Length contraction and time dilation in relativity

[Kennedy]

Homework Statement

A graduate student discovers that an elementary particle produced in his experiment travels 0.250 mm through the lab at a speed of 0.950c before it decays (becomes another particle). What is the lifetime of the particle measured in its rest frame (proper lifetime). A) 8.77 x 10-13 s B) 7.99 x 10-13 s C) 2.74 x 10-13 s D) 2.24 x 10-13 s E) 2.81 x 10-12 s

Homework Equations

L = L0 / ϒ
t = ϒt0
γ = 1/(1 - (v^2/c^2))^(1/2)

The Attempt at a Solution

I found ϒ to be equal to 3.2025 which I believe is correct. Given in the problem, it says that the observer sees the particle travel through 0.250 mm. That would be considered to be in the frame that is "moving" with respect to the particle. To find the proper lifetime, I would need to find the distance observed by someone who is a rest with respect to the particle (usually I imagine someone riding on the particle). The distance they would observe would be "L0," because that is the length observed by a someone at rest. L0 = L(ϒ) = 0.250 x 10^-3)(3.2025) = 8 x 10^-4 m.

With that I took the speed of the particle to be 0.95c and I divided that by 8 x 10^-4 to get the time observed by someone at rest with respect to the particle. This gave me 2.81 x 10^-12, which is not correct.

I got the right answer by solving first for the time observed by the graduate student ("moving" with respect to the particle). That time would be considered to be in the primed frame, and then using the second equation listed above I solved for t0. So, v = 0.95c, d = 0.250 x 10^-3... t = 0.250 x 10^-3/(0.95c) = 8.77 x 10^-13 s. Then t0 = t/γ = 8.77 x 10^-13/3.2025 = 2.74 x 10^-13. Shouldn't I be getting the same answer no matter which way I decide to do it?

 

[Orodruin]

The distance they would observe would be "L0," because that is the length observed by a someone at rest. L0 = L(ϒ) = 0.250 x 10^-3)(3.2025) = 8 x 10^-4 m.

Which distance are you talking about here?

Shouldn't I be getting the same answer no matter which way I decide to do it?

Yes, as long as the way you are doing it is correct, you will get the same result.

 

[PeroK]

Homework Statement

A graduate student discovers that an elementary particle produced in his experiment travels 0.250 mm through the lab at a speed of 0.950c before it decays (becomes another particle). What is the lifetime of the particle measured in its rest frame (proper lifetime). A) 8.77 x 10-13 s B) 7.99 x 10-13 s C) 2.74 x 10-13 s D) 2.24 x 10-13 s E) 2.81 x 10-12 s

Homework Equations

L = L0 / ϒ
t = ϒt0
γ = 1/(1 - (v^2/c^2))^(1/2)

The Attempt at a Solution

I found ϒ to be equal to 3.2025 which I believe is correct. Given in the problem, it says that the observer sees the particle travel through 0.250 mm. That would be considered to be in the frame that is "moving" with respect to the particle. To find the proper lifetime, I would need to find the distance observed by someone who is a rest with respect to the particle (usually I imagine someone riding on the particle). The distance they would observe would be "L0," because that is the length observed by a someone at rest. L0 = L(ϒ) = 0.250 x 10^-3)(3.2025) = 8 x 10^-4 m.

With that I took the speed of the particle to be 0.95c and I divided that by 8 x 10^-4 to get the time observed by someone at rest with respect to the particle. This gave me 2.81 x 10^-12, which is not correct.

I got the right answer by solving first for the time observed by the graduate student ("moving" with respect to the particle). That time would be considered to be in the primed frame, and then using the second equation listed above I solved for t0. So, v = 0.95c, d = 0.250 x 10^-3... t = 0.250 x 10^-3/(0.95c) = 8.77 x 10^-13 s. Then t0 = t/γ = 8.77 x 10^-13/3.2025 = 2.74 x 10^-13. Shouldn't I be getting the same answer no matter which way I decide to do it?

If the distance traveled in the lab is $L$ and the lab time is $t$, then:

Method 1: The particle measures $L' = L/\gamma$, hence a time of $\tau = t' = \frac{L}{\gamma v}$

Method 2: The time in the lab is $t = L/v$ and the proper time of the particle is $\tau = t/\gamma = \frac{L}{\gamma v}$

So, yes you should get the same answer both ways.

Is there a moral here that numbers let you down when algebra doesn't?

 

[Kennedy]

Which distance are you talking about here?Yes, as long as the way you are doing it is correct, you will get the same result.

Here, I was talking about the distance observed by the particle.

