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consider you are in a train that moves with a speed of $c/4$. A bulb is hanged from the ceiling. A light beam leaves the bulb downward. Exactly under the bulb on the floor is a detector. According to the principle of relativity the light will not strike the detector in the reference frame of a non-moving observer (standing out of train), because the velocity of light must be constant (light should continue it's motion vertically). But obviously the observer on the train should see that the beam strikes the detector.

But both can not be correct. (The beam is whether detected by the detector or not)

Whats wrong?

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You claim that

According to the principle of relativity the light will not strike the detector in the reference frame of a non-moving observer

If the detector is on the floor of the train, then this is false. Both the observer on the train, and the observer outside of the train (who is stationary with respect to the Earth let's say), will see the light strike the detector. The person outside of the train will simply see the light travel along a "diagonal" path as illustrated in the image below. The left-hand image is what the train observer sees, and the right-hand imagine is what someone on the outside of the train sees.

enter image description here

The Math.

In the train, which we call frame $S'$, the spacetime trajectory of the light between when it leaves the ceiling at parameter value $\lambda = 0$ and when it hits the detector on the floor at parameter value $\lambda = d/c$ is given by is given by the following parameterized curve \begin{align} t'(\lambda) = \lambda, \qquad x'(\lambda) = 0, \qquad y'(\lambda) = d-c\lambda \end{align} where I have taken the $y'$-axis perpendicular to the floor, the $x'$-axis parallel to the direction of motion of the train, and we ignore motion in $z'$. Notice that since $t'(\lambda) = \lambda$ the parameter $\lambda$ here is really just time according to a train observer. To determine what the person outside of the train, which we call frame $S$ sees, we apply the Lorentz transformation: \begin{align} \begin{pmatrix} ct(\lambda) \\ x(\lambda) \\ y(\lambda) \\ \end{pmatrix} &= \begin{pmatrix} \gamma & \gamma\beta & 0 \\ \gamma\beta & \gamma & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} ct'(\lambda) \\ x'(\lambda) \\ y'(\lambda) \\ \end{pmatrix} = \begin{pmatrix} c\gamma\lambda \\ c\gamma\beta\lambda \\ d-c\lambda \\ \end{pmatrix} \end{align} where we have assumed that the train moves along the positive $x$-axis with speed $v = \beta c$. In other words, the path of the light in the frame $S$ is \begin{align} t(\lambda) = \gamma\lambda, \qquad x(\lambda) = \gamma \beta c\lambda, \qquad x(\lambda) = d-c\lambda \end{align} For convenience, we reparameterize this curve by defining $\mu = \gamma\lambda$ to give \begin{align} t(\mu) = \mu, \qquad x(\mu) = v\mu, \qquad y(\mu) = d- c\mu/\gamma \end{align} where we used $\beta c = v$ to simplify the $x$ component. The parameter $\mu$ is just the time as measured by an $S$ observer since $t(\mu) = \mu$. Notice, that the $x$-component of the trajectory now is nonzero, and in fact the light has velocity $v$ in the same direction as the train!

As a final check, let's make sure that the speed of the light is $c$ in both frames. In the train frame we have \begin{align} v'_\mathrm{light} = \sqrt{\frac{dx'}{d\lambda}^2 + \frac{dy'}{d\lambda}^2} = \sqrt{0+c^2} = c \end{align} and in the frame outside the train we have \begin{align} v_\mathrm{light} &= \sqrt{\frac{dx}{d\mu}^2 + \frac{dy}{d\mu}^2} = \sqrt{v^2 + \frac{c^2}{\gamma^2}} = c\sqrt{\beta^2+\frac{1}{\gamma^2}} = c\sqrt{\beta^2 +1-\beta^2}=c \end{align}

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  • $\begingroup$ What did you use to make the diagram? $\endgroup$ – Larry Harson Sep 8 '13 at 0:57
  • $\begingroup$ @LarryHarson I used google :) commons.wikimedia.org/wiki/File:Light-clock.png $\endgroup$ – joshphysics Sep 8 '13 at 1:46
  • $\begingroup$ Haha, I know all the secrets of you guys now: Inkscape, Google, and... Microsoft powerpoint :| $\endgroup$ – Larry Harson Sep 8 '13 at 2:26
  • $\begingroup$ @LarryHarson I must say though, I have been learning TikZ as of late, and I think it's really the way to go if you put in the time to learn it, especially for those who are not averse to a little math and programming. texample.net/tikz/examples/all $\endgroup$ – joshphysics Sep 8 '13 at 2:46
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    $\begingroup$ @user215721 It's the speed that's frame-invariant, not the velocity. In fact the Lorentz transformation I used in the math section can be derived only from speed invariance and the principle of relativity, so we see that speed invariance leads to a transformation that necessarily gives velocity non-invariance. $\endgroup$ – joshphysics Sep 8 '13 at 18:29

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