First of all, I would like to say that I am somewhat new to four-vector notation.

I have a function of a four-vector that I want to expand.

$$ A_\mu (\mathbf{x} + \mathbf{x}_0) = A_\mu (\mathbf{x}) + \mathbf{x}_0 \frac{\partial A_\mu (\mathbf{x})}{\partial \mathbf{x}} + \dots $$

where both $\mathbf{x}$ and $\mathbf{x}_0$ are four-vectors so that e.g. $\mathbf{x} = (ct,x,y,z)$

My question is then what this actually looks like? The second term, should I have an index on the derivative? Is the derivative term a sum of several terms?

Could anyone help me write this out more explicitly, so that I am sure that I don't leave any terms behind?


Consider a scalar field, $\phi$. If we make the active transformation, $x^\mu \to x^\mu-\epsilon^\mu$ the change in the scalar field, infinitesimally, is given by,

$$\phi \to \phi + \epsilon^\mu\partial_\mu \phi + \mathcal{O}(\epsilon^2)$$

In more explicit notation, we may write,

$$\phi\to \phi + \epsilon^0 \frac{\partial \phi}{\partial t}+\epsilon^1 \frac{\partial \phi}{\partial x}+\epsilon^2 \frac{\partial \phi}{\partial y}+\epsilon^3 \frac{\partial \phi}{\partial z}$$

Note $\epsilon^\mu$ is simply the magnitude of the translation in the $\mu$th direction. For a $1$-form such as $A_\nu$, we simply pick up an extra index on the field in our expansion. The above is simply a higher dimensional analogue of the usual Taylor expansion. To see this clearly, note,

$$f(x+h)= f(x)+hf'(x)+\frac{1}{2}h^2f''(x)+\mathcal{O}(h^3)$$

In quantum field theory, a field takes the place of $f$, and $h$ becomes a vector, in which case we pick up derivatives, as shown, with respect to each coordinate.

  • $\begingroup$ It is long ago for me, but shouldn't the time derivative and space derivatives have different signs? $\endgroup$ – M.Herzkamp Jun 19 '14 at 11:06
  • $\begingroup$ @M.Herzkamp: No, I don't think so, because as I stated the translation is ACTIVE as opposed to PASSIVE. $\endgroup$ – JamalS Jun 19 '14 at 11:31
  • $\begingroup$ @M.Herzkamp you can see this from the fact that $\partial_\mu=\frac{\partial}{\partial x^\mu}$, which clearly doesn't involve any opposite signs ;) $\endgroup$ – Danu Jun 19 '14 at 12:46

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