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First, let’s restate the problem clearly:

- A friend (the traveler) moves at a speed of \(0.8c\).  
- On Earth, we measure his flight time to be \(3\) days.  
- We want to find out how long **he** (the traveler) measures the trip to take, i.e., the **proper time**.

---

## The Time Dilation Formula

The key idea from Special Relativity is time dilation. The standard formula is:

\[
t_{\text{obs}} \;=\; \gamma \, t_{\text{proper}},
\]

where:

- \(t_{\text{obs}}\) is the time interval measured by an observer who sees the clock moving (Earth’s reference frame, in this problem).
- \(t_{\text{proper}}\) is the **proper time**, i.e., the time measured by the clock moving along with the traveler (the pilot’s reference frame).
- \(\gamma\) (the Lorentz factor) is given by
  \[
  \gamma \;=\; \frac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}.
  \]

Because \(t_{\text{obs}} > t_{\text{proper}}\), the traveler always measures **less** time passing than someone who watches them fly by at high speed.

---

## Identify “Whose Time is Which?”

- **On Earth:** We observe (or measure) the flight to last \(3\) days.  
  - This \(3\) days is \(t_{\text{obs}}\).
- **The traveler’s own watch:** We want to find \(t_{\text{proper}}\).

Since we know \(t_{\text{obs}} = 3\) days and \(v = 0.8c\), we can compute \(\gamma\):

\[
\gamma = \frac{1}{\sqrt{1 - (0.8)^2}}
        = \frac{1}{\sqrt{1 - 0.64}}
        = \frac{1}{\sqrt{0.36}}
        = \frac{1}{0.6}
        = \frac{5}{3}
        \approx 1.6667.
\]

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## Compute the Traveler’s Time

Rewriting the time-dilation formula to solve for \(t_{\text{proper}}\):

\[
t_{\text{obs}} = \gamma \, t_{\text{proper}}
\quad\Longrightarrow\quad
t_{\text{proper}} = \frac{t_{\text{obs}}}{\gamma}.
\]

Plugging in the numbers:

\[
t_{\text{proper}}
= \frac{3 \text{ days}}{1.6667}
\approx 1.8 \text{ days}.
\]

So **the correct result** is that the pilot’s own watch reads **\(1.8\) days**.

---

## Common Source of Confusion

Sometimes there’s a mix-up about who measures which time. If the problem wording were flipped—if it said, “The pilot measures 3 days on his watch; how long does Earth measure?”—we would do the inverse calculation and get a longer time on Earth’s clocks. That might be how the teacher ended up with a larger figure (like 5 days). However, **as the question stands**—Earth sees 3 days, so the traveler must see **less** than 3 days because of time dilation. That leads to about **1.8 days** for the traveler.

---

### Bottom line:

- **Earth** sees **3 days**.  
- **Traveler** sees **1.8 days**, not 5 days.

If your teacher got 5 days, it’s almost certainly because they accidentally used the “proper time” and “observed time” the wrong way around. In special relativity, the clock moving at high speed **always** records **less** elapsed time than the stationary observer’s clock.

Tue, 14 Jan 2025 22:28 GMT