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## 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 $c$!

## Do GPS satellites move in the sky?

Following on from my blog “Is Tim Peake getting younger or older?” , a bit of fun to work out whether time was passing more slowly or more quickly for Tim Peake in the ISS than it is for us on the ground, it got me thinking about the global positioning system (GPS) that so many of us use on a daily basis. Whether it is using a SATNAV in our car, or a GPS-enabled watch to measure how far and fast we have run, or using maps on a smartphone, GPS must be one of the most-used satellite developments of the last few decades.

As I blogged about here, communication satellites need to be at a particular height above the Earth’s surface so that they orbit the Earth in the same time that it takes the Earth to rotate. In addition to their altitude, communication satellites can only orbit the Earth about the equator, no other orientation will allow the satellite to hover in the same place relative to a location on Earth.

But what about the satellites used in GPS? What kind of an orbit are they in?

## The GPS satellites’ orbits

It turns out that the GPS satellites are not in a geo-stationary orbit, but are in fact in an orbit which leads to their orbiting the Earth exactly twice in each sidereal day (for a definition of sidereal day see my blog here).

The GPS system consists of 31 satellites in orbit around the Earth

We can work out what radius from the Earth’s centre this needs to be by remembering that the speed of orbit is given by

$v = \sqrt { \frac{ GM }{ r } } \text{ (1) }$
where $v$ is the speed of orbit, $G$ is the universal gravitational constant, $M$ is the mass of the Earth and $r$ is the radius of orbit from the centre of the Earth (not from its surface).

A sideral day is 23 hours and 56 minutes, which in seconds is $8.6160 \times 10^{4}$ seconds. So, half a sidereal day is $4.308 \times 10^{4}$ seconds. We will call this the period $T$. The speed of orbit, $v$ is related to the period via the equation
$v = \frac{ 2 \pi r }{ T }$
where $r$ is the radius of the orbit, the same $r$ as in equation (1), and $2 \pi r$ is just the circumference of a circle. So, squaring Equation (1), we can write
$v^{2} = \frac{ GM }{ r } = \left( \frac{ 2 \pi r }{ T } \right)^{2}$
So, in terms of $r$ we can write
$r^{3} = \frac{ G M T^{2} }{ 4 \pi^{2} } \rightarrow r = \sqrt[3]{ \frac{ G M T^{2} }{ 4 \pi^{2} } }, \; \text{ so } r = 26.555 \times 10^{6} \text{ m}$
In terms of height above the Earth’s surface, we need to subtract off the radius of the Earth, so the altitude, which I will call $a_{gps}$, is going to be
$a_{gps} = 26.555 \times 10^{6} - 6.371 \times 10^{6} = 20.184 \times 10^{6} \text{ m } \text{ or } \boxed{ 20.2 \text{ thousand kilometeres} }$

## Why are GPS satellites in this kind of an orbit?

As I didn’t know what kind of an orbit GPS satellites were in before I wrote this blog, the next obvious question is – why are they in an orbit which is exactly half a sidereal day? It is clearly not coincidental! To answer this question, we need to first of all discuss how GPS works.

GPS locates your position by measuring the time a signal takes to get to your GPS device from at least four satellites. Your device can identify from which satellites it gets a signal, and the system knows precisely the position of these satellites. By measuring the time the signals take to you reach you from each of the satellites, it is able to calculate how far each one is from you, and then by using triangulation it can work our your location. There are currently 31 satellites in the system, so often there are more than four visible to your GPS device. The current 31 satellites have all been launched since 1997, the original suite of 38 satellites launched between 1978 and 1997 are no longer in operation.

As I mentioned in my blog about geostationary satellites, a satellite in a geostationary orbit can only orbit above the Earth’s equator. This would clearly be no good for a GPS system, as all the satellites would lie to the south of someone in e.g. Europe or North America. As I said above, there are currently 31 operational satellites; the 31 are divided into 6 orbital planes. If there were 30 satellites this would be 5 in each orbit. The orbits are inclined at $55^{\circ}$ to the Earth’s equator. Each orbit is separated from the other one by 4 hours (equivalent to $60^{\circ}$) in longitude.

As one can see approximately 6 hours in right ascension to both the east and west of one’s location, this means that there will be at least 3 of the orbits above the horizon, and sometimes more. If there were 5 satellites in each orbit this would mean that each one would pass a particular latitude 4 hours before the next one. So, at any particular time there should be some satellites further north than one’s location and some further south, as well as some further east and some further west. This configuration allows for the necessary triangulation to obtain one’s location.

