# Which observer is correct?

1. Jan 16, 2006

### Oxymoron

A muon is created 3km above the Earth's surface heading downward at a speed of $0.98c$. It is able to survive $2.2\mu s$ in its own frame before it decays.

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

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

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

$$\gamma_{v} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{(0.98)^2c^2}{c^2}}} = 5.03$$

$$\Delta t = \gamma_{v}\Delta t_0 = 5.03 \times 2.2\times 10^{-6}s = 11.1\mu s$$

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

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

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!

$$L = \frac{L_0}{\gamma_{v}} = \frac{3km}{5.03} = 597m$$

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

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

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 $0.98c$, in the muons frame? If so, I should get the same answer right?

Forgive my elementary-ness, Ive only just begun studying this stuff!

Last edited: Jan 16, 2006
2. Jan 16, 2006

### George Jones

Staff Emeritus
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.

Regards,
George

3. Jan 16, 2006

### Staff: Mentor

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 $L_0/\gamma$ = 3,000/5.03 = 596m. So everyone agrees that the muon survives long enough to hit the earth.

Right!

Right. And you do: Both frames agree that the muon hits the earth.

Last edited: Jan 16, 2006
4. Jan 16, 2006

### robphy

..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."

5. Jan 16, 2006

### Staff: Mentor

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. Jan 16, 2006

### robphy

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. Jan 16, 2006

### pervect

Staff Emeritus
I agree. Using "getting the right answer" as a judgement about which aproach 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.