PET scanning · PET扫描
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| tracer/ˈtreɪsə/ | 示踪剂 | shì zōng jì |
| metabolism/məˈtæbəlɪzəm/ | 代谢 | dài xiè |
| positron emission tomography/ˈpɒzɪtrɒn ɪˈmɪʃn təˈmɒɡrəfi/ | 正电子发射断层扫描 | zhèng diàn zi fā shè duàn céng sǎo miáo |
| positron/ˈpɒzɪtrɒn/ | 正电子 | zhèng diàn zi |
| antiparticle/ˌæntɪˈpɑːtɪkl/ | 反粒子 | fǎn lì zi |
| annihilate/əˈnaɪəleɪt/ | 湮灭 | yān miè |
| energy/ˈenədʒi/ | 能量 | néngliàng |
| momentum/məʊˈmentəm/ | 动量 | dòngliàng |
| electron/ɪˈlektrɒn/ | 电子 | diàn zi |
Two photons, exactly back to back
- A PET scanner does not watch the tracer. It watches for two gamma photons arriving at opposite sides of the ring at the same instant.
- That pair must have come from a single point somewhere on the straight line joining the two detectors. One coincidence, one line.
- Collect a few million of those lines from every angle and the intersection is a three-dimensional map of where the tracer went.
- The whole method rests on one conservation law forcing the photons to be back to back. This lesson is why.
两个光子,恰好背对背
- PET 扫描仪并不"看"示踪剂。它守候的是同一瞬间到达探测环相对两侧的两个伽马光子。
- 这一对必定来自连接那两个探测器的直线上某一点。一次符合,一条线。
- 从各个角度收集几百万条这样的线,它们的交汇处就是示踪剂去向的三维图。
- 整套方法都建立在一条守恒定律迫使两个光子背对背这一点上。这一课讲为什么。
The tracer
- A tracer 示踪剂 is a substance containing radioactive nuclei that is introduced into the body, usually bound to a molecule such as glucose, and is absorbed by the tissue being studied.
- A tumour has a high metabolism 代谢, so it takes up more glucose-tagged tracer than the tissue around it, and shows up bright.
- For positron emission tomography 正电子发射断层扫描 (PET) the tracer must be a beta-plus emitter, so that it gives out a positron 正电子. Fluorine-18 on a glucose analogue is the standard one.
- The half-life is a compromise: short enough that the activity is high during the scan and the patient's dose afterwards is small, but not so short that it decays before reaching the tissue. Fluorine-18 has $110$ minutes, oxygen-15 only $2$, so oxygen-15 is made on the spot.
示踪剂
- 示踪剂(tracer)是含有放射性核素、被引入体内的物质,通常与葡萄糖之类的分子结合,并被所研究的组织吸收。
- 肿瘤的代谢(metabolism)旺盛,所以比周围组织摄取更多带葡萄糖标记的示踪剂,于是显得很亮。
- 正电子发射断层扫描(positron emission tomography,PET)要求示踪剂是 β⁺ 放射体,好放出正电子(positron)。标准的一种是标记在葡萄糖类似物上的氟-18。
- 半衰期是个折中:短到让扫描期间活度高、扫描后患者剂量小,但又不能短到还没到达组织就衰变完。氟-18 是 $110$ 分钟,氧-15 只有 $2$ 分钟,所以氧-15 要现场制备。
A PET tracer is: · PET示踪剂是：
It is a β$^{+}$ emitter on a molecule (like glucose) that active tissue absorbs more of. · 它是附着在分子（如葡萄糖）上的 β$^{+}$ 发射体，活跃组织吸收更多此类物质。
Why is a tracer's half-life chosen as a compromise? Select all · 所有 that apply. · 为什么示踪剂的半衰期要选择折衷方案？选择 所有 适用项。
Fluorine-18 has 110 minutes; oxygen-15 has only 2 and must be made on the spot. Leaving activity in the patient afterwards is the thing to avoid, not to aim for. · 氟-18 有 110 分钟；氧-15 只有 2 分钟，必须在现场制备。避免在扫描后让患者体内残留活性是关键，而不是追求这一点。
Annihilation
- When a particle meets its antiparticle 反粒子 they annihilate 湮灭: their mass becomes electromagnetic energy. Mass-energy and momentum are both conserved.
- The four-mark explanation: the positron travels a few millimetres and meets an electron 电子 in the tissue; the pair annihilates, mass converting to energy; this appears as two gamma-ray photons; because momentum 动量 must be conserved and the pair had almost none, the photons travel in opposite directions with equal momenta.
- A single photon could not conserve momentum. That sentence is what the fourth mark is for, and it is the reason the whole technique works.
