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## Antimatter and Dirac’s Equation

Yesterday I introduced Paul Dirac, number 10 in “The Guardian’s” list of the 10 best physicists. I mentioned that his main contributions to physics were (i) predicting antimatter, which he did in 1928, and (ii) producing an equation (now called the Dirac equation) which describes the behaviour of a sub-atomic particle such as an electron travelling at close to the speed of light (a so-called relativistic theory). This equation was also published in 1928.

## The Dirac Equation

In 1928 Dirac wrote a paper in which he published what we now call the Dirac Equation.

This is a relativistic form of Schrödinger’s wave equation for an electron. The wave equation was published by Erwin Schrödinger two years earlier in 1926, and describes how the quantum state of a physical system changes with time.

### The Schrödinger eqation

The various terms in this equation need some explaining. Starting with the terms to the left of the equality, and going from left to right, we have $i$ is the imaginary number, remember $i = \sqrt{-1}$. The next term $\hbar$ is just Planck’s constant divided by two times pi, i.e. $\hbar = h/2\pi$. The next term $\partial/\partial t \text{ } \psi(\vec{r},t)$ is the partial derivative with respect to time of the wave function $\psi(\vec{r},t)$.

Now, moving to the right hand side of the equality, we have
$m$ which is the mass of the particle, $V$ is its potential energy, $\nabla^{2}$ is the Laplacian. The Laplacian, $\nabla^{2} \psi(\vec{r},t)$ is simply the divergence of the gradient of the wave function, $\nabla \cdot \nabla \psi(\vec{r},t)$.

In plain language, what the Schrödinger equation means “total energy equals kinetic energy plus potential energy”, but the terms take unfamiliar forms for reasons explained below.

## 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

## 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).

The blue ball $B$ moves solely in the y-direction in reference frame $S$, and the red ball $R$ moves solely in the y$^{\prime}$-direction in reference frame $S^{\prime}$. Ball $B$ starts by moving in the positive y-direction in reference frame $S$ with a velocity $u_{0}$, and ball $R$ 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 $R$ 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 $B$ will move vertically downwards in the negative y-direction, with a velocity $-u_{0}$. Ball $R$ 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 $B$ and $R$ looks like the diagram on the right of Figure 1. In $S^{\prime}$, it is ball $R$ which moves vertically, and ball $B$ which moves in both the $x^{\prime}$ and $y^{\prime}$ directions.

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

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

To calculate the velocity of ball $R$ 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 $R$’s velocities $u_{x}$ in the x-direction and $u_{y}$ in the y-direction in frame $S$.

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

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

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

We can see that $u(R)_{x}$ will not change, and $u(R)_{y}$ 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 $B$ and $R$ 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

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

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

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

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

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

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

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

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 $B$ and ball $R$ 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 $R$ needs to be greater than the mass of ball $B$ as seen in frame $S$, as the speed of ball $R$ in the y-direction in frame $S, |u(R)_{y}| = u_{0} / \gamma < u_{0}$.

## 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 $B$, the momentum in the x-direction is still zero, both before and after the collision. For ball $R$, we will now write the momentum in the x-direction, both before and after the collision, as

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

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

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

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

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

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

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 $B$ in the y-direction as being

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

And, for ball $R$ we can write

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

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

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

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

For ball $B$, we will write

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

where

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

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

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

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

For ball $R$, we will write

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

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

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

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

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

so

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

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

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

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

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

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

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

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

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

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

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

Comparing this to Equ. (7), the equation for $p(B)_{y}$, 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 $c$!

## Imaging the Galaxy’s supermassive blackhole – part 1

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

## The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity $v_{esc}$ when that velocity is equal to the speed of light $c$. That is

$v_{esc} = c = \sqrt{ \frac{2GM }{ R } }$

where $M$ is the mass of the blackhole, $G$ is the universal gravitational constant, and $R$ is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius $R_{s}$). So, we can write

$R_{s} = \frac{ 2GM }{ c^{2} }$

Putting in a mass of 4.5 million solar masses, we find

$R_{s} = 1.33 \times 10^{10} \text{ metres}$

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
$\theta = 6.17 \times 10^{-9} \text{ degrees}$ (remember to double $R_{s}$ to get the diameter of the event horizon). This is the same as

$\boxed{ \theta = 22.22 \text{ micro arc seconds} }$

Converting this to radians, we get

$\theta ( \text{in radians}) = 1.08 \times 10^{-10}$

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

$\theta ( \text{in radians}) \approx 4 \times 10^{-10}$

## The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

$\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }$

where $D$ is the diameter of the telescope and $\lambda$ is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of $50 \times 10^{-4} \text{ radians }$ at various wavelengths.

