# Video: Determining the Relative Velocity Required for Apparent Simultaneity of Measured Events

Two astronomical events are observed from Earth to occur at a time of 0.30 s apart and a distance separation of 2.0 × 10⁹ m 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?

06:24

### 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, we’ll assume that the speed of light 𝑐 is exactly 3.00 times 10 to the eighth meters per second. Let’s start by highlighting some of the vital information given to us. We’re told that in the Earth frame of reference, the two events occur a time of 0.30 seconds apart. We’ll call that value 𝑡 sub 𝑒 for the time in the Earth’s reference frame. In that same reference frame, the events are separated by a distance of 2.0 times 10 to the ninth meters. We’ll call that distance 𝑑 sub 𝑒. 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. We’ll call that speed 𝑣 sub 𝑠 for the speed of the spacecraft.

Let’s 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 we’ve 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 we’ve called 𝑣 sub 𝑠. In the frame of reference of the spacecraft, the two events are simultaneous for a given 𝑣 sub 𝑠 that will solve for. Relative to the Earth, we’re told how far the events are separated from one another in time 𝑡 sub 𝑒 and in distance 𝑑 sub 𝑒.

If we look carefully at our problem statement, we see that we’ve been told the time value in both of our reference frames, the Earth’s frame and the spaceship’s frame. In the Earth’s frame, the time separation of the events is 𝑡 sub 𝑒 which is 0.30 seconds. In the spaceship’s frame, the events happen at the same time. In other words, the time separation is zero seconds. So there’s the time passed experienced in the Earth’s 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 𝑡 prime, the time past experienced in one reference frame, is equal to 𝑡, the time passed experienced in another inertial reference frame, minus the speed of the object, 𝑣, divided by 𝑐 squared multiplied by the displacement of the object in the second reference frame, the same as for 𝑡. All this is divided by the square root of one minus 𝑣 squared over 𝑐 squared.

If we apply this Lorentz relationship to our scenario, then 𝑡 prime becomes 𝑡 sub 𝑠, the time passed experienced in the spaceship’s reference frame which equals 𝑡 sub 𝑒 minus 𝑣 sub 𝑠, the speed of the spaceship, divided by the speed of light squared times 𝑑 sub 𝑒, the distance between the events as observed from the Earth, all divided by the square root of one minus the spaceship’s speed squared over 𝑐 squared. We’re told that in the spaceship’s reference frame, these two events happen at the same time. That means that 𝑡 sub 𝑠 is equal to zero. So as we saw for 𝑣 sub 𝑠, the spaceship’s 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 𝑡 sub 𝑒 minus 𝑣 sub 𝑠 over 𝑐 squared times 𝑑 sub 𝑒 is equal to zero.

We’ll now rearrange this equation to solve for 𝑣 sub 𝑠. Let’s start by adding 𝑣 sub 𝑠 divided by 𝑐 squared times 𝑑 sub 𝑒 to both sides of the equation. This cancels out the term 𝑣 sub 𝑠 times 𝑑 sub 𝑒 divided by 𝑐 squared on the left-hand side of our equation. If we now multiply both sides of the equation by 𝑐 squared divided by 𝑑 sub 𝑒, then on the right-hand side of our equation, 𝑐 squared cancels out as does 𝑑 sub 𝑒 leaving us with 𝑣 sub 𝑠 by itself. So 𝑣 sub 𝑠, the speed of the spaceship, is equal to 𝑐 squared times 𝑡 sub 𝑒 divided by 𝑑 sub 𝑒.

𝑡 sub 𝑒 and 𝑑 sub 𝑒 are given in our problem statement and 𝑐, we’ll 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 spaceship’s speed, 𝑣 sub 𝑠, 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. That’s how fast the spaceship needs to be moving so that these two events appear simultaneous in the reference frame of the spaceship.