# Decay of a pion into a muon and a neutrino

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 31.

[Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of ${c}$ in the equations rather than setting ${c=1}$). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware\ldots ]

We can use the conservation of relativistic energy and momentum to analyze the interaction of elementary particles. For example, a pion at rest can decay into a muon and a neutrino. Conservation of energy and 3-momentum require

 $\displaystyle E_{\pi}=m_{\pi}c^{2}$ $\displaystyle =$ $\displaystyle E_{\mu}+E_{\nu}\ \ \ \ \ (1)$ $\displaystyle \mathbf{p}_{\pi}=0$ $\displaystyle =$ $\displaystyle \mathbf{p}_{\mu}+\mathbf{p}_{\nu} \ \ \ \ \ (2)$

We can use the relation

$\displaystyle E^{2}-p^{2}c^{2}=m^{2}c^{4} \ \ \ \ \ (3)$

to relate energy and momentum. Assuming the neutrino is massless (it isn’t quite, but it’s close) we have

$\displaystyle E_{\nu}=cp_{\nu} \ \ \ \ \ (4)$

while for the muon

$\displaystyle E_{\mu}=c\sqrt{p_{\mu}^{2}+m_{\mu}^{2}c^{2}} \ \ \ \ \ (5)$

so

$\displaystyle m_{\pi}c=\sqrt{p_{\mu}^{2}+m_{\mu}^{2}c^{2}}+p_{\nu} \ \ \ \ \ (6)$

But ${p_{\nu}=-p_{\mu}}$ from 2 so

 $\displaystyle m_{\pi}c$ $\displaystyle =$ $\displaystyle \sqrt{p_{\mu}^{2}+m_{\mu}^{2}c^{2}}-p_{\mu}\ \ \ \ \ (7)$ $\displaystyle \sqrt{p_{\mu}^{2}+m_{\mu}^{2}c^{2}}$ $\displaystyle =$ $\displaystyle m_{\pi}c+p_{\mu}\ \ \ \ \ (8)$ $\displaystyle p_{\mu}$ $\displaystyle =$ $\displaystyle \frac{m_{\mu}^{2}-m_{\pi}^{2}}{2m_{\pi}}c\ \ \ \ \ (9)$ $\displaystyle E_{\mu}$ $\displaystyle =$ $\displaystyle \frac{m_{\mu}^{2}+m_{\pi}^{2}}{2m_{\pi}}c^{2} \ \ \ \ \ (10)$

where the last line follows from 5.

The velocity of the muon can be found from

 $\displaystyle E_{\mu}$ $\displaystyle =$ $\displaystyle p^{0}c\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m_{\mu}c^{2}}{\sqrt{1-u^{2}/c^{2}}}\ \ \ \ \ (12)$ $\displaystyle \frac{m_{\mu}^{2}+m_{\pi}^{2}}{2m_{\pi}}c^{2}$ $\displaystyle =$ $\displaystyle \frac{m_{\mu}c^{2}}{\sqrt{1-u^{2}/c^{2}}}\ \ \ \ \ (13)$ $\displaystyle u$ $\displaystyle =$ $\displaystyle c\sqrt{1-\frac{4m_{\pi}^{2}m_{\mu}^{2}}{\left(m_{\pi}^{2}+m_{\mu}^{2}\right)}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m_{\pi}^{2}-m_{\mu}^{2}}{m_{\pi}^{2}+m_{\mu}^{2}}c \ \ \ \ \ (15)$

# Conservation of four-momentum implies the geodesic equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Box 22.1.

The stress-energy tensor obeys the conservation of four-momentum

$\displaystyle \nabla_{j}T^{ij}=0 \ \ \ \ \ (1)$

We can show that the geodesic equation actually follows from this conservation condition. For the case of ‘dust’ (a fluid whose constituent particles are locally at rest with one another), the stress-energy tensor is

$\displaystyle T^{ij}=\rho_{0}u^{i}u^{j} \ \ \ \ \ (2)$

where ${\rho_{0}}$ is the dust’s density in its own rest frame and ${u^{i}}$ is its four-velocity measured in the observer’s frame. In this case

 $\displaystyle \nabla_{j}T^{ij}$ $\displaystyle =$ $\displaystyle \nabla_{j}\left(\rho_{0}u^{i}u^{j}\right)\ \ \ \ \ (3)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle u^{i}\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}u^{j}\nabla_{j}u^{i} \ \ \ \ \ (4)$

Note that ${\rho_{0}}$ is not necessarily a constant so its gradient will, in general be non-zero.

From the equation

$\displaystyle g_{ij}u^{i}u^{j}=-1 \ \ \ \ \ (5)$

we get

 $\displaystyle \nabla_{k}\left(g_{ij}u^{i}u^{j}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle g_{ij}u^{i}\nabla_{k}u^{j}+g_{ij}u^{j}\nabla_{k}u^{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle 2g_{ij}u^{i}\nabla_{k}u^{j}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle g_{ij}u^{i}\nabla_{k}u^{j}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (9)$

The second line follows from the fact that the absolute gradient of the metric tensor is zero (so there’s no term ${u^{i}u^{j}\nabla_{k}g_{ij}}$ in the product rule expansion). The third line comes from swapping the bound indices ${i}$ and ${j}$ in the second term in line 2, and then using the symmetry of the metric tensor (${g_{ij}=g_{ji}}$).

