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duplicates list edited from Convention of tensor indices, Why is not ${(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu$?, Inverse and Transpose of Lorentz Transformation to Convention of tensor indices, Why is not ${(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu$?, Derivation of Inverse Lorentz Transformation in Index Notation, Inverse and Transpose of Lorentz Transformation

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edited Aug 21 at 17:34

Qmechanic ♦

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duplicates list edited from Convention of tensor indices, Inverse and Transpose of Lorentz Transformation to Convention of tensor indices, Why is not ${(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu$?, Inverse and Transpose of Lorentz Transformation

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edited Aug 21 at 17:32

Qmechanic ♦

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duplicates list edited from Convention of tensor indices to Convention of tensor indices, Inverse and Transpose of Lorentz Transformation

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edited Aug 21 at 17:31

Qmechanic ♦

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Post Closed as "Duplicate" by Qmechanic♦

Convention of tensor indices

occurred Aug 21 at 17:29

edited tags; edited title

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edited Aug 21 at 17:26

Qmechanic ♦

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A confusing tensor index calculation in quantum field (also in GR)

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}$$$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}.$$

Because we have

But there exists a subtlety, if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down, we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu},$$

where the RHS is usually defined as the transpose of the matrix form of $\Lambda$, but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general. How does this happen?

Even more confusingly, we have

This implies the matrix equation $\Lambda^T \eta \Lambda=\eta$, which is indeed how we define the Lorentz group mathematically, as the elements in $\mathbf{O}(1,3)$ preserving the inner product with respect to the metric $\eta$.

A confusing tensor index calculation in quantum field (also in GR)

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}$$

Because we have

But there exists a subtlety, if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down, we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu},$$

where the RHS is usually defined as the transpose of the matrix form of $\Lambda$, but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general. How does this happen?

Even more confusingly, we have

This implies the matrix equation $\Lambda^T \eta \Lambda=\eta$, which is indeed how we define the Lorentz group mathematically, as the elements in $\mathbf{O}(1,3)$ preserving the inner product with respect to the metric $\eta$.

A confusing tensor index calculation

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}.$$

Because we have

But there exists a subtlety, if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down, we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu},$$

where the RHS is usually defined as the transpose of the matrix form of $\Lambda$, but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general. How does this happen?

Even more confusingly, we have

This implies the matrix equation $\Lambda^T \eta \Lambda=\eta$, which is indeed how we define the Lorentz group mathematically, as the elements in $\mathbf{O}(1,3)$ preserving the inner product with respect to the metric $\eta$.

added 29 characters in body; edited title

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edited Aug 21 at 17:18

DanielC

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A confusing tensor index calculation in quantum field (also in GR)

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}$$

Because we have

But there exists a subtlety,if if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down,we we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu}$$$$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu},$$

Wherewhere the RHS is usually defined as the transpose of the matrix form of $\Lambda$,but but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general.How How does this happenshappen?

Even more confusingly,we we have

WhichThis implies the matrix equation $\Lambda^T \eta \Lambda=\eta$,which which is indeed how we define the Lorentz group mathematically,as as the elements in O(1;3)$\mathbf{O}(1,3)$ preserving the inner product with respect to the metric $\eta$.

A confusing tensor index calculation in quantum field(also in GR)

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}$$

Because we have

But there exists a subtlety,if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down,we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu}$$

Where the RHS is usually defined as the transpose of the matrix form of $\Lambda$,but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general.How this happens?

Even more confusingly,we have

Which implies the matrix equation $\Lambda^T \eta \Lambda=\eta$,which is indeed how we define Lorentz group mathematically,as the elements in O(1;3) preserving the inner product with respect to metric $\eta$

A confusing tensor index calculation in quantum field (also in GR)

We define Lorentz transformation as $\Lambda$ satisfying:$$\eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda^{\mu}_{\space\space\alpha}\Lambda^{\nu}_{\space\space\beta}$$

And its inverse$$(\Lambda^{-1})^{\mu}_{\space\space\nu}:=\eta^{\mu\beta}\eta_{\nu\alpha}\Lambda^{\alpha}_{\space\space\beta}$$

Because we have

But there exists a subtlety, if we use the two $\eta$ in the definition of $\Lambda^{-1}$ rising index $\beta$ up and lowing index $\alpha$ down, we have $$(\Lambda^{-1})^{\mu}_{\space\space\nu}=\Lambda_{\nu}^{\space\space\mu},$$

where the RHS is usually defined as the transpose of the matrix form of $\Lambda$, but we know that $\Lambda^{-1}=\Lambda^T$ is not right in general. How does this happen?

Even more confusingly, we have

This implies the matrix equation $\Lambda^T \eta \Lambda=\eta$, which is indeed how we define the Lorentz group mathematically, as the elements in $\mathbf{O}(1,3)$ preserving the inner product with respect to the metric $\eta$.