1. Feb 6, 2013

### chill_factor

lets say we have a covariant 4-gradient ∂$_{μ}$ = (1/c ∂/∂t, ∇). How do I actually apply this to a function, say x^2 where x is an arbitrary 4-vector (ct,-x,-y,-z)?

2. Feb 6, 2013

### Lavabug

What is x^2? Is it the scalar product of x*x or the contravariant vector ((ct)^2, -x^2, -y^2, -z^2)? If it is the latter, just contract:

$\partial_\mu x^2^\mu$

If it is the former, I suppose you would apply it just like an ordinary 3-gradient on a scalar function.

3. Feb 6, 2013

### chill_factor

that is what is bothering me as well. I assumed it was the scalar product, since it seems to me that you need a scalar function to apply a gradient operator, but in an example problem, the 4-gradient was applied to a 4-vector.

In an example I'm looking at, they tell me ∂$_{μ}$ x$^{μ}$ = 1 where x$^{μ}$ = (ct,-x,-y,-z). How did they get this?

4. Feb 6, 2013

### Lavabug

Assuming your indices are in the right place, that inner product should give -2, not 1.

You can apply a gradient to a vector, all the time in fluid dynamics. It produces a tensor.

If it is instead ∂$_{μ}$ x$_{μ}$, using the minkowski metric with signature +,-,-,-, you could get something that looks like (1,1,1,1). If somehow that "1" you have in your example is supposed to be a vector with just 1's, this is what they did, otherwise I have no clue.

Last edited: Feb 6, 2013
5. Feb 6, 2013

### chill_factor

thank you greatly. i believe i figured it out.

6. Feb 6, 2013

### Lavabug

What was it, out of curiosity?

7. Feb 6, 2013

### WannabeNewton

That is the divergence. The gradient is not being involved at all. $\partial _{\mu }x^{\mu } = \partial _{t}x^{t} + \triangledown \cdot \vec{x}$.

8. Feb 6, 2013

### dextercioby

No, $\partial_{\mu}x^{\mu} = 4$.