Returning to 4, we can multiply through by ${g_{il}u^{l}}$ and get

 $\displaystyle g_{il}u^{l}u^{i}\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}g_{il}u^{l}u^{j}\nabla_{j}u^{i}$ $\displaystyle =$ $\displaystyle -\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}u^{j}\left(g_{il}u^{l}\nabla_{j}u^{i}\right)\ \ \ \ \ (10)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -\nabla_{j}\left(\rho_{0}u^{j}\right) \ \ \ \ \ (11)$

where we used 5 on the first term on the LHS of the first line, and 9 on the second term of the RHS of the first line (with indices suitably relabelled). Therefore

$\displaystyle \nabla_{j}\left(\rho_{0}u^{j}\right)=0 \ \ \ \ \ (12)$

Substituting this back into 4 we get

$\displaystyle u^{j}\nabla_{j}u^{i}=0 \ \ \ \ \ (13)$

The absolute gradient of a four-vector can be written in terms of Christoffel symbols as

$\displaystyle \nabla_{j}u^{i}=\partial_{j}u^{i}+\Gamma_{kj}^{i}u^{k} \ \ \ \ \ (14)$

so we get

 $\displaystyle u^{j}\nabla_{j}u^{i}$ $\displaystyle =$ $\displaystyle u^{j}\left(\partial_{j}u^{i}+\Gamma_{kj}^{i}u^{k}\right)\ \ \ \ \ (15)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{\partial x^{j}}{\partial\tau}\frac{\partial u^{i}}{\partial x^{j}}+\Gamma_{kj}^{i}u^{k}u^{j}\ \ \ \ \ (16)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{du^{i}}{d\tau}+\Gamma_{kj}^{i}u^{k}u^{j}\ \ \ \ \ (17)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{d^{2}x^{i}}{d\tau^{2}}+\Gamma_{kj}^{i}\frac{dx^{j}}{d\tau}\frac{dx^{k}}{d\tau} \ \ \ \ \ (18)$

This is just the geodesic equation, so we see that (for dust, anyway; the result is generally true for fluids but is harder to prove) conservation of four-momentum implies the geodesic equation.

# Local flat coordinate systems; four-momentum of photons

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 12; Problem 12.3.

Sometimes it’s useful to use a local coordinate system to do calculations. If the spacetime metric (such as the Schwarzschild metric) is smooth over a region of space (that is, it has no singularities such as division by zero and no sudden jumps), then we can define a local metric that is essentially flat. A simple example is that of the surface of the Earth. Although the Earth is spherical, a locally flat 2-d coordinate system works well for distances of, say, a few miles.

In 4-d spacetime, we can define a locally flat coordinate system with the four mutually orthogonal basis vectors

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \left[1,0,0,0\right]\ \ \ \ \ (1)$ $\displaystyle \mathbf{o}_{x}$ $\displaystyle =$ $\displaystyle \left[0,1,0,0\right]\ \ \ \ \ (2)$ $\displaystyle \mathbf{o}_{y}$ $\displaystyle =$ $\displaystyle \left[0,0,1,0\right]\ \ \ \ \ (3)$ $\displaystyle \mathbf{o}_{z}$ $\displaystyle =$ $\displaystyle \left[0,0,0,1\right] \ \ \ \ \ (4)$

and the usual flat space metric from special relativity ${\eta_{ij}}$.

If we know some four vector ${\mathbf{A}}$ in this flat coordinate system, we can find its components by taking the scalar product with each of the basis vectors. That is

 $\displaystyle \mathbf{o}_{x}\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle \eta_{ij}o_{x}^{i}A^{j}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{xx}A^{x}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{x} \ \ \ \ \ (7)$

The second line follows because the only non-zero component of ${\mathbf{o}_{x}}$ is ${o_{x}^{x}}$ so ${i=x}$, and then the only non-zero component of the metric ${\eta_{xj}}$ is ${\eta_{xx}=+1}$. By the same argument, we get the other components:

 $\displaystyle \mathbf{o}_{t}\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle -A^{t}\ \ \ \ \ (8)$ $\displaystyle \mathbf{o}_{y}\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle A^{y}\ \ \ \ \ (9)$ $\displaystyle \mathbf{o}_{z}\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle A^{z} \ \ \ \ \ (10)$

This might seem trivial but the important point is that since the scalar product is invariant, if we work out the components in one coordinate system, then if we can find the basis vectors ${\mathbf{o}_{i}}$ in another coordinate system, their scalar products with ${\mathbf{A}}$ in that coordinate system must yield the same numerical results. This is often an easier way of finding the components of ${\mathbf{A}}$ in other coordinate systems (as opposed to using the general tensor transformation formula).

