# B Yet another question about Simultaneity

1. Jun 2, 2017

### MartinWyckmans

This is my train of thought, where did I go wrong?

Say a sensor is moving at a speed of c-1 m/s to the right. Somewhere to the left of the sensor there's a wall with a small opening. 2 balls fall down behind the wall and the distance between the balls is 1 meter, and when they pass the small opening the ray of light that gets reflected by the ball gets sent through the opening. The ray of light moves to the right, towards the sensor, with a speed of c. Taking the sensor as point of view, the light travels towards the sensor with 1 m/s. If the second ray of light coming from the second ball hits the sensor 1 second later than the first sensor, then the rays of light began travelling at the same time and the balls would fall simultaneously. In other words, if $$\Delta t = \Delta x / (c-v)$$ with $$\Delta t$$ the time difference measured in the sensor, then the actions happen simultaneously.
And say the sensor is moving towards the wall/balls/ray of light, then the formula would be $$\Delta t = \Delta x / (c+v)$$
My teacher dismissed my question almost immediately because he said (c+v) is impossible, but I'm not saying the sensor is travelling with a speed that's (c+v). Is it still wrong?

I don't know if I've made myself very clear, so maybe an analogy works better.

Say there're 3 ancient cities; Sparta, Athens and Rome. One day, a messager from Athens visits Sparta, saying the city has been sacked. Two days later, a messager from Rome visits Sparta, saying Rome has been sacked. The Spartans are fearful and wonder whether the same empire sacked both cities. They know that the distance between Sparta and Athens is 200 kilometers and the distance between Sparta and Rome is 600 kilometers. The messagers are trained to run 200 kilometers each day. The time between the two messagers was 2 days. Calculating the time difference with the formula: $$\Delta t = \Delta x / v$$ $$2 = 400 / 200$$ and we see that the time difference is indeed 2 days.

Does this not apply in special relativity? I know that getting the right velocity v would be very difficult to really use, but is this how it theoretically works?

2. Jun 2, 2017

### Vitro

Welcome to PF!
In which frame of reference? Distance is relative.
The light travels toward the sensor at speed c, the speed of light is c in every inertial frame of reference.

3. Jun 2, 2017

### Staff: Mentor

No, it doesn't. Velocities add differently in relativity; see here:

If you plug in c and c minus 1 m/sec in the formula, you will see that it gives the result c, which means that light moves at c relative to the sensor from the sensor's point of view.

4. Jun 2, 2017

### MartinWyckmans

Thanks both of you, I'm beginning to better understand it now!