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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 \( T' = \Lambda T \Lambda^T \).
  • \( \Lambda \) is the Lorentz transformation matrix.
  • \( \Lambda^T \) indicates its transpose.
Rank-two tensors are useful to represent measurable quantities like the electromagnetic field tensor. They differ from lower-rank tensors in terms of complexity and their interactions with other physical entities.
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:
  • \( a' = \Lambda a \)
This expression shows how the components of a four-vector combine space and time, ensuring that the laws of physics hold true regardless of the observer's velocity. An example of a four-vector is the four-momentum, which combines the relativistic energy and momentum of a particle. They reflect the isotropic nature of the spacetime continuum. Any object expressed as a result, like vector \( b \) in the exercise, inherently follows the transformation rule to remain consistent in different inertial frames.
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 \( u \).
  • This eliminates the need for the cumbersome \( \sum \), implied by the repetition of index.
  • Speeds up calculations in complex tensor equations.
This notation is particularly powerful in relativity and field theory, where tensors represent diverse physical quantities. The ability to simplify tensor equations is one of the reasons for its widespread use in advanced physics and mathematics.

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Most popular questions from this chapter

If one defines a variable mass \(m_{\text {var }}=\gamma m\), then the relativistic momentum \(\mathbf{p}=\gamma m \mathbf{v}\) becomes \(m_{\text {var }} \mathbf{v}\) which looks more like the classical definition. Show, however, that the relativistic kinetic energy is not equal to \(\frac{1}{2} m_{\mathrm{var}} v^{2}\)

(a) By exchanging \(x_{1}\) and \(x_{2}\), write down the Lorentz transformation for a boost of velocity \(V\) along the \(x_{2}\) axis and the corresponding \(4 \times 4\) matrix \(\Lambda_{\mathrm{B} 2}\). ( \(\mathbf{b}\) ) Write down the \(4 \times 4\) matrices \(\Lambda_{\mathrm{R}+}\) and \(\Lambda_{\mathrm{R}-}\) that represent rotations of the \(x_{1} x_{2}\) plane through \(\pm \pi / 2,\) with the angle of rotation measured counterclockwise. (c) Verify that \(\Lambda_{\mathrm{B} 2}=\Lambda_{\mathrm{R}-} \Lambda_{\mathrm{B} 1} \Lambda_{\mathrm{R}+},\) where \(\Lambda_{\mathrm{B} 1}\) is the standard boost along the \(x_{1}\) axis, and interpret this result.

When a radioactive nucleus of astatine 215 decays at rest, the whole atom is torn into two in the reaction $$^{215} \mathrm{At} \rightarrow^{211} \mathrm{Bi}+^{4} \mathrm{He}$$ The masses of the three atoms are (in order) \(214.9986,210.9873,\) and \(4.0026,\) all in atomic mass units. (1 atomic mass unit \(=1.66 \times 10^{-27} \mathrm{kg}=931.5 \mathrm{MeV} / c^{2}\).) What is the total kinetic energy of the two out coming atoms, in joules and in MeV?

A low-flying earth satellite travels at about 8000 m/s. What is the factor \(\gamma\) for this speed? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (as measured by the latter)? What is the percent difference?

Prove the following useful result, called the zero-component theorem: Let \(q\) be a four-vector, and suppose that one component of \(q\) is found to be zero in all inertial frames. (For example, \(q_{4}=0\) in all frames.) Then all four components of \(q\) are zero in all frames.
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