For visible light, assuming $\lambda = 550 \text{ nanometres}$

$D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }$

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

$D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }$

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

$D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }$

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!

## Evidence for the Galaxy’s supermassive black hole

On Tuesday, I blogged about the theoretical work done in the early 1970s by Martin Rees, and others, which proposed that there may be a supermassive black hole at the centre of our Galaxy and most spiral galaxies.

## What about the observational evidence?

In the early 1980s two teams set about observing the orbits of stars near Sgr A*. The two teams, working separately, were at UCLA and The Max Planck Institute For Extra-terrestrial Physics (MPE). The team at UCLA is known as the UCLA Galactic Center Group, the team at the MPE doesn’t have a snazzy name, but their website can be found here. Gradually, over many years, each of the two teams has determined the orbits of several dozens of stars, and hence have been able to use the laws of gravity to determine the mass of the enclosed mass which the stars are orbiting.

Below is an image of Sgr A* taken by the MPE team using the NACO near-infrared camera on the VLT with adaptive optics. The entire image is only 30 arc seconds across.

A combined H, K and L-band near infrared image of the Galactic Centre obtained by the NACO camera on the VLT using adaptive optics. This image is from the MPE website.

Here is a paper, published in 2009, entitled “Monitoring Stellar Orbits Around the Massive Black Hole in the Galactic Center”, published by the MPE group in The Astrophysical Journal. Here is a link to the paper.

This paper, published in The Astrophysical Journal in 2009, is one of several showing overwhelming evidence for a supermassive blackhole at the centre of the Milky Way galaxy.

In this paper, entitled “The Galactic Center massive black hole and nuclear star cluster”, Reinhard Genzel (the director of the MPE) and colleagues summarise the evidence from their studies of their being a supermassive blackhole at the centre of the Milky Way, with a calculated mass of about 4.4 million solar masses. Here is a link to the paper.

In a paper entitled “The Galactic Center massive black hole and nuclear star cluster”, Genzel etal. summarise their finding that the Galaxy harbours a massive black hole with a mass of about 4.4 million solar masses.

The UCLA group published this paper “Measuring Distance and Properties of the Milky Way’s Central Supermassive Black Hole With Stellar Orbits”, in 2008 in The Astrophysical Journal (here is a link to the paper). In it,they calculate the mass of the supermassive black hole to be 4.5 million solar masses, with an error of plus or minus 0.4 million solar masses.

Ghez etal. (2008) find the  mass of the supermassive black hole to be 4.5 million solar masses, slightly higher than the MPE group, but well within the errors of the two groups’ measurements.

The mass of this black hole is about 4.45 million times the mass of the Sun (the two groups calculate different masses, with the UCLA group calculating about 4.5 million solar masses, the MPE group about 4.4 million solar masses). Let us assume it is 4.5 million solar masses, just to round up to the nearest half a million solar masses.

As some of you may know, blackholes are observable in certain ways. They clearly affect the orbit of nearby objects (this is how the UCLA and MPE teams have garnered the evidence for the supermassive blackhole), but also the accretion disk which usually surrounds a blackhole has very hot gas spiralling into the blackhole. This very hot gas emits radiation as a blackbody, so most of it comes out in the X-ray part of the spectrum due to the very high temperature of several millions of Kelvin.

But, a blackbody will also radiate at other wavelengths (see my blog here to remind yourself of the shape of a blackbody curve), so such accretion disks will also radiate visible light, infrared light, and even radio emission. The question then arises, is it possible to observe the accretion disk way in towards the event horizon of the Galaxy’s supermassive blackhole?

I will answer that question next week.

## The Galaxy’s supermassive black hole

There is now overwhelming evidence that our Galaxy harbours a supermassive black hole at its centre. Not only that, but the Hubble Space Telescope has discovered that all spiral galaxies harbour supermassive black holes at their centres, and the mass of that black hole is directly proportional to the mass of the galaxy in which it resides. The reasons for this are still unclear.

