### Video Transcript

Two astronomical events are observed from Earth to occur at a time of 0.30 seconds apart and the distance separation of 2.0 times 10 to the ninth meters from each other. How fast must a spacecraft travel from the site of one event toward the other to make the events occur at the same time when measured in the frame of reference of the spacecraft?

In this problem, weโll assume that the speed of light ๐ is exactly 3.00 times 10 to the eighth meters per second. Letโs start by highlighting some of the vital information given to us. Weโre told that in the Earth frame of reference, the two events occur a time of 0.30 seconds apart. Weโll call that value ๐ก sub ๐ for the time in the Earthโs reference frame. In that same reference frame, the events are separated by a distance of 2.0 times 10 to the ninth meters. Weโll call that distance ๐ sub ๐. We want to solve for the speed of the spacecraft such that in the frame of reference of the moving spacecraft, the events occur simultaneously. Weโll call that speed ๐ฃ sub ๐ for the speed of the spacecraft.

Letโs begin our solution by drawing a diagram of the situation. We have an observer based on Earth and this observer sees things through a particular reference frame that weโve drawn here in three dimensions. The observer sees two events happening in space. These two events are points on the path of another reference frame. A spacecraft travels on a straight line distance from the first to the second event with some speed weโve called ๐ฃ sub ๐ . In the frame of reference of the spacecraft, the two events are simultaneous for a given ๐ฃ sub ๐ that will solve for. Relative to the Earth, weโre told how far the events are separated from one another in time ๐ก sub ๐ and in distance ๐ sub ๐.

If we look carefully at our problem statement, we see that weโve been told the time value in both of our reference frames, the Earthโs frame and the spaceshipโs frame. In the Earthโs frame, the time separation of the events is ๐ก sub ๐ which is 0.30 seconds. In the spaceshipโs frame, the events happen at the same time. In other words, the time separation is zero seconds. So thereโs the time passed experienced in the Earthโs reference frame and then a different time passed experienced in the spaceship frame.

There is a mathematical relationship that connects the time passed in one frame to the time passed in another inertial reference frame. This relationship for time is called a Lorentz transformation. This equation says that ๐ก prime, the time past experienced in one reference frame, is equal to ๐ก, the time passed experienced in another inertial reference frame, minus the speed of the object, ๐ฃ, divided by ๐ squared multiplied by the displacement of the object in the second reference frame, the same as for ๐ก. All this is divided by the square root of one minus ๐ฃ squared over ๐ squared.

If we apply this Lorentz relationship to our scenario, then ๐ก prime becomes ๐ก sub ๐ , the time passed experienced in the spaceshipโs reference frame which equals ๐ก sub ๐ minus ๐ฃ sub ๐ , the speed of the spaceship, divided by the speed of light squared times ๐ sub ๐, the distance between the events as observed from the Earth, all divided by the square root of one minus the spaceshipโs speed squared over ๐ squared. Weโre told that in the spaceshipโs reference frame, these two events happen at the same time. That means that ๐ก sub ๐ is equal to zero. So as we saw for ๐ฃ sub ๐ , the spaceshipโs speed, we only need to address the numerator of this fraction. If the numerator is zero, then the whole fraction itself is zero. Therefore, we can write that ๐ก sub ๐ minus ๐ฃ sub ๐ over ๐ squared times ๐ sub ๐ is equal to zero.

Weโll now rearrange this equation to solve for ๐ฃ sub ๐ . Letโs start by adding ๐ฃ sub ๐ divided by ๐ squared times ๐ sub ๐ to both sides of the equation. This cancels out the term ๐ฃ sub ๐ times ๐ sub ๐ divided by ๐ squared on the left-hand side of our equation. If we now multiply both sides of the equation by ๐ squared divided by ๐ sub ๐, then on the right-hand side of our equation, ๐ squared cancels out as does ๐ sub ๐ leaving us with ๐ฃ sub ๐ by itself. So ๐ฃ sub ๐ , the speed of the spaceship, is equal to ๐ squared times ๐ก sub ๐ divided by ๐ sub ๐.

๐ก sub ๐ and ๐ sub ๐ are given in our problem statement and ๐, weโll treat as the speed of light of 3.00 times 10 to the eighth meters per second. When we enter these values in for these symbols, we see that the spaceshipโs speed, ๐ฃ sub ๐ , is 3.00 times 10 to the eighth meters per second squared times 0.30 seconds divided by 2.0 times 10 to the ninth meters which is equal to a speed of 1.4 times 10 to the seventh meters per second. Thatโs how fast the spaceship needs to be moving so that these two events appear simultaneous in the reference frame of the spaceship.