Question

1\. Theorem: A transformation which transforms a metric \(g_{\mu r} \mathrm{~d} x^{n} \mathrm{~d} x^{\nu}\) with constant coefficients into a metric \(g_{\mu^{\prime} v^{\prime}} \mathrm{d} x^{\mu^{\prime}} \mathrm{d} x^{\gamma^{\prime}}\) with constant coefficients must be linear. Fill in the details of the following proof: \(g_{\mu v} \mathrm{~d} x^{\mu} \mathrm{d} x^{\nu}=g_{\mu^{\prime} v^{\prime}} \mathrm{d} x^{\mu^{\prime}} \mathrm{d} x^{\nu^{\nu}}=g_{\mu^{\prime} v^{\prime}} p_{\mu}^{\mu^{\prime}} p_{v}^{v^{\prime}} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}\), so \(g_{\mu \nu}=g_{\mu^{\prime} v^{\prime}} p_{\mu \nu}^{\|^{\prime}} p_{v}^{v^{\prime}}\) Differentiating with respect to \(x^{\phi}\) we get \(g_{\mu^{\prime} v} p_{\mu d} p_{v}^{v^{*}}+g_{\mu^{*} v} p_{\mu}^{\sigma_{v}} p_{v_{\sigma}}^{v^{*}}=0\) (i). Interchange \(\mu\) and \(\sigma\) to form equation (ii). Again, in (i) interchange and \(\sigma\) to form (iii). Subtract (iii) from the sum of (i) and (ii). Then \(2 g_{\mu^{*} v^{\prime}} p_{\mu \sigma}^{k^{*}} p_{v}^{v^{*}}=0\). Transvect first by \(p_{\sigma^{*}}^{v}\) and then by \(g^{v^{*} \sigma^{*}}\) and obtain pus \(=0\). This proves linearity.

Short answer

Question: Explain the steps involved in proving that a transformation that changes a metric with constant coefficients into another metric with constant coefficients must be linear. Answer: The proof involves the following steps: 1. Understand the Given Theorem and Proof: The theorem states that a transformation must be linear if it transforms a metric with constant coefficients into another metric with constant coefficients. We have to observe the given proof and provide explanations for each step. 2. Express the Transformed Metric in Terms of the Original Metric: The transformed metric can be expressed by using the transformation matrix relating the original and transformed coordinates. 3. Differentiate Both Sides with Respect to a Coordinate: Differentiate both sides of the equation with respect to a coordinate to obtain a new equation involving the original and transformed metrics. 4. Perform the Interchange of Indices: Interchange indices in the new equation to form two additional equations. 5. Subtract and Transvect to Show Linearity: Subtract one of the additional equations from the sum of the other two equations, then perform transvection on the resulting equation. If the term inside the parenthesis vanishes, this shows a vanishing curvature tensor, proving that the transformation is linear.

Step-by-step solution

Understand the Given Theorem and Proof

The theorem states that a transformation that transforms a metric \(g_{\mu\nu}\) with constant coefficients into a metric \(g_{\mu^\prime\nu^\prime}\) with constant coefficients must be linear. The given proof uses a transformation matrix \(p\), and starts with the transformation of the metric \(g_{\mu\nu}\). We need to observe the steps of the proof and fill in the missing details.

Express the Transformed Metric in Terms of the Original Metric

According to the proof, the transformed metric can be expressed as \(g_{\mu\nu}=g_{\mu^\prime\nu^\prime}p_{\mu}^{\mu^\prime}p_{\nu}^{\nu^\prime}\). This is derived from the equation \(g_{\mu\nu}dx^\mu dx^\nu=g_{\mu^\prime\nu^\prime}dx^{\mu^\prime}dx^{\nu^\prime}=g_{\mu^\prime\nu^\prime}p_{\mu}^{\mu^\prime}p_{\nu}^{\nu^\prime}dx^\mu dx^\nu\). The transformation matrix \(p\) relates the original coordinates \(x^\mu\) to the transformed coordinates \(x^{\mu^\prime}\).

Differentiate Both Sides with Respect to \(x^\phi\)

We now differentiate both sides of the equation \(g_{\mu\nu}=g_{\mu^\prime\nu^\prime}p_{\mu}^{\mu^\prime}p_{\nu}^{\nu^\prime}\) with respect to \(x^\phi\) to obtain \(g_{\mu^\prime\nu}p_{\mu\nu}^{\mu^\prime}p_{\nu}^{\nu^\prime}+g_{\mu^\prime\nu}p_{\mu}^{\mu^\prime}p_{\nu\phi}^{\nu^\prime}=0\). This is equation (i) in the given proof.