The idea of supermassive black holes driving the prodigious energy output at the centre of some galaxies was first proposed by Edwin Salpeter in a 1964 paper. Salpeter is probably better known for his work on the initial mass function of star formation, but in this paper (follow this link to read it), Salpeter proposed that supermassive black holes may be the energy source behind the then newly discovered quasars.

In a 1964 paper, Edwin Salpeter (possibly more famous for his work on the initial mass function of star-formation) was the first to propose supermassive black holes as the energy source of the newly-discovered quasars (or QSOs).

In 1971, Donald Lynden-Bell and Martin Rees wrote an important paper entitled “On quasars, dust and the galactic centre”, (follow this link to the paper). It was the first paper to suggest that our own Galaxy, the Milky Way, may harbour a supermassive black hole at its centre.

The possibility that our Milky Way harboured a supermassive blackhole at its centre was first proposed by Donald Lynden-Bell and Martin Rees in 1971.

Another important paper entitled “Accretion onto Massive Black Holes” was written in 1973 by Pringle, Rees and Pacholczyk (follow this link), who considered the observable effects that matter accreting onto a (super)massive black hole would have.

In a 1973 paper, Pringle, Rees and Packolczyk considered the observable effects of the accretion of matter onto a supermassive black hole.

Pringle etal. draw two main conclusions, the second of which is possibly the more important; that material falling onto a (super)massive black hole will emit a huge amount of radiation.

The two main conclusions of the Pringle etal. (1973) paper.

In a 1974 review article in The Observatory entitled “Black Holes”, Martin Rees further stated the arguments for supermassive black holes at the centres of galaxies.

In a 1974 review article in The Observatory, Martin Rees wrote that “a black hole might lurk in the centres of most galaxies.”. 35-40 years later, he was shown to be correct.

He stated (my highlight)

If we regard quasars as hyperactive galactic nuclei, then a black hole might lurk in the centres of most normal galaxies.

How prescient were these words!

Later in the same year, radio astronomers Bruce Balick and Robert Brown discovered a compact radio source in the constellation Sagittarius. They announced their result in a paper entitled “Intense Sub-Arcsecond Structure in the Galactic Center” (here is a link to the paper).

In 1974, Bruce Balick and Robert Brown used the Very Large Array of radio telescopes in New Mexico to discover a compact radio source at the centre of the Milky Way. We now call this source Sagittarius A*

Using the Very Large Array of radio telescopes in New Mexico, Balick and Brown found a sub-arcsecond radio source at both 2695 MHz (11cm) and 8085 MHz (3.7cm). We now call this source Sagittarius A*, and it is believed to be where the Galaxy’s supermassive black hole resides. Here is their image obtained at 2695 MHz (at right, the image at left is the base-line coverage in the u-v plane, the interferometry plane of the array).

Bruce Balick and Robert Brown discovered a sub-arcsecond radio source in the constellation Sagittarius. We now call this source Sagittarius A*. Their discovery was made at 2695 MHz (which corresponds to a wavelength of 11 centimetres)

Since these discovery observations, Sagittarius A* (or Sgr A* as it is often known) has been observed at many other wavelengths (but not in visible light, the dust extinction is too great). For example, here is a combined infrared and X-ray image.

A composite infrared and X-ray image of Sagittarius A*

And, here are some images taken by the SPIRE camera on the Herschel Space Observatory at 250, 350 and 500 microns.

Images of Sagittarius A* taken by the SPIRE camera on the Herschel Space Telescope. The observations are at (from left to right) 250, 350 and 500 microns.

Later this week I will blog about the observational evidence for this compact object being a supermassive black hole.

## Is there life on Mars? (part 1)

As those of you following my blog will know, I am currently on a cruise around New Zealand, giving astronomy talks. One of my six talks is about our current understanding of whether there is (or was) life on Mars. I try to only talk about objects which are visible during the cruise, and Mars is currently visible in the evening sky, albeit a lot fainter than it was in May when it was at opposition.

One of the talks I am giving on this cruise is our current understanding of whether there is (or was) life on Mars.

The question of whether there is life on Mars, or whether there ever has been in its history, is a fascinating one. I thought I would do a series of blogs to explore the question. But, I have to begin by saying that ANY search for life beyond Earth is predicated by our understanding of life on Earth. The only thing, it would seem, required by all forms of life which we have found on earth is water. Extremophiles show that life can exist without oxygen, without light, at high pressure, in radioactive environments; in fact in all sorts of environments which humans would find impossible. But, none of the life so far found on Earth can exist without water.

