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My son asked me the following question which I can't answer in a simple way.

A frame $O'$ has speed $v$ relative to $O$.

A photon travelling along the $x'$-axis in frame $O'$ over a distance $L'$ needs a time $t' = \frac{L'}{c}$. The speed of light is $c = \frac{L'}{t'}$.

For the observer in $O$ the photon covers a distance $L$ in a time $t$. The speed of light is $c = \frac{L}{t}$

For the observer in $O$, there is a time dilation: $t = \frac{t'}{\sqrt{1 - \frac{v^2}{c^2}}}$. So $t > t'$

From $\frac{L}{t} = \frac{L'}{t'}$ and $t > t'$ follows $L > L'$

Is this not a contradiction with the Lorentz contraction which says $L < L'$?

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    $\begingroup$ Advice: You can use MathJaX here to typeset formulae correctly. $\endgroup$
    – ACuriousMind
    Feb 27, 2015 at 14:24
  • $\begingroup$ you can't be sure if t>t' unless you know v. if v=c, then it makes no sense. If v<<c, then (v/c)^2 is approximately one, so t=t'. $\endgroup$
    – GRrocks
    Feb 27, 2015 at 14:34
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    $\begingroup$ Not a typical "My son asked ..." question. $\endgroup$
    – pfnuesel
    Feb 27, 2015 at 14:41
  • $\begingroup$ My son is 18 and in school they learn a little about relativity (speed of light is constant, Lorentz contraction, time dilatation, not the Lorentztransformations). I "translated" his question for this forum. $\endgroup$
    – léo
    Feb 27, 2015 at 14:53
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    $\begingroup$ Im not sure about a template, but I learned it by typesetting a couple of lecture notes for myself using LaTeX. Texmaker is my personal favorite editor, just play around with it. $\endgroup$
    – DK2AX
    Feb 27, 2015 at 19:44

6 Answers 6

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All clocks obey the time dilation formula, therefore all durations measured by a clock obey the time dilation formula.

All time differences definitely do not obey the time dilation formula.

To figure out the time that it takes for something to travel some distance you might use two clocks. In that case you are not measuring the time with a clock, you are calculating a time from the readings of two clocks.

To figure out the time that it takes for something to travel some distance you might use one clock, like a coach timing a 100 m sprint. In that case you are not measuring the time with a clock, you are measuring the time that the travel seems to take, and then calculating a "correct" result from that time.

If you attach a clock on the traveling thing, you get some duration measured by a clock, and you also know the velocity of the clock. In this case it's appropriate to use the time dilation formula.

Time dilation applies to clocks, not time calculations.

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  • $\begingroup$ I think you are right. Two points: 1) Time dilation is only true for events happening in ONE POINT. That was not clear for me! Now I understand also why a photon has to go up and down or back and forth in a lightclock ... to come back at the starting point. So we need ONE CLOCK (starting point) + MIRROR. 2) the distance the photon travels for O is not L, but L + the distance covered by O' (photon toward mirror) or L - distance covered by O' (return photon ). Taking into account these two points, I can solve the problem, got the formula for time dilation and L is Lorentcontracted! $\endgroup$
    – léo
    Feb 27, 2015 at 23:25
  • $\begingroup$ I approve this approach :) How about this: L is a distance moving at speed v, it is a contracted distance. Photon passes that distance slowly, this slowness is related to time dilation. Passing speed is c-v. Distance left for photon to pass shrinks at rate c-v. $\endgroup$
    – stuffu
    Feb 28, 2015 at 10:04
  • $\begingroup$ The problem is that the passing speed can't be c-v, because the speed of the photon is always c! $\endgroup$
    – léo
    Feb 28, 2015 at 13:39
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Given that you imposed $t= \gamma t'$, the obtained relation between $L$ and $L'$ is $$ \tag 1 L = \gamma L'.$$ Your reasoning is now, if I understand, that (1) is inconsistent with the phenomenon of length contraction because $L$ is the distance measured in the reference frame $\mathcal O$ and $L'$ is the one measured in the reference frame $\mathcal O'$ moving at a velocity $v'$ with respect to $\mathcal O$.