One event, one straight line
湮灭
- 当一个粒子遇到它的反粒子(antiparticle),二者湮灭(annihilate):它们的质量变成电磁能。质能和动量都守恒。
- 四分的解释:正电子走出几毫米后在组织中遇到一个电子(electron);这一对湮灭,质量转化为能量;能量表现为两个伽马光子;由于动量(momentum)必须守恒而这一对原本几乎没有动量,两个光子只能朝相反方向飞出,动量大小相等。
- **单个光子无法守恒动量。**这句话正是第四分的所在,也是整套技术能成立的原因。
一次事件,一条直线
PET scan route · PET扫描流程
Follow positron emission to a ring of detected photons. · 追踪正电子发射到光子环状探测器的过程。
When the emitted positron meets an electron, they: · 当发射的正电子遇到电子时，它们：
Matter and antimatter annihilate: their mass turns into two photons. · 物质与反物质湮灭：其质量转化为两个光子。
The two annihilation photons travel in opposite directions. · 两个湮灭光子向相反方向传播。
The pair was almost at rest, so to conserve momentum the photons go back-to-back. · 由于这对粒子几乎静止，为了守恒动量，光子必须背对背运动。
Put the four-mark explanation of annihilation in PET in order. · 按顺序排列PET湮灭的四分解释。
The last step is the one that earns the fourth mark: a single photon could not conserve momentum, which is exactly why the technique works. · 最后一步是获得第四分的步骤：单个光子无法守恒动量，这正是该技术有效的原因。
The energy of each photon
- Energy conservation shares the pair's rest energy 能量 between two identical photons:
- So each photon carries $m_e c^2 = 8.2\times10^{-14}\ \text{J} = 0.51\ \text{MeV} = 511\ \text{keV}$.
- Every PET photon has this same energy, which is what lets the scanner reject anything that is not $511\ \text{keV}$ as scatter or background.
每个光子的能量
- 能量守恒把这一对的静止能量(energy)平分给两个相同的光子:
- 所以每个光子携带 $m_e c^2 = 8.2\times10^{-14}\ \text{J} = 0.51\ \text{MeV} = 511\ \text{keV}$。
- 每一个 PET 光子的能量都是这个值,这正是扫描仪能把不是 $511\ \text{keV}$ 的信号当作散射或本底剔除掉的依据。
Each photon from electron–positron annihilation has an energy of (in keV): · 每个来自电子-正电子湮灭的光子能量为（单位 keV）：
Each carries $m_e c^{2} \approx 0.511\ \text{MeV} = 511\ \text{keV}$. · 每个携带 $m_e c^{2} \approx 0.511\ \text{MeV} = 511\ \text{keV}$。
Each annihilation photon carries an energy of ____ keV. · 每个湮灭光子携带 ____ keV 的能量。
That is m_e c^2, the rest energy of ONE particle. The pair's total is 1.02 MeV. Writing 2 m_e c^2 per photon is the standard error. · 那是 m_e c^2，即 ONE 粒子的静止能量。这对粒子的总能量为 1.02 MeV。每光子写 2 m_e c^2 是标准错误。
Worked example: the annihilation photons
- Find the total energy released when a slow-moving positron and electron annihilate, and the wavelength of each photon.
- Total: $E = 2m_e c^2 = 2(9.11\times10^{-31})(3.00\times10^{8})^2 = 1.64\times10^{-13}\ \text{J} = 1.02\ \text{MeV}$.
- Shared equally: $8.2\times10^{-14}\ \text{J}$ each.
- $\lambda = \dfrac{hc}{E} = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{8.2\times10^{-14}} = 2.4\times10^{-12}\ \text{m}$, a gamma ray.
- If the particles had been moving at $4.9\times10^{7}\ \text{m/s}$, each would add only $1.1\times10^{-15}\ \text{J}$, about $1\%$ of its rest energy. The rest energy dominates, which is why every annihilation gives very nearly $511\ \text{keV}$.
例题:湮灭光子
- 求一个慢速正电子与一个电子湮灭时放出的总能量,以及每个光子的波长。
- 总能量:$E = 2m_e c^2 = 2(9.11\times10^{-31})(3.00\times10^{8})^2 = 1.64\times10^{-13}\ \text{J} = 1.02\ \text{MeV}$。
- 平分:每个 $8.2\times10^{-14}\ \text{J}$。
- $\lambda = \dfrac{hc}{E} = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{8.2\times10^{-14}} = 2.4\times10^{-12}\ \text{m}$,是伽马射线。
- 若这两个粒子原本以 $4.9\times10^{7}\ \text{m/s}$ 运动,各自也只添上 $1.1\times10^{-15}\ \text{J}$,约为其静止能量的 $1\%$。静止能量占主导,这就是每次湮灭都给出几乎恰好 $511\ \text{keV}$ 的原因。
An annihilation photon has energy 8.2e-14 J. What is its wavelength, in units of 1e-12 m? · 一个湮灭光子的能量为 8.2e-14 J。其波长是多少（单位为 1e-12 m）？
lambda = hc/E = 2.4e-12 m, firmly in the gamma range. Every PET photon has this same energy, which lets the scanner reject anything else as scatter. · lambda = hc/E = 2.4e-12 m，明确处于伽马射线范围内。每个PET光子都具有相同的能量，这使得扫描仪能够排除其他散射信号。
Detecting a coincidence
- The two photons leave the body in opposite directions and strike a ring of detectors around the patient.
