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## First ever asteroid from another solar system detected

In late October 2017, astronomers announced the first ever discovery of an asteroid (or comet?) coming into our Solar System from another stellar system. The object was first spotted on 19 October by the University of Hawaii’s Pan-STARRS telescope, during its nightly search for near-earth objects. Based on its extreme orbit and its rapid speed, it was soon determined that the object has come into our Solar System from somewhere else, and this makes it the first ever asteroid/comet with an extra-solar origin to have been discovered. Originally given the designation A/2017 U1, the International Astronomical Union (IAU) have now renamed it 1I/2017 U1, with the I standing for “interstellar”.

The object, given the designation A/2017 U1, was deemed to be extra-solar in origin from an analysis of its motion.

In addition to its strange trajectory, observations suggest that the object also has quite an unusual shape. It is very elongated, being ten times longer than it is wide. It is thought to be at least 400 metres long but only about 40 metres wide. This was determined by the rapid and dramatic changes in its brightness, which can only be explained by an elongated object tumbling rapidly.

The object has also been given the name Oumuamua (pronounced oh MOO-uh MOO-uh), although this is not its official name (yet).  This means “a messenger from afar arriving first” in Hawaiian. In other respects, it seems to be very much like asteroids found in our own Solar System, and is the confirmation of what astronomers have long suspected, that small objects which formed around other stars can end up wandering through space, not attached to any particular stellar system.

## The Only Living Boy in New York – Simon & Garfunkel (song)

A couple of weeks ago I was in New York City. Although I have been to NYC many times since I first visited it in October 1985, this was the first time I had spent the night in Manhattan.

On the evening of Saturday 17th November I booked a ticket to go up the Empire State Building to take night-time photos. The cost of a ticket to the 80th floor is US$32, and I’d say that it’s good value. But, what is not is the extra$20 to go up to the 102nd floor. I had forgotten from my last time that this is not worth the extra money. Hopefully I’ll remember next time.

I was on the observation deck of the Empire State (86th floor) for about 2 hours taking photos and videos. I’ll post some of them over the next few weeks. Then, at about 11:30pm I walked to Time Square. I stopped to get a coffee and warm up a bit so got to Time Square at about 12:30am. The place was heaving, hundreds of people were milling around and many of the shops were open.

At about 1:30am I caught the subway to Brooklyn as I wanted to get a photo of Manhattan with the Brooklyn Bridge in the foreground. I took my photos from the Brooklyn Bridge Park, then walked back to Manhattan over the Brooklyn Bridge.

By this time it was lashing it down with rain and I was soaked. The rain had seeped through my winter coat and my sweater and trousers were pretty much wet through. But, as I walked over the bridge I caught sight of the Statue of Liberty illuminated (it was about 3am by this time). So I decided to walk to Battery Park to take photos of her at night.

When I got to Battery Park and set up my camera I discovered that my camera battery was dead from all the long exposure photos I’d been taking. So I got my spare battery out of my backpack, only to find that it too was dead. So, I didn’t get any photos of Lady Liberty at night. It was now 4am.

I then walked back to my hotel which was in the Little Italy part of Manhattan. The rain was still lashing it down, and by now my phone was getting damp leading to Google maps misbehaving. The app kept on going haywire every minute or two, so I couldn’t use it to guide me back from Battery Park to Little Italy. Instead I just tried to figure it out, and it took me two hours!

I collapsed into my bed at 6am, having spent nearly 12 hours wandering around nighttime Manhattan taking photos.

When I was walking across the Brooklyn Bridge at about 3am this great Simon & Garfunkel song kept playing in my head.

## The Only Living Boy in New York

This song appears on Simon & Garfunkel’s last album Bridge Over Troubled Water. Written, of course, by Paul Simon, the “Tom” in the lyrics refers to Art Garfunkel. When they were teenagers in Queens they released a single and briefly called themselves”Tom & Jerry”.

By 1970 Simon and Garfunkel were arguing and about to go their separate ways. Garfunkel decided to have a go at acting, he appears in the movie Catch 22. Simon is wishing him the best for his part in the movie Garfunkel is filming in Mexico.

As is usual with Paul Simon, the song’s lyrics are exquisite.