But (1) is not a contradiction, because the length contraction applies to the distance between two space points when measured in different reference frames, which is not the case at hand. Indeed, following the way you defined them,

  • $L$ is the distance travelled by light in the time $t$, in the reference $\mathcal O$. In other words, $L$ is the space-point reached by the light when the clock of $\mathcal O$ marks the time $t$.
  • $L'$ is the distance that light has travelled, according to $\mathcal O'$, at the moment in the reference $\mathcal O'$ in which the clock of $\mathcal O$ marks the time $t$, or equivalently at the moment in which the clock of $\mathcal O'$ marks the time $t'$. This follows from your saying that $t$ and $t'$ are related by $t=\gamma t'$.

So what (1) is telling you is that the distance travelled by light in the $\mathcal O$ frame in a time $t$ is greater than the distance travelled by light in the $\mathcal O'$ frame in the time $t'$ at which the clock of $\mathcal O$ marks the time $t$.

This is not a contradiction because you are comparating two quantities which are not describing a frame-independent space-time event.

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Be careful. L and L' are the lenghts of the paths followed by the photon. The contraction of lenghts due to relativistic effects is an effect that you experience when you can measure simoultaneously the lenght of something.

Take into account the following situation: http://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity

As you can see an observer in relative motion will experience a longer distance covered by the light ray, and so atime dilatation. The lenght contraction of a solid bar for instance in two different systems needs the concept of measure at the same time at both the extrema.

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No contradiction, because time dilation and length contraction, even if dilating and contracting, are working in the same sense:

For this purpose I recommend to use an example which is simpler than yours (I think it is not useful to consider the light ray beside O and O'. If you are asking a question try first to reduce it as much as possible).

Example: A spaceship is travelling near light speed at 0,99 c from Earth to an Exoplanet 100 light years away (Earth coordinates). The Lorentz factor γ for 0,99c is 7,09. For the spaceship, due to length contraction (with division by the Lorentz factor) the distance is reduced to 14,10 light years (spaceship coordinates). At a speed of 0,99c, the travel will take the spaceship 14,24 years (spaceship coordinates).

And it is only now that we apply time dilation: For the observers (on Earth), the time of the spaceship clock (14,24 years) has to be multiplied with 7,09, giving 101 years. Now look how exactly time dilation and length contraction were applied:

Length contraction is the contraction of the proper distance (Earth frame) as observed by the spaceship (spaceship frame).

Time dilation is the dilation of proper time (space ship frame) as observed by Earth (Earth frame).

We used the proper distance in Earth coordinates and proper time in spaceship coordinates. If you want you can say that in this example time and space are both contracted for the spaceship, and they are both dilated for Earth.

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In the $O$ frame in time $t$,the light will travel a distance $ct$.

For an interval $t$ in the $0$ frame,time interval in $0'$ will be $t'=t/γ$. In this time interval light will move a distance of $ct'$ in the $0'$ frame.

Now as we can see $ct$ is greater than $ct'$, means distance travelled by light in the $0$ frame is more(as $t$ is greater than $t'$?)

Is this is contradiction? Ofcourse not. This is the path travelled by the photon not a stick that should get contracted along the motion.

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Suppose you are observer $O$, and I am observer $O'$.

Suppose you, I, and the photon all cross paths at a time we can both agree to call $0$ and a place we can both agree to call home.

Eventually, the photon passes Joe's diner, which I say is $L$ miles from home, and --- if we choose units so $c=1$ --- it does so at time $t=L$.

According to you, Joe's diner is $L'=(L+vt)/\sqrt{1-v^2}$ miles from home, and the photon passes it at time $t'=(t+Lv)/\sqrt{1-v^2}$.

Now use the fact that $L=t$, plug this into the formulas for $L'$ and $t'$, and you'll see that $L'=t'$. In other words, you'll agree that the photon is traveling at speed $L'/t'=1=L/t$.

The calculations are a little more involved if you, I and the photon never cross paths, but they still work out to $L/t=L'/t'=1$, and no new ideas are needed.

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