The orbits are inclined at $55^{\circ}$ to the equator and separated by 4 hours (equivalent to $60^{\circ}$) in right ascension, as this diagram attempts to show

## Is the time-dilation effect due to SR or GR more important for these satellites?

We already showed in this blog that, for the International Space Station, the time-dilation due to Special Relativity (SR) has a greater effect on the passage of time than the time-dilation due to General Relativity (GR). What about for the GPS satellites?
The speed of orbit for the GPS satellites at a radius of $26.555 \times 10^{6}$ from the Earth’s centre is, using Equation (1),
$v = \sqrt{ \frac{ GM }{ r } } = 3.873 \times 10^{3} \text{ m/s}$
As we showed in my blog about Tim Peake, the speed of someone on the Earth’s surface relative to the centre of the Earth is $v_{se} = 463.35 \text{ m/s}$, so the relative speed between a GPS satellite and someone on the Earth’s surface is given by
$v = 3.873 \times 10^{3} - 463.35 = 3.410 \times 10^{3} \text{ m/s}$
Compare this to the value for the ISS, which was $7.4437 \times 10^{3}$, it is less than half the speed.

This value of $v$ leads to a time dilation factor $\gamma$ in SR of
$\gamma = \frac{ 1 }{ \sqrt{ 0.9999999999} } \approx 1$
which means that the time dilation due to SR is negligible.
The time dilation due to GR is given by (see my blog here on how to calculate this)
$\left( 1 - \frac{ gh }{ c^{2} } \right) = (1 - 2.2 \times 10{-9}) = 0.9999999978$, or 22 parts in $10^{10}$. Compare this to the ISS, where it was about 1 part in $10^{11}$. Clearly the GR effect for GPS satellites is greater, by about a factor of 5, than it was for the ISS. But, conversely, the SR time-dilation effect has become negligible.

To conclude, the time dilation for GPS satellites is nearly entirely due to General Relativity, and not due to Special Relativity. Time is passing more quickly for the clocks on the GPS satellites than it is for us on Earth, the converse of what we found for the ISS, which is in a much lower orbit.

Because the timings required for GPS to work are so precise, the time dilation effect due to GR needs to be taken into account, and is one of the best pieces of evidence we have that time dilation in GR actually does happen.

## Is Tim Peake getting younger or older?

As anyone who hasn’t been living under a rock knows, the International Space Station (ISS) orbits the Earth with (typically? always?) six astronauts on board. It has been doing this for something like the last fifteen years. One of the astronauts currently on board is the Disunited Kingdom’s first Government-funded astronaut, Tim Peake.

The first British person to go into space was Helen Sharman, but she went into space in a privately funded arrangement with the Russian Space Programme in 1991. Other British-born astronauts have gone into space through having become naturalised Americans, and going into space with NASA. But, Tim Peake has gone to the ISS as part of ESA’s space programme, and his place is due to Britain’s contribution to ESA’s astronaut programme. So, he is the first UK Government-funded astronaut, which is why there has been so much fuss about it in these lands.

Official NASA portrait of British astronaut Timothy Peake. Photo Date: August 28, 2013.

Anyway, I digress. This blog is not about Tim Peake per se, or about the ISS really. I wanted this blog to be about whether Tim Peake is getting older or younger whilst in orbit. Of course everyone is getting older, including Tim Peake, as ‘time waits for no man’ as the saying goes. What I really mean is whether time is passing more or less quickly for Tim Peake (and the other astronauts) in the ISS compared to those of us on Earth.

As some of you might now, when an astronaut is in orbit he is in a weaker gravitational field, as the Earth’s gravitational field drops off with distance (actually as the square of the distance) from the centre of the Earth. Time will therefore pass more quickly for Tim Peake than for someone on the Earth’s surface due to this effect. This is time dilation due to gravity, a general relativity (GR) effect.

But, there is also another time dilation, the time dilation due to one’s motion relative to another observer, the time dilation in special relativity (SR). Because Tim Peake is in orbit, and hence moving relative to someone on the surface of the Earth, this means that time will appear to move more slowly for him as observed by someone on Earth. Interestingly (at least for me!), the SR effect works in the opposite sense to the GR effect.

Which effect is greater? And, how big is the effect?

