Special Relativity - "Simultaneity"

In summary, an experimenter triggers two lights simultaneously, one at the origin of their reference frame and one at x = 23.8 km. An observer moving with speed 0.347c in the positive x direction also views the flashes. Using the equation \Delta t = \gamma (\Delta t' + \frac{v \Delta x'}{c^2}), the time interval between the flashes, according to the observer, is 2.935 \times 10^{-5} s or 29.35 \mu s.
  • #1
FaraDazed
347
2

Homework Statement


An experimenter triggers two lights simultaneously, which produces a
large flash located at the origin of his reference frame and a small one at x = 23.8 km.
An observer, who moves with speed 0.374c in the positive x direction, also views the
flashes. What is the time interval in seconds between them, according to her (the observer)?

Homework Equations


[itex]
\Delta t = \gamma (\Delta t' + \frac{v \Delta x'}{c^2})
[/itex]

The Attempt at a Solution


I have use the equation above and got an answer. However, if I am correct, I do not fully understand it, and the question also asks us to clearly explain each step of our solution, so would appreciate some advice/help with that. And if I am not correct then I would appreciate if someone could point out what I did wrong and any feedback.

I used the above equation, and because of the way the question is worded I assume that t' is zero, i.e. according to the experimenter there is no time between the flashes so then the above equation becomes...
[tex]
\Delta t = \gamma (\frac{v \Delta x'}{c^2}) \\
\Delta t = \frac{c^{-2}v \Delta x'}{\sqrt{1-\frac{v^2}{c^2}}} \\
\Delta t = \frac{(3 \times 10^8)^{-2}(0.347 \times 3 \times 10^8) (23800)}{\sqrt{1-\frac{(0.347 \times 3 \times 10^8)^2}{(3 \times 10^8)^2}}} = 2.935 \times 10^{-5} s = 29.35 \mu s
[/tex]

Thanks :)
 
Last edited:
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  • #2
Looks good, although that Δx should be a Δx'.

Chet
 
  • #3
Chestermiller said:
Looks good, although that Δx should be a Δx'.

Chet
Thanks :) Ah yes, I typed it up wrong once and then copied and pasted it through the calculation :oops:. I've gone back and changed it now :).
 

1. What is the concept of simultaneity in special relativity?

Simultaneity in special relativity refers to the idea that events that appear to occur at the same time for one observer, may not appear simultaneous for another observer moving at a different velocity. This is due to the relativity of time and space, as predicted by Einstein's theory.

2. How does the relativity of simultaneity affect our perception of time?

The relativity of simultaneity means that time is not absolute, and can be experienced differently by different observers. This challenges the traditional notion of a universal "now", and suggests that time is relative to an observer's frame of reference.

3. Can two events occur simultaneously in different frames of reference?

Yes, two events can appear simultaneous in one frame of reference, but not in another. This is because the perception of simultaneity is relative to an observer's frame of reference and their relative velocity.

4. How is the concept of simultaneity related to the speed of light?

The speed of light is a fundamental constant in the universe and is the same for all observers, regardless of their frame of reference. This means that the measurement of time and space is relative to the speed of light, and the concept of simultaneity is affected by an observer's relative motion to the speed of light.

5. What are some real-life applications of the relativity of simultaneity?

The relativity of simultaneity has been proven to be true in many experiments, and has important implications for technologies such as GPS navigation and particle accelerators. It also helps to explain phenomena such as time dilation and the twin paradox.

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