Chapter 15: Problem 96
Prove that if \(T\) and \(a\) are respectively a four-tensor and a four-vector, then \(b=T \cdot a=T G a\) is a four-vector; that is, it transforms according to the rule \(b^{\prime}=\Lambda b\)
Short Answer
Expert verified
Yes, the product is a four-vector; it obeys the transformation rule.
Step by step solution
01
Understanding the Problem
We need to prove that the product of a four-tensor \(T\) and a four-vector \(a\) results in a four-vector \(b = T \cdot a\). We also need to show that \(b\) transforms according to the four-vector transformation rule, \(b^{\prime} = \Lambda b\). Here, \(\Lambda\) is the Lorentz transformation matrix.
02
Defining the Four-Tensor and Four-Vector
A four-vector \(a\) transforms under Lorentz transformation \(\Lambda\) as \(a' = \Lambda a\). A rank-two four-tensor \(T\) transforms as \(T' = \Lambda T \Lambda^T\), where \(\Lambda^T\) is the transpose of \(\Lambda\).
03
Expressing the Product
The product \(b = T \cdot a\) is given by a tensor-vector multiplication. The elements of \(b\) are expressed as \(b^\mu = T^{\muu} a_u\), where the Einstein summation convention is implied over the index \(u\).
04
Applying Lorentz Transformation to the Product
Under Lorentz transformation, the product becomes \(b'^\mu = T'^\mu_{\,\sigma} a'^\sigma\). Substituting the transformation properties, we have \(b'^\mu = \Lambda^\mu_{\,\rho} T^{\rho\tau} \Lambda^\tau_{\,u} a^u\).
05
Simplifying the Transformation Expression
Using associativity, the expression simplifies to \(b'^\mu = \Lambda^\mu_{\,\rho} (T^{\rho\tau} a_\tau) = \Lambda^\mu_{\,\rho} b^\rho\). This matches the transformation rule for four-vectors, \(b'^\mu = \Lambda^\mu_{\,\rho} b^\rho\), confirming that \(b\) is a four-vector.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Four-tensor
In the realm of relativity, four-tensors extend the concept of tensors into four dimensions. They allow us to understand how physical quantities transform under the changes of reference frames, a key aspect of Einstein's theory of relativity.
Four-tensors can have various ranks, which denote their array structure and complexity.
Commonly discussed is the rank-two four-tensor. It has components represented by two indices, like in the exercise where tensor transposes as \( T' = \Lambda T \Lambda^T \).
Four-tensors can have various ranks, which denote their array structure and complexity.
Commonly discussed is the rank-two four-tensor. It has components represented by two indices, like in the exercise where tensor transposes as \( T' = \Lambda T \Lambda^T \).
- \( \Lambda \) is the Lorentz transformation matrix.
- \( \Lambda^T \) indicates its transpose.
Four-vector
Four-vectors are vital constructs in the study of relativity. They combine both spatial and temporal components into a single mathematical entity. These vectors are essential as they make the mathematical representation of physical laws invariant under Lorentz transformations.
A four-vector's transformation under a Lorentz transformation \(\Lambda\) is given by:
A four-vector's transformation under a Lorentz transformation \(\Lambda\) is given by:
- \( a' = \Lambda a \)
Einstein summation convention
The Einstein summation convention streamlines expressions involving tensor calculus by implying a sum over indices appearing twice in a term. Instead of explicitly writing out the sum, the repeated index suggests a summation, making equations less cumbersome and easier to manage.
In our exercise, when expressing the product \( b = T \cdot a \), the components \( b^\mu = T^{\mu u} a_u \), automatically sum over the repeated index \( u \).
In our exercise, when expressing the product \( b = T \cdot a \), the components \( b^\mu = T^{\mu u} a_u \), automatically sum over the repeated index \( u \).
- This eliminates the need for the cumbersome \( \sum \), implied by the repetition of index.
- Speeds up calculations in complex tensor equations.