Tensor equation in Dirac's 1975 book

[exmarine]

Dirac has equation 3.4 as:

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]

 

[dauto]

Dirac has equation 3.4 as:

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]

 

[dauto]

Let me open it some more

[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]

 

[exmarine]

? Wouldn't the first term, for example, be:

[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.

 

[Bill_K]

Consider the special case of the identity transformation, where all the new coordinates are the same as the old ones. That is, x^0' = x⁰, x^1' = x¹, x^2' = x², and x^3' = x³. Plug this into your equation, and I think you'll see that the result is 1, not 4.