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Poynting theorem

  1. Oct 16, 2012 #1
    I have to admit I don't really understand this theorem fully. As far as I understand it says that the work done on a volume of charge is equal to the change in the field energy inside the volume plus the energy density leaving the boundary. I guess that makes sense but then I did a calculating with an infinite coaxial cable where a current runs down the outer cylinder and comes back along the other. In this case there is a magnetic and electric field between the two cylinders. So you find an expression for the poynting vector which is nonzero which must mean that for every volume there is energy leaving it. I don't really understand this since the situation is completely symmetric in time since the currents are stationary. That doesn't look like energy is leaving any volume at any point.
  2. jcsd
  3. Oct 16, 2012 #2
    Poynting energy theorem can be expressed a number of ways, but is normally shown with LHS as dW/dt - the total rate-of-change of energy occurring within some volume owing to strictly EM causes. There are then four terms on the RHS - see e.g. boxed eq'n at http://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=0CGgQFjAJ&url=http%3A%2F%2Fwww.hep.man.ac.uk%2Fu%2Frmj%2FPHYS30441%2FPoyntings%2520Theorem.pdf&ei=si19UOGzJMyjiAeKsYHYBg&usg=AFQjCNF4l5ZbmzTTTqoZuL1oumuuqBB00Q&cad=rja
    [A more comprehensive set of alternate expressions: [http://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&ved=0CDQQFjAC&url=http%3A%2F%2Fwww.egr.uh.edu%2Fcourses%2Fece%2Fece3317%2FSectionWilton%2FClass%2520Notes%2FNOTES%2Fnotes%252014%25203317.pdf&ei=si19UOGzJMyjiAeKsYHYBg&usg=AFQjCNG9TJvQY83S1GiDiyXweKFuBMbn0w&cad=rja]
    Assuming zero resistivity in coax cable itself, there is an axially invariant radial E field and similarly azimuthal B field acting between inner and outer conductors, so Poynting vector ~ E×B is also axially uniform. This states power is flowing at a steady rate along the cable - power entering any region of cable equals power leaving. Just what we want.
    If finite resistivity is factored in, one has Poynting vector components normal to coax axis - representing waste heat that first enters cable, then leaves via radiation/conduction/convection. This will also be reflected in a corresponding slight decrease in axial Poynting vector power flow from source to load.
    Last edited: Oct 16, 2012
  4. Oct 16, 2012 #3
    But why does it say power in = power out. If you look at a small volume the poynting vector just states that a finite amount of energy is leaving the boundary surface of that volume per time? How do you see that an equal amount is entering?
  5. Oct 16, 2012 #4
    For dissipationless coax, you know that Poynting vector is entirely axial in direction and is uniform in magnitude wrt axial displacement. Create an imaginary boundary surface normal to axis at say position A along axis, and another further on at position B. What do we have? power dW/dt enters boundary surface at A, and same amount exits at surface B. Net change is zero. There is power flow through the enclosed volume, but no accumulation or loss of energywithin the volume. Poynting theorem does not require a net increase or decrease within a given volume - only that there is a definite relationship between the various possible contributors to EM power. All that is consistent with a source/generator connected at one end of coax delivering power at a steady rate to a load/resistor connected to the other end of coax - with no loss in-between.
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