- #1

- 241

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x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=g[itex]^{\lambda}_{\nu}[/itex]

Shouldn't that have a 4 on the right side?

x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=(4?)g[itex]^{\lambda}_{\nu}[/itex]

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In summary, the Dirac equation, which is represented by x^{\lambda}_{,\mu}x^{\mu}_{,\nu}=g^{\lambda}_{\nu}, does not have a 4 on the right side. When expanded, the equation shows that the first term, for example, is 1 and all off-diagonal terms are 0. This is consistent with the special case of the identity transformation.

- #1

- 241

- 10

x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=g[itex]^{\lambda}_{\nu}[/itex]

Shouldn't that have a 4 on the right side?

x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=(4?)g[itex]^{\lambda}_{\nu}[/itex]

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- #2

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- 201

exmarine said:

x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=g[itex]^{\lambda}_{\nu}[/itex]

Shouldn't that have a 4 on the right side?

x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=(4?)g[itex]^{\lambda}_{\nu}[/itex]

Nope. let me open the expression for you

[itex]\Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\delta ^ \lambda _ \nu = g ^ \lambda _ \nu [/itex]

- #3

- 1,948

- 201

[itex]

\Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\Sigma _ \mu \delta ^ \lambda _ \mu \delta ^ \mu _ \nu =\delta ^ \lambda _ \nu = g ^ \lambda _ \nu [/itex]

- #4

- 241

- 10

[itex]\frac{\partial x^{0}}{\partial x^{0'}}[/itex][itex]\frac{\partial x^{0'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{1'}}[/itex][itex]\frac{\partial x^{1'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{2'}}[/itex][itex]\frac{\partial x^{2'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{3'}}[/itex][itex]\frac{\partial x^{3'}}{\partial x^{0}}[/itex]=4?

And all the off-diagonals be 0 of course.

- #5

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