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Tuesday, October 18, 2016

Deriving Dirac's Lagrangian

To derive Dirac's Lagrangian, we begin with the Dirac field's adjoint spinor (bar-psi) and spinor (psi). Note that each spinor contains right-handed (psi-R) and left-handed (psi-L) fields. (Right-handed means the spin and momentum of the field particles are in the same direction. Left-handed means spin and momentum are opposite.)

Let's put the fields into an algebraic form:

Next, we multiply them together:

On the right side of the equal sign the first two terms are each zero and add to zero. Here's why:

The mixed terms don't equal zero. Here's why:

The first two terms that equal zero won't equal zero if we multiply them both by Dirac's matrix.

Now we can set those first two terms equal to the mixed terms:

On the left side, we take the derivative of the fields and multiply by -i, Planck's constant, and the speed of light (c). That gives us kinetic energy. On the right side we multiply by mass (m) and c^2. That gives us potential energy.

The Lagrangian is kinetic energy minus potential energy, so we subtract the potential energy from both sides to get the Lagrange (L):

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