How do I differentiate this Scalar Field?

Click For Summary

Homework Help Overview

The discussion revolves around differentiating a scalar field within the context of general relativity, specifically focusing on the computation of covariant derivatives and the relationship between the Christoffel symbols and the scalar field.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of covariant derivatives and their application to scalar fields versus vector fields. There is an attempt to clarify the distinction between partial derivatives and covariant derivatives in this context.

Discussion Status

Some participants have provided insights into the correct interpretation of the second covariant derivative and its application to scalar fields. There is an ongoing exploration of the implications of these derivatives and the role of Christoffel symbols.

Contextual Notes

Participants are questioning the assumptions regarding the differentiation of scalar fields and the definitions of covariant derivatives, indicating a need for clarity on the underlying mathematical framework.

unscientific
Messages
1,728
Reaction score
13

Homework Statement



(a) Find the christoffel symbols (Done).
(b) Show that ##\phi## is a solution and find the relation between A and B.[/B]

2013_B5_Q2.png

Homework Equations

The Attempt at a Solution


Part(b)
\nabla_\mu \nabla^\mu \phi = 0
I suppose for a scalar field, this is simply the normal derivative:
\frac{\partial^2 \phi}{\partial \eta^2} + \frac{\partial^2 \phi}{\partial r^2} = 0

Starting with the ##\eta## component:
\frac{\partial \phi}{\partial \eta} = exp() \left[ B + (A+B\eta)(-ic|k|) \right]
\frac{\partial^2 \phi}{\partial \eta^2} = exp() \left[ -2ic|k|B - |k|^2c^2(A+B\eta) \right]

Now for the ##r## component:
\frac{\partial \phi}{\partial r} = exp() \left[ A+B\eta \right](\vec k \cdot \hat r)
\frac{\partial^2 \phi}{\partial r^2} = exp() \left[ -(A+B\eta)(\vec k \cdot \hat r)^2 \right]

Equating both real parts doesn't work; It gives ##(\vec k \cdot \hat r)^2 = |k|^2 c^2##..
 
Physics news on Phys.org
What is ##\nabla_\mu## here? I would have thought they'd be covariant derivatives, not partial derivatives. (?)

Even though a covariant derivative of a scalar field is just the partial derivative, the 2nd covariant derivative is acting on a vector...
 
strangerep is correct, the second covariant derivative is not acting on a scalar, but on a covariant vector. I suggest rewriting it as
$$
\nabla^\mu \nabla_\mu \phi = g^{\mu\nu} \nabla_\mu \nabla_\nu \phi
$$
and apply the rules for covariant derivatives from there.
 
Orodruin said:
strangerep is correct, the second covariant derivative is not acting on a scalar, but on a covariant vector. I suggest rewriting it as
$$
\nabla^\mu \nabla_\mu \phi = g^{\mu\nu} \nabla_\mu \nabla_\nu \phi
$$
and apply the rules for covariant derivatives from there.

I know that for a vector field, ##\nabla_\alpha V^\mu = \partial_\alpha V^\mu + \Gamma^\mu_{\alpha \gamma} V^\gamma##. Does this apply to scalar fields as well? Or in the case of a scalar field, does it simply reduce to a normal derivative?

I have also found that
\nabla_\alpha \nabla_\beta V^\mu = \partial_\alpha \partial_\beta V^\mu + \left(\partial_\alpha \Gamma^\mu_{\beta \nu} \right)V^\nu + \Gamma^\mu_{\beta \nu} \left( \partial_\alpha V^\nu \right) + \Gamma^\mu_{\alpha \nu} \left[ \partial_\beta V^\nu + \Gamma^\nu_{\beta \epsilon} V^\epsilon \right] + \Gamma^\nu_{\alpha \beta} \left[ \partial_\nu V^mu + \Gamma^\mu_{\nu \epsilon} V^\epsilon \right]
 
unscientific said:
I know that for a vector field, ##\nabla_\alpha V^\mu = \partial_\alpha V^\mu + \Gamma^\mu_{\alpha \gamma} V^\gamma##. Does this apply to scalar fields as well? Or in the case of a scalar field, does it simply reduce to a normal derivative?

No, for scalar fields you have ##\nabla_\mu \phi = \partial_\mu \phi##. However, ##\nabla_\mu \phi## is a covector field and as such its covariant derivative ##\nabla_\nu\nabla_\mu \phi## will involve a Christoffel symbol.
 
  • Like
Likes   Reactions: unscientific

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K