#3 Given time shift of aircraft, calculate speed.
Time difference = 4.00E-09 seconds, c= 3.00E+08
Distance 3.00E+06 meters in earth frame
a. Earth time interval is longer, stationary observer.
b. Pilot sees 3000 km as length contracted.
Both observers measure same value for v
t(earth) = 3000/v
t(pilot) = (3000)sqrt(1-v^2/c^2)/v
3000/v(1 - sqrt(1-v^2/c^2)) = 4ns
Use first two terms of binomial expansion:
sqrt(1-a) approx= 1 - a/2

#3 已知飞机的时间偏移，计算速度。时间差=4.00E-09秒，c=3.00E+08。在地球框架下，距离为3.00E+06米。a. 地球时间间隔更长，静止观察者。b. 飞行员看到3000公里为长度收缩。两个观察者测量的v相同。t(earth)=3000/v，t(pilot)=3000/(sqrt(1-v^2/c^2)*v)。3000/(v*(1 - sqrt(1-v^2/c^2)))=4ns。使用二项式展开的前两项：sqrt(1-a)近似=1 - a/2。

3000000(v/2c²) = 4 ns v = 239.7 m/s

6. Proper lifetime of particle
Proper lifetime means lifetime in rest frame of particle
v =.99c with respect to lab frame
Lab frame distance = 0.001 m
Particle frame distance = 0.000141
Two frames agree on the 0.99c velocity
Particle frame time = distance/velocity = 4.75 E-13 sec

粒子的固有寿命粒子在静止框架中的寿命是指以0.99c相对于实验室框架的速度运动时的寿命实验室框架距离 = 0.001 m 粒子框架距离 = 0.000141 两个框架同意0.99c的速度粒子框架时间 = 距离/速度 = 4.75 E-13 秒

8. Muon and pion
a. Muon proper lifetime = 2.20E-06 seconds
speed = .99c
Distance before decay = 652.96 meters in muon frame
The relativity factor is = 7.0888
In the lab frame, this would be 4629 meters in lab frame.

muon的固有寿命 = 2.20E-06 秒速率 = 0.99c 在muon参考系中衰变前的距离 = 652.96 米相对论因子是 = 7.0888 在实验室参考系中，这将是 4629 米在实验室参考系中。

b. Charged pion lifetime = 2.60E-08 seconds
speed = .99c
relativity factor = 7.0888
Lab frame distance = 54.7 meters
9. Twin paradox
a. Earthbound twin determines time to be shorter in travelers frame.
b. In the absence of acceleration, the traveler would measure the earthbound twin's time to be shorter.
c. The acceleration is in the travelers frame, so the earth frame is considered to be the inertial frame.
The traveling twin would be younger.

b. 带电π介子的寿命 = 2.60E-08 秒速率 = 0.99c 相对论因子 = 7.0888 实验室参考系距离 = 54.7 米 9. 双生子悖论 a. 地面双生子确定在旅行者参考系中时间更短。 b. 在没有加速度的情况下，旅行者会测量地面双生子的时间更短。 c. 加速度存在于旅行者参考系中，因此地球参考系被认为是惯性参考系。旅行者会更年轻。

13. Momentum and speed of a 1 GeV proton.
From the relativistic kinetic energy expression, the speed must be 0.875 c From the relativistic momentum expression, its momentum is then = 9.06E-19 kg m/s
At this speed, the non-relativistic momentum expression would give 4.38E-19 kg m/s
which would be in error by 51%

Another approach is to use pc =sqrt(E^2 - (m0c^2)^2) and express p in MeV/c
Must note that the E in this expression includes both KE and the rest mass energy.
pc=sqrt((1GeV +.938GeV)^2-(.938 GeV)^2)) = 1.696 GeV
The momentum can then be expressed as p=1.70 GeV/c

13. 1 GeV质子的动量和速率。从相对论动能表达式可知，其速率必须为0.875 c。从相对论动量表达式可知，其动量为9.06E-19 kg m/s。在该速率下，非相对论动量表达式会给出4.38E-19 kg m/s，这将产生51%的误差。另一种方法是使用pc = sqrt(E^2 - (m0c^2)^2)并以MeV/c表示动量。必须注意，此表达式中的E包括了动能和静止质量能量。pc = sqrt((1GeV + 0.938GeV)^2 - (0.938 GeV)^2) = 1.696 GeV。动量可以表示为p = 1.70 GeV/c。

14. Radius of curvature of 1 MeV electron in 1 Tesla magnetic field
Force equation mv^2/r = evB, so r = mv/eB, but relativistic momentum must be used
From the relativistic kinetic energy expression, the velocity must be= 0.941 c
for the kinetic energy to be 1 MeV.
From the relativistic momentum relationship, p = 7.59E-22 kg m/s
e= 1.60E-19 C
Resulting radius = 4.74E-03 m

14. 1 MeV电子在1特斯拉磁场中的曲率半径力方程 mv²/r = evB，所以 r = mv/(eB)，但必须使用相对论动量。从相对论动能表达式可知，当动能为1 MeV时，速度必须为0.941 c。从相对论动量关系式可知，p = 7.59E-22 kg m/s，e= 1.60E-19 C。结果半径为4.74E-03 m

