Relativistic Kinematics
Special relativity was first divised by Albert Einstein in 1905. It describes objects moving at speeds near the speed of light,
where classical equations fail to accurately describe the physics. Relativistic kinematics is the kinematics of objects
described through special relativity, and is the basis for
particle physics.
There are two important postulates of special relativity:
- The laws of physics are the same in all intertial frames of reference,
- The speed of light in vacuum, \(c\), is the same in all frames of reference.
An inertial frame of reference is one with no linear or angular acceleration: one in which objects travel at a constant velocity
\(\underline{v}\) in a straight line, unless acted upon by an external force.
Four-vectors are vectors with four components. In addition to the three spatial components of a normal vector, they have an
additional time-like zeroth component. A four-vector \(a\) is written:
\[a=[a^0,\ \underline{a}]=[a^0,\ a^1,\ a^2,\ a^3].\]
The square of a four-vector is given by:
\[a^2=(a^0)^2-\underline{a}^2=(a^0)^2-(a^1)^2-(a^2)^2-(a^3)^2.\]
This can be written more compactly using the Einstein summation convention, in which repeated indices are implied to be
summed over. Using \(\mu\) as the dummy index, the square is:
\[a^2=a^{\mu}a_{\mu}\]
When \(\mu\) is in the superscript, the four-vector is contravariant (a column vector); when it is in the subscript, it is
covariant (a row vector). The metric tensor, \(g_{\mu\nu}\), can be used to convert from one to the other:
\[a_{\mu}=g_{\mu\nu}a^{\nu}.\]
Using the convention of negative space-like components in the first expression for \(a^2\) above, the metric tensor is given by:
\[g_{\mu\nu}=\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{bmatrix}.\]
This matrix is its own inverse, meaning \(g_{\mu\nu}=g^{\mu\nu}\). Converting from a contravariant four-vector to a covariant
four-vector (or vice versa) means changing the signs of the spatial components.
The sum of two four-vectors is the sum of their components, and the dot product is the product of the zeroth components minus
the dot product of the three-vectors:
\[a+b=[a^0+b^0,\ \underline{a}+\underline{b}]\]
\[a\cdot b=a^0b^0-\underline{a}\cdot\underline{b}\]
There are two important frames of reference in relativistic kinematics:
- The Lab Frame: The reference frame of the observer. In scattering experiments, the target particle is at rest in the
lab frame.
- The Center of Mass Frame: The reference frame in which the center of mass between two (or more) moving particles is
at rest.
Both are inertial frames of reference, meaning an event in one frame can be converted into an event in another by applying a
Lorentz transformation.
An event occuring at time and location \((ct, x, y, z)\) in reference frame \(S\) will, in a reference frame \(S^{\prime}\)
moving with speed \(v\) with respect to \(S\), be given by the Lorentz transformations (the inverse transforms are on the
right):
| \(ct^{\prime}=\gamma(ct-\beta x)\) |
\(ct=\gamma(ct^{\prime}-\beta x^{\prime})\) |
| \(x^{\prime}=\gamma(x-\beta ct)\) |
\(x=\gamma(x^{\prime}-\beta ct^{\prime})\) |
| \(y^{\prime}=y\) |
\(y=y^{\prime}\) |
| \(z^{\prime}=z\) |
\(z=z^{\prime}\) |
The Lorentz factors \(\beta\) and \(\gamma\) are given by:
\[\beta=\frac{v}{c},\]
\[\gamma=\frac{1}{\sqrt{1-\beta^2}}.\]
The Lorentz transformation can be carried out by the linear operator \(\Lambda\):
\[\Lambda=\begin{bmatrix}
\gamma & -\gamma\beta & 0 & 0\\
-\gamma\beta & \gamma & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}.\]
The position four-vector in frame \(S^{\prime}\) is then written \(x^{\prime}=\Lambda x\), and the \(\mu\)-th component is
given by \(x^{\prime\mu}=\Lambda^{\mu}_{\nu}x^{\nu}\).
A quantity is said to be Lorentz-invariant if it remains the same under Lorentz transformation, meaning the quantity is the same
in all inertial reference frames. The square of any four-vector is Lorentz-invariant.
