2
$\begingroup$

I have a dilemma concerning my understanding of Special Relativity. Maybe I am understanding or calculating something wrong and would hear so.

The problem is based on muons created in the upper atmosphere by cosmic rays. Basically, what happens when we time the difference between such a muon and light created the same place at the same time.

Since the muon travels at 0.994c, it arrives 301 nanoseconds later than the light. However this is from observation on earth. From the muon's observation, the point of creation and the destination on the earth's surface are moving at 0.994c and the distance is contracted from 15 Km to 1638 m, so the light arrives 33 nanoseconds earlier.

Further, from the muon's observation, a clock on earth will experience time dilation and should only record 3.6 nanoseconds of time. So how much time does the earth clock record?

Detailed Description: In practice, identifying one muon and whether it would not decay in the journey would be problematic. Since we are using observations from the muon, replace it with a spaceship traveling at the (now arbitrary) velocity of 0.994c. If you do not like crashing the spaceship into the surface of the earth, or worry about general theory effects from the earth or sun, move the thought experiment to space, say the midpoint between Sol and Alpha Centauri.

A point m, is 15.0 Km above a point on the surface of the earth, e. At point me is a space station, Muse.

A space ship traveling at 298 m/μs constant velocity passes point m and then point e.

When the spaceship passes point m, the Muse Space Station turns on a light. A photographic plate at point e collects light only from the Muse Space Station. This will be our clock, and we will measure how long an exposure to light is indicated by the photographic plate at the time where the space ship reaches point e. (Perhaps when the space ship passes we close a shutter on the plate.)

The light takes 50.03461 μs to reach point e. The spaceship takes 50.03461 μs to reach point e, 301 nanoseconds later. So the photographic plate is exposed to light for 301 ns.

However, from the space ship's reference, the space ship is not moving, but points m and e are moving at 298 m/μs constant velocity. When point m reaches the space ship, it turns on a light.

Because of length contraction predicted by special relativity, the distance between points m and e is 1637 m. The light from point m takes 5.49540 μs to reach point e. Point e reaches the ("stationary") space ship after 5.46254 μs. So the photographic plate is exposed to light for 32.9 ns.

So how much has the exposure to light darkened the photographic plate? As much as expected for 301 ns or 33 ns?

Further, since from the space ship's observation, the photographic plate at point e is moving at a speed close to c. So it should experience time dilation according to special relativity.

If instead of leaving the light on when point m passes the space station, the light at point m was only turned on for 1 nanosecond, the clock timing that nanosecond would be slower observed from the space station, and would be on for 9 ns. Then the 9 ns of light would only expose the plate at e as much as for 1 ns of light, because of the time dilation at point e. So, observed from the space ship for every 9 units of time that the plate at point e is exposed to light, it only reacts as much as for one unit of time. So the 33 ns of light when point e reaches the space ship, the plate should only show 3.6 ns of exposure.

Since we are comparing times in microseconds, and subtracting to get a time difference in nanoseconds, to get four digit precision in our result, we need to start with and carry seven digit precision in our thought experiment, even when the input was arbitrary.

Earth Observation:
Distance of segment em:         15 000.000 000 m
Speed of muon space ship:          298.000 000 m/μs
Speed of Light:                    299.792 458 m/μs
Time for space ship to reach e      50.335 57 μs
Time before light reaches e         50.034 61 μs
Time of exposure of p-plate            301.0 ns


Muon Space Ship Observation:
Distance of segment em:          1 637.628 062 m
Rel Speed of e & m to muon space ship:
                                   298.000 000 m/μs
Speed of Light:                    299.792 458 m/μs
Time for space ship to reach e       5.495 40 μs
Time before light reaches e          5.462 54 μs
Time of exposure of p-plate             32.9 ns

How much has the plate darken from exposure to light over time?

$\endgroup$
  • $\begingroup$ References to muons is extraneous. The reason for selecting the values for distance and speed are based on hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html where it describes a relativistic muon-frame observer. In the case of muons created from cosmic rays, muon decay is slowed down due to time dilation. However, to the same results from the muon-frame, so a muon would explain why after reaching earth so many muons still have not decayed, length contraction is applied to the distance of segment em. $\endgroup$ – Marlin Pierce Jul 16 '14 at 18:26
  • $\begingroup$ Have you noticed that this page (tables by the 4th and 5th graphics) show not only time dilatation, but also distance dilatation. Not exactly what SR teaches. $\endgroup$ – bright magus Jul 17 '14 at 9:19
  • $\begingroup$ The page explains both the earth observation (which observes muons experiencing time dilation) and observation from a moving muon (which observes earth experiencing length contraction). The 4th and 5th graphic compare the two, so for length the contracted value for the muon 2 Km is compared to the uncontracted value for earth 15 Km. $\endgroup$ – Marlin Pierce Jul 17 '14 at 12:25
  • $\begingroup$ Time-dilatation and length (distance) contraction are supposed to be observed from one frame of reference. They are both denoted by the primed variables, which belong to one ref. frame. Otherwise, they are both dilated (from one frame of reference) or both contracted (from the other). But this is not what Special Relativity predicts. $\endgroup$ – bright magus Jul 17 '14 at 12:29
0
$\begingroup$

So how much has the exposure to light darkened the photographic plate? As much as expected for 301 ns or 33 ns ?