## Time dilation due to SR – slowing it down for Tim Peake

As I showed in this blog, the time dilation due to SR can be calculated using the equation

$t^{\prime} = \gamma t \text{ where } \gamma = \frac{ 1 }{ \sqrt{ ( 1 -v^{2}/c^{2} )} }$

If he is in orbit at an altitude of 500km (I guessed at this amount, according to Wikipedia it is 400km, but it does not alter the argument which ensues) then his distance from the centre of the Earth (assuming a spherical Earth) is $6.371 \times 10^{6} + 500 = 6.3715 \times 10^{6}$ metres. The centripetal force keeping him in orbit is provided by the force of gravity, and in this blog I showed that the centripetal force $F_{c}$ is given by

$F_{c} = \frac{mv^{2} }{r}$

where $m$ is the mass of the object in orbit, $v$ is its velocity and $r$ is the radius of its orbit.This centripetal force is being provided by gravity, which we know is

$F_{g} = \frac{ GMm }{ r^{2} }$

where $G$ is the universal gravitational constant, and $M$ is the mass of the Earth. Putting these two equal to each other

$\frac{ mv^{2} }{ r } = \frac{ GMm }{ r^{2} } \rightarrow v^{2} = \frac{GM}{r}$

Putting in the values we have for the ISS, where $r=6.3715 \times 10^{6}, G=6.67 \times 10^{-11}$ and $M= 5.97237 \times 10^{24}$, we find that

$v^{2} = 6.2522 \times 10^{7} \rightarrow v = 7.907(067129) \times 10^{3} \text { m/s} = \boxed{ 7.907(067129) \text{ km/s} }$

But, this is the motion relative to the centre of the Earth. People on the surface of the Earth are also moving about the centre, as the Earth is spinning on its axis. But, we cannot calculate this speed as we have done above; people on the surface are not in orbit, but on the Earth’s surface. For something to stay e.g. 1 metre above the Earth’s surface in orbit it would have to move considerably quicker than the rotation rate of the Earth.

The Earth turns once every 24 hours, so for someone on the equator they are moving at

$v_{se} = \frac{ 2 \pi \times 6.3715 \times 10^{6} }{ 24 \times 3600 } = 463.348(5554) \text { m/s}$

where $v_{se}$ refers to the speed of someone on the surface of the Earth. Someone at other latitudes is moving less quickly, at the poles they are not moving at all relative to the centre of the Earth. The speed of someone on the surface will go as $v_{se} \cos (\theta)$ where $\theta$ is the latitude. This is why we launch satellites as close to the Earth’s equator as is feasible; we maximise $v_{se}$ and thus get the benefit of the speed of rotation of the Earth at the launch site to boost the rocket’s speed in an easterly direction.

The difference in speeds between the ISS and someone at the equator on the surface of Earth is therefore

$7.907(067129) \times 10^{3} - 463.348(5554) = \boxed { 7.443(718574) \times 10^{3} \text { m/s} }$

Referring back to my blog on time dilation in special relativity that I mentioned at the start of this section, this means that the time dilation factor $\gamma$, using this value of $v$, is

$\gamma = \frac{ 1 }{ \sqrt{(1 - (v/c)^{2})} } = \frac{ 1 }{ 0.9999999997 }$
(where $c$ is, of course, the speed of light).
This value of $\gamma$ is equal to unity to 3 parts in $10^{10}$, so it would require Tim Peake to orbit for about $3 \times 10^{9}$ seconds for the time dilation factor to amount to 1 second. $3 \times 10^{9}$ seconds is just over 96 years, let us say 100 years.

## The time dilation due to GR – speeding it up for Tim Peake

For GR, the time dilation works in the other sense, it will run more slowly for those of us on the Earth’s surface; we experience gravitational time dilation which is greater than that experienced by Tim Peake. In this blog here, I derived from the principle of equivalence the time dilation due to GR, and found

$\Delta T_{B} = \Delta T_{A} \left( 1 - \frac{ gh }{ c^{2} } \right)$

where, in this case, $\Delta T_{B}$ would be the rate of time passing on the Earth’s surface, $\Delta T_{A}$ the rate of time passing on the ISS,  $g = 9.81$ (the acceleration due to gravity at the Earth’s surface) and $h$ is the height of the orbit, which we have assumed (see above) to be 500 km = $500 \times 10^{3}$.

Plugging in these values we get that

$\frac{ \Delta T_{B} }{ \Delta T_{A} } = 1 - 5.45 \times 10^{-11}$

So, the GR effect is about one part in $10^{11}$ (100 billion). In six months, the number of seconds that Tim Peake will be in orbit is about $1.6 \times 10^{7}$ seconds, so a factor of about 10,000 less than for the GR effect to amount to 1 second. Tim Peake would need to be in orbit for about 5,000 years for the GR effect to amount to 1 second of difference!