The problem then is to find ${\mathbf{o}_{i}}$ in the other coordinate system. If we take this system to be the global system with the Schwarzschild metric, we can work out this transformation. First, we need to align the axes in the two systems. We’ll take the ${x}$, ${y}$ and ${z}$ axes in the local system to be aligned with the ${\phi}$,${\theta}$ and ${r}$ axes in the general system. To get started, suppose the observer (who has the local, flat system) is at rest in the general system. Then his four velocity ${\mathbf{u}^{\prime}}$ as measured in the general system must have all its spatial components equal to zero. Using the Schwarzschild metric, we must have

 $\displaystyle \mathbf{u}^{\prime}\cdot\mathbf{u}^{\prime}$ $\displaystyle =$ $\displaystyle g_{tt}u^{\prime t}u^{\prime t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)\left(u^{\prime t}\right)^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (13)$ $\displaystyle u^{\prime t}$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)^{-1/2} \ \ \ \ \ (14)$

In the observer’s own local frame, because the metric is flat and the observer is not moving relative to himself, ${\mathbf{u}=\left[1,0,0,0\right]}$. That is, in the local frame, ${\mathbf{u}=\mathbf{o}_{t}}$. Therefore, ${\mathbf{u}^{\prime}}$ is the transformed version ${\mathbf{o}_{t}^{\prime}}$ of the time basis vector, and

$\displaystyle \mathbf{o}_{t}^{\prime}=\left[\left(1-\frac{2GM}{r}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (15)$

What about transforming the spatial basis vectors? We know that in the flat system ${\mathbf{o}_{i}\cdot\mathbf{o}_{j}=\eta_{ij}}$ so the same must be true in the general system (since these are scalar products). That is, it must also be true that ${\mathbf{o}_{i}^{\prime}\cdot\mathbf{o}_{j}^{\prime}=\eta_{ij}}$. If we’ve aligned the axes in the two systems as stated above, then ${\mathbf{o}_{x}^{\prime}}$ must have zero components along the ${\theta}$ and ${r}$ directions, so we must have (where the components are listed in the order ${t}$, ${r}$, ${\theta}$, ${\phi}$):

$\displaystyle \mathbf{o}_{x}^{\prime}=\left[o_{x}^{\prime t},0,0,o_{x}^{\prime\phi}\right] \ \ \ \ \ (16)$

We must also have

 $\displaystyle \mathbf{o}_{t}^{\prime}\cdot\mathbf{o}_{x}^{\prime}$ $\displaystyle =$ $\displaystyle g_{tt}\left(1-\frac{2GM}{r}\right)^{-1/2}o_{x}^{\prime t}+g_{\phi\phi}\times0\times o_{x}^{\prime\phi}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (18)$

from which we get ${o_{x}^{\prime t}=0}$. Finally, we must also have

 $\displaystyle \mathbf{o}_{x}^{\prime}\cdot\mathbf{o}_{x}^{\prime}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}\left(o_{x}^{\prime\phi}\right)^{2}=1\ \ \ \ \ (19)$ $\displaystyle o_{x}^{\prime\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r\sin\theta} \ \ \ \ \ (20)$

Therefore

$\displaystyle \mathbf{o}_{x}^{\prime}=\left[0,0,0,\frac{1}{r\sin\theta}\right] \ \ \ \ \ (21)$

By the same process, we can work out the other two vectors:

 $\displaystyle \mathbf{o}_{y}^{\prime}$ $\displaystyle =$ $\displaystyle \left[0,0,-\frac{1}{r},0\right]\ \ \ \ \ (22)$ $\displaystyle \mathbf{o}_{z}^{\prime}$ $\displaystyle =$ $\displaystyle \left[0,\sqrt{1-\frac{2GM}{r}},0,0\right] \ \ \ \ \ (23)$

where the minus sign for ${\mathbf{o}_{y}^{\prime}}$ is because we’re taking ${y}$ to be in the ${-\theta}$ direction in order to get a right-handed coordinate system.

Having worked out the basis vectors in the general system, if we’ve calculated the the components of ${\mathbf{A}}$ in the local, flat system as above, we can use the invariance of the scalar product to work out the components of ${\mathbf{A}}$ in the general system.

Note that it’s not true that in the Schwarzschild (or any non-flat) metric, ${A^{\prime i}=\pm\mathbf{o}_{i}^{\prime}\cdot\mathbf{A}^{\prime}}$. For example, suppose that ${\mathbf{A}^{\prime}=\left[1,0,0,0\right]}$. Then

$\displaystyle -\mathbf{o}_{t}^{\prime}\cdot\mathbf{A}^{\prime}=-g_{tt}\left(1-\frac{2GM}{r}\right)^{-1/2}A^{\prime t}=\left(1-\frac{2GM}{r}\right)\left(1-\frac{2GM}{r}\right)^{-1/2}=\left(1-\frac{2GM}{r}\right)^{1/2} \ \ \ \ \ (24)$

That is, the transformations derived above apply only when we start with a locally flat space and then transform to some other metric.

