# Relativity of Simultaneity and lightening

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1. Jan 19, 2017

### WWCY

1. The problem statement, all variables and given/known data
You are standing at x = 9.0 km and your assistant is standing at x = 3.0 km. Lightning bolt 1 strikes at x = 0 and lighting Bolt #2 strikes at x = 12.0 km. You see the flash from Bolt #2 at t = 10 μs and the flash from Bolt #1 at t = 50 μs. According to your assistant, were the lightning strikes simultaneous? If not, which occurred first, and what was the time difference between the two?

2. Relevant equations
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3. The attempt at a solution
Drawing a ct/t graph it is clear that the observer sees both flashes simultaneously (while the bolts strike at different times in my frame of reference). The solution manual states however, that the observer will say that the bolts do not strike at the same time as he/she is standing in 'my' frame of reference. My questions are:

1. In the case of the assistant does seeing the flashes =/= measuring the strike? How would one be able to see a simultaneous flash but measure a non-simultaneous strike?

2. Why does the assistant share the same frame of reference with 'me'? Is it because of the equation x = x' +vt where v (assistant w.r.t 'me') = 0 and therefore x = x'? (Where x is 'my' frame and x' is the assistant's frame)

2. Jan 19, 2017

### BvU

You both calculate the (x,t) of the flashes in the same frame of reference: your relative speed is 0.
So
They do, but NOT at 10 c.q. 50 $\mu$s
So
Indeed, No. He calculates $\ t_{\rm strike} = t_{\rm observe} - {\rm distance}/c \$ just like you do.

Last edited: Jan 19, 2017
3. Jan 19, 2017

### WWCY

I'm sorry but I'm not quite sure i get what you mean, if the assistant sees the flash simultaneously, how does he go on and deduce that the strikes do not occur simultaneously?

4. Jan 19, 2017

### BvU

He calculates, just like you do.

5. Jan 19, 2017

### WWCY

Is this because he sees a simultaneous flash at different distances and concludes that they couldn't have happened at the same time due to constant c?

As for:
This means that v = 0, x' always = x, and therefore both frames s' and s are the same and coinciding, right?

Apologies if this comes across as obvious to you, I'm barely beginning to wrap my head around these concepts.

6. Jan 19, 2017

### BvU

Correct. You calculate $t_2= 0$ and $t_1 = 20 \mu$s. From that you found that your assistant observes both flashes at the same time, namely at $t=30 \mu s$. From that, your assistant calculates the same $t_1$ and $t_2$ as you.

Very important for your understanding for the upcoming relativity curriculum:
Events have a time and a place in 'spacetime'. Independent observers, even when in different frames of reference, must all agree on the spacetime between events. Not on the difference in time and/or place separately.
Observers in the same inertial frame of reference must all agree on time and place in their own frame.​

Completely correct. We always assume that clocks in the same frame are synchronized perfefctly.
No apologies. I well remember the shattering experience of first contact with relativity as a first-year student (long ago ).

7. Jan 19, 2017

### WWCY

Thanks a lot! However there is one more thing that still really bugs me.

Since me and the assistant stand on different points of the x axis, how does this affect (or not affect) our frames of reference? If i drew a frame for the assistant with his origin on x = 3, and drew one for myself with origin on x = 9, why are our frames of reference still the same and is there a mathematical proof to this?

I know I'm missing something but I'm not sure what it is...

8. Jan 19, 2017

### BvU

You and your assistant can agree on a common frame of reference via a simple coordinate transformation that is constant in time and place. That you indeed need to do that if you want to communicate about events is not really relevant. For the scenario in this exercise you do need to agree.

9. Jan 19, 2017

### WWCY

I believe that I'm starting to understand now, thank you for the help and patience, appreciate it :)