 

[Kennedy]

If the distance traveled in the lab is $L$ and the lab time is $t$, then:

Method 1: The particle measures $L' = L/\gamma$, hence a time of $\tau = t' = \frac{L}{\gamma v}$

Method 2: The time in the lab is $t = L/v$ and the proper time of the particle is $\tau = t/\gamma = \frac{L}{\gamma v}$

So, yes you should get the same answer both ways.

Is there a moral here that numbers let you down when algebra doesn't?

Wait, but didn't I do it right by method one? Where did I go wrong?

 

[PeroK]

Wait, but didn't I do it right by method one? Where did I go wrong?

I don't know. Maybe a $10^{-4}$ should've been $10^{-5}$ somewhere. And the rest is rounding errors.

I'd just put $L$ and $v$ into a spreadsheet and let it calculate $\gamma$ and $L/v\gamma$.

 

[Orodruin]

Here, I was talking about the distance observed by the particle.

Which distance observed by the particle? Please be more specific.

 

[Kennedy]

Which distance observed by the particle? Please be more specific.

I don't really know what more to say. That is the distance that I calculated that the particle sees itself travelling.

 

[Orodruin]

I don't really know what more to say. That is the distance that I calculated that the particle sees itself travelling.

This is what I thought and it is a very common misconception that you need to throw out of the window as soon as possible. The particle does not see itself traveling anywhere. In the particle's rest frame, it is the lab that is moving and the lab (and therefore the distance between the position where in the lab the particle is created to where it decays) is length contracted in the particle's rest frame. This is why you get a result that differs by an order of magnitude (you are confusing which length is contracted where) as $\gamma^2 \simeq 10$.

Edit: It is always a bit confusing, in particular when you are learning, to figure out which way length contraction goes. Whenever in doubt, do the full analysis using the Lorentz transformations!

 

[PeroK]

I don't really know what more to say. That is the distance that I calculated that the particle sees itself travelling.

Ah, so in method 1, you multiplied by $\gamma$? To explain my method 1 in post #3:

The particle is created at one point in the lab. It lives for a short time and decays at another point in the lab. The distance between these two points in the lab frame is $L$. In the particle's frame the lab is moving at $v$, so in this frame this distance is contracted to $L' = L/\gamma$. And, in the particle frame, the time it takes something of length $L'$ to move past at speed $v$ is $L'/v$.

But, really, if you do things algebraically you would see immediately that in one method you have $L/v\gamma$ and in another you have $L\gamma/v$.

 

[Kennedy]

This is what I thought and it is a very common misconception that you need to throw out of the window as soon as possible. The particle does not see itself traveling anywhere. In the particle's rest frame, it is the lab that is moving and the lab (and therefore the distance between the position where in the lab the particle is created to where it decays) is length contracted in the particle's rest frame. This is why you get a result that differs by an order of magnitude (you are confusing which length is contracted where) as $\gamma^2 \simeq 10$.

Edit: It is always a bit confusing, in particular when you are learning, to figure out which way length contraction goes. Whenever in doubt, do the full analysis using the Lorentz transformations!

I still don't understand. Why can't I use the rest length, and divide by the speed of the particle to get the lifetime of the particle measure by a clock traveling at the same speed as the particle?

 

[Orodruin]

I still don't understand. Why can't I use the rest length, and divide by the speed of the particle to get the lifetime of the particle measure by a clock traveling at the same speed as the particle?

You can, but you are going about it in the wrong way. The relevant length is the distance from production point to decay point in the lab. The rest frame of this length is the lab frame, not the particle rest frame. Again, when in doubt, use the full Lorentz transformation.

 

[Kennedy]

You can, but you are going about it in the wrong way. The relevant length is the distance from production point to decay point in the lab. The rest frame of this length is the lab frame, not the particle rest frame. Again, when in doubt, use the full Lorentz transformation.

We were never taught how to use the full Lorentz transformation. Why is the lab frame the rest frame? It's the one that's moving with respect to the particle?

 

[PeroK]

We were never taught how to use the full Lorentz transformation. Why is the lab frame the rest frame? It's the one that's moving with respect to the particle?

You were given a measurement of $0.250mm$ explicitly in the lab frame. This measurement is not in the particle frame.

 

[Kennedy]

You were given a measurement of $0.250mm$ explicitly in the lab frame. This measurement is not in the particle frame.

Yes, so since it's not in the particle frame, is it not the moving frame? This is what I understand, because we're trying to find the particle's lifetime when measured with respect to itself.

 

[Orodruin]

Imagine placing a rod at rest in the lab frame such that the particle is at one end when it is created and at the other end when it decays. Call the length of this rod in the lab frame is $L_0$, since it is the rest frame of the rod. In the particle rest frame, this rod is moving with speed $v = 0.95c$ and is therefore length contracted to $L = L_0/\gamma$, but the particle must still be created when the first end of the rod passes it and decay when the second end passes it. The time this takes is $L/v = L_0/(\gamma v)$.