Perform the Interchange of Indices

We need to perform the interchange of indices in equation (i) as suggested in the given proof. First, interchange \(\mu\) and \(\sigma\) to form equation (ii). Then, interchange \(\nu\) and \(\sigma\) to form equation (iii).

Subtract and Transvect to Show Linearity

Subtract equation (iii) from the sum of equation (i) and equation (ii) to obtain \(2g_{\mu^\prime\nu^\prime}p_{\mu\sigma}^{\mu^\prime}p_{\nu}^{\nu^\prime} = 0\). To prove linearity, we need to show that \(p\) has a vanishing curvature tensor. Transvect the resulting equation first by \(p_{\sigma^\prime}^{\nu}\) and then by \(g^{\mu^\prime\sigma^\prime}\) and obtain \(p_{\sigma^\prime}^{\nu}g^{\mu^\prime\sigma^\prime}2g_{\mu^\prime\nu^\prime}p_{\mu\sigma}^{\mu^\prime}p_{\nu}^{\nu^\prime} = 0\). Since the transformed metric \(g_{\mu^\prime\nu^\prime}\) is non-degenerate, the term inside the parenthesis must vanish, which shows the vanishing curvature tensor. Therefore, the theorem is proved, and the transformation is linear.

Key concepts

Linear Transformation

In the realm of special relativity, understanding how different observers perceive the same event is critical. The concept of a linear transformation is at the heart of this understanding. A linear transformation involves converting a set of coordinates into a new set, while preserving the straight-line relationship between points and their coordinates.

A transformation is considered linear if the new coordinates \( x^{\mu'} \) of a point can be written as a linear combination of its old coordinates \( x^\mu \) using a transformation matrix \( p \) such that \( x^{\mu'} = p_\mu^{\mu'} x^\mu \). This concept is pivotal because it ensures that the equations of motion remain consistent under different inertial frames, which is a fundamental aspect of special relativity.

In the exercise provided, we see the transition from one metric tensor to another while maintaining constant coefficients, implying a linear transformation. By following the steps given in the solution, we can verify this transformation's linearity, as it maintains the form of the metric despite changing the observer's frame of reference.

Transformation Matrices

The use of transformation matrices is an elegant mathematical tool to facilitate the conversion of coordinates from one inertial frame to another in special relativity. These matrices contain information on how space and time coordinates change under certain transformations, especially Lorentz transformations.

In our example, the matrix \( p \) holds the coefficients that relate the old and new metric's components. Each element of this matrix, \( p_\mu^{\mu'} \) and \( p_u^{u'} \) maps the original coordinates \( x^\mu \) and \( x^u \) to new ones. The metric tensor in the new coordinate system \( g_{\mu'u'} \) can then be written as a product of the original metric \( g_{\muu} \) and the transformation matrix elements, illustrating how distances and intervals are perceived in the transformed frame.

Constant Coefficient Metrics

The concept of constant coefficient metrics concerns the simplicity and uniformity of the spacetime fabric. When the metric tensor, which encodes the geometry of spacetime, has constant coefficients, it indicates that the spacetime is flat and unchanging. This is typical of Minkowski space, the stage upon which special relativity plays out.

Within the context of the exercise, constant coefficient metrics allow us to assert that the transformation in question preserves the linearity of spacetime. This means that the transformed inertial frames, which is universally true in special relativity. The proof states that transforming from one metric with constant coefficients to another necessarily involves a linear transformation because the constancy of coefficients suggests a direct proportional change, which is a hallmark of linearity.

Differentiation of Tensor Components

The action of differentiation of tensor components is particularly significant in the study of transformations in relativity. It's a way to understand how the components of a tensor change with respect to variations in coordinates. In our case, it means examining how the metric tensor's components vary when we perturb the coordinates slightly.

The exercise directs us to differentiate the equation \( g_{\muu} = g_{\mu'u'} p_\mu^{\mu'} p_u^{u'} \) with respect to \( x^\phi \). Through this process, we see that if the metric tensor components indeed have constant coefficients, the derivative with respect to any coordinate should yield zero. By performing these differentiation steps and following the conditions set by the theorem, it reinforces the proof that our transformation retains linearity, as the metric components - unchanged by differentiation - sustain their constant value.