We can now apply this method to the specific four-vector which is ${\mathbf{p}}$, the four-momentum of a photon. As usual, we start with the four-momentum of a particle with rest mass and seek a form that makes no mention of ${m}$ or ${\tau}$, the proper time. We get

 $\displaystyle p^{i}$ $\displaystyle =$ $\displaystyle m\frac{dx^{i}}{d\tau}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\frac{dx^{i}}{dt}\frac{dt}{d\tau}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle me\left(1-\frac{2GM}{r}\right)^{-1}\frac{dx^{i}}{dt} \ \ \ \ \ (27)$

where we’ve used the definition of ${e}$.

We still need to get rid of ${m}$, but when defining ${e}$ we discovered that it is the energy per unit mass at ${r=\infty}$, so ${me}$ is the total energy ${E}$ of the object at infinity. This is carried over to photons by just defining their four-momentum as

$\displaystyle p^{i}=E\left(1-\frac{2GM}{r}\right)^{-1}\frac{dx^{i}}{dt} \ \ \ \ \ (28)$

For a photon moving in the equatorial plane, we therefore get

 $\displaystyle p^{t}$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)^{-1}\frac{dt}{dt}=E\left(1-\frac{2GM}{r}\right)^{-1}\ \ \ \ \ (29)$ $\displaystyle p^{r}$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)^{-1}\frac{dr}{dt}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm E\left(1-\frac{2GM}{r}\right)^{-1}\left(1-\frac{2GM}{r}\right)\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm E\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}}\ \ \ \ \ (32)$ $\displaystyle p^{\theta}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (33)$ $\displaystyle p^{\phi}$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)^{-1}\frac{d\phi}{dt}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)^{-1}\frac{1}{r^{2}}\left(1-\frac{2GM}{r}\right)b\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E\frac{b}{r^{2}} \ \ \ \ \ (36)$

For ${p^{r}}$ and ${p^{\phi}}$ we’ve used the photon equations of motion.

Now suppose we want the velocity components of the photon as measured by our observer in his local, flat frame. The components are, as measured in the flat local frame:

 $\displaystyle v^{i}$ $\displaystyle =$ $\displaystyle \frac{p^{i}}{p^{t}}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mathbf{o}_{i}\cdot\mathbf{p}}{-\mathbf{o}_{t}\cdot\mathbf{p}} \ \ \ \ \ (38)$

Because we’ve managed to express the components in terms of scalar products, we can evaluate them in any frame, so since we know the components of the ${\mathbf{o}_{i}}$s and ${\mathbf{p}}$ in the general frame, we can do the scalar product in that frame. We must remember to use the Schwarzschild metric in calculating the scalar products in the general frame, of course! So we get (remember ${\theta=\frac{\pi}{2}}$)

 $\displaystyle -\mathbf{o}_{t}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle -g_{tt}\left(1-\frac{2GM}{r}\right)^{-1/2}E\left(1-\frac{2GM}{r}\right)^{-1}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)\left(1-\frac{2GM}{r}\right)^{-1/2}\left(1-\frac{2GM}{r}\right)^{-1}\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{r}\right)^{-1/2}\ \ \ \ \ (41)$ $\displaystyle \mathbf{o}_{x}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}\frac{1}{r\sin\theta}E\frac{b}{r^{2}}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\frac{1}{r\sin\theta}E\frac{b}{r^{2}}\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E\frac{b}{r}\ \ \ \ \ (44)$ $\displaystyle \mathbf{o}_{y}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (45)$ $\displaystyle \mathbf{o}_{z}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle \pm g_{rr}\left(1-\frac{2GM}{r}\right)^{1/2}E\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}}\ \ \ \ \ (46)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm\left(1-\frac{2GM}{r}\right)^{-1}\left(1-\frac{2GM}{r}\right)^{1/2}E\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}}\ \ \ \ \ (47)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm E\left(1-\frac{2GM}{r}\right)^{-1/2}\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}} \ \ \ \ \ (48)$

We therefore get

 $\displaystyle v_{x}$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)^{1/2}\frac{b}{r}\ \ \ \ \ (49)$ $\displaystyle v_{y}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (50)$ $\displaystyle v_{z}$ $\displaystyle =$ $\displaystyle \pm\sqrt{1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}} \ \ \ \ \ (51)$

The magnitude of the velocity is then

 $\displaystyle \sqrt{v_{x}^{2}+v_{z}^{2}}$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}+1-\left(1-\frac{2GM}{r}\right)\frac{b^{2}}{r^{2}}\ \ \ \ \ (52)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (53)$

Thus the photon’s speed is always 1 to the observer in the local frame, which is a relief, since photons must always have speed 1.

# Schwarzschild metric: four-momentum of a photon

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapters 9; Problem 9.6.

This is a first example of the use of the time component of the Schwarzschild metric. This metric is, for a spherical mass ${M}$:

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

Suppose we have an observer at a Schwarzschild radius ${R}$ from the centre of a star of mass ${M}$, and this observer watches a photon move radially outward. The observer measures the energy of the photon to be ${E}$. We can use this to calculate the four-momentum of the photon.