## Conclusions

In conclusion, the SR effect on how quickly time is passing for Tim Peake is about 3 parts in 10 billion, in the sense that it passes more slowly for Tim Peake. The GR effect is even smaller, about one part in 100 billion, but in the sense that time is passing more quickly for him. The SR ‘time slowing down’ effect is greater than the GR time ‘passing more quickly effect’, by roughly a factor of 300.

Tim Peake is therefore actually ageing more slowly by being in orbit than if he were on Earth. But, he would need to orbit for nearly 100 years for this difference to amount to just 1 second! And, none of this of course takes into account the detrimental biological effects of being in orbit, which are probably not good to anyone’s longevity!

## Time dilation in General Relativity

A student asked me last week if I could explain the difference between time dilation in Special Relativity (SR) and that in General Relativity (GR), so here is my attempt at doing so. Time dilation in SR comes about when something travels near the speed of light, and is due to the Lorentz transformations which ensure that experiments in any inertial frame are indistinguishable from each other.

I have already derived the Lorentz Transformations from first principles in this blog, and these equations are at the heart of SR, and show why time dilation occurs when one travels near the speed of light. In this blog here, I worked through some examples of time dilation in SR. But, what about time dilation in GR?

## How does time dilation come about in GR?

As I have already explained in this blog here, Einstein’s principle of equivalence tells us that whatever is true for acceleration is true for a gravitational field. So, to see how gravity affects time the easiest way is to consider how time would be affected in an accelerating rocket.

We will consider a rocket in empty space, away from any gravitational fields, which is accelerating with an acceleration $g$. We will have two people in the rocket, Alice and Bob. Alice is at the top end of the rocket, the nose end. Bob is at the bottom end of the rocket, where the tail is. Alice sends two pulses of light, one at time $t=0$, and the second one at a time $t= \Delta \tau_{A}$ later. They are received at the back of the rocket by Bob; the first pulse is received when the time is $t=t_{1}$, and the second one when the time is $t=t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between flashes as measured by Bob.

This is illustrated in the figure below.

We can see how time dilation comes about in GR by considering a rocket accelerating in empty space with an acceleration $g$, and a light flashing from Alice at the front-end of the rocket and being received at the back-end by Bob.

We will set it up so that Bob’s position at time $t=0$ when the first flash is emitted by Alice is $z_{B}(0)=0$, and so his position at any other time is given by
$z_{B}(t) = \frac{1}{2}gt^{2} \text{ (Equ. 1) }$

(this just comes from Newton’s 2nd equation of motion $s=ut + \frac{1}{2}at^{2}$, see my blog here which derives those equations).

The position of Alice will just be Bob’s position plus the distance between them, which we will call $h$ (the height of the rocket), so
$z_{A}(t) = h + \frac{1}{2}gt^{2} \text{ (Equ. 2) }$

We will assume that the first pulse takes a time of $t=t_{1}$ to travel from Alice to Bob. The second pulse is emitted by Alice at a time $\Delta \tau_{A}$ after the first pulse, this is the time interval between each light pulse that Alice sends. This second pulse is received by Bob at a time of $t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between pulses as measured by Bob using a clock next to him.

When the first pulse leaves Alice her position is $z_{A}(0)$, which from equation (2) is h, as she is at the top of the rocket. When Bob receives the pulse at time $t=t_{1}$ his position will be $z_{B}(t_{1})$ which, from equation (1) is $z_{B}(t_{1}) = \frac{1}{2} gt_{1}^{2}$. So, the distance travelled by the pulse is going to be
$z_{A}(0) - z_{B}(t_{1})= ct_{1} \text{ (Equ. 3) }$

as the speed of light is $c$ and it travels for $t_{1}$ seconds. Because the rocket is accelerating, the distance travelled by the second pulse will not be same (as it would be if the rocket were moving with a constant velocity). The distance travelled by the second pulse will be less, and is given by
$z_{A}(\Delta \tau_{A}) - z_{B}(t_{1} + \Delta \tau_{B}) = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 4) }$

We can use Equations (1) and (2), which give expressions for $z_{B} \text{ and } z_{A}$ as a function of $t$, to put in the values that $z_{A} \text{ and } z_{B}$ would have when $t = \Delta \tau_{A}$ for $z_{A}$ and $(t_{1} + \Delta \tau_{B})$ for $z_{B}$ respectively.

