- #1
LCSphysicist
- 644
- 162
- Homework Statement
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- Relevant Equations
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i am having some hard time thinking about this problem:
It is basically this:
Imagine a bulb and a receptor distant L from each other (at the same axis x) inside a room, the roof of the room is at a height d from the bulb and receptor. Now you are at a train moving horizontally, parallel to the x axis, with speed v. You are looking to the room. What time does it take in your reference frame to a photon emitted by the bulb to reach the receptor, in such way that the photon should reflect at the roof, and not be direct emitted along the closest distance between them.
The problem is that i am having two different answers, using different approach:
(1) The horizontal distance will be contract, so the time it take is $$t' = \frac{2\sqrt{(L/2\gamma)^2 + d^2}}{c}$$
(2) Using lorentz transformation, the time it take is $$t' = \gamma t - \gamma \beta \Delta x/c = \gamma (t -\Delta x \beta /c) = \gamma (2\sqrt{(L/2)^2 + d^2}/c - L \beta / c)$$
Now $$|\beta / c| << 1$$, so $$t' \approx \frac{2 \gamma \sqrt{(L/2)^2 + d^2}}{c} $$.
Since i should choose just one answer, i would say that the first one use more solid arguments, but still i can not say definitely if the first is in fact the right answer, or in another words, i am not able to refute the second answer.
So which one is right? what argument the wrong answer used? Thank you.
It is basically this:
Imagine a bulb and a receptor distant L from each other (at the same axis x) inside a room, the roof of the room is at a height d from the bulb and receptor. Now you are at a train moving horizontally, parallel to the x axis, with speed v. You are looking to the room. What time does it take in your reference frame to a photon emitted by the bulb to reach the receptor, in such way that the photon should reflect at the roof, and not be direct emitted along the closest distance between them.
The problem is that i am having two different answers, using different approach:
(1) The horizontal distance will be contract, so the time it take is $$t' = \frac{2\sqrt{(L/2\gamma)^2 + d^2}}{c}$$
(2) Using lorentz transformation, the time it take is $$t' = \gamma t - \gamma \beta \Delta x/c = \gamma (t -\Delta x \beta /c) = \gamma (2\sqrt{(L/2)^2 + d^2}/c - L \beta / c)$$
Now $$|\beta / c| << 1$$, so $$t' \approx \frac{2 \gamma \sqrt{(L/2)^2 + d^2}}{c} $$.
Since i should choose just one answer, i would say that the first one use more solid arguments, but still i can not say definitely if the first is in fact the right answer, or in another words, i am not able to refute the second answer.
So which one is right? what argument the wrong answer used? Thank you.