In special relativity, for an observer at rest the observer’s four-velocity is ${u^{i}=\left[1,0,0,0\right]}$ so the scalar product of the observer’s four-velocity with another object’s momentum (as measured by the observer) is

 $\displaystyle \mathbf{p}\cdot\mathbf{u}_{obs}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}u^{j}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}u^{t}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E \ \ \ \ \ (5)$

since the time component of an object’s four-momentum is its energy. Since this is a tensor equation, it should be true in curved space-time as well. In the Schwarzschild metric, an observer at rest has

$\displaystyle u^{t}=\left[\left(1-\frac{2GM}{R}\right)^{-1/2},0,0,0\right] \ \ \ \ \ (6)$

Therefore, we get

 $\displaystyle \mathbf{p}\cdot\mathbf{u}_{obs}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}u^{j}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{R}\right)p^{t}\left(1-\frac{2GM}{R}\right)^{-1/2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p^{t}\left(1-\frac{2GM}{R}\right)^{1/2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E\ \ \ \ \ (10)$ $\displaystyle p^{t}$ $\displaystyle =$ $\displaystyle E\left(1-\frac{2GM}{R}\right)^{-1/2} \ \ \ \ \ (11)$

For a photon, ${\mathbf{p}\cdot\mathbf{p}=0}$, and for a photon moving in the radial direction ${p^{\theta}=p^{\phi}=0}$ so

 $\displaystyle \mathbf{p}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle g_{ij}p^{i}p^{j}\ \ \ \ \ (12)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{R}\right)E^{2}\left(1-\frac{2GM}{R}\right)^{-1}+\left(1-\frac{2GM}{R}\right)^{-1}\left(p^{r}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -E^{2}+\left(1-\frac{2GM}{R}\right)^{-1}\left(p^{r}\right)^{2}\ \ \ \ \ (14)$ $\displaystyle p^{r}$ $\displaystyle =$ $\displaystyle E\sqrt{1-\frac{2GM}{R}} \ \ \ \ \ (15)$

Thus the photon’s four-momentum in the Schwarzschild basis is

$\displaystyle \mathbf{p}=\left[E\left(1-\frac{2GM}{R}\right)^{-1/2},E\sqrt{1-\frac{2GM}{R}},0,0\right] \ \ \ \ \ (16)$

# Electromagnetic field tensor: conservation of mass

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.6.

As we’ll study in more detail a bit later, the electric and magnetic fields can be combined into a single tensor known as the electromagnetic field tensor ${F^{ij}}$:

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

We can see from its definition that this tensor is anti-symmetric, that is, that ${F^{ij}=-F^{ji}}$. For any anti-symmetric tensor we can show that

$\displaystyle F^{ij}\eta_{ia}\eta_{jb}u^{a}u^{b}=0 \ \ \ \ \ (2)$

In this equation, ${\eta_{ij}}$ is the metric tensor in flat space and ${u^{a}}$ is the four-velocity, but in fact the formula is valid for any tensors ${\eta}$ and ${u}$, provided that ${F}$ is anti-symmetric. The proof involves a bit of index-switching.

 $\displaystyle F^{ij}\eta_{ia}\eta_{jb}u^{a}u^{b}$ $\displaystyle =$ $\displaystyle -F^{ji}\eta_{ia}\eta_{jb}u^{a}u^{b}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -F^{ij}\eta_{ja}\eta_{ib}u^{a}u^{b}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -F^{ij}\eta_{jb}\eta_{ia}u^{b}u^{a} \ \ \ \ \ (5)$

In the second line, we swapped the dummy indexes ${i}$ and ${j}$, and in the third line we swapped ${a}$ and ${b}$. The result shows that the original quantity is equal to its negative, which means it must be zero.

In terms of ${F^{ij}}$, the electric and magnetic (Lorentz) force laws for a charge ${q}$ can be combined into a single equation:

$\displaystyle \frac{dp^{i}}{d\tau}=qF^{ij}\eta_{ja}u^{a} \ \ \ \ \ (6)$

where ${u^{a}=\gamma\left[1,v_{x},v_{y},v_{z}\right]}$ is the four-velocity.

For example, if ${i=1}$ we get

 $\displaystyle \frac{dp^{1}}{d\tau}$ $\displaystyle =$ $\displaystyle q\gamma\left(E_{x}+v_{y}B_{z}-v_{z}B_{y}\right) \ \ \ \ \ (7)$

In the non-relativistic limit, ${\gamma\rightarrow1}$ and this is the ${x}$ component of the force law ${\frac{d\mathbf{p}}{dt}=q\mathbf{E}+q\mathbf{v}\times\mathbf{B}}$. We’ll explore some of the other properties of this tensor later.

Since the square of the four-momentum of a particle is the negative of its mass squared (${\mathbf{p}\cdot\mathbf{p}=\gamma^{2}m^{2}\left(-1+v^{2}\right)=-m^{2}}$), this should be conserved for a charged particle moving in an electromagnetic field. (Its total momentum is, of course, not conserved since the fields exert a force on the particle.)