Substituting from Equations (1) and (2) into Equation (3) we have
$z_{A}(0) = h, \; \; z_{B}(t_{1}) = \frac{1}{2}gt_{1}^{2}$

which makes equation (3) become
$h - \frac{1}{2}gt_{1}^{2} = ct_{1} \text{ (Equ. 5) }$

Doing the same kind of substitution into equation (4) we have
$z_{A}(\Delta \tau_{A}) = h + \frac{1}{2}g \left( \Delta \tau_{A} \right)^{2} \rightarrow h$

$z_{B}(t_{1} + \Delta \tau_{B}) = \frac{1}{2}g(t_{1} + \Delta \tau_{B})^{2} \rightarrow \frac{1}{2}gt_{1}^2 +gt_{1} \Delta \tau_{B}$

assuming that we can ignore terms in $(\Delta \tau_{A})^{2} \text{ and } (\Delta \tau_{B})^{2}$

Substituting these expressions into equation (4) gives
$h - \frac{1}{2}gt_{1}^{2} - gt_{1} \Delta \tau_{B} = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 6) }$

We now subtract equation (6) from (5) to give
$gt_{1} \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B} \text{ Equ. (7) }$

Re-arranging equation (5) as $\frac{1}{2}gt_{1}^{2} +ct^{1} -h$ and using the quadratic formula to find $t_{1}$ we can write that
$t_{1} = \frac{ -c \pm \sqrt{ c^{2} + 2gh } }{ g } \rightarrow \frac{ -c + \sqrt{ c^{2} + 2gh } }{ g }$

(we can ignore the negative solution because the time is always positive). We will next use the binomial expansion to write
$\sqrt{ c^{2} + 2gh } \approx c ( 1 + \frac{gh}{ c^{2} } )$

(where we have ignored terms in $\left( \frac{2gh}{c^{2}} \right)^{2}$ and higher in the Binomial expansion), and so we can write for $t_{1}$
$t_{1} \approx \left( \frac{ -c + c \left( 1 + \frac{gh}{ c^{2} } \right) }{ g } \right) \rightarrow gt_{1} = c \left( \frac{gh}{ c^{2} } \right)$

Substituting this expression for $gt_{1}$ into equation (7) we now have
$c \left( \frac{gh}{ c^{2} } \right) \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B}$

We can cancel the $c$ in each term and bringing the terms in $\Delta \tau_{B}$ onto one side and the term in $\Delta \tau_{A}$ on the other side we have
$\Delta \tau_{B} \left( 1 + \frac{ gh }{ c^{2} } \right) = \Delta \tau_{A}$

and so
$\Delta \tau_{B} = \frac{ \Delta \tau_{A} }{ \left( 1 + \frac{ gh }{ c^{2} } \right) }$

and using the binomial expansion for $(1 + gh/c^{2})^{-1}$ (and ignoring terms in $\left( \frac{ gh }{ c^{2} } \right)^{2}$ and higher), we can finally write
$\boxed{ \Delta \tau_{B} = \Delta \tau_{A} \left( 1 - \frac{ gh }{ c^{2} } \right) \text{ (Equ. 8) } }$

Because $\frac{gh}{c^{2}}$ is always positive, this means that $\Delta \tau_{B}$ is always less than $\Delta \tau_{A}$, or to put it another way the time interval as measured by Bob at the back-end of the rocket will always be less than the time interval measured by Alice where the light pulses were sent. This means that Bob will measure time to be going at a slower rate than Alice, Bob’s time will be dilated compared to Alice.

From the principle of equivalence, whatever is true for acceleration is true for gravity, so if we now imagine the rocket stationary on the Earth’s surface, with the top end in a weaker gravitational field than the bottom end, we can see that a gravitational field will also lead to pulses arriving at Bob being measured closer together than where they were emitted by Alice. So, gravity slows clocks down!

A very important difference between time dilation in SR and time dilation in GR is that the time dilation in GR is not symmetrical. In SR, both observers in their respective inertial frames think it is the other person’s clock which is running slow. In GR, both Alice and Bob will agree that it is Bob’s clock which is running slower than Alice’s clock.

In a future blog I will do some calculations on this effect in different situations, but as you can see from Equation (8), the size of the dilation depends on the acceleration $g$ and the difference in height between $A \text{ and } B$. I will also discuss whether it is time dilation due to GR or time dilation due to SR which affect the satellites which give us GPS the more, as both effects have to be taken into account to get the accuracy we seek in the GPS position.

