Problem 76
Question
(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-v t\) and \(t^{\prime}=t\) . In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{2}}=0$$ This has a different form than the wave equation in \(S\) . Hence the Galiean transformation violates the first relativity postulate that all physical laws have the same form in all inertial reference frames. (Hint: Express the derivatives \(\partial / \partial x\) and \(\partial / \partial t\) in terms of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S\) : $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)
Step-by-Step Solution
VerifiedKey Concepts
Galilean Transformation
- \( x' = x - vt \)
- \( t' = t \)
However, this approach works well when dealing with low speeds, much slower than the speed of light. At these speeds, the addition of velocities seems simple, and the temporal aspect requires no alteration. But when speeds approach the speed of light, such straightforward assumptions falter. The transformation does not accommodate the relativistic effects necessary for light-speed scenarios, leading to discrepancies in equations such as the wave equation for electromagnetic waves.
Lorentz Transformation
- \( x' = \gamma(x - vt) \)
- \( t' = \gamma(t - \frac{vx}{c^2}) \)
- where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
Using Lorentz transformations, the wave equation maintains its form across frames, proving the particle's behavior remains consistent. This confirms that laws of physics, such as the wave equation, sustain uniformity across all inertial frames when critical speed thresholds are approached.
Speed of Light
This invariance is a key part of Einstein's postulates of special relativity. In our context, properly applying Lorentz transformations rather than Galilean ensures the wave equation remains unchanged across inertial frames. The consistent form of the wave equation vindicates the constancy of \( c \) in every frame.
This constancy elucidated by the wave equation's form implies that in both frame \( S \) and \( S' \), light maintains its speed \( c \). Therefore, the universe's "speed limit" applies everywhere and every time, highlighting how relativity reshapes our understanding of speed and time.
Inertial Frames
The need for consistent laws across inertial frames leads us to inspect transformations like Galilean and Lorentz more closely. Especially at speeds closer to light, discrepancies arise if purely Galilean approaches are used. Lorentz transformations emerge necessary to preserve the law's nature across various inertial frames.
- All inertial frames are equivalent in their description of physical processes.
- The Lorentz transformation ensures identical physical laws across frames moving at constant, high velocities.