Tom, get your plane right on time
I know your part’ll go fine
Fly down to Mexico
Da-n-da-da-n-da-n-da-da a
And here I am
The only living boy in New York

I get the news I need on the weather report
Oh, I can gather all the news I need on the weather report
Hey, I’ve got nothing to do today but smile
Da-n-do-da-n-do-da-n-do
Here I am
The only living boy in New York

Half of the time we’re gone but we don’t know where
And we don’t know where
Here I am
Half of the time we’re gone, but we don’t know where
And we don’t know where

Tom, get your plane right on time
I know that you’ve been eager to fly now
Hey, let your honesty shine, shine, shine
Like it shines on me
The only living boy in New York
The only living boy in New York

Here is a video of this beautiful song. Enjoy!

## The origin of the elements

A couple of weeks ago this fascinating version of the periodic table of the elements was the NASA Astronomy Picture of the Day (APOD). Most people have seen the periodic table of the elements, it is shown on the wall of most high school chemistry classrooms. But, what is totally fascinating to me about this version is it shows the origin of each element.

It has been a long process of several decades to understand the origin of the elements. In fact, we have not totally finished understanding the processes yet. But, we do know the story for most elements. All the hydrogen in the Universe was formed in the big bang. This is true for nearly all the helium too. A small amount of the 25% or so of helium in the Universe has been created within stars through the conversion of hydrogen into helium. But, not much has been created this way because most of that helium is further converted to carbon.

The only other element to be formed in the big bang is lithium. About 20% of the lithium in the Universe was formed in the big bang, the rest has been formed since,

Together, hydrogen and helium comprise 99% of the elements in the Universe by number (not by mass).

Where Your Elements Came From – from the NASA Astronomy Picture Of the Day (APOD) 24 October 2017.

I have decided to use this fascinating table as the basis for a series of blogs over the next few weeks to explain each of the 6 processes in these six boxes

## Forever Young – Bob Dylan (song)

The Bob Dylan song about which I am going to blog about today is his 1973 song “Forever Young”, another one of his songs included on his Vevo channel. It appears on his 1974 album Planet Waves in two different versions, a slow and a fast version. It is the slow version which is in the video included here. As an aside, Planet Waves is the only studio album which Dylan released through Asylum Records. Apart from the live album Before The Flood which is his next album after Planet Waves, all his other albums have been with Columbia Records.

“Forever Young” was recorded by Dylan in November 1973. The slow version runs for 4m57s and is the 6th track on Planet Waves, the last track on the first side of the record. The fast version (which is a shorter track at 2m49s) is the 7th track on the album, the first track on the second side of the record. Dylan first performed “Forever Young” live in January 1974 and his most recent live performance of it was in November 2011. He has performed it live a remarkable 493 times as of my writing this.

“Forever Young” was recorded by Bob Dylan in November 1973 and appears on his 1974 album Planet Waves. There are two version of the song on the album, a slow version and a fast version.

You may be familiar with a 1988 Rod Stewart song by the same name. Confusingly, it is not a cover version in the traditional sense, but bears such a remarkable similarity to Dylan’s song in both melody and some of the lyrics that Stewart agreed to share his royalties with Dylan (presumably to avoid a lawsuit).

The inspiration for the song was Dylan’s eldest son Jesse who was born in 1966. Dylan wrote “Forever Young” as a lullaby to his young son, and over the years it has been covered by many artists.

May God bless and keep you always
May your wishes all come true
May you always do for others
And let others do for you
May you build a ladder to the stars
And climb on every rung
May you stay forever young
Forever young, forever young
May you stay forever young

May you grow up to be righteous
May you grow up to be true
May you always know the truth
And see the lights surrounding you
May you always be courageous
Stand upright and be strong
May you stay forever young
Forever young, forever young
May you stay forever young

May your hands always be busy
May your feet always be swift
May you have a strong foundation
When the winds of changes shift
May your heart always be joyful
May your song always be sung
May you stay forever young
Forever young, forever young
May you stay forever young

Here is the official Vevo video of this great song. Enjoy!

## Derivation of E=mc2

There are quite a few ways to derive Einstein’s famous equation $E=mc^{2}$. I am going to show you what I consider to be the simplest way.  Feel free to comment if you think you know of an easier way.

We will start off with the relationship between energy, force and distance. We can write

$dE = F dx \text{ (1) }$

Where $dE$ is the change in energy, $F$ is the force and $dx$ is the distance through which the object moves under that force.  But, force can also be written as the rate of change of momentum,

$F = \frac{dp}{dt}$

Allowing us to re-write Equation (1) as

$dE = \frac{dp}{dt}dx \rightarrow dE = dp \frac{dx}{dt} = vdp \text{ (2) }$

Remember that momentum $p$ is defined as

$p =mv$

In classical physics, mass is constant. But this is not the case in Special Relativity, where mass is a function of velocity (so-called relativistic mass).