## Einstein’s General Relativity – part 1 – the Principle of Equivalence

Next year, 2015, marks the centennial of Einstein’s theory of gravity, what we now call the General theory of Relativity (or just “General Relativity” – “GR”). It is widely recognised as one of the greatest achievements in science, and when Arthur Eddington validated one of its predictions in 1919 Einstein was catapulted to the status of an international star. It is often said that, whereas Einstein’s 1905 special theory of relativity (or “special relativity”) would have been thought of by someone else had Einstein not come up with it, general relativity was so far ahead of its time that we may still be waiting for it if it were not for Einstein’s unparalleled genius.

A portrait of Albert Einstein from around the period that he started developing his theory of gravity, General Relativity.

As it turns out, the development of Einstein’s new theory of gravity was not an easy one. Over the course of several blogs I will trace this tortuous path, which took the best part of ten years, mainly because he had to learn the mathematics of curved space and Tensor calculus to be able to express his ideas in equations. Today I will discuss the beginnings of GR, and in particular what we now call Einstein’s “principle of equivalence”, which he thought of in 1907.

## Einstein’s 1905 Special theory of Relativity

I have already blogged about Einstein’s ground-breaking Special theory of Relativity here. Just to recap, based on two assumptions

1. There is no experiment one can do to distinguish between one inertial (non-accelerating) frame of reference and another
2. The speed of light is constant in all inertial (non-accelerating) frames of reference

Einstein was able to show that these two postulates require that strange things happen to space and time when one travels an appreciable fraction of the speed of light. Lengths get shorter, and time passes more slowly. One of the other consequences of this theory is that Einstein predicted that no information can travel faster than the speed of light.

Einstein soon realised, after he had developed his theory, that Newton’s theory of gravity was in violation of special relativity because it violates both of the postulates on which special relativity is based. In Newton’s theory of gravity, the gravitational force between two objects acts instantaneously. So, according to Newton, if the Sun were to disappear, we would instantly notice its absence (the Earth would move in a straight line rather than continue in its orbit).

Secondly, you could have two inertial (non-accelerating) frames of reference in two different gravitational fields (e.g. one on the surface of the Earth and the other on the surface of the Moon), and a simple experiment like the swinging of a pendulum would yield a different result. This is because the force of gravity (which, along with the length of the pendulum’s string, determines its period of motion) would be different in the two places.

## Einstein’s “happiest thought”

In 1907 Einstein was still working in obscurity in the Patent Office in Bern. Although his special theory of relativity had been published two years before, it was yet to have received much attention. It wasn’t until 1908 that he would get his first academic appointment. In his largely boring patent clark job, Einstein had allowed his mind to wander just as he had done leading up to his miraculous year of 1905. This time, it was in pondering how he could fit Newton’s theory of gravity into his own special relativity. One day he had what he would later refer to as the “happiest thought of my life”. In a lecture on the origins of general relativity which he gave at Glasgow University in June 1933 (“The Origins of the General Theory of Relativity”), he expressed this 1907 thought as

If a person falls freely he will not feel his own weight

Very few of us have experienced free-fall, but most of us have been in a lift (elevator). Right at the start, when the lift starts moving, we temporarily feel heavier and our stomach may feel as if it is sinking. When we slow down at the top of the lift’s travel we temporarily experience the opposite, we feel lighter and our stomach may feel as if it is about to hit our diaphragm!

What Einstein realised is that, if a person were in a lift and the cable were to snap so that the lift fell freely towards the Earth, that person would feel weightless whilst the lift was falling. Their feet would come away from the floor of the lift, and if they took e.g. coins out of their pocket, those coins would not fall towards the floor of the lift but instead would appear to “float” next to the person.

Einstein realised in 1907 that being in a lift (elevator) which is falling freely would feel the same as being in empty space – you would feel weightless.

Einstein next illustrated his absolute genius – he went from this idea, which is fairly specific, to the much more general principle of equivalence – which states that:

there is no experiment you can do to distinguish between the effects of a uniform gravitional field and that of uniform acceleration

Einstein’s “happiest thought” led to his principle of equivalence, which simply states that being in a uniform gravitational field feels the same as accelerating in empty space. They cannot be distinguished from each other. The consequences of this idea are profound and far reaching.

