Muon Decay Example from Morin's Book: Issues Explained

[realanswers]

TL;DR Summary

Why is time dilation calculated as $$t = \gamma T$$ and not $$t = \gamma(T - x^{'} v/c^2)$$ ?

From the Lorentz transformation equations we know that $$t = \gamma(t^{'} - x^{'} v/c^2)$$

but for the Muon decay example where the setup is as follows :
"Assume for simplicity that a certain muon is created at a height of 50 km, moves straight downward, has a speed v = .99998 c, decays in exactly T = 2 · 10−6 seconds, and doesn’t collide with anything on the way down.11 Will the muon reach the earth before it decays? "

This particular example is from Morin's book. The solution is that $$t = \gamma T$$ but why is it not $$t = \gamma(T - x^{'} v/c^2)$$ ?

Is the idea that x' (where the muon decays in its reference frame) is 0 ? because consider a reference that is offset from the muon by a distance d perpendicular in direction of velocity (so it is beside it) and is moving at the same velocity as the muon. Then x' = d and v is the speed of the muon. but then it is a different answer than that in Morin's book. Why is this the case?

 

[Dale]

You certainly could consider a non-zero $x’$. It does not make any difference to the outcome.

 

[realanswers]

You certainly could consider a non-zero $x’$. It does not make any difference to the outcome.

Can you explain more? It would differ by the $x' v/c^2$ How does that "not make any difference" ?

Also does this mean it is true that they consider $x' = 0$ to obtain their result?

Would appreciate a more verbose, fully-fledged answer!

 

[malawi_glenn]

Use Delta t' and Delta x' instead. Clearly Delta x' = 0

 

[realanswers]

Use Delta t' and Delta x' instead. Clearly Delta x' = 0

why must I use the difference between the two events rather than just position in a reference frame?

 

[TSny]

For the muon example, you are considering a time interval between two events.

event #1: creation of the muon
event #2: decay of the muon

Thus, the time interval in the earth frame is $$\Delta t = t_2 - t_1 = \left[\gamma \left(t'_2 - x'_2 v/c^2 \right) \right] - \left[\gamma \left(t'_1 - x'_1 v/c^2\right) \right] $$
In the muon frame, the two events occur at the same place. So, $x'_2 = x'_1$. Therefore, the above equation reduces to $$\Delta t = \gamma \left(t'_2 - t'_1\right) = \gamma \Delta t'$$.

[Edit] Here, I followed your sign choice for the Lorentz transformation: $t = \gamma \left( t' - x'v/c^2 \right)$. This corresponds to the primed frame moving in the negative x direction relative to the unprimed frame. It is more common to have the primed frame move in the positive x direction relative to the unprimed frame. Then, $t = \gamma \left( t' + x'v/c^2 \right))$. As a simple exercise, show that it doesn't make any difference for this problem.

 

[malawi_glenn]

why must I use the difference between the two events rather than just position in a reference frame?

How else would you measure life time?

 

[PeterDonis]

From the Lorentz transformation equations

For a single event. But the muon decay example involves two events: the event at which the muon is created in the upper atmosphere, and the event at which the muon reaches the Earth's surface. The key quantity you are trying to calculate is the timelike interval between those events, so you can compare that with the muon half-life. You can't do that just by looking at the Lorentz transformation equations.

The formula you refer to, $t = \gamma T$, is a shortcut for comparing the muon's proper time $T$ to the coordinate time elapsed in the Earth frame. But if you're not sure why that shortcut works, the best thing to do is to do the full computation I described above. $T$ will be the timelike interval I described (which is the proper time along the muon's worldline from creation to Earth's surface); $t$ is the coordinate time in the Earth frame that that interval corresponds to. Note that to compute that coordinate time, once you know the interval $T$, you need to ignore the spatial motion of the muon in the Earth frame, since you are not trying to do a full Lorentz transformation, you are just trying to compute a coordinate time interval in the Earth frame.

 

[realanswers]

For a single event. But the muon decay example involves two events: the event at which the muon is created in the upper atmosphere, and the event at which the muon reaches the Earth's surface. The key quantity you are trying to calculate is the timelike interval between those events, so you can compare that with the muon half-life. You can't do that just by looking at the Lorentz transformation equations.

The formula you refer to, $t = \gamma T$, is a shortcut for comparing the muon's proper time $T$ to the coordinate time elapsed in the Earth frame. But if you're not sure why that shortcut works, the best thing to do is to do the full computation I described above. $T$ will be the timelike interval I described (which is the proper time along the muon's worldline from creation to Earth's surface); $t$ is the coordinate time in the Earth frame that that interval corresponds to. Note that to compute that coordinate time, once you know the interval $T$, you need to ignore the spatial motion of the muon in the Earth frame, since you are not trying to do a full Lorentz transformation, you are just trying to compute a coordinate time interval in the Earth frame.

Thanks for the enlightening answer. I do want to know however, why does the short cut work?

Moreover, you seem to have a great handle on the subject and I would like to gain the insight you have, what book on special relativity do you recommend? The chapters in Morin's and Taylor's classical mechanics seem to fall short so far....

 

[Ibix]

Thanks for the enlightening answer. I do want to know however, why does the short cut work?

Algebraically, for the reason @TSny gave. In the frame where a clock is at rest it measures $t$ and its $x$ is constant. You want to know a difference in $t'$ coordinates, and the contribution of $x$ to both is the same, so drops out of the difference.

More fundamentally, it must work that way because otherwise we'd be writing physical laws that were different depending on where you are. That doesn't appear to be the way the world works, so successful physical theories don't either.

It is much safer to always write down coordinates and then use the Lorentz transforms on them, because there are whole classes of errors you can't make if you do that.

 

[robphy]

Draw a position-vs-time diagram ( a spacetime diagram ).