Your approach is wrong because it would imply the following argumentation: Place a rod at rest in the particle rest frame such that (in that frame) the particle is at one end (and remains at that end, since neither rod nor particle moves in that frame). Also place a marker at rest in the lab frame such that the particle is created when the marker passes one end of the rod and decays when the marker reaches the particle. Since the rod is at rest in the particle rest frame, you may call its proper length $L_0$ and say that it takes the marker a time $L_0/v$ to reach the particle from the time the particle was created. So far, nothing is wrong. You continue to relate this length $L_0$ to the length of the rod $L = L_0/\gamma$ in the lab frame. This is also correct. You then say that $L$ must be equal to the distance in lab frame that the particle is from the marker when it is created. This is not correct. While the marker passing the front of the rod is simultaneous with the particle's creation in the particle rest frame, it is not simultaneous with the particle creation in the lab frame. This is called relativity of simultaneity and is a crucial ingredient in understanding relativity. Not discussing it properly in popular science and introductory level courses is probably the one main cause of misconceptions about special relativity and "paradoxes" such as the twin paradox. (See my signature ...)

 

[Kennedy]

You then say that LLL must be equal to the distance in lab frame that the particle is from the marker when it is created. This is not correct. While the marker passing the front of the rod is simultaneous with the particle's creation in the particle rest frame, it is not simultaneous with the particle creation in the lab frame.

I don't understand what you're saying here. Is this beyond the realm of a first year physics course? Would you be able to show a video explaining this, or is it not worth my time to know this, and to just do it the other way? I have a final coming up soon and I don't recall ever talking about anything like this.

 

[PeroK]

Yes, so since it's not in the particle frame, is it not the moving frame? This is what I understand, because we're trying to find the particle's lifetime when measured with respect to itself.

When I described method 1, I imagined it exactly as @Orodruin described above.

The student's measuring device is at rest in his frame, hence moving in the particle frame hence contracted in that frame.

 

[Kennedy]

When I described method 1, I imagined it exactly as @Orodruin described above.

The student's measuring device is at rest in his frame, hence moving in the particle frame hence contracted in that frame.

But wouldn't it have been measure by the student in the lab frame? So, he's at rest with respect to the measuring device, and so he doesn't measure a contracted length?

 

[Orodruin]

Unfortunately, I am pretty sure that many university teachers do not understand relativity of simultaneity properly, much less teach it properly. Even if they do understand it, it is too often swept under the carpet. However, relativity of simultaneity is a fundamental issue that is completely necessary to understand what is going on in relativity. In many cases, such as this one, it is important to know how it works in order to apply length contraction in the correct manner. For the purposes of this type of problems, it is clearly simpler to use the argumentation with time dilation instead.

I also find it rather haphazard to teach length contraction and time dilation and expect students to apply them properly without going through the underlying assumptions first and how they relate to the Lorentz transformations. I would go so far as saying that you really cannot understand SR without understanding both Lorentz transformations and relativity of simultaneity. Now, whether that is part of what is expected on your finals or not, I cannot say.

I don't understand what you're saying here.

Exactly which part of the statement do you have problems with? Without knowing that it is difficult to elaborate further.

 

[Orodruin]

But wouldn't it have been measure by the student in the lab frame? So, he's at rest with respect to the measuring device, and so he doesn't measure a contracted length?

The student does not measure a contracted length - however, the length that he does measure is contracted in the rest frame of the particle.

 

[Kennedy]

Unfortunately, I am pretty sure that many university teachers do not understand relativity of simultaneity properly, much less teach it properly. Even if they do understand it, it is too often swept under the carpet. However, relativity of simultaneity is a fundamental issue that is completely necessary to understand what is going on in relativity. In many cases, such as this one, it is important to know how it works in order to apply length contraction in the correct manner. For the purposes of this type of problems, it is clearly simpler to use the argumentation with time dilation instead.

I also find it rather haphazard to teach length contraction and time dilation and expect students to apply them properly without going through the underlying assumptions first and how they relate to the Lorentz transformations. I would go so far as saying that you really cannot understand SR without understanding both Lorentz transformations and relativity of simultaneity. Now, whether that is part of what is expected on your finals or not, I cannot say.Exactly which part of the statement do you have problems with? Without knowing that it is difficult to elaborate further.

I just don't understand what event is not happening simultaneously. I understand that the decaying of the particle does not happen at the same time, but that's a given, because it's exactly what they're asking of me in the problem.

 

[Kennedy]

The student does not measure a contracted length - however, the length that he does measure is contracted in the rest frame of the particle.

Right, so then I measure the contracted length as measured by the particle, in the particle frame, and it seems logical to me to use that length and divide by the speed of the particle to get the time. But apparently being logical doesn't work when talking about special relativity.

