# I forgot what this is called and what is is written out.

1. Aug 22, 2011

### LostConjugate

$$\partial_u \partial^u$$

Isn't this like a box symbol? How is it written out again? You can write it out in 2 dimensions x,t no need to include all 3 spacial dimensions.

2. Aug 22, 2011

### WannabeNewton

$\partial _{\mu }\partial ^{\mu } = \square ^{2}$ or $\square$. Both notations are used (with and without the 2). As in, $\square ^{2}A^{\mu } = \partial ^{2}_{t}A^{\mu } - \triangledown ^{2}A^{\mu }$ in natural units.

3. Aug 22, 2011

### G01

It's the D'Alembert Operator, essentially the 3+1 generalization of the Laplacian. Using a space-negative metric:

$$\partial_{\mu}\partial^{\mu}=\Box=\frac{\partial^2}{\partial t^2}-\nabla^2$$

http://en.wikipedia.org/wiki/D'Alembertian

4. Aug 22, 2011

### LostConjugate

I am confused about why it is not written $$\partial_u \partial_u$$

I thought an upper index on a derivative implied that $$\partial x^u$$ is in the numerator and not the denominator.

5. Aug 22, 2011

### WannabeNewton

$\partial _{\mu }\partial _{\mu }$ is simply the second partial derivative with respect to a single component $\mu$ and won't give you the divergence which is $\partial ^{\mu }\partial _{\mu }$ (notice that here you are summing over all the possible values of $\mu$, not just with respect to a single component like the other form). Don't try to write $\partial ^{\mu }$ as you would the right hand side of $\partial _{\mu } = \frac{\partial }{\partial x^{\mu }}$, it won't help. Just think of it as $\partial ^{\mu } = g^{\mu \nu }\partial _{\nu }$.

6. Aug 22, 2011

### LostConjugate

Oh, right. Thanks everyone.