$m = \frac{ m_{0} }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } \text{ (3) }$

where $m_{0}$ is defined as the rest mass (the mass of an object as measured in a reference frame where it is stationary).

Assuming that both $m \text{ and } v$ can change, we can therefore write

$dp =mdv + vdm$

This allows us to write Equ. (2) as

$dE = vdp = v(mdv + vdm) = mvdv + v^{2}dm \text{ (4) }$

Differentiating Equ. (3) with respect to velocity we get

$\frac{dm}{dv} = \frac{d}{dv} \left( \frac{ m_{0} }{ \sqrt{ (1 - v^{2}/c^{2}) } } \right) = m_{0} \frac{d}{dv} (1 - v^{2}/c^{2})^{-1/2}$

Using the chain rule to differentiate this, we have

$\frac{dm}{dv} = m_{0} \cdot - \frac{1}{2} (1 - v^{2}/c^{2})^{-3/2} \cdot (-2v/c^{2}) = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-3/2} \text{ (5) }$

But, we can write

$(1 - v^{2}/c^{2})^{-3/2}$ as $(1-v^{2}/c^{2})^{-1/2} \cdot (1-v^{2}/c^{2})^{-1}$

This allows us to write Equ. (5) as

$\frac{dm}{dv} = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-1} \cdot (1 - v^{2}/c^{2})^{-1/2}$

From the definition of the relativistic mass in Equ. (3), we can rewrite this as

$\frac{dm}{dv} = \frac{ m v }{ c^{2} }(1-v^{2}/c^{2})^{-1}$

Which is

$\frac{dm}{dv} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}} - \frac{ v^{2}}{c^{2} } \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}-v^{2}}{c^{2}} \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}-v^{2}} \right)$

$\frac{dm}{dv} = \frac{ m v }{ (c^{2}-v^{2}) } \text{ (6) }$

So we can write

$c^{2}dm - v^{2}dm = mvdv$

Substituting this expression for $mvdv$ into Equ. (4) we have

$dE = vdp = vd(mv) = mvdv + v^{2}dm = c^{2}dm - v^{2}dm + v^{2}dm$

So

$dE = c^{2} dm$

Integrating this we get

$\int_{E_{0}}^{E} dE = c^{2} \int_{m_{0}}^{m} dm$

So

$E - E_{0} = c^{2} ( m - m_{0} ) = mc^{2} - m_{0}c^{2}$

$E - E_{0} = mc^{2} - m_{0}c^{2}$

This tells us that an object has rest mass energy $E_{0} = m_{0}c^{2}$ and that its total energy is given by

$\boxed{ E = mc^{2} }$

where $m$ is the relativistic mass.

## Derivation of relativistic mass

I have taught special relativity for many years, but every time I teach it I present the result that mass changes as a function of velocity as a consequence of the modified version of Newton’s 2nd law.

As almost everyone knows, Newton’s 2nd law says that

$F=ma$

where $F$ is the force applied, $m$ is the mass, and $a$ is the acceleration felt by the body. In Newtonian mechanics, mass is invariant, but a consequence of special relativity is that nothing can travel faster than the speed of light $c$. This raises the conundrum of why can’t we keep applying a force to a body of mass $m$, causing it to continue accelerating and to ultimately increase its velocity to one greater than the speed of light?

The answer is that Newton’s 2nd law is incomplete. Einstein showed that mass is also a function of velocity, and so we should write

$m = \gamma m_{0} \text{ (1) }$

Where $\gamma = \frac{ 1 }{ \sqrt{ (1 - V^{2}/c^{2}) } }$ is the so-called Lorentz factor and$m_{0}$ is the rest mass (also known as the invariant mass or gravitational mass), the mass an object has when it is at rest relative to the observer. Hence we can argue that, as we approach the speed of light, the applied force goes into changing the mass of the body, rather than accelerating it, leading to a modified version of Newton’s 2nd law

$F = \gamma m_{0} a$

where both velocity and/or mass change as a force is applied. But, because of the fact that $\gamma \approx 1$ until $V \approx c/2$ (see Figure 1), very little increase in mass occurs until $V$ has reached appreciable values.

