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Special relativity

Can somebody please explain the following question to me please. I really don't understand why they used the half life to find the time taken to travel between the two detectors and why the detectors and lab don't have the same frame of reference B874EC5E-2BD1-45A4-8A0C-A14A46001E6C.jpg.jpeg 71EE0AB3-5EDB-4F76-B940-EA0578290B02.jpg.jpeg E6CF030D-ADA3-40E3-9006-29F4104B456B.jpg.jpeg
Original post by MissMathsxo
Can somebody please explain the following question to me please. I really don't understand why they used the half life to find the time taken to travel between the two detectors and why the detectors and lab don't have the same frame of reference B874EC5E-2BD1-45A4-8A0C-A14A46001E6C.jpg.jpeg 71EE0AB3-5EDB-4F76-B940-EA0578290B02.jpg.jpeg E6CF030D-ADA3-40E3-9006-29F4104B456B.jpg.jpeg


Okay. Here's an idea of the train of thinking; if it doesn't help, please say and I can elucidate the issue better... either ways, here goes:

From the first point where the mesons pass, you start a clock. By the end, in your frame, you record a time, and for that set velocity the mesons have, a distance, and so in that time you see a fraction of them disappear.

The mesons are at relativistic speeds, and so, in effect, time will be slower as you observe them with respect to yours: you can derive this quite simply w/ a light clock and pythagoras for many examples, alongside other derivations. It's a simple way of saying it, but you will just observe their clock moving slower than yours. So, for 3/4 of them to disappear, there must have been two half lives past: Is it in your frame, or their frame? If their time is moving slower than ours for example, then the half life we record will be longer for them, after all their time is slowed, and in effect their decay is slowed, correct?

SO! That's where this question goes. You need to apply a transformation for the half lives they pass. 3/4 are gone, so 0.25 remains, 25% of them, and so two half lives have passed. Not in your frame, but in theirs, in their frame 75% are gone, and so we observe that 75% have poofed.

Do you know a transformation that can do this?
Original post by Callicious
Okay. Here's an idea of the train of thinking; if it doesn't help, please say and I can elucidate the issue better... either ways, here goes:

From the first point where the mesons pass, you start a clock. By the end, in your frame, you record a time, and for that set velocity the mesons have, a distance, and so in that time you see a fraction of them disappear.

The mesons are at relativistic speeds, and so, in effect, time will be slower as you observe them with respect to yours: you can derive this quite simply w/ a light clock and pythagoras for many examples, alongside other derivations. It's a simple way of saying it, but you will just observe their clock moving slower than yours. So, for 3/4 of them to disappear, there must have been two half lives past: Is it in your frame, or their frame? If their time is moving slower than ours for example, then the half life we record will be longer for them, after all their time is slowed, and in effect their decay is slowed, correct?

SO! That's where this question goes. You need to apply a transformation for the half lives they pass. 3/4 are gone, so 0.25 remains, 25% of them, and so two half lives have passed. Not in your frame, but in theirs, in their frame 75% are gone, and so we observe that 75% have poofed.

Do you know a transformation that can do this?

Thank you for your response
I get the idea behind the question, I got others like it correct but surely the detector and the observer are in the same frame of reference which is a different one to the particles?
Original post by MissMathsxo
Thank you for your response
I get the idea behind the question, I got others like it correct but surely the detector and the observer are in the same frame of reference which is a different one to the particles?


Detector and observer are in the same frame, that's true, so that shouldn't cause a problem. The detector gets the lab frame time, not the time in the frame of the actual mesons.
Would you be able to explain what the first step in the mark scheme is trying to do please? Is the half life in the particles or our reference frame?
Oh or have I just completely understood the question and the first part shows the half life stuff in the frame of reference of the mesons and therefore how much has actually decayed and the second part shows the half life in the observer frame if you do not take into account the fact that the half life would be different for the two references? That would explain why I got the same answer for both as I was including special relativity theory ?
Original post by MissMathsxo
Would you be able to explain what the first step in the mark scheme is trying to do please? Is the half life in the particles or our reference frame?


Original post by MissMathsxo
Oh or have I just completely understood the question and the first part shows the half life stuff in the frame of reference of the mesons and therefore how much has actually decayed and the second part shows the half life in the observer frame if you do not take into account the fact that the half life would be different for the two references? That would explain why I got the same answer for both as I was including special relativity theory ?


I don't have the full question, so I can't answer it completely, well I can based on the MS but still :-;

Provided you can understand the question and you got the right answer, i.e. 58 for the time it took in the observers frame and understand that this is derived from the lorentz transform from the frame of the mesons, I'd be happy.
Original post by Callicious
I don't have the full question, so I can't answer it completely, well I can based on the MS but still :-;

Provided you can understand the question and you got the right answer, i.e. 58 for the time it took in the observers frame and understand that this is derived from the lorentz transform from the frame of the mesons, I'd be happy.


Yeah I think I understand it now. I kept getting this time and 2 half life's for both because I misunderstood the question. Thank you for your help

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