Why Does Tensor Contraction Yield Zero in This Calculation?

  • Thread starter teddd
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In summary: The reason why is that when you contract the second factor, you're actually multiplying all the terms by the same denominator. This is why all the terms end up being zero.
  • #1
teddd
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Homework Statement


Hello guys, hope you'll help me out with this!
I'm asked to calculate [tex]g^{\alpha\beta}g^{\sigma\rho}(g_{\alpha\sigma}g_{\beta\rho}-g_{\alpha\rho}g_{\beta\sigma})[/tex]
where g is the metric tensor on a n-dimensional manifold but I can't get to the right result, i keep on getting zero! (i know that's wrong although i don't know the exact solution -sorry- but it should depend on the dimension of the manifold)


Homework Equations



none

The Attempt at a Solution



Well, i contracted firs with respect to [itex]g^{\sigma\rho}[/itex] and i end up with [tex]g^{\alpha\beta}(g_{alpha\beta}+g_{alpha\beta}-g_{alpha\beta}-g_{alpha\beta})[/tex]which obviously vanish.
I must be missing something!
 
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  • #2
Are you so sure it's zero? Let's look at the first factor:
[tex]
g^{\alpha\beta}g^{\sigma\rho}g_{\alpha\sigma}g_{ \beta\rho}=g^{\alpha\beta}(g^{\sigma\rho}g_{\alpha\sigma})g_{\beta\rho}=g^{\alpha\beta}\delta^{\rho}_{\alpha}g_{\beta\rho}=g^{\rho\beta}g_{\beta\rho}=\delta^{\rho}_{\rho}=n
[/tex]
What is the other factor?
 
  • #3
Thanks for the answer hunt_mat but... I keep on getting zero!

That's becaouse contracting the second factor like you did for the first i get
[tex]-g^{\alpha\beta} g^{\sigma\rho}g_{\alpha\rho}g_{\beta\sigma}=-g^{\alpha\beta} (g^{\sigma\rho}g_{\alpha\rho})g_{\beta\sigma}=-g^{\alpha\beta} \delta^{\sigma}_{\alpha}g_{\beta\sigma}=-g^{\sigma\beta}g_{\beta\sigma}=-n[/tex]

which added to the first factor gives zero!
They seems equal to me, becaouse when i get the kronecker delta it doesent matter which index it has, it will be a mute index anyhow! (maybe the error is here?)where am i mistaking?
 
  • #4
I don't think that you're making a mistake at all, I think the answer really is zero.
 

FAQ: Why Does Tensor Contraction Yield Zero in This Calculation?

What are tensors and why are they used in scientific research?

Tensors are mathematical objects that describe the relationships between physical quantities in a multidimensional space. They are used in scientific research because they allow for a more precise and comprehensive understanding of complex systems and phenomena.

What is contracting tensors and how is it used?

Contracting tensors involves summing over repeated indices in a tensor equation. This operation is used to simplify and manipulate tensor equations, making them easier to solve and interpret.

What are some common applications of contracting tensors in scientific research?

Contracting tensors is commonly used in the fields of physics, engineering, and mathematics. It is particularly useful in mechanics, electromagnetism, and general relativity for solving complex equations and describing physical phenomena.

How does contracting tensors relate to the concept of symmetry?

Contracting tensors can uncover hidden symmetries in a system by reducing the number of independent components in a tensor equation. This allows for a more elegant and concise description of the system's properties and behavior.

Are there any limitations or challenges when using contracting tensors?

Contracting tensors can become increasingly complex and computationally demanding when working with higher order tensors. It also requires a strong understanding of tensor algebra and can be difficult to visualize in higher dimensions.

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