Lorentz factor depend on speed, but to measure the speed, we need to know the distance traveled and the times it takes to get the ratio. But according to special relativity theory, for two equivalent observer that moving relative to each other, each of them will see the other experiencing time dilation and lorentz contraction. So distance measured By one observer become smaller when seen from other observer by lorentz factor. And also time measured By one observer become longer when seen from other observer. And speed as measured By one observer must by multiplied By square of lorentz factor when calculated By other observer.

Update: With their average life time and speed, moun can reach a distance only 0.66 km, but when using time dilation formula it can reach distance 10 km, their average life span increase according ovserver on earth. But from moun frame it is not his time that dilated, but the distance to earth that become shorter. So with their average life span, they can cover a shorter distance compared the distance measured By observer on earth. So in this case, proper time is measured in moun frame but proper distance measured from earth frame. But for the case measuring speed like I mentioned in above equation, both the distance and time is proper according to one frame, and we need to calculated the same distance and time that being dilated and contracted By lorentz factor from other frame reference. So it is become multiplied By square of lorentz factor. (Note: this is verified By experiment)

  • $\begingroup$ Please note that the update you added to your question does not itself contain any questions. Nor does the original content. The only question I can see here is that found in the title. I am not entirely clear on why muons are specifically brought into this, but if you want answers to properly address more specific issues than just what was asked in the title, you should include more detailed questions $\endgroup$ – Jim Apr 12 '17 at 15:51
  • $\begingroup$ Because I found that content on the book, Introduction to Modern Physics by Arthuer Beiser. I need to reconcile it with with my question, to make clear what I ask. Because that what I understand about time dilation. $\endgroup$ – Mohammad Fajar Apr 12 '17 at 19:57

Here is the short answer: on a spacetime diagram, both the earth-frame and the muon frame use a Minkowski-right-triangle to determine the velocity as ratio of the spacelike-leg to the timelike-leg. These triangles share a common side (namely, the muon's worldline from birth to death). Although these triangles are different, they are Minkowski-similar and thus determine the same relative speed.

In this analysis, I am using the results from special relativity to show that corresponding measurements from each frame are consistent with a common relative speed.

I'll start with a problem with simpler numbers, then apply it to your muon case.

First, consider a "muon" with a lifetime of 6 seconds, travelling at V=0.8c (so $\gamma=\frac{1}{\sqrt{1-V^2}}=5/3$). [Since I'll use rapidities later: if $v=\tanh\theta$, then $\gamma=\cosh\theta=\cosh(\rm arctanh(0.8))=5/3$ and $\gamma V=\sinh\theta=\sinh(\rm arctanh(0.8))=4/3=(5/3)(4/5)$.]

Here's a spacetime diagram on rotated graph paper. Time runs upwards.

muon-simple on rotated graph paper

Suppose there was a mountain where the muon was created at the peak and decayed when it reached it ground. We draw the worldlines of the peak and ground levels and of the muon.

In the earth frame,
consider Minkowski-right-triangle BPD, with Minkowski-right-angle at P.
In the earth frame, event P is simultaneous with the death event D.
The muon's 6-sec life is the timelike-hypotenuse BD of this triangle.
The adjacent side is the timelike-leg BP.
So, in the earth frame, the muon lives $(BP)=(BD)\cosh\theta=(6\rm sec)\gamma=(10\rm sec)$ [time-dilation]
and travels a distance $(PD)=V(BP)=0.8(10\rm sec)=(8\rm lightsec)$.
[Note this is the same as $(PD)=(BD)\sinh\theta=(BD)\cosh\theta\tanh\theta=(BD)\gamma v$.]

In the muon frame,
we must consider Minkowski-right-triangle BDX, with Minkowski-right-angle at D.
In the muon frame, event X is simultaneous with the death event D.
DX is parallel to the muon's spatial axis and
is thus Minkowski-perpendicular to the muon's timelike axis along BD.

The muon's 6-sec life is the timelike-leg BD of this triangle.
The timelike hypotenuse is BX.
So, according to the muon frame,
a clock at the peak ticked-off hypotenuse $(BX)=\frac{(BD)}{\cosh\theta}=\frac{(6\rm sec)}{\gamma}=(3.6\rm sec)$ [time-dilation]
and is now a distance $(DX)=V(BP)=0.8(6\rm sec)=(4.8\rm lightsec)$ away.

Note there is a third Minkowsk-right-triangle PDX, with Minkowski-right-angle at P. In this triangle, the Minkowski angle at D is the angle between the spatial-axes of the earth and muon frame, which is equal to the angle between their timelike worldlines. Thus, in this triangle, the hypotenuse $(DX)=\frac{(PD)}{\cosh\theta}=\frac{(8\rm lightsec)}{\gamma}=4.8$---this is length-contraction.
So, from this, the muon could say that, using triangle BDX,
"the peak has moved $(4.8\rm lightsec)$ in $(6\rm sec)$, with speed $\frac{(DX)}{(BD)}=\frac{(4.8\rm lightsec)}{(6\rm sec)}=0.8c$".

