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An observer standing by the railroad tracks sees two bolts of lightning strike the ends of a 5.0 × 10²-m-long train simultaneously, at the instant the middle of the train passes him at a speed of 50 m/s. Find the time between the lightning strikes as measured by a passenger seated in the middle of the train.
An observer standing by the railroad tracks sees two bolts of lightning strike the ends of a 5.0-times-10-to-the-two-meter-long train simultaneously, at the instant the middle of the train passes him at a speed of 50 meters per second. Find the time between the lightning strikes as measured by a passenger seated in the middle of the train.
In this problem statement, we’re told that the train is 5.0 times 10 to the two meters long, we’ll call that 𝑥, and that the train is moving forward at a speed of 50 meters per second. We’ll call that 𝑣. We want to find the time between the lightning strikes as measured by a passenger seated in the middle of the train. We’ll call that 𝑡 prime.
As we work through 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 begin our solution by drawing a diagram of the scenario. In this situation, we have a train car moving past a platform. The instant of time we’re considering is when the center of the car is right in front of an observer standing on the platform.
According to this observer, two bolts of lightning strike either end of the car at the same time simultaneously. In the frame of reference of the person in the train car, however, the bolts do not strike simultaneously and in fact arrive at different times with that time difference measured as 𝑡 prime.
Knowing the length of the car and the speed at which it moves, 50 meters per second, we want to solve for the observed time difference between strikes for the passenger sitting in the middle of the train car. In this scenario, we have two different reference frames, the frame of the observer on the platform and the frame of the observer in the car.
To transition between them, we’ll use what’s called the Lorentz transformations. These are equations used to translate between one inertial frame of reference to another. These equations will let us solve for all four coordinate variables 𝑥, 𝑦, 𝑧, and 𝑡 in our two different reference frames.
We’ll define our primed coordinates, those with a prime sign above them, as the reference frame relative to the observer inside the train car and our unprimed coordinates as those relative to the observer on the platform. Looking through these transformations, we see that the final one, 𝑡 prime, in terms of the time 𝑡 on the platform looks like a close fit for what we wanna solve for.
If we apply this relationship to our scenario, then we see that 𝑡, the time between the events that is the strikes of lightning in the frame of reference of the platform observer, is zero because that observer sees them happening simultaneously. 𝑣 is given. That’s the speed of the train car. And 𝑥 is also given. That’s the distance between the two events.
Now what about 𝛾? We can recall that 𝛾, also known as the Lorentz factor, is equal to one divided by the square root of one minus 𝑣 squared divided by 𝑐 squared. We can plug in this expression for 𝛾 to our equation for 𝑡 prime. Looking back at our diagram, if we choose to let motion to the left opposite the direction of the train, the motion in the positive direction, then that means 𝑣 in our equation for 𝑡 prime is negative.
We can now plug-in for 𝑣, 𝑥, and 𝑐 in our equation for 𝑡 prime. When we plug in negative 50 meters per second for 𝑣, 5.0 times 10 to the two meters for 𝑥, and 3.00 times 10 to the eighth meters per second for 𝑐 and enter these numbers into our calculator, we find that to, two significant figures, 𝑡 prime is equal to 2.8 times 10 to the negative 13th seconds. That’s how much time separates the bolts of lightning according to an observer sitting in the middle of the train car.
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