For 33 ns. That is the time that the clock inside the ship-muon indicates. We measure 301 ns of exposure, but a much shorter time was experienced by the moving photographic plate. This question is tricky also because the light will be red-shifted by Doppler effect, but lets forget about this detail.

These questions about "what the other observer would observe that I observe" are complicated. I suggest to follow the approach of the invariants: there are quantities that, measured by any observer, give the same result. One of them is the proper-distance between events. An event is any time and position (in this example, height) coordinate. There are only two important events in this problem: the muon enters the atmosphere at 15 km above the ground (lets called it $E_1$), and the muon reaching the ground ($E_2$). The specific coordinates given to the events depend on the observer. For example, for the muon, assuming that it sets its clock to zero when it enters the atmosphere:

$E_1: (t,z)_M=(0 s,z_0)$

Is $z_0=0$ for the muon? well, it could be $z_0=15$ km, or $z_0=100$ km, it really does not matter. What it matters is that for the event 2, for the muon:

$E_2: (t,z)_M=(t_M,z_0)$,

that is, both events, for the muon, happen at its same spatial position. What about us from the ground? well, lets assume we set our clock to zero when the muon reach the top of the atmosphere (but it does not matter really):

$E_1: (\tau,\zeta)_{\rm Earth}=(0,15~{\rm km}),\quad E_2: (\tau,\zeta)_{\rm Earth}=(\tau_M,0~{\rm km})$.

I am using Greek letters for my coordinates. Now, special relativity tells us that the proper distance between these events is the same as measured by any observer. The proper distance is the squared difference of the time (times $c^2$) minus the spatial distance squared. That is: $$c^2(t_M-0)^2-(z_0-z_0)^2=c^2(0-\tau_M)^2-(15~{\rm km}-0)^2\tag{1}$$ And that's it. Forget about that one will be contracted measured by the other, and one time is dilated, etc...Lets divide Eq. (1) by $c^2\tau_M^2$: $$ \left(\frac{t_M}{\tau_M}\right)^2=1-\left(\frac{15~{\rm km}}{c\tau_M}\right)^2$$ but we know that $(15~{\rm km}/\tau_M)=0.994 c$, and therefore you have it, the time dilation. Since $\tau'_M=301~{\rm ns}$, therefore $t'_M=301~{\rm ns}\,\sqrt{1-0.994^2}=33~{\rm ns}$. I use a $'$ to denote the fact that we are not talking of the first event at 15 km from the ground, but from a much closer distance. The time dilation factor is the same.

Ok, lets solve the original problem that has three events:

  1. Muon enters the atmosphere at 15 km of height.
  2. The ray of light that the muon emitted toward the ground reaches the ground.
  3. The muon reaches the ground.

Again, latin letters for the muon coordinates and greek for the ground observer. For the first event: \begin{align} E_1: &(0,0) \quad\text{ for the muon.}\\ &(0,15~{\rm km}) \quad\text{ for the ground observer.} \end{align}
For the second event: \begin{align} E_1: &(t_2,z_2) \quad\text{ for the muon.}\\ &(15{\rm km}/c,0) \quad\text{ for the ground observer.} \end{align} For the third event: \begin{align} E_1: &(t_3,0) \quad\text{ for the muon.}\\ &(15{\rm km}/0.994 c,0) \quad\text{ for the ground observer.} \end{align}

For the invariant equation between $E_1$ and $E_2$ we get $$z_z=ct_2,$$ which is important since light speed is the same for everybody.

For the invariant equation between $E_1$ and $E_3$ we get $$t_3=\frac{15~{\rm km}}{0.994 c}\sqrt{1-0.994^2},$$ which is the equation we got at the beginning.

For the invariant equation between $E_2$ and $E_3$ we get $$(15~{\rm km})^2\left(\frac{1}{0.994}-1\right)^2=c^2t_3(t_3-2t_2)$$.