We have

 $\displaystyle \frac{d\left(\mathbf{p}\cdot\mathbf{p}\right)}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{d}{d\tau}\left(\eta_{ij}p^{i}p^{j}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta_{ij}\left[\frac{dp^{i}}{d\tau}p^{j}+p^{i}\frac{dp^{j}}{d\tau}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\eta_{ij}\frac{dp^{i}}{d\tau}p^{j}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2q\eta_{ij}F^{ik}\eta_{ka}u^{a}p^{j}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2qmF^{ik}\eta_{ij}\eta_{ka}u^{a}u^{j}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

In the third line, we used the fact that ${\eta_{ij}=\eta_{ji}}$ and swapped ${i}$ and ${j}$ in the second term. The fourth line uses 6 and the last line uses 2.

# Four-momentum conservation: electron-photon collision

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.10.

Here’s another example of using the invariance of the scalar product under Lorentz transformations. This time we have a photon colliding with a stationary electron (in the lab frame) and producing an extra electron-positron pair (so we have 2 electrons and a positron after the collision; the photon gets absorbed so it disappears). What is the minimum energy of the incoming photon (in the lab frame) to achieve this?

First, we can look at the problem in the centre of mass (COM) frame. There, the electron travels at some speed ${v}$ (the photon, of course, travels at ${v=1}$) and after the collision all 3 particles are at rest. From conservation of momentum we must have

$\displaystyle \left(\mathbf{p}_{e}+\mathbf{p}_{\gamma}\right)^{2}=-\left(3m\right)^{2} \ \ \ \ \ (1)$

where ${\mathbf{p}_{e}}$ is the momentum of the electron and ${\mathbf{p}_{\gamma}}$ is the momentum of the photon before the collision, and ${m}$ is the mass of an electron or positron.

Multiplying this out, we get

 $\displaystyle p_{e}^{2}+p_{\gamma}^{2}+2\mathbf{p}_{e}\cdot\mathbf{p}_{\gamma}$ $\displaystyle =$ $\displaystyle -m^{2}-0+2\mathbf{p}_{e}\cdot\mathbf{p}_{\gamma}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -9m^{2}\ \ \ \ \ (3)$ $\displaystyle \mathbf{p}_{e}\cdot\mathbf{p}_{\gamma}$ $\displaystyle =$ $\displaystyle -4m^{2} \ \ \ \ \ (4)$

Since the scalar product on the LHS is invariant, it must also be true in the lab frame. There, ${\mathbf{p}_{e}^{\prime}=\left[m,0\right]}$ since the electron is at rest, and ${\mathbf{p}_{\gamma}^{\prime}=\left[E,-E\right]}$. Therefore

 $\displaystyle \mathbf{p}_{e}^{\prime}\cdot\mathbf{p}_{\gamma}^{\prime}$ $\displaystyle =$ $\displaystyle -mE\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{p}_{e}\cdot\mathbf{p}_{\gamma}=-4m^{2}\ \ \ \ \ (6)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle 4m \ \ \ \ \ (7)$

This gives us the minimum energy of the photon in the lab frame.

We can work out the wavelength of this photon by using ${E=h/\lambda}$ where Planck’s constant has the value ${h=1240\times10^{-9}\;\mbox{eV m}}$ and the electron mass is ${m=0.51\times10^{6}\mbox{ eV}}$. We get

 $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \frac{h}{E}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1240\times10^{-9}}{4\times0.51\times10^{6}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.08\times10^{-13}\mbox{m} \ \ \ \ \ (10)$

This is well into the gamma ray region of the electromagnetic spectrum.

# Four-momentum conservation: electron-electron collision

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.9.

The invariance of the scalar product under Lorentz transformations can be useful in working out energies of particles in collisions. For example, suppose we have one electron colliding with another electron (where the second electron is at rest in the lab) and producing a new electron-positron pair (so after the collision there are 4 particles: 3 electrons and a positron). What is the minimum energy of the incoming electron (in the lab frame) to achieve this?

First, we can look at the problem in the centre of mass (COM) frame. There, the two electrons travel towards each other at the same speed ${v}$ and after the collision all 4 particles are at rest. From conservation of momentum we must have

$\displaystyle \left(\mathbf{p}_{1}+\mathbf{p}_{2}\right)^{2}=-\left(4m\right)^{2} \ \ \ \ \ (1)$

where ${\mathbf{p}_{i}}$ is the momentum of electron ${i}$ before the collision, and ${m}$ is the mass of an electron or positron.