Thus, triangle BPD is Minkowski-similar to triangle BDX.

Now, I'll use the same basic diagram with realistic values for the muon, as suggested by you.
Let's use a lifetime of $(BD)=2.2\mu sec$ and a velocity $V=\tanh\theta=0.999c$ so that $\gamma=\cosh(\rm arctanh(0.999))=22.36627204$ and $\gamma V=\sinh(\rm arctanh(0.999))=22.34390576$.

Thus, the earth measures a muon speed of $$\displaystyle\frac{14746.97780 \rm m}{49.20579849\rm \mu sec}=0.999c.$$

Thus, the muon measures an earth speed of $$\displaystyle\frac{659.3399997\rm m}{2.2\rm \mu sec}=0.999c.$$

muon-real on rotated graph paper

Here are some updates to address issues raised by the OP in the comments.

  • (To address the "there are no observer in muon frame" comment [even though earlier the same commenter said "Moun observer using his clock"].) It is implicitly assumed that there are observers along timelike-worldlines and that all observers have identical measuring instruments (or else it makes no sense to compare measurements made in different reference frames.. or at least the issue is unnecessarily complicated). I believe this is the point of view for Einstein-type thought-experiments.
    So, there is an implicit observer in the muon frame of reference.

    There might arise the issue of calibration between reference frames.
    So, I offer a calibration procedure in the spirit of the Bondi k-calculus where one inertial observer uses a radar measurement (using her clock and light signals) to determine the calibration of the clock of another inertial observer.

    A time $T$ after meeting, the Peak sends a light signal to the Muon, which is received by the Muon-observer a time $T'=kT$ after their meeting (where $k$ is not yet determined). The reflected-light-signal is received by the Peak at a time $kT'=k(kT)$, where we have used the same value of $k$ in accordance with the Relativity Principle.
    Since the Peak measured times $T$ and $k^2T$, the square root of that ratio is $k$, which is now measured. From the diagram, $T=2$ and $k^2T=18$ implies that $k=3$... so we find that $kT=6$ has elapsed for the muon along BD, in agreement with the first diagram in my answer. I can now construct 6 light-clock diamonds (with lightlike sides) along BD. Bondi radar measurement

  • In my example above, the Peak-observer determined that his 10th-tick was simultaneous with the Muon-observer's 6th-tick and the Muon's 6th-tick was simultaneous withe the Peak's 3.6th-tick. Note the ratio of my ticks to the other-observer's ticks (10/6)=(6/3.6) are the same, in fact, this is the time-dilation factor $\gamma$.
    Indeed, this is consistent with the earlier statement that
    triangle BPD is Minkowski-similar to triangle BDX.

    For completeness (to address the "So this is not fair" comment), I construct Minkowski-congruent triangles so that each observer can conduct the same experiment: "compare my clock reading with a distant clock reading at events declared by me to be simultaneous".
    Compare the Peak's measurement with triangle BPD with the Muon-obsersver's measurement with triangle BP'D': Peak says "10 for me, 6 for the other" while Muon says "10 for me, 6 for the other". Similarly, compare the Muon's measurement with triangle BDX with the Peak's measurement with triangle BD'X': Muon says "6 for me, 3.6 for the other" while Peak says "6 for me, 3.6 for the other".
    This is the symmetry of time-dilation.

Symmetry of time-dilation


Suppose we have your two observers travelling towards each other, and you are standing exactly halfway between them. So in your frame observer $A$ is moving at some velocity $+v$ and observer $B$ is moving at some velocity $-v$:

Symmetrical motion

The motion is symmetrical about your position because the velocities of the two observers $A$ and $B$ are equal and opposite.

Now suppose $A$ and $B$ measure each other's velocities, but get different answers. The problem is that this would break the symmetry because how could you tell who would measure the higher velocity when the situation is symmetrical? Therefore $A$ and $B$ must measure each other's velocities to be equal and opposite, that is they must agree on their relative velocity.


In Alice's frame, Bob is traveling at velocity $v$. In Bob's frame, Alice is traveling at velocity $w$. Then the velocity addition formula (which follows from the Lorentz transformation) tells you that in Alice's frame, Alice is traveling at velocity $$v+w\over 1+vw$$

But Alice considers herself stationary, so the above expression must equal zero, i.e. $w=-v$.

  • 1
    $\begingroup$ I think this is a circular answer. The velocity addition equation is derived from the Lorentz transformations, and they assume the relative velocity is the same for both observers. So you have ended up proving what you originally assumed. $\endgroup$ – John Rennie Apr 9 '17 at 10:08

protected by Qmechanic Apr 10 '17 at 5:05

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