Combining the three equations we get \begin{align} t_1&=0~{\rm \mu s} &\tau_1&=0~{\rm \mu s} \\ t_2&=2.74~{\rm \mu s} &\tau_1&=50.035~{\rm \mu s}\\ t_3&=5.51~{\rm \mu s} &\tau_1&=50.336~{\rm \mu s}\\ \end{align}

So, why is the difference between $t_2$ and $t_3$ not 33 ns? Well, this illustrates the sort of confusion you can get unless you track systematically the events. If the 301 ns were the exposure time measured by the ground observer, then the exposure time measured by the muon would have been 33 ns. However, in this design, we cannot say that simultaneously to the arrival of the light to the ground the exposure starts. Simultaneously for who? This needs to be specified. If we assume that the exposure starts with the arrival of the muon-plate at the atmosphere and ends with its arrival to the ground, according to us ground observers it would have been exposed for $50.3~{\rm\mu s}$, while if we recover the plate at ground level, we would measure an exposure equivalent to only $5.5~{\rm\mu s}$. Shorter. By a factor of $\sim10$. That is it.

$\endgroup$
  • $\begingroup$ In your last equation block, it should be 15 Km over c * tau[M]. It's not c squared since the whole term is squared. $\endgroup$ – Marlin Pierce Jul 16 '14 at 20:45
  • $\begingroup$ What you are saying is that the 301 ns measured from the earth observation is time dilated. However, observed from earth and the photographic plate, the photographic plate is motionless and does not experience time dilation, and so we should expect it to be exposed for 301 ns. In hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html the earth frame observer sees time dilation of the muon decay but not of the earth observation. $\endgroup$ – Marlin Pierce Jul 16 '14 at 20:53
  • $\begingroup$ From the muon observation, earth, and the photographic plate experience time dilation. You correctly calculated the time dilation magnitude. However, in my original question I had (incorrectly) calculated the muon's observation of the plate's exposure as 33 ns. When in my self-answer I calculate it as 2,756 ns the time dilation gives us 301 ns. $\endgroup$ – Marlin Pierce Jul 16 '14 at 20:56
  • $\begingroup$ I am basically questioning when you applied the time dilation. On the hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html web site, the muon observation show time dilation of earth, but the earth observation has no time dilation. $\endgroup$ – Marlin Pierce Jul 16 '14 at 20:59
  • $\begingroup$ Earth and the photographic plate experience time dilation respect to the other. But I find these sort of statements confusing, that is why I suggest to stick to the invariants. $\endgroup$ – Enredanrestos Jul 16 '14 at 21:04
1
$\begingroup$

Because of length contraction predicted by special relativity, the distance between points m and e is 1637 m. The light from point m takes 5.49540 μs to reach point e. Point e reaches the ("stationary") space ship after 5.46254 μs. So the photographic plate is exposed to light for 32.9 ns

1637m is after length contraction in the reference frame of the space ship with point e having velocity close to but < c. This transformed distance only applies to objects in the reference frame of the ship, definitely excluding photons on the light beam. In fact you don't apply length contraction to light itself. Try out the formula for objects moving at c.

Also, where the photographic plate darkening is concerned, if you are in the same reference plane as the plate, then the clock you are really interested in is the clock next to the plate. The ship has velocity 0.994c toward you/clock/plate, the light has velocity c, and the total distance travelled by each is 15km.

$\endgroup$
  • $\begingroup$ I'm applying the length contraction to the segment em. Instead of two points, m and e, replace that with a 15 Km long ship. From the short ship's reference the 15 Km ship contracted to a length of 1637m. The light, unaffected by the transformation, travels at velocity c across the 1637m. So I am applying the transformation to an object, and not light. $\endgroup$ – Marlin Pierce Jul 16 '14 at 18:08
  • $\begingroup$ What you suggest, apply length contraction to objects but not light, is exactly what I did. I followed the example for a muon experiment (link added to question comments) which explains the observation of reduced muon decay for muons traveling at near light speed. $\endgroup$ – Marlin Pierce Jul 16 '14 at 18:28
1
$\begingroup$

Trying to defend the application of SR in my question, I discovered my mistake. From the observation framework of the photographic plate, it's a simple race between a muon speed space ship and light. Not so to the other observer.

from the muon speed space ship's observation, the space ship is stationary and points m and e are approaching at 0.994c. When the light source at point m passes the (stationary) space ship, it turns on a light which approaches the photographic plate at velocity c. However, it's not a fair race because the photographic plate is approaching the light source at a speed of 0.994c.

Thus the light and the photographic plate close on each other at a speed of 597.792458 m/μs, nearly twice the speed of light. They meet after 2.73946 μs.

Since it takes 5.49540 μs for the photographic plate to reach the space ship, the plate is exposed for 2,756 nanoseconds. Due to time dilation, the photographic plate only experiences 301 ns of exposure, which matches the earth observation.

Muon Space Ship Observation:
Distance of segment em:          1 637.628 062 m
Rel Speed of e & m to muon space ship:
                                   298.000 000 m/μs
Speed of Light:                    299.792 458 m/μs
Time for e to reach space ship       5.495 40 μs
Time before light reaches e          2.739 46 μs
Time of exposure of p-plate          2,756 ns
Actual p-plate exposure from time dilation
                                       301 ns
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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