The variation of $\gamma$ (the Lorentz factor) as a function of the speed $V$. Until $V \approx c/2, \; \gamma$ is very close to unity

However, I have always found this an inadequate explanation of the relativistic mass, as it does not derive it but rather argues for its necessity. So, as I’m teaching special relativity again this year, I decided a few weeks ago to see if I could find a way of deriving it from a simple argument. After several weeks of hunting around I think I have found a derivation which is robust and easy to understand. But, in my searching I came across several “derivations” which were nothing more than circular arguments, and also some derivations which were simply incorrect.

## Two balls colliding

The best explanation that I have found to derive the relativistic mass is to use the scenario of two balls colliding. Although it would be possible, in theory to have the balls moving in any direction, we are going to make things a lot easier by having the balls moving in the y-direction, but with the two reference frames $S \text{ and } S^{\prime}$ moving relative to each other with a velocity $V$ in the x-direction. Also, the balls are going to have the same rest mass, $m_{0}$, as measured in their respective frames $S$ and $S^{\prime}$ (the rest mass of each ball can be measured by each observer in their respective reference frames when they are at rest in their respective frames).

Ball $A$ moves solely in the y-direction in reference frame $S$, and ball $B$ moves solely in the y$^{\prime}$-direction in reference frame $S^{\prime}$. Ball $A$ starts by moving in the positive y-direction in reference frame $S$ with a velocity $u_{0}$, and ball $B$ starts moving in the negative y$^{\prime}$-direction in reference frame $S^{\prime}$ with a velocity $-u_{0}$ in frame $S^{\prime}$.

Reference frame $S^{\prime}$ is moving relative to frame $S$ at a velocity $V$ in the positive x-direction. So, as seen in $S$, the motion of ball $B$ appears as shown in the left of Figure 2. That is, it appears in $S$ to move both in the negative y-direction and the positive x-direction, and so follows the path shown by the red arrow pointing downwards and to the right.

At some moment the two balls collide. After the collision, as seen in $S$, ball $A$ will move vertically downwards in the negative y-direction, with a velocity $-u_{0}$. Ball $B$ moves upwards (positive y-direction) and to the right (positive x-direction), as shown by the red arrow in the diagram on the left of Figure 1.

In reference frame $S^{\prime}$ the motions of balls $A$ and $B$ looks like the diagram on the right of Figure 1. In $S^{\prime}$, it is ball $B$ which moves vertically, and ball $A$ which moves in both the $x^{\prime}$ and $y^{\prime}$ directions.

Two balls colliding. Ball $A$ (in blue) moves solely in the y-direciton as seen in frame $S$, ball $B$ (in red) moves solely in the y-direction in frame $S^{\prime}$.

## The velocity of ball $B$ in $S$

To calculate the velocity of ball $B$ as seen in $S$, we have to use the Lorentz transformations for velocity. As we showed in this blog here, if we have an object moving with a velocity $u^{\prime}$ in $S^{\prime}$ which is moving relative to $S$ with a velocity $V$, then the velocity $u$ in frame $S$ is given by

$u = \frac{ u^{\prime} + V }{ \left( 1 + \frac{ u^{\prime}V }{ c^{2} } \right) } \text{ (2) }$

This equation is true when the velocity is in the x$^{\prime}$-direction, and the frames are moving relative to each other in the x-direction. So we are going to re-write Equ. (2) as

$u_{x} = \frac{ u^{\prime}_{x} + V }{ \left( 1 + \frac{ u^{\prime}_{x}V }{ c^{2} } \right) } \text{ (3) }$

However, if the velocity of an object is in the y$^{\prime}$-direction, rather than the x$^{\prime}$-direction, then we need a different expression. We can derive it from going back to our equations for the Lorentz transformations

The Lorentz transformations

This time we write

$dy = dy^{\prime}$

and

$dt = \gamma \left( dt^{\prime} + \frac{ dx^{\prime}V }{ c^{2} } \right)$

So

$\frac{ dy }{ dt } = \frac{ dy^{\prime} }{ \gamma \left( dt^{\prime} + \frac{ dx^{\prime}V }{ c^{2} } \right) }$

Dividing each term in the right-hand side by $dt^{\prime}$, we get

$\frac{ dy }{ dt } = \frac{ dy^{\prime}/dt^{\prime} }{ \gamma \left( dt^{\prime}/dt^{\prime} + \frac{ dx^{\prime}V }{ dt^{\prime}c^{2} } \right) }$

$u_{y} = \frac{ u^{\prime}_{y} }{ \gamma \left( 1 + \frac{ u^{\prime}_{x}V }{ c^{2} } \right) } \text{ (4) }$

Equations (3) and (4) allow us to work out the components of ball $B$’s velocities $u_{x}$ in the x-direction and $u_{y}$ in the y-direction in frame $S$.

