Using 4-Gradient to Apply Function to 4-Vector

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In summary: This is because the covariant derivative of a 4-vector x^{\mu} is given by \partial_{\mu}x^{\mu} = (1/c) \partial x^{0}/\partial t + \partial x^{i}/\partial x^{i} where i = 1,2,3. In this case, x^{0} = ct and x^{i} = -x,-y,-z, so the expression simplifies to \partial_{\mu}x^{\mu} = (1/c) \partial(ct)/\partial t + \partial(-x)/\partial x + \partial(-y)/\partial y + \partial(-z)/\partial z = (1
  • #1
chill_factor
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lets say we have a covariant 4-gradient ∂[itex]_{μ}[/itex] = (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)?
 
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  • #2
chill_factor said:
lets say we have a covariant 4-gradient ∂[itex]_{μ}[/itex] = (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:

[itex]\partial_\mu x^2^\mu[/itex]

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

[itex]\partial_\mu x^2^\mu[/itex]

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 ∂[itex]_{μ}[/itex] x[itex]^{μ}[/itex] = 1 where x[itex]^{μ}[/itex] = (ct,-x,-y,-z). How did they get this?
 
  • #4
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 ∂[itex]_{μ}[/itex] x[itex]_{μ}[/itex], 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.
 
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  • #5
Lavabug said:
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 ∂[itex]_{μ}[/itex] x[itex]_{μ}[/itex], 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.
 
  • #6
What was it, out of curiosity?
 
  • #7
chill_factor said:
In an example I'm looking at, they tell me ∂[itex]_{μ}[/itex] x[itex]^{μ}[/itex] = 1
That is the divergence. The gradient is not being involved at all. [itex]\partial _{\mu }x^{\mu } = \partial _{t}x^{t} + \triangledown \cdot \vec{x}[/itex].
 
  • #8
No, [itex] \partial_{\mu}x^{\mu} = 4 [/itex].
 

FAQ: Using 4-Gradient to Apply Function to 4-Vector

1. What is a 4-gradient?

A 4-gradient is a mathematical tool used in the field of special relativity to describe the change of a quantity or function over time and space. It is represented by a vector with four components, each representing the change in a specific direction.

2. How is a 4-gradient used to apply a function to a 4-vector?

A 4-vector is a mathematical object with four components that represent a physical quantity in space and time. To apply a function to a 4-vector, the 4-gradient is used to describe the rate of change of the function in each direction of the 4-vector. This allows us to calculate the value of the function at any point along the 4-vector.

3. What are the advantages of using a 4-gradient to apply a function to a 4-vector?

Using a 4-gradient allows us to easily calculate the value of a function at any point along a 4-vector, even in the complex space-time of special relativity. It also provides a more comprehensive description of the function's change over time and space compared to traditional methods.

4. Can a 4-gradient be used in other fields of science?

Yes, 4-gradients are not limited to special relativity and can be applied in other fields such as quantum mechanics, electromagnetism, and fluid dynamics. They are a powerful mathematical tool for describing the change of a function in complex systems.

5. Are there any limitations to using 4-gradients to apply functions to 4-vectors?

While 4-gradients are a useful tool, they have limitations in describing the behavior of complex systems. They do not take into account certain factors such as quantum effects and can become difficult to interpret in highly curved space-time. Additionally, they may not be applicable in non-relativistic scenarios.

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