## The first mention of what would become “General Relativity”

Einstein was under pressure from his German editor to write up a review of his principle of special relativity, and so in late 1907 he wrote an article entitled “Über das Relativitätsprinzip und die aus demselben gezogenen Folgerungen”
(On the Relativity Principle and the Conclusions Drawn from It) which appeared on the 4th of December 1907 in the journal Jahrbuch der Radioaktivität. In a section of this review article he included some ideas as to what would happen if he were to generalise his special theory of relativity to include the effects of gravity. He noted a few consequences (without going into the details as he had yet to work them out) – gravity would alter the speed of light and hence cause clocks to run more slowly (i.e. gravity would slow down time). He even postulated that generalising special relativity to include gravity may explain the drift in the perihelion of Mercury’s orbit, something which had been confusing astronomers for several decades.

## Gravity bends light

One of the more celebrated predictions of Einstein’s general theory of relativity is that gravity should bend light. As I mentioned above, in 1919 this was shown to be the case by England’s foremost theoretical astrophysicist of the day, Arthur Eddington. I will go into the details of what he measured in another blog in this series on general relativity, but to finish this part one I will explain how gravity bends light in Einstein’s model.

To understand how this happens, we have to go back to the principle of equivalence. Remember, this states that whatever is true inside a lift which is accelerating in empty space is also going to be true for a lift which is stationary in a uniform gravitational field.

Imagine that a beam of light enters the lift horizontally on the left hand side of the lift. Because the lift is accelerating, rather than follow a straight path across the lift, it will appear to follow a curve (actually a parabola), and it will exit at a lower point on the right hand side than where it entered (this is exactly the same kind of path as a ball would follow if it is projected horizontally from a platform e.g. 200m above the Earth’s surface).

Through the principle of equivalence, if a beam of light crossing an accelerating lift will follow a curve, so will a beam of light crossing a stationary lift which is in a gravitational field. So, gravity should bend light!

Light traversing a lift which is accelerating will appear to bend (in fact it will follow a parabolic path). Because of the principle of equivalence, light should be similarly affected by gravity.

As Einstein developed the mathematics of his general theory he was able to work out precisely how much a given gravitational field should bend light, and his predicted amount was found to be true for the Sun in a celebrated experiment in 1919 by Arthur Eddington.

In part two of this blog I will discuss some of the mathematical obstacles Einstein faced in bringing his general theory of relativity to fruition.

## The 10 best physicists – no. 4 – Albert Einstein

At number 4 in The Guardian’s list of the ten best physicists is Albert Einstein.

The fact that Einstein is at number 4 in this list of “ten best physicists” leads me to believe that this list is not in any particular order. I can’t imagine there would be anyone who would not place Einstein alongside Newton as being one of the two greatest physicists in history.

Quite simply, Einstein revolutionised our understanding of Physics and of the Universe. He overturned the absoluteness of Newtonian mechanics, replacing it with his theories of relativity. He was also instrumental in our understanding of the World at the atomic scale. His name is now synonymous with genius, we all understand what a phrase like “he’s not an Einstein” means.

## Einstein’s brief biography

Albert Einstein was born in Ulm (in present day Germany) in 1879. His father Hermann and his mother Pauline (née Koch) were also both German. In 1880 the family moved to Munich where Einstein’s father and uncle started a company manufacturing electrical equipment running on direct current electricity. In 1894 this company failed, when alternating current became the standard worldwide for distributing electricity. The family moved to Italy, but Einstein stayed in Munich to finish his studies.

In 1896, having obtained top marks in his mathematics and physics exams, Einstein enrolled on a 4-year mathematics and physics teaching diploma course at Zurich Polytechnic. He graduated in 1900. During his time at Zurich Polytechnic he became romantically involved with a classmate, MIleva Mariƈ. They married in 1903, by which time Einstein was working as a patent clark in the Swiss Patent Office.

In 1905 Einstein had his annus mirablis (miraculous year). He completed his PhD at the end of April from the University of Zurich, and then proceeded to publish five important papers, four of which are generally recognised as being great works of physics in their own right. I will discuss these five papers below, but by 1908 his work from 1905 had gained enough attention that he was appointed a lecturer at the University of Bern.

Einstein was only at Bern for one year, in 1909 he was offered a lecturing position at the University of Zurich, and then in 1911 was offered a full Professorship at the Charles-Ferdinand University in Prague. In 1914 he was appointed the Director of the Kaiser Wilhelm Institute for Physics and made a Professor at the Humboldt University in Berlin, arguably Germany’s most prestigious university.

