# Tensor equation in Dirac's 1975 book

• exmarine
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.

#### 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$^{\lambda}_{\nu}$

exmarine said:
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

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

Let me open it some more

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

? 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.

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