Using 4-Gradient to Apply Function to 4-Vector

[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)?

 

[Lavabug]

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)?

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.

 

[chill_factor]

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.

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?

 

[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.

 

[chill_factor]

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).

thank you greatly. i believe i figured it out.

 

[Lavabug]

What was it, out of curiosity?

 

[WannabeNewton]

In an example I'm looking at, they tell me ∂$_{μ}$ x$^{μ}$ = 1

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

 

[dextercioby]

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