 

[Orodruin]

I just don't understand what event is not happening simultaneously.

That what event is happening simultaneously as what other event? The following is true:

• In the particle's rest frame, the particle creation is simultaneous with the marker passing the other end of the rod.
• In the lab frame, the particle creation is not simultaneous with the marker passing the other end of the rod.
• In both frames, the marker passes the particle at the same time that the particle decays. This is because this is a single event in spacetime.

Note that simultaneity for events that are separated depends on the frame. Hence relativity of simultaneity.

 

[PeroK]

But wouldn't it have been measure by the student in the lab frame? So, he's at rest with respect to the measuring device, and so he doesn't measure a contracted length?

Of course not. But, if the student has a special ruler exactly $0.250mm$ long, then it is only this length in his reference frame. In the particle frame, for example, that ruler is a lot shorter.

 

[Orodruin]

Right, so then I measure the contracted length as measured by the particle, in the particle frame, and it seems logical to me to use that length and divide by the speed of the particle to get the time. But apparently being logical doesn't work when talking about special relativity.

Yes it does. But you did it the other way around. You claimed that the length measured in the lab frame was the contracted one, which is not the case. The measured length is a length in the lab frame. This is length is contracted in the particle rest frame and therefore you obtain $L/(\gamma v)$, not $L\gamma/v$.

 

[Kennedy]

That what event is happening simultaneously as what other event? The following is true:

• In the particle's rest frame, the particle creation is simultaneous with the marker passing the other end of the rod.
• In the lab frame, the particle creation is not simultaneous with the marker passing the other end of the rod.
• In both frames, the marker passes the particle at the same time that the particle decays. This is because this is a single event in spacetime.

Note that simultaneity for events that are separated depends on the frame. Hence relativity of simultaneity.

I don't understand the analogy of the marker either. That extra piece is confusing me. That's the main confusion here. Is the marker always moving with respect to the particle?

 

[Orodruin]

I don't understand the analogy of the marker either. That extra piece is confusing me. That's the main confusion here. Is the marker always moving with respect to the particle?

The marker is at rest in the lab frame and therefore moving at speed $v$ relative to the particle. If you are confused by the marker, keep in mind that your argumentation is essentially equivalent to using such a marker and making the (false) assumption that the marker passing the first end of the rod is simultaneous to the particle creation in all frames.

 

[Kennedy]

Yes it does. But you did it the other way around. You claimed that the length measured in the lab frame was the contracted one, which is not the case. The measured length is a length in the lab frame. This is length is contracted in the particle rest frame and therefore you obtain $L/(\gamma v)$, not $L\gamma/v$.

Why wouldn't it be the contracted one? The particle is moving at a fraction of the speed of light in the lab, so wouldn't the moving frame be the lab frame? I think we've agreed on that. The moving frame is the lab frame. L' = 0.250 x 10 x^-3m. Then I calculate LLL, which is 0.250 x 10^-3(λ)?

 

[Orodruin]

Alternatively, imagine that the rod is at rest in the particle rest frame and that in the lab frame the other end of the rod passes the position where the particle eventually will decay at the same time as the particle is created. Since the rod is length contracted, it will indeed be longer in the particle's rest frame. However, in the particle's rest frame, the particle creation is not simultaneous to the event that the other end of the rod passes the position in the lab frame where the particle will eventually decay. Therefore the length of the rod divided by the velocity of the lab frame will not give the correct time.

Why wouldn't it be the contracted one? The particle is moving at a fraction of the speed of light in the lab, so wouldn't the moving frame be the lab frame?

I am going to stop you right there because this is a fundamental misunderstanding of relativity. There is no way you can identify a particular frame as a "moving" frame. All movement is relative, in classical mechanics as well as in special relativity. Relative to the lab frame, the particle is moving. Relative to the particle's rest frame, the lab is moving.

 

[Kennedy]

Alternatively, imagine that the rod is at rest in the particle rest frame and that in the lab frame the other end of the rod passes the position where the particle eventually will decay at the same time as the particle is created. Since the rod is length contracted, it will indeed be longer in the particle's rest frame. However, in the particle's rest frame, the particle creation is not simultaneous to the event that the other end of the rod passes the position in the lab frame where the particle will eventually decay. Therefore the length of the rod divided by the velocity of the lab frame will not give the correct time.I am going to stop you right there because this is a fundamental misunderstanding of relativity. There is no way you can identify a particular frame as a "moving" frame. All movement is relative, in classical mechanics as well as in special relativity. Relative to the lab frame, the particle is moving. Relative to the particle's rest frame, the lab is moving.

I'm very very very confused, but yes I understand that all movement is relative. That was a poor choice of words on my part.