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, $\partial_{\mu}x^{\mu} = 4$.