In 1919 Einstein became a celebrity when Sir Arthur Eddington verified one of the key predictions of Einstein’s theory of gravity, his General Theory of Relativity. Newspaper headlines around the World acclaimed him, and from that year on his life was never the same. In 1921 he was awarded the Nobel Prize in Physics for his work on the photoelectric effect.

In 1933, with Hitler coming to power in Germany, Einstein seized on the opportunity of a visit to the United States to “defect”. He never returned to Germany, and with many academic job offers to choose from he settled at the Institute for Advanced Study at Princeton University in New Jersey, USA. It is there he saw out the rest of his career, dying in 1955.

## Einstein’s contributions to Physics

It is difficult to know where to start in discussing Einstein’s contributions to Physics. As I said above, alongside Newton, Einstein stands as one of the two most important people in Physics. Time Magazine decided to make Einstein their “Person of the Century”, placing above all the 20th Century’s great statesmen. Why?

Einstein changed our perception of our Universe. He will be best remembered for his two theories of relativity. In 1905, as the third paper in his annus mirablis, he published a paper entitled “On the Electrodynamics of Moving Bodies”, a fairly innocuous title. But the contents were far from innocuous, it asked us to accept the idea that if two observers are moving relative to each other they will always measure the speed of light to be the same. This has far reaching consequences. It means that time passes differently depending on how you are moving, it means that lengths will be measured differently, and it means that the inertial mass of an object will change as its speed increases. Later in the same year, Einstein wrote a paper entitled “Does the inertia of a body depend upon its energy content?”, and this paper introduced what has become the most famous equation in Physics, $E=mc^{2}$, showing that mass and energy are equivalent. These two papers together describe the ideas of what we now call Special Relativity.

Before publishing his paper on the electrodynamics of moving bodies, Einstein published a paper which would win him the 1921 Nobel Prize. This was a paper on the photoelectric effect, a phenomenon which had been known about for several decades but for which no-one had a successful explanation. The photoelectric effect is when electrons are released from the surface of certain metals when a light is shone upon the metal’s surface. The experimental results of this phenomenon could not be explained with the accepted “wave theory” of light. Einstein extended the idea of “light quanta”, which had first been suggested by Max Planck in 1900 to explain blackbody radiation. Einstein suggested that light could be thought of as individual quanta of energy (what we now called photons), with the energy of each photon given by $E=h\nu$ where $h$ is Planck’s constant and $\nu$ is the frequency of the light. This simple but revolutionary idea explained the results of the photoelectric effect perfectly, and was the beginnings of having to think of all sub-atomic phenomena as a combination of waves and particles, something which is at the heart of Quantum Mechanics.

In 1905 Einstein also published a paper explaining Brownian motion as the jostling of pollen grains by molecules in the liquid in which the grains are suspended. This was one of the first direct proofs of the existence of atoms. With the five papers Einstein published in 1905 he was destined to be an important physicist. But what propelled him to being considered the greatest physicist next to Newton was what came next, his General Theory of Relativity.

Einsteins’ General Theory of Relativity is his re-working of Newton’s theory of gravity, which had been around for over two hundred years and which most physicists felt was as correct a theory as any that had been developed. But Einstein realised that Newton’s way of thinking about gravity was incompatible with his earlier Special Theory of Relativity. Over a period of some seven years, from 1908 to 1915, Einstein worked out the mathematical details of his theory, and when it was finally published it was even more revolutionary in some ways than his Special Theory of Relativity. It asked us to think of gravity not as a force, but as a bending of space-time. Masses cause the space-time continuum to deform, and in this deformed continuum objects move along the shortest path. I am not going to go into too much detail here, but will explain many of the most important ideas and predictions of General Relativity in series of future blogs. Suffice it to say that, to date, his new way of thinking about gravity has withstood every test to which it has been put, and is considered one of the most important theories in physics, as well as one of the most beautiful.

Einstein made many other contiributions to 20th Century Physics, including the Einstein coefficients which enable us to work out how long electrons will spend in different energy levels in atoms, and Bose-Einstein Statistics, which is how particles which are indistinguishable from each other behave in a statistical way.

It is difficult to imagine where Physics would be today if it weren’t for Einstein’s contributions. Some people have argued that, whereas his Special Theory of Relativity was just waiting to be discovered (indeed, in a talk in 1904 the French mathematician/physicist Henri Poincaré suggested the idea of the relativity of time), his General Theory was so revolutionary that it required someone with Einstein’s genius to even think of it, and that no other physicist before or since could have thought of it. Einstein’s name towers over Physics like no one else, except of course for Isaac Newton, whom I will talk about in a few weeks’ time.