-1
$\begingroup$

The fact that tan 90° is undefined indicates a key issue in the problem. Does this mean it is impossible to observe rain falling at 90 degree?

When rain is observed to be falling at 90° with the y-axis: $$\tan \theta = \frac{1}{0}$$ This means,

  • velocity of rain with respect to ground is zero
  • velocity of man w.r.t to ground 1 which implies velocity = 1x = x (velocity can be anything)
$\endgroup$
2
  • 1
    $\begingroup$ You should start by stating what you would consider as "horizontal enough". What angle tolerance do you have? $\endgroup$ Commented May 15 at 6:40
  • $\begingroup$ @naturallyInconsistent Unfortunately if the maximum angle were given, it would make it a homework question. $\endgroup$ Commented May 15 at 9:01

2 Answers 2

7
$\begingroup$

If you are running horizontally and the rain is falling with any vertical component (as it must have, otherwise we would not call it "falling"), both with respect to the ground, then indeed in both Newtonian mechanics and special relativity will you never see the rain fall horizontally.

In Newtonian mechanics, the vertical component of the rain velocity is unaffected by the change of reference frame from ground to runner, so the velocity will retain its vertical component.

In special relativity, both velocity components transform because the time coordinate changes between frames. However, this transformation can never make the vertical velocity component zero, it can only rescale it by a multiplicative factor.

Hence in both cases, the velocity will have a non-zero velocity component in the runner's frame of reference.

$\endgroup$
4
  • 3
    $\begingroup$ There is a perpendicular velocity transformation law. Not that it matters much, but it is not exactly unaffected. $\endgroup$ Commented May 15 at 6:40
  • $\begingroup$ The condition that the rain is falling vertically is not necessary. If the velocity vector of rain has any non-zero vertical component then this will be unaffected by adding the horizontal velocity vector of the runner, so the rain is never horizontal in the runner's reference frame. $\endgroup$
    – gandalf61
    Commented May 15 at 7:26
  • $\begingroup$ @gandalf61 rewritten. $\endgroup$ Commented May 15 at 7:49
  • 1
    $\begingroup$ @naturallyInconsistent rewritten. $\endgroup$ Commented May 15 at 7:55
-1
$\begingroup$

$$\tan(\theta)=\frac{v_x}{v_y}$$ $\theta$ is the angle made with the $y$ axis, $v_x$ is the speed of the man (alternatively, the horizontal speed of rain from the man's view point) and $v_y$ is the vertical (falling) speed of the rain. If $\theta \to 90°$, $v_x \to \infty$. In other words, the man has to travel at an enormous speed. Practically, $\frac{v_x}{v_y}$ is always finite - no matter how large - so, $\theta=90°$ is not possible.

$\endgroup$
1
  • $\begingroup$ There is no train in the original question! $\endgroup$ Commented May 15 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.