$u(B)_{x} = \frac{ 0 + V }{ \left( 1 + \frac{ 0 \cdot V }{ c^{2} } \right) } = V \text{ (5) }$

$u(B)_{y} = \frac{ -u_{0} }{ \gamma \left( 1 + \frac{ 0 \cdot V }{ c^{2} } \right) } = \frac{ -u_{0} }{ \gamma } \text{ (6) }$

After the collision, the velocity of ball $A$ becomes $u(A) = -u_{0}$. What about ball $B$?

We can see that $u(B)_{x}$ will not change, and $u(B)_{y}$ after the collision will be $- \frac{ + u_{0} }{ \gamma }$.

## The momentum before and after the collision

We are now going to look at the momentum of balls $A$ and $B$ before and after the collision, as seen in frame $S$. We will start off by assuming that the mass is constant for both balls, that is that $m=m_{0}$ for both balls, despite the two reference frames moving relative to each other.

If we do this, we can write that the momentum in the x-direction before the collision is given by

$(p(A)_{x} + p(B)_{x})_{i} = 0 + m_{0}V = m_{0}V$

The momentum after the collision in the x-direction is given by

$(p(A)_{x} + p(B)_{x})_{f} = 0 + m_{0}V = m_{0}V$

So, momentum is conserved in the x-direction. But, what about in the y-direction? Before the collision, the momentum is given by

$(p(A)_{y} + p(B)_{y})_{i} = + m_{0}u_{0} + m_{0} \left( \frac{ -u_{0} }{ \gamma } \right) =m_{0}u_{0} - \frac{ m_{0}u_{0} }{ \gamma }$

After the collision, the momentum in the y-direction is given by

$(p(A)_{y} + p(B)_{y})_{f} = m_{0}(-u_{0}) + m_{0} \left( \frac{ +u_{0} }{ \gamma } \right) = -m_{0}u_{0} + \frac{ m_{0}u_{0} }{ \gamma }$.

If we assume that momentum is conserved, we can write

$m_{0}u_{0} - \frac{ m_{0}u_{0} }{ \gamma } = -m_{0}u_{0} + \frac{ m_{0}u_{0} }{ \gamma } \rightarrow 2m_{0}u_{0} = \frac{ 2m_{0}u_{0} }{ \gamma } \rightarrow \gamma = 1$

So, if we assume that the mass of both ball $A$ and ball $B$ in frame $S$ is $m_{0}$, the momentum in the y-direction is only conserved if $\gamma =1$. But, $\gamma$ is only equal to unity when the relative velocity $V$ between the two frames is zero; in other words when the two frames are not moving relative to each other! If $V \neq 0$ and mass is constant, momentum will not be conserved.

In physics, the conservation of momentum is considered a law, it is believed to always hold. In order for momentum to be conserved, we can qualitatively see that the mass of ball $B$ needs to be greater than the mass of ball $A$ as seen in frame $S$, as the speed of ball $B$ in the y-direction in frame $S, |u(B)_{y}| = u_{0} / \gamma < u_{0}$.

## Allowing the mass to change

We have just shown above that, if we assume both masses are invariant, momentum will only be conserved in the y-direction in the trivial case where the two frames are stationary relative to each other. So, let us now assume that, if $V \neq 0$, we have to allow the masses to change.

We will assume that mass is a function of speed. For ball $A$, the momentum in the x-direction is still zero, both before and after the collision. For ball $B$, we will now write the momentum in the x-direction, both before and after the collision, as

$p(B)_{x} = m(B) u(B)_{x} = m(B) V$

What about in the y-direction? For ball $A$, before the collision we can write

$p(A)_{y} = m(A) u(A)_{y} = m(A) u_{0}$

Where $m(A)$ is the mass of ball $A$ in frame $S$ which is affected by its velocity in frame $S$, which is $u_{0}$.

