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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

One way to set up the system of synchronized clocks in a frame \(\mathcal{S}\), as described at the beginning of Section 15.4, would be for the chief observer to summon all her helpers to the origin \(O\) and synchronize their clocks there, and then have them travel to their assigned positions very slowly. Prove this claim as follows: Suppose a certain observer is assigned to a position \(P\) at a distance \(d\) from the origin. If he travels at constant speed \(V\), when he reaches \(P\) how much will his clock differ from the chief's clock at \(O\) ? Show that this difference approaches 0 as \(V \rightarrow 0\).

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

Let \(\Lambda_{\mathrm{B}}(\theta)\) denote the 4 \(\times 4\) matrix that gives a pure boost in the direction that makes an angle \(\theta\) with the \(x_{1}\) axis in the \(x_{1} x_{2}\) plane. Explain why this can be found as \(\Lambda_{\mathrm{B}}(\theta)=\Lambda_{\mathrm{R}}(-\theta) \Lambda_{\mathrm{B}}(0) \Lambda_{\mathrm{R}}(\theta)\) where \(\Lambda_{\mathrm{R}}(\theta)\) denotes the matrix that rotates the \(x_{1} x_{2}\) plane through angle \(\theta\) and \(\Lambda_{\mathrm{B}}(0)\) is the standard boost along the \(x_{1}\) axis. Use this result to find \(\Lambda_{\mathrm{B}}(\theta)\) and check your result by finding the motion of the spatial origin of the frame \(\mathcal{S}\) as observed in \(\mathcal{S}^{\prime}\).

The pion \(\left(\pi^{+} \text {or } \pi^{-}\right)\) is an unstable particle that decays with a proper half-life of \(1.8 \times 10^{-8}\) s. (This is the half-life measured in the pion's rest frame.) (a) What is the pion's half-life measured in a frame \(\mathcal{S}\) where it is traveling at \(0.8 c ?\) (b). If 32,000 pions are created at the same place, all traveling at this same speed, how many will remain after they have traveled down an evacuated pipe of length \(d=36 \mathrm{m} ?\) Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive. (c) What would the answer have been if you had ignored time dilation? (Naturally it is the answer (b) that agrees with experiment.)

(a) What is a mass of \(1 \mathrm{MeV} / c^{2}\) in kilograms? ( \(\mathbf{b}\) ) What is a momentum of \(1 \mathrm{MeV} / c\) in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} ?\)

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