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

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\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 spatial dimensions.
 
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\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.
 
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
 
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.
 
\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 }.
 
WannabeNewton said:
\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 }.

Oh, right. Thanks everyone.
 
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