Lie derivative of covariant vector

In summary, the conversation discusses deriving the equation L_v(u_a)=v^b \partial_b u_a + u_b \partial_a v^b using the equations L_v(w^a)=v^b \partial_b w^a - w^b \partial_b v^a and L_v(f)=v^a \partial_a f. The attempt at a solution involves using Einstein notation and changing the order of the indexes in the last term, which is allowed in this notation.
  • #1
zardiac
16
0

Homework Statement


Derive [itex]L_v(u_a)=v^b \partial_b u_a + u_b \partial_a v^b[/itex]


Homework Equations


[itex]L_v(w^a)=v^b \partial_b w^a - w^b \partial_b v^a[/itex]

[itex]L_v(f)=v^a \partial_a f[/itex] where f is a scalar.

The Attempt at a Solution


In the end I get stuck with something like this,
[itex]L_v(u_a)w^a=v^b u_a \partial_b w^a -u_a v^b \partial_b w^a +v^b w ^a \partial_b u_a +u_a w^b \partial_b v^a [/itex]
Which makes me think I am on the right way, but I end up with
[itex]L_v(u_a)w^a=v^b w ^a \partial_b u_a +u_a w^b \partial_b v^a [/itex]

Which is what I want if I can change place of a and b in the last term only. But I am not allowed to do that just like that right?
 
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  • #2
zardiac said:
Which is what I want if I can change place of a and b in the last term only. But I am not allowed to do that just like that right?

Of course you are. There's no significance to what letter you use to represent the sums. [itex] u^a v_a = u^b v_b = u^\Upsilon v_\Upsilon [/itex] all mean exactly the same thing.
 
  • #3
Really? I guess I am not used to the einstein notation contention. How can you see what indexes are free and which ones are dummy?
I thought since I was to determine [itex]L_v(u_a)[/itex] then a was a free index, and thus if I change in one of the terms I have to change in all of the terms. (Which would not lead to the correct result in this case)
 

Related to Lie derivative of covariant vector

1. What is the definition of the Lie derivative of a covariant vector?

The Lie derivative of a covariant vector is a mathematical operation that determines how the vector changes along a specified direction, without altering the underlying coordinate system. It is denoted by the symbol ℓ and is a measure of the rate of change of the vector with respect to a specific vector field.

2. How is the Lie derivative of a covariant vector different from the covariant derivative?

The Lie derivative and the covariant derivative are two different mathematical operations. While the Lie derivative measures the change of the vector along a specified direction, the covariant derivative measures the change of the vector with respect to a coordinate system. Additionally, the Lie derivative does not require the use of a metric tensor, while the covariant derivative does.

3. What are the applications of the Lie derivative of a covariant vector?

The Lie derivative of a covariant vector has various applications in fields such as differential geometry, general relativity, and fluid mechanics. It is used to study the behavior of vector fields and their transformations, which can provide insights into the dynamics of physical systems.

4. How is the Lie derivative of a covariant vector calculated?

The Lie derivative of a covariant vector is calculated using the Lie bracket, a mathematical operation that measures the difference between two vector fields. The Lie derivative is then expressed as a combination of the Lie bracket and the vector field along which the derivative is being calculated.

5. Can the Lie derivative of a covariant vector be extended to higher dimensions?

Yes, the concept of the Lie derivative can be extended to higher dimensions, including tensors of higher order. This allows for the calculation of the Lie derivative of more complex objects, such as tensors that represent physical quantities like stress or strain.

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