Multiplying this out, we get

 $\displaystyle p_{1}^{2}+p_{2}^{2}+2\mathbf{p}_{1}\cdot\mathbf{p}_{2}$ $\displaystyle =$ $\displaystyle -m^{2}-m^{2}+2\mathbf{p}_{1}\cdot\mathbf{p}_{2}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -16m^{2}\ \ \ \ \ (3)$ $\displaystyle \mathbf{p}_{1}\cdot\mathbf{p}_{2}$ $\displaystyle =$ $\displaystyle -7m^{2} \ \ \ \ \ (4)$

Since the scalar product on the LHS is invariant, it must also be true in the lab frame. There, ${\mathbf{p}_{1}^{\prime}=\left[m,0\right]}$ since the first electron is at rest, and ${\mathbf{p}_{2}^{\prime}=\left[E,p_{x}\right]}$. Therefore

 $\displaystyle \mathbf{p}_{1}^{\prime}\cdot\mathbf{p}_{2}^{\prime}$ $\displaystyle =$ $\displaystyle -mE\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{p}_{1}\cdot\mathbf{p}_{2}=-7m^{2}\ \ \ \ \ (6)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle 7m \ \ \ \ \ (7)$

This gives us the minimum energy of the incoming electron in the lab frame.

To work out the velocity of the incoming electron, we note that

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \gamma m\ \ \ \ \ (8)$ $\displaystyle \gamma$ $\displaystyle =$ $\displaystyle 7=\frac{1}{\sqrt{1-v^{2}}}\ \ \ \ \ (9)$ $\displaystyle 1-v^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{49}\ \ \ \ \ (10)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{4\sqrt{3}}{7} \ \ \ \ \ (11)$

# Electron-positron collision must produce at least 2 photons

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.8.

When an electron and positron collide, they can decay into a pair of photons as we’ve seen. However, 2 photons is the minimum number of photons that can be produced. To see why a single photon is not possible, re-examine the solution we had earlier. In that post, we started with the electron and positron moving towards each other, one with velocity ${v}$ and the other with velocity ${-v}$. We then transformed to the frame in which the electron is at rest. We can just as easily reverse the process, and start with the electron’s frame. Let’s say that in the electron’s frame, the positron is moving with speed ${-v^{\prime}}$ towards it. Then the total momentum before the collision is

$\displaystyle \mathbf{p}^{\prime}=\left[m,0\right]+\gamma^{\prime}\left[m,-v^{\prime}\right] \ \ \ \ \ (1)$

where ${\gamma^{\prime}=1/\sqrt{1-\left(v^{\prime}\right)^{2}}}$.

We can now transform to a frame in which the two particles are moving towards each other with the same speed ${v}$. That is, we want the four-velocities to be

 $\displaystyle \mathbf{u}_{e}$ $\displaystyle =$ $\displaystyle \left[\gamma m,v\right]\ \ \ \ \ (2)$ $\displaystyle \mathbf{u}_{p}$ $\displaystyle =$ $\displaystyle \left[\gamma m,-v\right] \ \ \ \ \ (3)$

From our earlier analysis, we see that this works if

 $\displaystyle v^{\prime}$ $\displaystyle =$ $\displaystyle \frac{2v}{1+v^{2}}\ \ \ \ \ (4)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{1-\sqrt{1-\left(v^{\prime}\right)^{2}}}{v^{\prime}} \ \ \ \ \ (5)$

Thus it’s always possible to find ${v}$ such that the particles are moving towards each other with the same speed, and in that frame, the total momentum is

$\displaystyle \mathbf{p}=\left[2\gamma m,0\right] \ \ \ \ \ (6)$

Since the ${x}$ component of a photon’s momentum is always ${\pm1}$, there is no way this momentum can be conserved after the collision if only a single photon is emitted. However, it’s always possible to produce 2 photons moving in opposite directions, since then the ${x}$ component of momentum can add up to zero, as required.

# Doppler effect and four-momentum

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.7.

The Doppler effect can be derived from the Lorentz transformation of momentum for a photon. The energy and wavelength of a photon are related by Planck’s formula

$\displaystyle E_{0}=\frac{h}{\lambda_{0}} \ \ \ \ \ (1)$

where ${h}$ is Planck’s constant and ${\lambda_{0}}$ is the wavelength.

In the rest frame of the photon’s source, we have

$\displaystyle \mathbf{p}_{0}=\left[E_{0},E_{0}\right]=E_{0}\left[1,1\right] \ \ \ \ \ (2)$

If we now transform to a frame moving at speed ${v}$ away from the source, we get

 $\displaystyle \mathbf{p}_{v}$ $\displaystyle =$ $\displaystyle E_{0}\gamma\left[1-v,1-v\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}\sqrt{\frac{1-v}{1+v}}\left[1,1\right] \ \ \ \ \ (4)$

Thus the transformed energy is ${E_{v}=E_{0}\sqrt{\frac{1-v}{1+v}}}$ and the new wavelength is

$\displaystyle \lambda_{v}=\sqrt{\frac{1+v}{1-v}}\lambda_{0} \ \ \ \ \ (5)$

which is the usual Doppler red-shift formula.

