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Chapter 35: Problem 56

Show that momentum and energy transform from one inertial frame to another as \(p_{x}^{\prime}=\gamma\left(p_{x}-v E / c^{2}\right) ; p_{y}^{\prime}=p_{y}\) \(p_{z}^{\prime}=p_{p} ; E^{\prime}=\gamma\left(E-v p_{x}\right) .\) Hint: Look at the derivation for the space-time Lorentz transformation.

Short Answer

Expert verified
Based on our analysis and solution, we have demonstrated that the momentum and energy transform from one inertial frame to another according to the given equations by applying the space-time Lorentz transformation. We defined the four-momentum vector, applied the Lorentz transformation matrix, and compared our results to the given equations to verify the relationships between the inertial frames.

Step by step solution

01

Review the Lorentz Transformation

The Lorentz transformation is a set of equations that relates the space-time coordinates \((t, x, y, z)\) of an event in one inertial frame (S) to the space-time coordinates \((t', x', y', z')\) in another inertial frame (S'). The frames are moving relative to each other with a constant velocity v along the x-axis. The transformation equations are: 1. \(t^{\prime} = \gamma\left(t - \frac{v x}{c^2}\right)\) 2. \(x^{\prime} = \gamma\left(x - vt\right)\) 3. \(y^{\prime} = y\) 4. \(z^{\prime} = z\) where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor.
02

Define the Four-Momentum Vector

In special relativity, the four-momentum vector is defined as a four-dimensional vector with components: $$ P = \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix} $$ with \(E = \gamma m c^2\) being the relativistic energy and \((p_x, p_y, p_z)\) the relativistic momentum. The four-momentum transforms according to the Lorentz transformation for contravariant vectors in Minkowski space: $$ P^{\mu\prime} = \Lambda^{\mu}{}_{\nu} P^\nu \\ P^\prime = \begin{pmatrix} E^{\prime}/c \\ p_x^{\prime} \\ p_y^{\prime} \\ p_z^{\prime} \end{pmatrix} $$ where \(\Lambda\) is the Lorentz transformation matrix.
03

Evaluate the Lorentz Transformation for the Four-Momentum Vector

Now, we evaluate the Lorentz transformation applied to the four-momentum vector: $$ \begin{pmatrix} E^{\prime}/c \\ p_x^{\prime} \\ p_y^{\prime} \\ p_z^{\prime} \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma v/c & 0 & 0 \\ -\gamma v/c & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix} $$ By multiplying the matrices we obtain: 1. \(E^{\prime}/c = \gamma(E/c - v p_x/c^2)\) 2. \(p_x^{\prime} = \gamma(p_x - vp_z)\) 3. \(p_y^{\prime} = p_y\) 4. \(p_z^{\prime} = p_z\)
04

Simplify the Transformations and Compare with the Given Equations

Now we need to simplify the results and compare them with the given equations for momentum and energy. 1. Multiplying both sides of the first equation by c gives \(E^{\prime} = \gamma\left(E-v p_x\right)\) 2. The second equation is \(p_x^{\prime} = \gamma\left(p_x - v E / c^2\right)\) 3. The third equation is \(p_y^{\prime} = p_y\) 4. The fourth equation is \(p_z^{\prime} = p_z\) Comparing these equations with the given ones, we see that they match. Thus, we have successfully shown that momentum and energy transform from one inertial frame to another according to the given equations by analyzing the space-time Lorentz transformation.

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