# Proving Lorentz Metric on Real Type (1,0;1,0) Tensors in Wald Ch. 13

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• ergospherical
In summary, Wald is asking for a Lorentz metric on a vector spacecontaining the real elements of the tensorvector space ##Y## of type ##(1,0;1,0)##. The basis ##t^{AA'} =\dfrac{1}{\sqrt{2}}(o^A \bar{o}^{A'} + \iota^A \bar{\iota}^{A'})##, ##x^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{\iota}^{A'} + \iota^A \bar{o}^{A'})##, ##y^{AA'
ergospherical
In ch. 13, pg.349 of Wald it's asked to prove that ##g_{AA'BB'} = \epsilon_{AB} \bar{\epsilon}_{A'B'}## is a Lorentz metric on ##V## (containing the real elements of the vector space ##Y## of type ##(1,0;1,0)## tensors). Given the basis ##t^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{o}^{A'} + \iota^A \bar{\iota}^{A'})##, ##x^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{\iota}^{A'} + \iota^A \bar{o}^{A'})##, ##y^{AA'} = \dfrac{i}{\sqrt{2}}(o^A \bar{\iota}^{A'} - \iota^A \bar{o}^{A'})##, ##z^{AA'} = \dfrac{1}{\sqrt{2}}(o^A \bar{o}^{A'} - \iota^A \bar{\iota}^{A'})##, where ##o_A \iota^A = 1## by definition. I need to show that these are Minkowski orthonormal. So for example\begin{align*}
\epsilon_{AB} \bar{\epsilon}_{A'B'} t^{AA'} t^{BB'} &= \dfrac{1}{2} o_B o_{B'} o^B \bar{o}^{B'} + \dfrac{1}{2} o_B \bar{o}_{B'} \iota^B \bar{\iota}^{B'} + \dfrac{1}{2} \iota_B \bar{\iota}_{B'} o^B \bar{o}^{B'} + \dfrac{1}{2} \iota_B \bar{\iota}_{B'} \iota^B \bar{\iota}^{B'}
\end{align*}should be equal to ##1##. Since ##o_B \iota^B = 1## it also follows that
\begin{align*}
1 = o_B \iota^B = \epsilon_{AB} o^A \iota^B = - \epsilon_{BA} o^A \iota^B = -o^A \iota_A
\end{align*}therefore the middle two terms are both equal to ##\dfrac{1}{2}##. What I can't see is what to write for ##\iota_A \iota^A## and ##o_A o^A##; they should be zero (right...?), but why?

Last edited:
##o
ergospherical said:
What I can't see is what to write for ##\iota_A \iota^A## and ##o_A o^A##; they should be zero (right...?), but why?
##o_A o^A=\epsilon_{BA}o^B o^A=0## The two form ##\epsilon## is antysymmetric. Everything is orthogonal to itself.

ergospherical
martinbn said:
##o_A o^A=\epsilon_{BA}o^B o^A=0## The two form ##\epsilon## is antysymmetric. Everything is orthogonal to itself.
ahhh course, thanks.

## 1. What is the Lorentz Metric?

The Lorentz Metric is a mathematical concept used in the theory of relativity. It describes the geometry of spacetime and is represented by a matrix with four dimensions.

## 2. What is a Real Type (1,0;1,0) Tensor?

A Real Type (1,0;1,0) Tensor is a mathematical object that has one contravariant and one covariant index. It is used to represent physical quantities in the theory of relativity.

## 3. How is the Lorentz Metric proven on Real Type (1,0;1,0) Tensors?

The Lorentz Metric is proven on Real Type (1,0;1,0) Tensors by using the properties of tensors and the Lorentz transformation. This involves manipulating the components of the tensor and showing that they satisfy the Lorentz Metric equation.

## 4. What is the significance of proving the Lorentz Metric on Real Type (1,0;1,0) Tensors?

Proving the Lorentz Metric on Real Type (1,0;1,0) Tensors is important because it provides a mathematical foundation for the theory of relativity. It allows us to accurately describe the geometry of spacetime and make predictions about the behavior of physical systems.

## 5. Are there any real-world applications for the Lorentz Metric on Real Type (1,0;1,0) Tensors?

Yes, the Lorentz Metric on Real Type (1,0;1,0) Tensors is used in various fields such as astrophysics, cosmology, and particle physics. It is essential for understanding the behavior of objects moving at high speeds and in the presence of strong gravitational fields.

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