Another way of getting this is to consider the scalar product ${-\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}}}$ where ${\mathbf{u}_{\mbox{obs}}}$ is the four-velocity of an observer and ${\mathbf{p}}$ is the four-momentum of a passing object. Since the scalar product is invariant under a Lorentz transformation, we can work it out in the rest frame of the observer, where ${\mathbf{u}_{\mbox{obs}}=\left[1,0,0,0\right]}$. In this case, we get

 $\displaystyle -\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}}$ $\displaystyle =$ $\displaystyle -\left[E,p_{x},p_{y},p_{z}\right]\cdot\left[1,0,0,0\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E \ \ \ \ \ (7)$

That is, the scalar product is the energy of the object as seen by the observer.

In the case of the photon above, if we work out this product in the rest frame of the source, then in that frame ${\mathbf{u}_{\mbox{obs}}=\left[\gamma,\gamma v\right]}$ and ${\mathbf{p}=\mathbf{p}_{0}=E_{0}\left[1,1\right]}$ so

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -\mathbf{p}\cdot\mathbf{u}_{\mbox{obs}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E_{0}\left(\gamma v-\gamma\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}\sqrt{\frac{1-v}{1+v}} \ \ \ \ \ (10)$

# Four-momentum conservation: a trip to Alpha Centauri

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.6.

As a rather fanciful example of using the conservation of four-momentum, suppose we have a spaceship of total mass (including fuel) ${M}$, initially at rest on Earth. The fuel consists of matter/anti-matter which when mixed, produces photons that are ejected out of the back of the ship. If the ship burns enough fuel to accelerate to ${v=0.95}$, then travels to some star system such as Alpha Centauri, then decelerates to zero for a landing, then, after some time at its destination it reverses the trip by again accelerating to ${v=0.95}$, returning to Earth, and decelerating to rest, what is its final mass as a fraction of its initial mass?

We can assume that all motion takes place along the ${x}$ axis, and treat the problem in the Earth’s frame. Then the initial momentum is

 $\displaystyle \mathbf{p}_{0}$ $\displaystyle =$ $\displaystyle \left[M,0\right] \ \ \ \ \ (1)$

After accelerating, the combined momentum of the ship + ejected photons is

$\displaystyle \mathbf{p}_{1}=\gamma m_{1}\left[1,0.95\right]+E_{1}\left[1,-1\right] \ \ \ \ \ (2)$

where ${\gamma=1/\sqrt{1-v^{2}}=3.2}$, ${E_{1}}$ is the energy of the ejected photons and ${m_{1}}$ is the mass of the ship after burning the fuel needed to accelerate.

By the conservation of momentum, we have ${\mathbf{p}_{1}=\mathbf{p}_{0}}$ so

 $\displaystyle \gamma m_{1}+E_{1}$ $\displaystyle =$ $\displaystyle M\ \ \ \ \ (3)$ $\displaystyle 0.95\gamma m_{1}-E_{1}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Adding these 2 equations, we get

$\displaystyle m_{1}=\frac{M}{1.95\gamma}=0.16M \ \ \ \ \ (5)$

Now to decelerate the ship, we eject the photons ahead of the ship, and we start with a momentum of ${\gamma m_{1}\left[1,0.95\right]}$. After deceleration, the mass is now ${m_{2}}$ and the ship is at rest, so we must have

$\displaystyle \mathbf{p}_{2}=\left[m_{2},0\right]+E_{2}\left[1,1\right] \ \ \ \ \ (6)$

Again, conservation of momentum requires ${\mathbf{p}_{2}=\gamma m_{1}\left[1,0.95\right]}$, so

 $\displaystyle \gamma m_{1}$ $\displaystyle =$ $\displaystyle m_{2}+E_{2}\ \ \ \ \ (7)$ $\displaystyle 0.95\gamma m_{1}$ $\displaystyle =$ $\displaystyle E_{2} \ \ \ \ \ (8)$

Subtracting these equations we get

 $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle 0.05\gamma m_{1}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{0.05}{1.95}M \ \ \ \ \ (10)$

On the return trip, we go through exactly the same procedure, except we now start with a mass ${m_{2}}$ rather than ${M}$. Thus on the return to Earth, the ship’s mass will be

 $\displaystyle m_{E}$ $\displaystyle =$ $\displaystyle \frac{0.05}{1.95}m_{2}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{0.05}{1.95}\right)^{2}M \ \ \ \ \ (12)$

Thus the ship’s initial mass is

$\displaystyle M=1521m_{E} \ \ \ \ \ (13)$

Virtually all the initial mass is fuel.

Incidentally, if we do this calculation for an arbitrary velocity ${v}$, we get

 $\displaystyle m_{1}$ $\displaystyle =$ $\displaystyle \frac{M}{\left(1+v\right)\gamma}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-v}{1+v}}M\ \ \ \ \ (15)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle \frac{1-v}{1+v}M\ \ \ \ \ (16)$ $\displaystyle m_{E}$ $\displaystyle =$ $\displaystyle \left(\frac{1-v}{1+v}\right)^{2}M \ \ \ \ \ (17)$

Thus each acceleration or deceleration multiplies the previous mass by a factor of ${\sqrt{\frac{1-v}{1+v}}}$.