For ball $B$ as seen in frame $S$ we can write that the momentum in the y-direction before the collision is given by

$p(B)_{y} = m(B) u(B)_{y} = m(B) \cdot \left( \frac{ - u_{0} }{ \gamma } \right) = \frac{ -m(B)u_{0} }{ \gamma }$

Where $\gamma = 1/(1-V^{2}/c^{2})$, the Lorentz factor due to the relative velocity $V$ between $S$ and $S^{\prime}$.

After the collision, we can write the momentum for ball $A$ in the y-direction as being

$p(A)_{y} = m(A) u(A)_{y} = -m(A) u_{0}$

And, for ball $B$ we can write

$p(B)_{y} = m(B) u(B)_{y} = \frac{ +m(B)u_{0} }{ \gamma }$

Equating the momentum in the y-direction before and after the collision, we have

$m(A) u_{0} - \left( \frac{ m(B) u_{0} }{ \gamma } \right) = -m(A) u_{0} + \left( \frac{ m(B) u_{0} }{ \gamma } \right)$

$\rightarrow 2m(A) u_{0} = 2 \left( \frac{ m(B) u_{0} }{ \gamma } \right) \rightarrow m(A) =\frac{ m(B) }{ \gamma }$

For ball $A$, we will write

$m(A) = \gamma_{A} m_{0}$

where

$\gamma_{A} = \frac{ 1 }{ \sqrt( 1 - u^{2}(A)/c^{2} ) } = \frac{ 1 }{ \sqrt( 1 - u_{0}^{2}/c^{2} ) }$

(that is, $\gamma_{A}$ depends on the speed of ball $A$ in frame $S$, and that speed is $u_{0}$).

So, the momentum of ball $A$ in the y-direction is given by

$p(A)_{y} = m(A)u(A)_{y} \rightarrow \boxed {p(A)_{y} = \frac{ m_{0} u_{0} }{ \sqrt( 1 - u_{0}^{2}/c^{2} ) } \text{ (7) } }$

For ball $B$, we will write

$m(B) = \gamma_{B} m_{0}$

Where $\gamma_{B} = 1/\sqrt{ (1 - u^{2}(B)/c^{2} ) }$ depends on the speed $u(B)$ of ball $B$ as seen in frame $S$. (Note: the mass does not depend on just the y-component of ball $B$‘s speed (as is often incorrectly stated), it depends on its total speed).

To calculate the value of $u(B)$ we note that it is made up of the x-component $u(B)_{x}$ and the y-component $u(B)_{y}$. But, $u(B)_{x} = V$, and we showed above that $u(B)_{y} = -u_{0}/ \gamma$, where this $\gamma = 1/\sqrt{ (1 - V^{2}/c^{2}) }$.

Using Pythagoras to calculate $u(B)$, we have

$u(B)^{2} = V^{2} + u_{0}^{2}/\gamma^{2} = V^{2} + u_{0}^{2}(1 -V^{2}/c^{2})$

so

$u(B)^{2} = u_{0}^{2} + V^{2}( 1 - u_{0}^{2}/c^{2} )$

Using this value of $u(B)$ we can write

$\gamma_{B} = \frac{ 1 }{ \sqrt{ ( 1 - u(B)^{2}/c^{2} )} } = \frac{ 1 }{ \sqrt{ ( 1 - u_{0}^{2}/c^{2} - V^{2}/c^{2} + V^{2}u_{0}^{2}/c^{4} ) } }$

But, the terms $( 1 - u_{0}^{2}/c^{2} - V^{2}/c^{2} + V^{2}u_{0}^{2}/c^{4} )$ can be factorised as

$( 1 - u_{0}^{2}/c^{2} - V^{2}/c^{2} + V^{2}u_{0}^{2}/c^{4} ) = (1 - u_{0}^{2}/c^{2})(1 - V^{2}/c^{2})$

And so we can write

$\gamma_{B} = \frac{ 1 }{ \sqrt{ ( 1 - u_{0}^{2}/c^{2} ) } \sqrt{ (1 - V^{2}/c^{2}) } }$

But, $1/\sqrt{ (1 - V^{2}/c^{2}) } = \gamma$, so we can write

$\gamma_{B} = \gamma \cdot \frac{ 1 }{ \sqrt{ ( 1 - u_{0}^{2}/c^{2} ) } }$

This means that we can write the momentum for ball $B$ in the y-direction as

$p(B)_{y} = m(B) u(B)_{y} = \gamma_{B} m_{0} u(B)_{y}$

$p(B)_{y} = \gamma \cdot \frac{ 1 }{ \sqrt{ ( 1 - u_{0}^{2}/c^{2} ) } } \cdot m_{0} \cdot \frac{ u_{0} }{ \gamma }$

$\boxed{ p(B)_{y} = \frac{ m_{0}u_{0} }{ \sqrt{ ( 1 - u_{0}^{2}/c^{2} ) } } \text{ (8) } }$

Comparing this to Equ. (7), the equation for $p(A)_{y}$, we can see that they are equal, as required.

