Killing Equation: Why Set g_μν,ρ V^ρ = 0?

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In summary, the conversation discusses the Killing Equation and the term ##g_{\mu \nu, \rho} V^\rho##, which is often set to zero. However, one user questions the motivation for setting this term to zero and asks for more details. Another user also questions if the term should actually be written as ##g_{\mu \nu;\rho} V^\rho## instead. The original poster then reveals that they made a mistake in their derivation by equating ##g_{\mu \nu} V^\nu{}_{, \ \rho} = V_{\mu, \ \rho}##. The conversation ends with a reminder to always include all relevant information when asking a question.
  • #1
kent davidge
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When deriving the Killing Equation, one ends up with a term ##g_{\mu \nu, \rho} V^\rho## where ##V## is the Killing Vector. This terms is often set to zero. Why?
I see no motivation for demanding that the derivative of the metric coefficients vanish.
 
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  • #2
Can you give more details?
 
  • #3
martinbn said:
Can you give more details?
What kind of details? I tried to put everything into the opening post.
 
  • #4
kent davidge said:
What kind of details? I tried to put everything into the opening post.
You did not though. You said ”This term is usually set to zero. Why?” without any reference to a text actually doing so or reciting an actual derivation doing so.
 
  • #5
Are you sure that they didn't write ##g_{\mu \nu;\rho} V^\rho##? The covariant derivative (indicated by semicolon, rather than comma) of ##g## is zero.
 
  • #6
stevendaryl said:
Are you sure that they didn't write ##g_{\mu \nu;\rho} V^\rho##?
It's me. I'm deriving it. However I get ##\mathcal L g _{\sigma \rho} = \nabla_\sigma V_\rho + \nabla_\rho V_\sigma + 2 V_\mu \Gamma^\mu{}_{\sigma \rho} + g_{\sigma \rho, \kappa} V^\kappa##.
 
  • #7
kent davidge said:
It's me. I'm deriving it. However I get ##\mathcal L g _{\sigma \rho} = \nabla_\sigma V_\rho + \nabla_\rho V_\sigma + 2 V_\mu \Gamma^\mu{}_{\sigma \rho} + g_{\sigma \rho, \kappa} V^\kappa##.
So why are you withholding the rest of your derivation?
 
  • #8
Orodruin said:
So why are you withholding the rest of your derivation?
because I was not wanting to bore you with it. but never mind, i found my fault. i was equating ##g_{\mu \nu} V^\nu{}_{, \ \rho} = V_{\mu, \ \rho}## right away.
 
  • #9
kent davidge said:
because I was not wanting to bore you with it

As you see, this is not a good strategy. When you ask a question about something you did, you need to include all of what you did. You can't assume that just the part you think you need to include is relevant, because if you already knew what was relevant and what was not, you wouldn't have made a mistake and you wouldn't have needed to post a question in the first place. Please keep that in mind for the future.

Since you say you have found your mistake, this thread is closed.
 
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Likes jim mcnamara and Orodruin

1. What is the Killing Equation?

The Killing Equation is a mathematical equation used in the field of differential geometry to calculate symmetries of certain mathematical objects, such as manifolds. It is used to determine whether a vector field on a manifold is a symmetry of the metric tensor of that manifold.

2. What is the significance of gμν,ρ and Vρ in the Killing Equation?

The gμν,ρ term represents the components of the metric tensor, which is used to measure distances and angles on a manifold. The Vρ term represents the components of the vector field, which describes the direction and magnitude of a vector at each point on the manifold. These two terms are essential in determining the symmetries of the manifold using the Killing Equation.

3. Why is it important to set gμν,ρ Vρ = 0 in the Killing Equation?

Setting gμν,ρ Vρ = 0 ensures that the vector field V is a symmetry of the metric tensor g. This means that the vector field preserves the distances and angles on the manifold, making it a useful tool in studying the geometry of the manifold.

4. How is the Killing Equation used in physics?

In physics, the Killing Equation is used in the study of general relativity, where it is used to determine the symmetries of the space-time manifold. These symmetries can then be used to find solutions to Einstein's field equations, which describe the curvature of space-time.

5. Are there any limitations to the Killing Equation?

While the Killing Equation is a powerful tool in studying the symmetries of manifolds, it does have some limitations. It can only be applied to manifolds with a symmetric metric tensor, and it does not account for the effects of matter on the geometry of the manifold. Additionally, it may not be applicable to certain non-compact manifolds.

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