The four-momentum is the most useful quantity in particle physics. It is defined by:
\[p=\left[\frac{E}{c},\ \underline{p}\right]=\left[\frac{E}{c},\ p_x,\ p_y,\ p_z\right].\]
The momentum vector in this case is the relativistic momentum, not the classical momentum. While in classical physics,
momentum is given by \(\underline{p}=m\underline{v}\), in relativity, it is calculated using the proper velocity,
\(\eta=\gamma v\), and is \(p=m\eta=\gamma mv\). The four-momentum is therefore:
\[p=\left[\gamma mc,\ \gamma mv_x,\ \gamma mv_y,\ \gamma mv_z\right].\]
The square of the four-momentum is Lorentz invariant, and is given by:
\[p^2=\frac{E^2}{c^2}-\underline{p}^2=m^2c^2\]
The dot product of two four-momenta is:
\[p_A\cdot p_B=\frac{E_AE_B}{c^2}-\underline{p}_A\cdot\underline{p}_B\]
In outdated notation, the relativistic mass of a particle was written \(m=\gamma m_0\), where \(m_0\) was the rest mass. The
distinction between \(m\) and \(m_0\) is now usually abandoned in favour of using the rest mass everywhere, calling it
\(m\).
The relativistic energy of a particle is:
\[E=\gamma mc^2.\]
Knowing that the relativistic momentum is given by \(p=\gamma mv\), the relativistic kinetic energy, \(T\), can be calculated:
\[\begin{align}
T & =\int F\mathrm{d}x\\
& =\int\left(\frac{\mathrm{d}p}{\mathrm{d}t}\right)\mathrm{d}x\\
& =\int\left(\frac{\mathrm{d}(\gamma mv)}{\mathrm{d}t}\right)\mathrm{d}x\\
& =\int\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)\mathrm{d}(\gamma mv)\\
& =\int v\mathrm{d}(\gamma mv).
\end{align}\]
Here, the quantity \(\mathrm{d}(\gamma mv)\) must be calculated, because both \(\gamma\) and \(v\) are functions of time, since
\(\gamma\) contains a \(v\) in it. Taking the proper derivatives and applying the chain rule, this comes out to:
\[\mathrm{d}(\gamma mv)=\frac{m\mathrm{d}v}{\left(1-\frac{v^2}{c^2}\right)^{3/2}}.\]
Performing the integral gives:
\[\begin{align}
T & =\int_0^v\frac{mv}{\left(1-\frac{v^2}{c^2}\right)^{3/2}}\mathrm{d}v\\
& =\left[\gamma mc^2\right]_0^v.
\end{align}\]
The relativistic kinetic energy is then:
\[T=(\gamma-1)mc^2.\]
The energy-mass relation is an equation that relates the energy, momentum, and mass of a particle. It is the complete and
modern version of Einstein's famous equation \(E=mc^2\). Rearranging the expressions for \(p^2\) above, it is given by:
\[E^2=\underline{p}^2c^2+m^2c^4.\]
For massless particles, like the photon, this reduces to \(E=pc\). For particles at rest, this reduces to \(E=mc^2\).
It can be easily shown that this is consistent with the relativistic energy given above, by substituting into the energy-mass
relation the expression for relativistic momentum, \(p=\gamma mv\):
\[\begin{align}
E^2 & =\gamma^2m^2v^2c^2+m^2c^4\\
& =m^2c^2(\gamma^2v^2+c^2)\\
& =m^2c^2(\gamma^2c^2)\\
& =\gamma^2m^2c^4.
\end{align}\]
Therefore,
\[E=\gamma mc^2,\]
as above. The third line was calculated using the definition of \(\gamma\) and simple algebra.
In the non-relativistic limit (NRL), \(|\underline{p}|c\ll mc^2\). With the energy-mass relation, this implies:
\[E\approx mc^2.\]
In the ultra-relativistic limit (URL), \(|\underline{p}|c\gg mc^2\). With the energy-mass relation, this implies:
\[E\approx |\underline{p}|c.\]
The Mandelstam variables are Lorentz-invariant variables.
\[s=(p_1+p_2)^2c^{-2}=(p_3+p_4)^2c^{-2}\]
\[t=(p_1-p_3)^2c^{-2}=(p_4-p_2)^2c^{-2}\]
\[u=(p_1-p_4)^2c^{-2}=(p_3-p_2)^2c^{-2}\]
They can be represented by the respective diagrams below:
The table below summarises what types of interactions each channel is relevant for:
| Channel |
Interaction Type(s) |
| \(s\)-channel |
\(AB\rightarrow AB\) |
| \(t\)-channel |
\(AB\rightarrow AB\) and \(AA\rightarrow BB\) |
| \(u\)-channel |
\(AA\rightarrow BB\) |
The center of mass energy is given by \(\sqrt{s}\), and the momentum transfer is given by \(\sqrt{t}\). Using basic algebra,
the sum of the variables can be shown to be:
\[s+t+u=m_1^2+m_2^2+m_3^2+m_4^2\]
Suggestions:
- To get the energy of a particle when its momentum is known (or vice versa), use \(E^2-\underline{p}^2c^2=m^2c^4\).