17. Calculated mass of alpha particle
mass of proton= 938.3 MeV in energy equivalents
mass of neutron= 939.6 MeV
Binding energy = 28.4 MeV
mass of alpha= 3727.4 MeV
mass of alpha= 6.64E-27 kg

计算出的α粒子质量等于质子质量=938.3 MeV的能量当量，中子质量=939.6 MeV，结合能=28.4 MeV，α粒子质量=3727.4 MeV，α粒子质量=6.64E-27 kg

18. Carbon-14 decay energy
From Appendix K of Rohlf
mass of carbon 13040.976 MeV
mass of nitrogen 13040.309 MeV
mass of electron 0.511 MeV
energy release 0.156 MeV

碳-14衰变释放的能量从Rohlf附录K中得知，碳的质量为13040.976 MeV，氮的质量为13040.309 MeV，电子的质量为0.511 MeV，能量释放为0.156 MeV

21. Electron KE=mass energy, find photon energy for photon with same momentum
Electron mass energy = 0.511 MeV
KE of electron = 0.511 MeV requires speed of    0.866 c
The momentum of this electron is then =    4.73E-22 kg m/s
The momentum of a photon = E/c, so E=pc =   1.42E-13 joules
    8.85E+05 eV

21. 电子的动能等于其静止能量，求具有相同动量的光子的能量。电子静止能量为0.511 MeV，电子动能为0.511 MeV，要求其速度为0.866 c。该电子的动量则为4.73E-22 kg m/s。光子的动量等于E/c，因此E=pc=1.42E-13 joules，即8.85E+05 eV。

22. Speed to red-shift to 1% of frequency
To shift frequency to 1% of its source frequency requires a recession
velocity of 0.9998 c

22. 将频率移至其源频率的1%需要退行速度为0.9998 c。

29. Z-zero decay into electron and positron
Mass energy of Z-zero = 91.2 GeV
Particle is at rest, so electron and positron must have equal but
oppositely directed momenta.
Energy of each is 45.6 GeV p=45.6 GeV/c
For this extreme relativistic case, the mass energy can be neglected
in calculating the momentum, and the photon expression p=E/c used.
With this approximation, p= 152.1014009 eV s/m
The required speed is so close to c that you can't take the square root
to calculate c.
The relativistic gamma factor is = 45,600MeV/.511MeV =
gamma = 89236.8
(use b and g for beta and gamma since Greek letters can't be depended upon.)
We can assess the departure from c with the fractional departure from c, 1 -v/c
The trick is to find an expression for 1-v/c = 1-b which we can evaluate.
Since g^2 = 1/(1-b^2) = 1/((1-b)(1+b)), then we can write
1-b = 1/((1+b)g^2) or approximately: 1-b = 1/2g^2
This gives 1-b= 6.28E-11

29. Z零衰变为电子和正电子。Z零的静止能量为91.2 GeV。粒子处于静止状态，因此电子和正电子必须具有大小相等但方向相反的动量。每个粒子的能量为45.6 GeV。p=45.6 GeV/c。对于这种极端相对论性情况，计算动量时可以忽略质量能量，使用光子表达式p=E/c进行近似。使用此近似值，p=152.1014009 eV s/m。所需速度非常接近c，以至于无法通过平方根计算c。相对论的伽马因子为=45,600 MeV/0.511 MeV=伽马=89236.8（使用b和g表示beta和gamma，因为希腊字母无法依赖。）我们可以评估与c的偏差，即1-v/c。关键在于找到一个表达式来计算1-v/c=1-b，我们可以评估它。由于g²=1/(1-b²)=1/((1-b)(1+b))，因此可以写成1-b=1/((1+b)g²)或近似为：1-b=1/2g²。这给出1-b=6.28E-11

36. Back-scattered 10GeV photon
a) High energy collisions like this need to be viewed in the center-of-mass
frame. The photon energy does not change in the collision as viewed in the
COM frame. Transforming back to the laboratory frame, the electron has
essentially all the energy. Take electron energy = 10 GeV, and the photon
energy can be approximated by:

36. 反散射10GeV光子 a) 这样的高能碰撞需要在质心系中观察。在质心系中，光子的能量在碰撞过程中不发生变化。转换回实验室系，电子几乎拥有全部的能量。取电子能量为10 GeV，光子能量可以近似表示为：

E' = mc² E/(mc² + 2E) ≈ mc²/2 = 0.511 MeV/2 = 0.26 MeV

b) Again the collision is examined in the COM frame. This time the photon has
essentially all the energy. The photon energy is about 10 GeV; the electron
energy is about mc^2

b) 再次在质心系中考察碰撞。这次光子几乎拥有全部的能量。光子能量约为10 GeV；电子能量约为mc²

Relativistic projectile problem.
Enterprise traveling at 0.8c as seen by external observer
Projectile fired at 0.6c as seen by Enterprise
Klingon vessel approaches at 0.7c in opposite direction as seen by external observer

相对论弹道问题。企业号以0.8c的速度运动，这是外部观察者看到的。企业号发射的弹丸以0.6c的速度运动，这是企业号观察到的。克林贡飞船以0.7c的速度朝相反方向接近，这是外部观察者看到的。

Projectile speed as seen by external observer =   0.9459 c
Approach speed of Klingons as seen by Enterprise =   -0.9615 c
Projectile speed as seen from Klingon ship =    0.9902 c