So, we have proved that, to conserve momentum, we need mass to be a function of speed, and specifically that

$\boxed{ m = \frac{ m_{0} }{ \sqrt{ (1 - u^{2}/c^{2}) } } }$

Where $u$ is the speed of the ball in a particular direction in frame $S$.

## How to add velocities in special relativity

As I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be $c$, irrespective of whether the observer is moving towards or away from the source of light. We can think of the speed of light as a cosmic speed limit, nothing can travel faster than it.

But, let us suppose that we have two reference frames $S$ and $S^{\prime}$ moving relative to each other with a speed of $v=0.9c$, 90% of the speed of light. Surely, if someone in frame $S^{\prime}$ fires a high-speed bullet at a speed of $u^{\prime}= 0.6c$, an observer in frame $S$ will think that the bullet is moving away from him at a speed of $u = v + u^{\prime} = 0.9c + 0.6c = 1.5c$, which seemingly violates the comic speed limit.

What have we done wrong?

We cannot simply add velocities, as we would do in Newtonian mechanics. In special relativity we have to use the Lorentz transformations to add velocities. How do we do this? Let us remind ourselves that the Lorentz transformations can be written as

The Lorentz transformations to go either from reference frame $S \text{ to } S^{\prime}$, or to go from $S^{\prime} \text{ to } S$.

## Calculating a velocity in two different reference frames

To calculate the velocity $u$ of some object moving with a velocity $u^{\prime}$ in reference frame $S^{\prime}$ we need to use these Lorentz transformations.

We start off by writing

$x = \gamma \left( x^{\prime} + vt^{\prime} \right) \text{ (1) }$

and

$t = \gamma \left( t^{\prime} + \frac{x^{\prime}v}{c^{2} } \right) \text{ (2) }$

We will now take the derivative of each term, so we have

$dx = \gamma \left( dx^{\prime} + vdt^{\prime} \right)$

and

$dt = \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right)$

We can now write $dx/dt = u$ (the velocity of the object as seen in frame $S$) as

$\frac{dx}{dt} = \frac{ \gamma \left( dx^{\prime} + vdt^{\prime} \right) }{ \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right) }$

The $\gamma$ terms cancel, and dividing each term on the right hand side by $dt^{\prime}$ gives

$\frac{dx}{dt} = \frac{ \left( dx^{\prime}/dt^{\prime} + vdt^{\prime}/dt^{\prime} \right) }{ \left( dt^{\prime}/dt^{\prime} + \frac{dx^{\prime}v}{c^{2} dt^{\prime} } \right) } = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) }$

$\boxed{ u = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) } }$

where $u^{\prime}$ was the velocity of the object in reference frame $S^{\prime}$.

Going back to our example of $v = 0.9c$ and $u^{\prime} = 0.6c$, we can see that the velocity $u$ as measured by an observer in reference frame $S$ will be

$u = \frac{ 0.6c + 0.9c }{ \left( 1 + \frac{ (0.6c \times 0.9c) }{ c^{2} } \right) } = \frac{ 1.5c }{ 1 + 0.54 } = \frac{ 1.5c }{ 1.54} = \boxed {0.974c}$, not $1.5c$ as we naively calculated.

## The constancy of the speed of light

What happens if a person in reference frame $S^{\prime}$ shines a light in the same direction as $S^{\prime}$ is moving away from $S$? In this case, $u^{\prime}=1.0c$. Putting this into our equation for $u$ we get

$u = \frac{ 0.6c + 1.0c }{ \left( 1 + \frac{ (0.6c \times 1.0c) }{ c^{2} } \right) } = \frac{ 1.6c }{ 1 + 0.6 } = \frac{ 1.6c }{ 1.6} = 1.0c$

So they both agree that the light is moving away from them with the same speed $c$!