- To get the velocity of a particle when its energy and momentum are known, use \(\underline{v}^2=\underline{p}^2c/E\).
- Exploit the invariance of the dot product \(p^2=m^2c^2\) for any real particle.
- If a problem is complicated in the lab frame, analyse it in the center of mass frame.
Center of Mass Energy
Show that \(\sqrt{s}\) is the center of mass energy.
Start by writing the four-momenta of particles A and B:
\[p_1=\left[\frac{E_1}{c},\ \underline{p}_1\right],\]
\[p_2=\left[\frac{E_2}{c},\ \underline{p}_2\right]=\left[\frac{E_2}{c},\ -\underline{p}_1\right].\]
The
Mandelstam variable \(s\) is given by:
\[s=(p_1+p_2)^2c^{-2}\]
The momenta can simply be added in the forms they were given in above:
\[p_1+p_2=\left[\frac{E_1+E_2}{c},\ \underline{p}_1-\underline{p}_1\right]=\left[\frac{E}{c},\ \underline{0}\right],\]
where \(E=E_1+E_2\) is the total energy. \(s\) is then:
\[s=\frac{E^2}{c^4}.\]
Rearranging for \(E\) gives:
\[\boxed{E=\sqrt{s}c^2}.\]
In natural units, where \(c=1\), \(\sqrt{s}\) is the center of mass energy.
Collision in the Lab Frame
Two particles A and B collide, where B is initially at rest, to produce particles C and D. Find the energy of A in the lab
frame.
Start by writing the four-momenta of particles A and B:
\[p_A=\left[\frac{E_A}{c},\ \underline{p}_A\right],\]
\[p_B=\left[m_Bc,\ \underline{0}\right].\]
The expression for \(p_A\) can give access to \(E_A\) in terms of the
Mandelstam variable
\(s\) after some evaluation. \(s\) is given by:
\[s=(p_A+p_B)^2c^{-2}=(p_A^2+p_B^2+2p_A\cdot p_B)c^{-2}.\]
The first two terms in the brackets are easily evaluated using
Suggestion (3):
\[p_A^2=m_A^2c^2,\]
\[p_B^2=m_B^2c^2.\]
The dot product is given by:
\[p_A\cdot p_B=E_Am_B.\]
Therefore, \(s\) is:
\[s=m_A^2+m_B^2+2E_Am_Bc^{-2}.\]
Rearranging for \(E_A\) gives:
\[\boxed{E_A=\frac{(s-m_A^2-m_B^2)c^2}{2m_B}}.\]
Collision in the Center of Mass Frame
Two particles A and B collide to produce particles C and D. Find the energy of A in the center of mass frame.
Start by writing the four-momenta of particles A and B:
\[p_A=\left[\frac{E_A}{c},\ \underline{p}_A\right],\]
\[p_B=\left[\frac{E_B}{c},\ \underline{p}_B\right]=\left[\frac{E_B}{c},\ -\underline{p}_A\right].\]
The momentum \(\underline{p}_B=-\underline{p}_A\) because the problem is being analysed in the center of mass frame. The
expression for \(p_A\) can give access to \(E_A\) in terms of the
Mandelstam variable \(s\)
after some evaluation. \(s\) is given by:
\[s=(p_A+p_B)^2c^{-2}=(p_A^2+p_B^2+2p_A\cdot p_B)c^{-2}.\]
The first two terms in the brackets are easily evaluated using
Suggestion (3):
\[p_A^2=m_A^2c^2,\]
\[p_B^2=m_B^2c^2.\]
The dot product is given by:
\[\begin{align}
p_A\cdot p_B & =p_A^0p_B^0-\underline{p}_A\cdot\underline{p}_B,\\
& =\frac{E_AE_B}{c^2}-\underline{p}_A\cdot(-\underline{p}_A),\\
& =\frac{E_AE_B}{c^2}+\underline{p}_A^2.
\end{align}\]
From rearranging the
energy-mass relation, \(\underline{p}_A^2\) is:
\[\underline{p}_A^2=\frac{E_A^2}{c^2}-m_A^2c^2.\]
This can be substituted into the dot product:
\[p_A\cdot p_B=\frac{E_AE_B}{c^2}+\frac{E_A^2}{c^2}-m_A^2c^2,\]
which can now be inserted into \(s\):
\[\begin{align}
s & =\left(m_A^2c^2+m_B^2c^2+2\left(\frac{E_AE_B}{c^2}+\frac{E_A^2}{c^2}-m_A^2c^2\right)\right)c^{-2},\\
& =m_B^2-m_A^2+2E_A^2c^{-4}+2E_AE_Bc^{-4}.
