theory in Minkowski space, the hypersurface integrals are always chosen to be 0.
 
  • #5
kakarukeys
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So is there a condition
[tex]T^{i0} = 0[/tex] at boundary?
 
  • #6
da_willem
599
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If you take a volume in which the charge is conserved that will mean there is no net charge flowing in or out of the boundary, wich is the last condition you mention (integrated over the surface).
 

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