- Recording two simultaneous arrivals, a coincidence, fixes the line the annihilation lay on. A single detection on its own tells you nothing about position.
- Comparing the two arrival times more precisely narrows the position along that line. This is time-of-flight PET.
- The photons must escape the body to be detected at all, which is why gamma rays are used rather than the positron itself: a positron travels only a few millimetres before it annihilates.
探测一次符合
- 两个光子朝相反方向离开身体,打在患者周围的探测器环上。
- 记录到两个同时到达的信号,即一次符合,就确定了湮灭所在的那条线。单独一次探测本身说明不了位置。
- 更精确地比较两个到达时间,可以把位置沿那条线进一步收窄。这就是飞行时间 PET。
- 光子必须能逃出体外才谈得上被探测,这就是用伽马射线而不是正电子本身的原因:正电子走不出几毫米就湮灭了。
Two photons detected at the same instant — a ____ — fix the line the annihilation happened on. · 同时探测到两个光子——这种____——确定了湮灭发生的直线位置。
Many such coincidence lines, from many angles, let a computer build a 3-D map of the tracer. · 许多这样的符合线，来自许多角度，让计算机构建示踪剂的3D地图。
A single detected gamma photon is enough to locate an annihilation event. · 单个被探测到的伽马光子不足以定位湮灭事件。
One detection gives no position at all. It takes two simultaneous arrivals on opposite sides, a coincidence, to fix the line the event lay on. · 单次探测完全无法确定位置。需要两侧同时到达的两个光子，即符合计数，才能确定事件所在的直线。
Building the image
- Many coincidences, from many angles, give many lines. The computer finds where the lines concentrate, and that is where the tracer is.
- The result is a three-dimensional map, not of anatomy but of tracer concentration, which means of metabolic activity.
- That is the difference from a CT scan worth stating: CT shows structure, PET shows function. Modern scanners often do both and overlay them.
重建图像
- 来自许多角度的许多次符合给出许多条线。计算机找出这些线密集交汇的地方,那里就是示踪剂所在。
- 结果是一幅三维图,画的不是解剖结构,而是示踪剂的浓度,也就是代谢活跃程度。
- 这是与 CT 的区别,值得写出来:CT 显示结构,PET 显示功能。现代扫描仪常常两者都做并叠加在一起。
Match each scan to what it shows. · 将每种扫描与其显示的内容匹配。
A coincidence gives a LINE, not a point; the point comes from where many lines cross. Modern scanners often overlay the two images. · 符合计数给出的是直线，而非点；点来自于多条直线的交汇处。现代扫描仪通常会将这两种图像叠加。
Marks that slip away
- The two photons go in opposite directions because momentum must be conserved. Say the conservation law, not just the direction.
- Each photon is $511\ \text{keV}$, which is $m_e c^2$, not $2m_e c^2$. The pair's total is $1.02\ \text{MeV}$.
- The tracer is a beta-plus emitter. A gamma emitter would not annihilate and could not be located by coincidence.
- The half-life is a compromise: short for a low dose, but long enough to reach the tissue.
- A coincidence gives a line, not a point. The point comes from where many lines cross.
容易丢掉的分
- 两个光子朝相反方向飞,是因为动量必须守恒。要说出守恒定律,不能只说方向。
- 每个光子是 $511\ \text{keV}$,即 $m_e c^2$,不是 $2m_e c^2$。一对的总和才是 $1.02\ \text{MeV}$。
- 示踪剂是 β⁺ 放射体。伽马放射体不会湮灭,也无法用符合来定位。
- 半衰期是折中:短以降低剂量,但要长到足以到达组织。
- 一次符合给出的是一条线,不是一个点。点来自许多条线的交汇。
You've got it
- a tracer is radioactive nuclei introduced into the body and taken up by the tissue studied; PET needs a beta-plus emitter such as fluorine-18
- a positron annihilates with an electron, and momentum conservation forces two gamma photons in opposite directions
- each photon carries $m_e c^2 = 0.51\ \text{MeV}$, a wavelength of about $2.4\times10^{-12}\ \text{m}$
- a coincidence at the detector ring fixes the line of the event, and many lines from many angles build a 3-D map of tracer concentration
你掌握了
- 示踪剂是引入体内、被所研究组织摄取的放射性核素;PET 需要氟-18 这样的 β⁺ 放射体
- 正电子与电子湮灭,而动量守恒迫使产生方向相反的两个伽马光子
- 每个光子携带 $m_e c^2 = 0.51\ \text{MeV}$,波长约 $2.4\times10^{-12}\ \text{m}$
- 探测环上的一次符合确定事件所在的线,来自多个角度的许多条线建立起示踪剂浓度的三维图