Tensor equation in Dirac's 1975 book

[exmarine]

Dirac has equation 3.4 as:

x$^{\lambda}_{,\mu}$x$^{\mu}_{,\nu}$=g$^{\lambda}_{\nu}$

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

x$^{\lambda}_{,\mu}$x$^{\mu}_{,\nu}$=(4?)g$^{\lambda}_{\nu}$

 

[dauto]

Dirac has equation 3.4 as:

x$^{\lambda}_{,\mu}$x$^{\mu}_{,\nu}$=g$^{\lambda}_{\nu}$

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

x$^{\lambda}_{,\mu}$x$^{\mu}_{,\nu}$=(4?)g$^{\lambda}_{\nu}$

Nope. let me open the expression for you


$\Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\delta ^ \lambda _ \nu = g ^ \lambda _ \nu$

 

[dauto]

Let me open it some more

$\Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\Sigma _ \mu \delta ^ \lambda _ \mu \delta ^ \mu _ \nu =\delta ^ \lambda _ \nu = g ^ \lambda _ \nu$

 

[exmarine]

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

$\frac{\partial x^{0}}{\partial x^{0'}}$$\frac{\partial x^{0'}}{\partial x^{0}}$+$\frac{\partial x^{0}}{\partial x^{1'}}$$\frac{\partial x^{1'}}{\partial x^{0}}$+$\frac{\partial x^{0}}{\partial x^{2'}}$$\frac{\partial x^{2'}}{\partial x^{0}}$+$\frac{\partial x^{0}}{\partial x^{3'}}$$\frac{\partial x^{3'}}{\partial x^{0}}$=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.