Who has the correct perspective on the muon's journey?

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In summary, the muon's lifetime is shorter in the observer's frame because the muon perceives the distance to the ground as being shorter.
  • #1
Oxymoron
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A muon is created 3km above the Earth's surface heading downward at a speed of [itex]0.98c[/itex]. It is able to survive [itex]2.2\mu s[/itex] in its own frame before it decays.

(1) The muon travels a distance of 647m before it decays in its frame

[tex]d = vt = (0.98\times 3\times 10^8m/s)(2.2\times 10^{-6}s) = 647m[/tex]

(2) To an observer on Earth the muon's lifetime is longer. To the Earth observer, [itex]11.2\mu s[/itex] have passed in the time it took for the muon to decay.

[tex]\gamma_{v} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{(0.98)^2c^2}{c^2}}} = 5.03[/tex]

[tex]\Delta t = \gamma_{v}\Delta t_0 = 5.03 \times 2.2\times 10^{-6}s = 11.1\mu s[/tex]

(3) To the Earth observer the muon now travels a distance of 3.25km

[tex]d = vt = (0.98\times 3 \times 10^8m/s)(11.1\times 10^{-6}s) = 3.25km[/tex]

From these calculations, the muon actually hits the Earth in the observer's frame on Earth. But in the muon's frame it doesn't travel far enough. Who is right? If this happened in real life, would the muon hit the ground or not (to an observer on the ground or to a muon).

Lets take this a little further.

Does the muon actually percieve the distance to the ground as being 3.5km? To the observer on the ground the distance is well and truly 3.5km. But length contraction says that length contracts in the direction of motion. So the observer measures the length to be 3.5km in his frame, but that is simply because he is not moving relative to the distance.

The muon, however, IS moving relative to the distance. In fact the muon thinks that the distance to the ground is 597m, NOT 3.5km!

[tex]L = \frac{L_0}{\gamma_{v}} = \frac{3km}{5.03} = 597m[/tex]

So according to the muon, the distance to the ground is only 597m, which it will travel through in a mere [itex]2.03\mu s[/itex].

[tex]t = \frac{L}{u} = \frac{597m}{0.98\times 3\times 10^8m/s} = 2.03\mu s[/tex]

Therefore, according to the muon it hits the Earth.

If all my reasoning is correct here, my initial problem of discerning who was right (did the muon hit or not?) was incomplete. I hadn't adjusted for length contraction for the moving object. I this the reason why I had my dilemma?

Is it safe to say that the muon is at rest and the Earth is moving toward it at [itex]0.98c[/itex], in the muons frame? If so, I should get the same answer right?

Forgive my elementary-ness, I've only just begun studying this stuff!
 
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  • #2
Oxymoron said:
A muon is created 3km above the Earth's surface heading downward at a speed of [itex]0.98c[/itex].

I strongly believe (Many people disagree with me!) that questions of this type should be handled using invariance of the spacetime interval, i.e., the metric.

Isolate the key events - here creation and decay of the muon - and use the metric to determine what happens.

See my comments (and comments by others) in https://www.physicsforums.com/showthread.php?t=106450" about a similar problem.

Regards,
George
 
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  • #3
Oxymoron said:
From these calculations, the muon actually hits the Earth in the observer's frame on Earth. But in the muon's frame it doesn't travel far enough. Who is right? If this happened in real life, would the muon hit the ground or not (to an observer on the ground or to a muon).
You have misinterpreted your first calculation: The 647m is how far the muon would travel in its lifetime according to the muon's frame. But in the muon's frame, the distance to the Earth is only [itex]L_0/\gamma[/itex] = 3,000/5.03 = 596m. So everyone agrees that the muon survives long enough to hit the earth.

If all my reasoning is correct here, my initial problem of discerning who was right (did the muon hit or not?) was incomplete. I hadn't adjusted for length contraction for the moving object. I this the reason why I had my dilemma?
Right!

Is it safe to say that the muon is at rest and the Earth is moving toward it at , in the muons frame? If so, I should get the same answer right?
Right. And you do: Both frames agree that the muon hits the earth.
 
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  • #4
George Jones said:
I strongly believe (Many people disagree with me!) that questions of this type should be handled using invariance of the spacetime interval, i.e., the metric.
Isolate the key events - here creation and decay of the muon - and use the metric to determine what happens.
See my comments (and comments by others) in https://www.physicsforums.com/showthread.php?t=106450" about a similar problem.
Regards,
George
..and a spacetime diagram should be drawn!

Then, many problems in special relativity reveal themselves to be a problem in [Minkowskian] geometry... often solvable with the intuition and analogues of methods learned in Euclidean geometry and trigonometry. In my opinion, this takes away some of the mystery and misunderstanding of the various "relativistic formulas".

"A spacetime diagram is worth a thousand words."
 
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  • #5
robphy said:
..and a spacetime diagram should be drawn!
...
"A spacetime diagram is worth a thousand words."
Absolutely!

I think true understanding of special relativity comes only with the mastery of the spacetime diagram and the invariant interval. Nonetheless, one should also be able to "explain" things in terms of the relativistic behavior of moving rods and clocks.
 
  • #6
Doc Al said:
Absolutely!
I think true understanding of special relativity comes only with the mastery of the spacetime diagram and the invariant interval. Nonetheless, one should also be able to "explain" things in terms of the relativistic behavior of moving rods and clocks.
I'll admit that I have had trouble thinking in terms of the "relativistic behavior of moving rods and clocks"... that's why I try to do problems invariantly first, then interpret physically. In other words, I put my faith in the mathematical reformulation of the physical problem, do some math, then reinterpret the physics. In some sense, I'm trying to use the geometry to develop my physical intuition. When I first appreciated the "geometric viewpoint" [in graduate school], relativity made so much more sense to me.
 
  • #7
George Jones said:
I strongly believe (Many people disagree with me!) that questions of this type should be handled using invariance of the spacetime interval, i.e., the metric.

I agree. Using "getting the right answer" as a judgement about which approach is best, the approach using the invariance of the space-time interval to solve SR problems seems to be the most reliable.

At the very least, it is an extremely simple "double check". You compute the Lorentz interval in one reference frame, and make sure that you get the same number in the second.
 

1. How do we determine which observer is correct?

The determination of which observer is correct depends on a variety of factors, including the position and perspective of the observer, the accuracy and reliability of their observations, and the validity of any supporting evidence or data. It is important to carefully consider all of these factors before making a determination.

2. What if two observers have conflicting observations?

In cases where two observers have conflicting observations, it is important to carefully review the evidence and consider alternative explanations. It may also be helpful to seek additional input from other experts in the field to reach a consensus.

3. Can an observer be considered "correct" if their observations are subjective?

In science, we strive to make objective and unbiased observations. However, some observations may still be considered "correct" if they are based on subjective experiences or interpretations. It is important to clearly communicate any potential subjectivity in observations and to support them with other evidence.

4. What is the role of scientific consensus in determining which observer is correct?

In many cases, scientific consensus plays a key role in determining which observer is correct. Consensus is reached through peer review and replication of experiments, and it indicates a widely accepted understanding within the scientific community.

5. How can we account for human error in determining which observer is correct?

Human error is a common occurrence in scientific observations and can greatly impact the accuracy and reliability of data. It is important for observers to carefully follow scientific protocols and for researchers to thoroughly review and validate data to minimize the impact of human error.

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