# Use of the Kronecker Delta in Translations

A translation in special relativity is, as I understand, a kind of Lorentz transformation given by:

$x^{\mu} \rightarrow \delta^{\mu'}_{\mu}(x^{\mu} + a^{\mu})$ where $\delta^{\mu'}_{\mu} = 1$ if and only if $\mu' = \mu$.

What I fail to understand is the use of the Kronecker delta here. Why would you need to multiply by $1$ if the two coordinates are the same and $0$ if they aren't? Isn't that the whole point of the transformation, i.e., if two coordinate systems are the same, why have a translation? In other words, what is the point of using the Kronecker delta and what purpose does it serve?

• $1\in\mathrm{SO}(1,d-1)$; the identity is just a particular example of a Lorentz matrix. You probably found this as a subcase of a Poincaré transformation $x\to\Lambda x+a$. Commented Oct 8, 2017 at 18:17

The most likely intended meaning of your equation is $x^{\mu'} = \delta^{\mu'}{}_\mu (x^\mu+a^\mu)$. In that case, the $\delta$ is important because $\mu$ and $\mu'$ are different indices, and you need something that can link one with the other. Of course, this is a bit of a technicality, and if you write $x^{\mu'} = x^\mu+a^\mu$ everybody will understand what you mean; however this kind of technicality sometimes helps keeping the indices straight when dealing with tricky expressions.

• To me the clearest is taking the expression off the index, so for example $\tilde{x}^\mu = x^\mu + a^\mu$ or $x'^\mu = x^\mu + a^\mu$. Commented Oct 8, 2017 at 18:42

I have a guess on why this notation is used, but if someone notices it is incorrect, please feel free to leave a comment. You are dealing with Special Relativity. In that setting fortunately you can take advantage of the structure of $\mathbb{R}^4$ to understand things better.

Let $x\in \mathbb{R}^4$ be some event. Consider the translation by $a\in\mathbb{R}^4$ namely $T_a : \mathbb{R}^4\to \mathbb{R}^4$ given by

$$T_a(x)=x+a.$$

Now look in components: you have that

$$T_a(x)^\mu=x^\mu+a^\mu.$$

What is the relation to your formula them? After all, it seems that $x^\mu \mapsto x^\mu + a^\mu$.

That is true. The point is notation. There is a notation which basicaly uses the following convention: let $(x^\mu)$ be a set of coordinates. The coordinates after the transformation are denoted then by $(x^{\mu'})$.

This is usually seem when dealing with the compatibility condition which defines tensors via components: some people write

$$T_{\mu'\nu'}=\dfrac{\partial x^\mu}{\partial x^{\mu'}}\dfrac{\partial x^\nu}{\partial x^{\nu'}}T_{\mu\nu}$$

understanding the $'$ indicates the things after the transformation. In my notation

$$x^{\mu'}=T_a(x)^{\mu'}$$

and then to make everything consistent, and leave the untransformed without $'$ you insert the $\delta^{\mu'}_\mu$.

$\mu$ can take values 0,1,2,3. So, The reason, there is a Kronecker delta is to identify the one to one transformation. Say, you are translating the system in $x^1$, then the change is done only in the $x^1$ component and not others.

It is there so that you don't have a situation when you have,

$x^2 = x^1 + a^1$

This implies, if you shift your system's $x^1$ coordinate along $x^1$ axis by $a^1$, then you get the $x^2$ coordinate of the system. But doesn't make any sense right?

So, the presence of the Kronecker delta makes sure, that your transformation along $x^1$ axis results is a transformation in $x^1$ axis and not something else.

You are entirely correct; the $\delta$ in this circumstance is entirely superfluous.

It was probably chosen to emphasize a common form for all Poincaré transformations, which can be understood as a pair of a translation and a Lorentz transformation. In this particular context it means "Translations are embedded in the Poincaré group as the pair $(T, I)$ where $T$ is the translation and the Lorentz transformation is the identity transformation." In other words, they are explicitly calling out, for ill-defined reasons, that the "Lorentz" part of the transform doesn't do anything but perhaps relabel an index.