\end{align}\]
From rearranging the
energy-mass relation, similar to \(\underline{p}_A^2\) above, \(E_B\)
can be written as:
\[\frac{E_B^2}{c^2}=m_B^2c^2+\underline{p}_B^2,\]
\[E_Bc^{-1}=\sqrt{m_B^2c^2+\underline{p}_B^2}.\]
From before \(\underline{p}_B=-\underline{p}_A\), which means \(\underline{p}_B^2\) can be replaced with the expression for
\(\underline{p}_A^2\):
\[E_Bc^{-1}=\sqrt{\frac{E_A^2}{c^2}+m_B^2c^2-m_A^2c^2}.\]
This can now be substituted into the expression for \(s\). After rearranging to isolate the square root:
\[2E_Ac^{-3}\sqrt{\frac{E_A^2}{c^2}+m_B^2c^2-m_A^2c^2}=s+m_A^2-m_B^2-2E_A^2c^{-4}.\]
Square both sides:
\[4E_A^2c^{-6}\left(\frac{E_A^2}{c^2}+m_B^2c^2-m_A^2c^2\right)=(s+m_A^2-m_B^2)^2+(2E_A^2c^{-4})^2-4E_A^2c^{-4}(s+m_A^2-m_B^2).\]
Multiply a factor of \(c^{-2}\) into the brackets on the left-hand side, and refactor both side:
\[\cancel{4E_A^4c^{-8}}+\cancel{4E_A^2c^{-4}(m_B^2-m_A^2)}=(s+m_A^2-m_B^2)^2+\cancel{4E_A^4c^{-8}}-4E_A^2c^{-4}s+\cancel{4E_A^2c^{-4}(m_B^2-m_A^2)}.\]
Simplify:
\[0=(s+m_A^2-m_B^2)^2-4E_A^2c^{-4}s.\]
Finally, rearrange for \(E_A\):
\[4E_A^2=\frac{(s+m_A^2-m_B^2)^2c^4}{s},\]
\[\boxed{E_A=\frac{(s+m_A^2-m_B^2)c^2}{2\sqrt{s}}}.\]
The problem is clearly much simpler in the lab frame.
Collision with Multiple Products
Two particles A and B collide, where B is initially at rest, to produce \(n\) C particles. Find the minimum energy of
particle A required to produce this collision.
?
Decay into Two Particles
Particle A, initially at rest, decays into particles B and C. Find the energies of particles B and C.
Because particle A is initially at rest, its four-momentum is:
\[p_A=\left[\frac{E_A}{c},\ \underline{0}\right].\]
The four-momenta of particles B and C are:
\[p_B=\left[\frac{E_B}{c},\ \underline{p}_B\right],\]
\[p_C=\left[\frac{E_C}{c},\ \underline{p}_C\right],\]
From conservation of momentum:
\[p_A=p_B+p_C,\]
\[p_C=p_A-p_B.\]
Using
Suggestion (3), square the expression for \(p_C\):
\[p_C^2=p_A^2+p_B^2-2p_Ap_B,\]
\[m_C^2c^2=m_A^2c^2+m_B^2c^2-2p_A\cdot p_B.\]
The dot product is given by:
\[\begin{align}
p_A\cdot p_B & =\frac{E_A}{c}\frac{E_B}{c}-\underline{p}_A\cdot\underline{p}_B,\\
& =\frac{E_AE_B}{c^2}-\underline{0}\cdot\underline{p}_B,\\
& =\frac{E_AE_B}{c^2},\\
& =m_AE_B.
\end{align}\]
The simplified form of the
energy-mass relation for a particle at rest was used to replace
\(E_A\) with \(m_Ac^2\). This can now be subsituted back into the expression for \(m_C^2c^2\):
\[m_C^2c^2=m_A^2c^2+m_B^2c^2-2m_AE_B.\]
Rearranging for \(E_B\) gives:
\[\boxed{E_B=\frac{(m_A^2+m_B^2-m_C^2)c^2}{2m_A}}.\]
The exact same steps can be followed for \(E_C\) to obtain:
\[\boxed{E_C=\frac{(m_A^2+m_C^2-m_B^2)c^2}{2m_A}}.\]