The time-dilation calculation is akin to finding the adjacent side of a right triangle in Euclidean geometry. ($\gamma$ is akin to cosine as a ratio of adjacent side to hypotenuse.)

Often a diagram will clarify the meanings of an equation and the algebraic symbols in that equation.

Many details are at my answer
https://physics.stackexchange.com/a/325582/148184

 

[malawi_glenn]

I would like to gain the insight you have

It comes with experience

 

[Dale]

why must I use the difference between the two events rather than just position in a reference frame?

Muons don’t decay at a specific time. They decay at a specific time after they were created.
That is a difference in time. The extra term subtracts out in the difference

 

[malawi_glenn]

Why do we need to know both the date of birth and today's date to figure out someone's age?
Similar question.

 

[vanhees71]

The muon example is nice, because it's so simple that you can explain it in many different ways. All you need to know is that a muon has an intrinsic parameter, it's (mean) lifetime. As an intrinsci parameter of the particle, it's defined in its rest frame. So let's call it $\tau$ (since "proper time" is usually labelled as $\tau$).

Now we have the situation that a muon is created at some place at some time and then it moves force-free, i.e., with constant velocity $v$ in a straight line until it decays from the point of view of the observer at rest on Earth. The question is, at which time, measured by the observer on Earth, after its creation it decays.

You can just answer this by saying that according to Earth the life-time is dilated by the Lorentz factor, i.e., it decays at $t=\gamma \tau$ after its creation at $t=0$.

Another way is to use the idea of world lines. In the rest frame of the muon the world line is $x(\lambda)=(\lambda,0)$ (setting the place of its creation at $x=0$. It's decay is at $(c \tau,0)$. For the observer on Earth the world line is given by the (inverse) Lorentz boost with the boost velocity $v=\beta c$, i.e., it's described by $x'(\lambda)=\gamma (\lambda, \beta \lambda)$, i.e., its decay occurs at $(\gamma c \tau,\beta \gamma c \tau)$, i.e., measured in the time of the observer on Earth it took a time $t=\gamma \tau$ to decay and in this time has travelled from $x=0$ to $x=v t$.

Another line of arguments starts with the question, why the muon can reach the earth from a place of creation at a distance $L>v \tau$. Of course, from the point of view of the observer on Earth again you can argue that the lifetime measured in terms of this observer's time, the muon can travel a distance $L'=v t=\gamma v \tau>v \tau$ during its lifetime.

From the muon's point of view the distance it needs the Earth to travel to it is only $L/\gamma$, i.e., the Earth can reach it if $L/\gamma \leq v \tau$ or $L \leq \gamma v \tau$.

So everything is always consistent, and no contradictions occur in whatever frame of reference you analyze the motion of the muon.

 

[realanswers]

It comes with experience

I am trying to learn special relativity well and I am finding it difficult to have good resources. So any concrete suggestions or recommendations would be helpful other than just "experience"!

 

[vanhees71]

Here's my attempt:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Note however that indeed you can learn physics only in an active way. Just to read a textbook or, even worse, just listen to lectures and/or Youtube videos won't help you. As a first step, you have to read the textbook and/or go over the content of lectures you listened to "with pencil and paper" and derive everything written or said there for yourself. After that you have to solve a lot of problems yourself to see, whether you really understood the material and can solve these problems.

 

[Dale]

After that you have to solve a lot of problems yourself to see, whether you really understood the material and can solve these problems.

I 100% “second” this.

I also recommend programming solutions. The process of programming requires you to dive deeply into the math and theory, and testing your code requires a good exploration of existing known solutions. And at the end then you have a tool you can easily use for questions that go outside of the typical homework examples.

 

[realanswers]

I 100% “second” this.

I also recommend programming solutions. The process of programming requires you to dive deeply into the math and theory, and testing your code requires a good exploration of existing known solutions. And at the end then you have a tool you can easily use for questions that go outside of the typical homework examples.

Yes I understand both your points. I am just looking for a good detailed exposition of special relativity that is at the advanced undergraduate/ graduate level.

 

[robphy]

Yes I understand both your points. I am just looking for a good detailed exposition of special relativity that is at the advanced undergraduate/ graduate level.

To make recommendations, it might help to learn of your previous physics, math, & relativity experience and preparation, preferences, and specific goals.

What specific textbooks have you looked at? And what is good and bad about each?

 

[Ibix]

I'm fond of Taylor and Wheeler's Spacetime Physics (free download via Taylor's website now). Others here seem to prefer Morin's Special Relativity for the Enthusiastic Beginner, which is downloadable for a tenner I think. Vanhees71 has linked his own notes, and @bcrowell wrote an SR book which is free for download from lightandmatter.com (I haven't read it, but his GR book was nice, as is his non-mathematical Relativity for Poets).

I recommend getting a selection. There'll always be something you don't understand in one book's presentation, and another may help.

300 Problems in Special and General Relativity by Blennow (@Orodruin) and Ohlsson is also a good resource, IMO. You don't know you know something until you can solve problems.

 

[vanhees71]

I 100% “second” this.

I also recommend programming solutions. The process of programming requires you to dive deeply into the math and theory, and testing your code requires a good exploration of existing known solutions. And at the end then you have a tool you can easily use for questions that go outside of the typical homework examples.

Of course, that's part of the game nowadays anyway, but not only programming in the numerical sense but also using computer-algebra systems for analytical calculations. A good "swiss-knife tool" for both is Python (including great visualization tools).

 

[PeterDonis]

I do want to know however, why does the short cut work?

I told you how to find that out: do the full computation and then look at what it says about the shortcut.

what book on special relativity do you recommend?

I learned SR from Taylor & Wheeler's Spacetime Physics.

Sean Carroll's online lecture notes on GR also have a good introductory chapter on SR.