As a consequence, all searches for life in our Solar System tend to begin with the search for water. Now, it may be that life beyond Earth could have evolved to exist without the need for water. I am no chemist, but I don’t think there is anything particularly unique about water in its chemistry which makes it impossible for living cells to use some other substance. Water is the only substance on Earth which can exist in all three forms naturally (solid, liquid and gas), so it does occupy an unique place in the environment found on Earth. But, on Titan for example, methane seems to exist in all three forms. Maybe life has evolved on Titan to metabolise using methane in the same way that life on Earth has evolved to metabolise using water. We don’t know.

So, I thought I would start this series of blogs with the big news in the 1890s, that Martians had built canals on the red planet!

## Schiaparelli and Martian ‘canali’

The Schiaparelli space probe which ESA sadly failed to land on Mars recently was named after Italian astronomer Giovanni Schiaparelli. In the late 1880s he reported seeing ‘canali’ on the surface of Mars. Although this means ‘channels’, it got mis-translated to ‘canals’, and led to a flurry of excitement that this was evidence of an intelligent civilisation on Mars.

The idea grew that Martians had built canals to transport water from the “wet” regions near the poles to the arid equatorial regions. The ice caps of Mars are easily visible through a small telescope, so astronomers had known for decades that Mars had ice caps which they assumed were similar to the ice caps on Earth.

Giovanni Schiaparelli’s map of ‘canali’ on Mars, from 1888.

One person who became particularly taken with this idea of canals on Mars was American Percival Lowell. Lowell came from a rich Bostonian family, and had enough personal wealth to build an observatory in Flagstaff, Arizona. He set about proving the existence of life on Mars, writing several books on the subject. He published Mars (1895), Mars and Its Canals (1906), and Mars As the Abode of Life (1908). But, by 1909 the 60-inch telescope at Mount Wilson Observatory had shown that the ‘canali’ were natural features, and Lowell was forced to abandon his ideas that intelligent life existed on Mars.

However, his Flagstaff Observatory was to go on and make important contributions to astronomy. In the 1910s Vesto Slipher was the first person to show that nearly all spiral nebulae (spiral galaxies as we now call them) showed a redshift, the first bit of observational evidence that the Universe is expanding. And, in 1930 Clyde Tombaugh discovered Pluto at Flagstaff Observatory.

In part 2 of this blog, next week, I will talk about the first space probes sent to Mars, and the first images taken of Mars by a space probe which successfully orbited the planet, Mariner 4.

## A cylinder rolling down a slope

The other week I was asked to explain how a cylinder (or ball) rolling down a slope differs from e.g. a ball being dropped vertically. It is an interesting question, because it illustrates some things which are not immediately obvious. We all know that, if you drop two balls, say a tennis ball and a cannon ball, they will hit the ground at the same time. This is despite their having very different masses (weights). Galileo supposedly showed this idea by dropping objects of different weights from the tower of Pisa (although he probably never did this, see our book Ten Physicists Who Transformed Our Understanding of Reality).

With a tennis ball and a cannon ball, they clearly have very different masses (weights), but will fall to the ground at the same rate. This fact, contrary to the teachings of Aristotle, was one of the key breakthroughs which Galileo made in our understanding of motion. But, what about if we roll the two balls down a slope? If we build a track to keep them going straight, will a tennis ball roll down a slope at the same rate as a cannon ball? The answer is no, and I will explain why.

## Rolling rather than dropping

When a ball rolls down a slope, it starts off at the top of the slope with gravitational potential energy. When it starts rolling down the slope, this gravitational potential energy gets converted to kinetic energy. This is the same as when the ball drops vertically. But, in the case of the ball dropping vertically, the kinetic energy is all in the form of linear kinetic energy, given by

$\text{ linear kinetic energy } = \frac{ 1 }{ 2 } mv^{2}$

where $m$ is the mass of the ball and $v$ is its velocity (which is increasing all the time as it falls and speeds up). The gravitational potential energy is converted to linear kinetic energy as the ball drops; by the time the ball hits the bottom of its fall all of the PE has been converted to KE.

