Contracting one index of a metric with the inverse metric

In summary, we use the completeness condition to write the dot products in terms of summations over all indices, and then use the orthogonality condition to simplify these expressions. This leads to the desired result of ##g_{\mu\nu}g^{\nu\lambda} = \delta^\lambda_\mu##.
  • #1
shinobi20
267
19
Homework Statement
Show ##g_{\mu\nu}g^{\nu\lambda} = \delta^\lambda_\mu## using the orthogonality and completeness condition.
Relevant Equations
##g_{\mu\nu} = \vec{e_\mu} \cdot \vec{e_\nu}##
##g^{\mu\nu} = \vec{e^\mu} \cdot \vec{e^\nu}##
##\vec{e_\mu} \cdot \vec{e^\nu} = \delta^\nu_\mu \quad \rightarrow \quad \sum_\alpha (e_\mu)_\alpha (e^\nu)_\alpha = \delta^\nu_\mu ~## (orthogonality)
##\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~## (completeness)
Since ##\nu## is contracted, we form the scalar product of the metric and inverse metric,

##g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) = \vec{e_\mu} \cdot (\vec{e_\nu} \cdot \vec{e^\nu}) \cdot \vec{e^\lambda} = \delta^\lambda_\mu##

I am not sure how to use the completeness condition here, can anyone point me in the proper direction?
 
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  • #2
shinobi20 said:
Homework Statement:: Show ##g_{\mu\nu}g^{\nu\lambda} = \delta^\lambda_\mu## using the orthogonality and completeness condition.
Relevant Equations:: ##g_{\mu\nu} = \vec{e_\mu} \cdot \vec{e_\nu}##
##g^{\mu\nu} = \vec{e^\mu} \cdot \vec{e^\nu}##
##\vec{e_\mu} \cdot \vec{e^\nu} = \delta^\nu_\mu \quad \rightarrow \quad \sum_\alpha (e_\mu)_\alpha (e^\nu)_\alpha = \delta^\nu_\mu ~## (orthogonality)
##\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~## (completeness)

Since ##\nu## is contracted, we form the scalar product of the metric and inverse metric,

##g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) = \vec{e_\mu} \cdot (\vec{e_\nu} \cdot \vec{e^\nu}) \cdot \vec{e^\lambda} = \delta^\lambda_\mu##

I am not sure how to use the completeness condition here, can anyone point me in the proper direction?
You have to be very explicit with all the steps. First, note that when you wrote
##g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) ##
it is confusing since you are using a dot symbol to have more than one meaning here. You should start by writing
##g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda}) ##
Now write these dot products in terms of summation over the other types of indices (that you wrote as [itex] \alpha,\beta [/itex]). Also, note that to be precise, the completeness relation is

## \vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \sum_\mu \quad (e_\mu)_\alpha (e^\mu)_\beta
= \delta_{\alpha\beta}~ ##
 
  • #3
nrqed said:
You have to be very explicit with all the steps. First, note that when you wrote
##g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) ##
it is confusing since you are using a dot symbol to have more than one meaning here. You should start by writing
##g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda}) ##
Now write these dot products in terms of summation over the other types of indices (that you wrote as [itex] \alpha,\beta [/itex]). Also, note that to be precise, the completeness relation is

## \vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \sum_\mu \quad (e_\mu)_\alpha (e^\mu)_\beta
= \delta_{\alpha\beta}~ ##

##g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda}) ##

##= \sum_\nu (\sum_\alpha (\vec{e_\mu})_\alpha (\vec{e_\nu})_\alpha) (\sum_\beta (\vec{e^\nu})_\beta (\vec{e^\lambda})_\beta)##

##= \sum_\alpha \sum_\beta (\vec{e_\mu})_\alpha (\sum_\nu (\vec{e_\nu})_\alpha (\vec{e^\nu})_\beta) (\vec{e^\lambda})_\beta##

##= \sum_\alpha \sum_\beta (\vec{e_\mu})_\alpha \delta_{\alpha \beta} (\vec{e^\lambda})_\beta##

##= \sum_\alpha (\vec{e_\mu})_\alpha (\vec{e^\lambda})_\alpha##

##= \delta^\lambda_\mu##
 
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Related to Contracting one index of a metric with the inverse metric

1. What does it mean to contract one index of a metric with the inverse metric?

Contracting one index of a metric with the inverse metric means to perform a mathematical operation where one index of a tensor (a mathematical object that represents a physical quantity) is summed over with the corresponding index of the inverse metric. This operation is commonly used in the study of general relativity and other branches of physics.

2. Why is it important to contract one index of a metric with the inverse metric?

Contracting one index of a metric with the inverse metric allows us to define a scalar quantity, which is a quantity that remains unchanged under coordinate transformations. This is important in physics as it allows us to describe physical quantities that are independent of the chosen coordinate system.

3. How is contracting one index of a metric with the inverse metric related to the curvature of space-time?

Contracting one index of a metric with the inverse metric is related to the curvature of space-time through the Einstein field equations, which describe how matter and energy are distributed in space-time and how this distribution affects the curvature of space-time. The curvature of space-time is a fundamental concept in general relativity and is essential in understanding the behavior of objects in the presence of gravity.

4. Can you give an example of contracting one index of a metric with the inverse metric in a real-life scenario?

One example of contracting one index of a metric with the inverse metric in a real-life scenario is in the calculation of the stress-energy tensor in general relativity. The stress-energy tensor is a mathematical object that describes the distribution of matter and energy in space-time, and it is obtained by contracting two indices of the energy-momentum tensor with the inverse metric.

5. Are there any practical applications of contracting one index of a metric with the inverse metric?

Contracting one index of a metric with the inverse metric has many practical applications in physics, particularly in the study of general relativity and other branches of theoretical physics. It is also used in computer simulations and numerical calculations to model and understand the behavior of physical systems. Additionally, it is essential in the development of theories and models that describe the fundamental laws of the universe.

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