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## Main Question or Discussion Point

Of course, a spin 0 field, by definition, has to be a scalar field under rotations. But what about boosts?

Compare to classical charge density [tex]\rho[/tex], which must satisfy

[tex]\frac{\partial\rho}{\partial t}+\nabla\cdot J=0[/tex]

That is, [tex]\rho[/tex] is the time-component of a four-vector under boosts. But under rotations it is a scalar and in that sense, [tex]\rho[/tex] has "spin 0".

One might say, well the Klein-Gordon equation satisfied by spin-0 fields is a

If your first reaction is to respond, "You don't seem to have a good grasp of what a scalar under boosts means," you're right. Please enlighten me.

Todd

Compare to classical charge density [tex]\rho[/tex], which must satisfy

[tex]\frac{\partial\rho}{\partial t}+\nabla\cdot J=0[/tex]

That is, [tex]\rho[/tex] is the time-component of a four-vector under boosts. But under rotations it is a scalar and in that sense, [tex]\rho[/tex] has "spin 0".

One might say, well the Klein-Gordon equation satisfied by spin-0 fields is a

*scalar*equation. That is, all the operations in the equation transform as a scalars, under rotations and boosts both. True, but that does not mean it has to act on scalars. Scalar operators can act on anything. For instance, each of the four components of the Dirac field independently must satisfy the Klein-Gordon equation.If your first reaction is to respond, "You don't seem to have a good grasp of what a scalar under boosts means," you're right. Please enlighten me.

Todd