Special Relativity , firework question

AI Thread Summary
The discussion revolves around the concepts of simultaneity and the calculation of event timing in the context of special relativity. It is established that the two firecracker explosions are not simultaneous despite their light reaching an observer at the same time, as the distances to each explosion differ. The calculations reveal that Firecracker 1 exploded at 1.0μs and Firecracker 2 at 2.0μs. A similar problem involving lightning bolts is also analyzed, emphasizing the importance of understanding the timing of events and the finite speed of light. The conclusion highlights that simultaneity is relative and depends on the observer's position.
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Homework Statement



You are standing at x=600m. Firecracker 1 explodes at x=0m. Firecracker 2 explodes at x=900m. The flashes from both explosions reach your eyes at t=3.0μs.
a. Are the two events simultaneous. Why?
b. at what time did each firecracker explode.?


Homework Equations


Im not sure if there is an exact equation.



The Attempt at a Solution



I know the answers are a) no, b) t1=1.0μs, t2=2.0μs



Can anyone please help me figure out how to work these questions, any help will be appreciated.
 
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The fact that the two flashes of light arrived at the same time doesn't make the two events simultaneous. For them to be simultaneous, they must have actually occurred at the same time. That is, if you calculate the time at which they happened, taking into account the finite speed of light, you have to find that they happened at the same time.

For part b, there are no tricks to it. Time elapsed is just distance divided by speed, as it was in Newtonian mechanics. You know the distances, and the speed of light is a constant in all inertial reference frames.
 
a. The distances between you and explosions are not equal(600m and 300m). However you receive the flashes on same time. It means the closer one must have occurred later.

b. velocity is c=3*10^8m/s. for the first explosion, time t= t1+d1/c, 3.0μs = t1 + 600/c
which gives t1 = 1.0μs; same calculation for second one..

Hope it helps.
 
Thanks ideasrule and q11we both your posts helped me a lot.
 
I have another similar problem except this time no time is given.

Person A is standing at x=9.0 km. Lighting bolt 1 strikes at x=0 km and lighting bolt 2 strikes at x = 12.0 km. A second person (person B) is standing at x=3.0km. Person A experiences the both lighting bolts at the same time.

question: What time difference does person B experience between 2 events.

My attempt:
fist i converted the two differences
Person B—Bolt 1= 3km to 3000m
Person B—Bolt 2 = 9km to 9000m

Then i divided each one by the speed of light.
3000m/300m∕μs= 10μs
9000m/300m∕μs= 30μs

Lastly i just added them together to get 40μs. The answer the paper gives me is 40μs I am just unsure if i did the problem right i might of just got lucky.
 
My equations show 40μs as well. However, I suspect that you actually just got lucky. Let's go through what is going on in this problem.

This problem (as you noticed) is different from the previous one because it does not include the time information. However, it does give you the critical piece of information regarding simultaneity by telling you that person A experienced both lightning bolts simultaneously.

You did right by evaluating the time light would take to go from each lightning bolt to each person. However, we must also take into account that lightning bolt 1 actually occurred first. Because lightning bolt 1 occurred before any other event happened, let's say that it happened at time t=0s. In that case, then lightning bolt 2 must have occurred at time t=20μs, in order for the light to reach person A simultaneously.

Next, ask yourself, "At what time will the light from lightning bolt 1 reach person B?"

Next, ask yourself"At what time will the light from lightning bolt 2 reach person B?"

The difference between these two times will yield the time difference. That was my strategy for solving this. You may have had an insight that I didn't, and so you might have taken a different, albeit equivalent, path to the answer.
 
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