I am a bit confused about the implications of length contractions;

For example, in the muon decay problem, we assume that the distance between the muon and the earth is contracted only in the frame of reference of the muon (I've heard justification saying this is because it is Earth's atmosphere) but suppose there were no atmosphere then in the reference frame of Earth shouldn't the length between them also be contracted?

What I don't get is if one object is moving at a speed relative to another object, shouldn't this movement affect the distance between them in the reference frames of both of them, since their movement is merely relative?

  • $\begingroup$ You are not entirely correct, but almost. :) If we assume object A as stationary than it sees the distance traveled by B as contracted, and not the other way round. But then, we can switch and assume object B as stationary, which should then see the distance traveled by A as contracted. See also my comments to David Z's answer. $\endgroup$ Nov 29 '14 at 11:35

Don't confuse time dilation and length contraction,(even if length contraction is a consequence of time dilation) because there is no length contraction for the distance between two reference frames. The length (or distance) must be found in one reference frame, and the observer must be in another reference frame, he may not be part of the reference frame of the distance.

By consequence, it is not the distance between muon and Earth which is contracted. It is the distance between the starting point A of the muon and Earth (if we suppose that the muon is traveling from A to Earth).

From the Earth frame the distance A-Earth is at its largest. For the muon frame, the distance A-Earth is contracting according to its relative velocity.

The example of David Z is the opposite example: There is a distance in the muons' frame (between two muons which are belonging to the same frame) which is observed from the Earth frame.

  • $\begingroup$ "... because there is no length contraction for the distance between two reference frames." David Z's answer shows the opposite. In his answer the muon sees Earth travel the distance of 500 m, which is contracted (down from 50,000 m). "... length contraction is a consequence of time dilation". This is just plain wrong - see my comments to David Z's answer: $t'$ calculated from the $x'/v$ equation is different to that calculated from $t\gamma$. $\endgroup$ Nov 29 '14 at 11:47

Short answer: Yes, length contraction is a symmetric transformation between reference frames. However... You may not like this answer, but I think it's far easier to regard muon decay via the concept of time dilation rather than length of contraction. Reason is, from the standpoint of the muon it's not moving (and yes the Eath is moving but it's not relevant to the muon's decay), and it's going to decay after some time t.

In the earth's frame, that interval of length t gets split up into part time T and part space L, and because of the "minus sign" in the metric it means that the muon seems to 'live longer' before it decays. In units where $c=1$, $$t^2 = T^2 - L^2,$$ so $$T^2 = t^2 + L^2.$$

(If I can figure out how to upload pictures from my iPad I'll show you a couple spacetime diagrams that more directly answer your question re length contraction. Posting what I've got now.)

  • $\begingroup$ A slightly re-framed question might be, if there is a muon on the surface of the Earth, and another muon is traveling to Earth high from the atmosphere, would they meet before either one of them decays completely? $\endgroup$ Nov 27 '14 at 19:47
  • $\begingroup$ @PeterG.Chang actually that's a good way to ask it. In fact the Earth and atmosphere are irrelevant; you can just think about two muons in space, one at rest and one moving (in a particular reference frame). You could edit your question to reflect this, if you like. $\endgroup$
    – David Z
    Nov 28 '14 at 5:41

Well, Lorentz transformation and whole special relativity gives us short qualitative answer - time dilation and length contraction are very similar. Sitting on Earth we see muons with slower decay rate, because of time dilatation. Being a muon, we calculate time in our own frame of reference - so to keep calculations clear, we need to contract length instead.

Muon's half life time is 2,2 microseconds. We can assume they travel roughly with the speed of light (0.9997 c). Watching them on Earth, their lifetime is increased, because their clock 'tick tocks' slower. Then if they are, suppose, 660 meters from us, in one second they will reach us.

Okay, but what about being a muon? Our clocks is running with exact speed of rest muon. Then, to reach observer in one second with same velocity, we have to travel shorter distance.

This is why Lorentz contraction and time dilatation are in fact very similar to each other.

  • $\begingroup$ hmmm But why don't we consider length contraction for the length between muon and Earth in the reference frame of Earth? $\endgroup$ Nov 27 '14 at 19:32
  • $\begingroup$ We don't have to - we already know the distance. If it will help, imagine it as already contracted and just measured. $\endgroup$ Nov 27 '14 at 19:33
  • $\begingroup$ But isn't motion relative? In the frame of Earth, muon is approaching at a speed close to light and in the frame of muon, Earth is approaching at the speed close to light; I don't understand why we use length contraction on one case and not the other $\endgroup$ Nov 27 '14 at 19:35
  • $\begingroup$ In fact, variables in Lorentz transformation are for measured values. $\endgroup$ Nov 27 '14 at 19:35
  • $\begingroup$ In case standing on Earth, we can put flag when muon starts and when muon ends - these flags will be motionless, only muon would move. We could change muon with particle with some structure, here we would observe Lorentz contraction for this structure. But distance between two flags won't change, they are motionless. $\endgroup$ Nov 27 '14 at 19:37

Length is contracted the other way, we just don't notice it because muons are pointlike particles (as far as anyone knows), and a length of zero is still zero no matter how much you contract it.

