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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

Newton's first law can be stated: If an object is isolated (subject to no forces), then it moves with constant velocity. We know that this is invariant under the Galilean transformation. Prove that it is also invariant under the Lorentz transformation. [Assume that it is true in an inertial frame \(\mathcal{S}\), and use the relativistic velocity-addition formula to show that it is also true in any other \(\mathcal{S}^{\prime} .\) ]

Show that the four-velocity of any object has invariant length squared \(u \cdot u=-c^{2}\)

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?

For any two objects \(a\) and \(b\), show that the scalar product of their four- velocities is \(u_{a} \cdot u_{b}=\) \(-c^{2} \gamma\left(v_{\mathrm{rel}}\right),\) where \(\gamma(v)\) denotes the usual \(\gamma\) factor, \(\gamma(v)=1 / \sqrt{1-v^{2} / c^{2}},\) and \(v_{\mathrm{rel}}\) denotes the speed of \(a\) in the rest frame of \(b\) or vice versa.

An excited state \(X^{*}\) of an atom at rest drops to its ground state \(X\) by emitting a photon. In atomic physics it is usual to assume that the energy \(E_{\gamma}\) of the photon is equal to the difference in energies of the two atomic states, \(\Delta E=\left(M^{*}-M\right) c^{2},\) where \(M\) and \(M^{*}\) are the rest masses of the ground and excited states of the atom. This cannot be exactly true, since the recoiling atom X must carry away some of the energy \(\Delta E .\) Show that in fact \(E_{\gamma}=\Delta E\left[1-\Delta E /\left(2 M^{*} c^{2}\right)\right] .\) Given that \(\Delta E\) is of order a few ev, while the lightest atom has \(M\) of order \(1 \mathrm{GeV} / c^{2},\) discuss the validity of the assumption that \(E_{\gamma}=\Delta E\)

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