91影视

Consider at t=0 , a bright beacon at the origin flashes, sending light uniformly in all directions. Anna is moving at speed v in the +x direction relative to the beacon and passes through the origin at t=0.

Write the formula of beacon鈥檚 reference frame is within an angle $胃$of the +x-axis.

${{\mathit{u}}^{\mathbf{\text{'}}}}_{\mathbf{x}}\mathbf{=}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{V}}{{\mathbf{c}}^{\mathbf{2}}}}$ 鈥︹�� (1)

Here, ${\mathbf{u}}_{\mathbf{x}}$ is certain velocity, v is moving speed and c is speed of light.

Length contraction happens in the same direction of motion as Anna travels in the x-direction with a velocity v, but the vertical component maintains the same value (I may refer here to problems${}^{38}$and ${}^{39}$of this chapter). The conclusion is that these items appear to have been at different angles. The following relation will be reached mathematically if we suppose that we are travelling with the same x-component of a light beam in frame${}^{{}^{\left(S^{\text{'}}\right)}}$.

Determine the beacon鈥檚 reference frame is within an angle $胃$of the +x-axis.

Substitute 0 for $u_x^{\text{'}}$and $c\cos \left(胃\right)$ for v into equation (1).

$$\begin{gathered} 0=\frac{c\cos \left(胃\right)-v}{1-{\displaystyle \frac{u_{x}v}{c^2}}} \\ c\cos \left(胃\right)-v=0 \\ c\cos \left(胃\right)=v \\ 胃={\cos }^{-1}\left(\frac{v}{c}\right) \end{gathered}$$

Remember that the denominator of the velocity transformation relation is always positive (for speeds less than c) and thus for a given velocity $u_x=v$, Anna will distinguish between positive and negative values of the velocity recorded in her reference frame.

To examine the limiting situations, let's now utilise the previous relation:

1. The angle $胃$will equal $\frac{\mathrm{蟺}}{2}=90掳$if $v=0$, or when this frame is motionless with regard to the bright beacon. By extension, both frames will include just positive light that is released in the positive x -direction, say on her right.

2. The angle will be zero when $v=c$, which denotes that Anna will be travelling at the speed of light. In other words, the window where Anna perceives the light is becoming smaller until it finally reduces to a single line. According to Anna, the light beam will be regarded as positive if it is travelling along the positive x-axis. The remainder of the beams, however, will appear to include a negative element.

In every frame of reference, the overall velocity's magnitude remains constant atc. The relative velocity between the observer and the light beam will remain constant, even for an observer travelling at the speed of light. Since Anna and the light are synchronised at$t=t^{\text{'}}=0$, the light will continue to move ahead of her.

Therefore, the phenomenon is known as the light beam maintaining constant speed across all frames of reference; it will continue to travel in front of Anna even if she moves at a speed of $v=c$.