If you had a beam of muons spaced a fixed distance apart (and yet somehow their "decay counters" only start at $50\text{ km}$ altitude), then it would be a different story. You could tell that length is contracted by looking at the spacing between consecutive muons. Suppose the muons are traveling at $0.99995c$, so $\gamma \approx 100$. Then, in the muons' rest frame:

  • they are spaced, say, $10\text{ km}$ apart
  • they take $2.2\ \mathrm{\mu s}$ to decay
  • an atmosphere of thickness $50\text{ km}/\gamma = 500\text{ m}$ is approaching them
  • that atmosphere will pass by them in $500\text{ m}/(0.99995c) = 1.67\ \mathrm{\mu s}$
  • the atmosphere will reach a new muon every $10\text{ km}/(0.99995c) = 33.4\ \mathrm{\mu s}$

In the Earth's rest frame:

  • the muons are spaced $10\text{ km}/\gamma = 100\text{ m}$ apart
  • they take $2.2\ \mathrm{\mu s}\times \gamma = 220\ \mathrm{\mu s}$ to decay
  • they are approaching an atmosphere of thickness $50\text{ km}$
  • they will pass through that atmosphere in $50\text{ km}/(0.99995c) = 167\ \mathrm{\mu s}$
  • a new muon will hit the atmosphere every $100\text{ m}/(0.99995c) = 0.334\ \mathrm{\mu s}$

You can check that all these numbers are consistent. In particular, in both frames, the muons last long enough to make it through the atmosphere. The time between impacts is shortest in the rest frame of the atmosphere and is dilated by the right factor of $\gamma = 100$ in the rest frame of the muons.

  • $\begingroup$ But the moving object's length has nothing to do with the contraction of space it is traveling through. The space should contract the same whether the muon is considered to be moving toward the Earth or whether the Earth is moving toward the muon. Also, you used two different methods to obtain the time for travel. So, if in the first case you calculate $t$ from $v$ and $x$ just the way you did in your second case (instead of calculating $t'$ from $v$ and $x'$) you will obtain $167\ \mathrm{\mu s}$. If you then calculate time dilatation $t'$, the muon is not going to make it to the Earth at all $\endgroup$ Nov 29 '14 at 8:48
  • $\begingroup$ Yes, my answer shows that space contracts the same way no matter which reference frame it is viewed from. In either case, objects moving at $v$ have their lengths contracted by $\gamma$. I don't know where you're getting the idea that I used two different methods of obtaining the time; I used $\Delta x/v$ in both cases. If you'd like to discuss this, please do so in Physics Chat and I'll be happy to explain why your reasoning doesn't work, but I won't get into the details here so as not to clutter up the comments. $\endgroup$
    – David Z
    Nov 29 '14 at 9:14
  • $\begingroup$ In the first case Earth travels 500 meters, while in the second, muon travels 50,000 meters. The distance should not change depending on which frame you assume as stationary, and therefore it is obvious that you used $\Delta x'$ for your calculations in the first case, while you used $\Delta x$ for your calculations in the second case. (And as I said - calculating $t'$ from $t$ gives a different number than calculating $t'$ from $x'$.) $\endgroup$ Nov 29 '14 at 9:38
  • $\begingroup$ @brightmagus again, I will be happy to explain why that's not correct in Physics Chat. $\endgroup$
    – David Z
    Nov 29 '14 at 11:47
  • $\begingroup$ Why don't you do it in your answer? It does not address OP's question at all, while being misleading in its claim that the objects length has anything to do with distance contraction. $\endgroup$ Nov 29 '14 at 11:59

Confusion about Length Contraction [...] if one object is moving at a speed relative to another object, shouldn't this movement affect [...]

Talk about "length contraction" (or "time dilation") is inherently confusing; it is improper and should be avoided.

In the typical "cosmic ray generated atmospheric muon" example we have the following unambiguous experimental facts:

  1. the muon "mean life" duration is $\approx~2.2\times 10^{-6}~\text s$, as it has always been found for samples of "free" muons,

  2. the duration of a clock which is (practically) at rest at the bottom of the atmosphere

    • from its indication (practically) simultaneous to the indication of an "air atom" having been hit by a cosmic ray proton and (at the end of a rapid decay chain) having emitted a muon

    • until its indication of being passed by this muon

    is $\approx \frac{1}{\sqrt{ \stackrel{~}{1 - v^2/c^2} }} \times 2.2\times10^{-6}~\text s$.

    From this relation we can determine that "$\text s$" here actually means the same unit duration in both measurements.
    But to call this duration of the clock on the Earth's surface also a "dilated duration of the muon" is plainly a misattribution.


  3. the distance between the described clock and "air atom" is typically $\approx~10~\text{km}$,

  4. the distance between the described muon and some (hypothetical) other muon which was "moving behind" at the same velocity (wrt. the clock on the Earth's surface), starting out from an "air atom" which had the same distance ($\approx~10~\text{km}$) from the clock, such that

    • the indication of the former muon having passed the clock was simultaneous to

    • the birth indication of the latter muon

    is $\approx \sqrt{ \stackrel{~}{1 - v^2/c^2} } \times 10~\text{km}$.

    From this relation we can determine that "$\text {km}$" here actually means the same unit distance in both measurements.
    But to call this (thought-experimental) distance between two muons also a "contracted distance between the clock and some particular piece of the atmosphere" is plainly a misattribution.


I think most of you are confusing what contracts because of motion. Not the distance between the muon and the Earth, nor the distance between the point in which the muon began its motion and its destination. Length contraction deals with the contraction "of the moving object" (that is the muon's length, if we could talk of it). If you could see an airplane or a train passing in front of you at relativistic velocity it would appear contracted. But the object. And moreover this effect only affects the way you "see" the object (the way it appears to your eyes) not its physical characteristiques, there's no structural deformation. Moving, solid bodies don't become really shorter. This effect is due to the fact that visual information is carried by photons and when the object to be seen has (almost) the same speed of what has to carry the visual information (light, photons) then there will be of course a lack of information, which will be the more greater the more distant the points of that object will be from the observer (photons have no time enough to deliver visual information). So the points at the extremities of the object will be lost when seeing it and the body will appear as contracted. I agree that the case of muons is rather explainable through time dilation.


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