If, instead, we roll a ball down a slope, the kinetic energy is in two forms, linear kinetic energy but also rotational kinetic energy, which is given by

$\text{ rotational kinetic energy } = \frac{ 1 }{ 2 } I \omega^{2}$

where $I$ is the ball’s moment of inertia, and $\omega$ is the ball’s angular velocity, usually measured in radians per second. The key point is that the the moment of inertia for the two balls in this example (a tennis ball and a cannon ball) have a different value, because the distribution of the mass in the two balls is different. For the tennis ball it is all concentrated in the layer of the rubber near the ball’s surface, with a hollow interior. For the cannon ball, the mass is distributed throughout the ball.

## Two cylinders rolling down a slope

Let us, instead, consider the case of two cylinders rolling down a slope. One is a solid cylinder, the other is a hollow one with all of its mass concentrated near the surface. We will make the two cylinders have the same mass; this can be done by making the material from which the hollow cylinder is made denser than the material for the solid cylinder. So, even though the material of the hollow cylinder is all concentrated near the surface of the cylinder, and there is a lot less of it, if it is denser it can have equal mass.

A solid cylinder on an inclined plane. We will make the mass of this solid cylinder the same as that of the hollow cylinder, by making it of less dense material. Although it will have the same mass $m$ and the same radius $R$, it will not have the same moment of inertia $I$.

We will start both cylinders from rest near the top of the slope, and let them roll down. We will observe what happens.

A hollow rolling down an inclined plane. We will make the hollow cylinder denser than the solid one, so that they both have the same mass $m$ and the same outer radius $R$. But, they will not have the same moment of inertia $I$.

When things are dropped, the rate at which they fall is independent of the mass, but when they roll the rate at which they roll is not indpendent of the moment of inertia. In particular, it is not independent of the distribution of mass in the rolling object. As this video shows, the solid cylinder rolls down the slope faster than the hollow one!

But, why??

## Why does the solid cylinder roll down quicker?

The reason that the solid cylinder rolls down faster than the hollow cylinder has to do with the way that the potential energy (PE) is converted to kinetic energy. Because the cylinder is rolling, some of the PE is converted to rotational kinetic energy (RKE), not just to linear kinetic energy (LKE). The only way that a cylinder can roll down a slope is if there is friction between the cylinder and the slope, if the slope were perfectly smooth the cylinder would slide and not roll.

The torque (rotational force) $\tau$ is related to the angular acceleration $\alpha$ in a similar way that the linear force $F$ is related to linear accelerate $a$. From Newton’s second law we know that $F = ma$ where $m$ is the mass of the object. The rotational equivalent of this law is

$\tau = I \alpha$

where $I$ is the moment of inertia. The moment of inertia $I$ is different for a hollow cylinder and a solid cylinder. For the solid cylinder it is given by

$I_{sc} = \frac{ 1 }{ 2 } mR^{2} = 0.5mR^{2}$

where $m$ is the mass of the cylinder and $R$ is the radius of the cylinder. For the hollow cylinder, the moment of inertia is given by

$I_{hc} = \frac{ 1 }{ 2 }m(R_{2}^{2} + R_{1}^{2})$

where $R_{2} \text{ and } R_{1}$ are the outer and inner radii of the annulus of the cylinder. We are going to make the hollow cylinder such that the inner 80% is hollow, so that $R_{1} = 0.8R_{2} = 0.8R$. We will make the mass $m$ of the two cylinders the same.

Thus, for the hollow cylinder, we can now write

$I_{hc} = \frac{ 1 }{ 2 }m(R^{2} + (0.8R)^{2}) = \frac{ 1 }{ 2 }mR^{2}(1+0.64) = \frac{ 1 }{ 2 }mR^{2}(1.64) = 0.82 mR^{2}$

The cylinder accelerates down the slope due to the component of its weight which acts down the slope. This component is $mg sin(\theta)$ where $g$ is the acceleration due to gravity and $\theta$ is the angle of the slope from the horizontal. To make the maths easier, we are going to set $\theta = 30^{\circ}$, as $sin(30) =0.5$.

Friction always acts in the opposite direction to the direction of motion, and in this case the friction $F_{f}$ is related to the torque $\tau$ via the equation

$\tau = F_{f}R \text{ (1)}$

so we can write

$F_{f}R = \tau = I \alpha \text{ (2)}$

where $\alpha$ is the rotational acceleration. Re-arranging this to give $F_{f}$, we have

$F_{f} = \frac{I \alpha}{ R }$

The force down the slope, $F (=ma)$ is just the component of the weight down the slope minus the frictional force $F_{f}$ acting up the slope.

$ma = mg\sin(30) - F_{f} = 0.5mg - \frac{ I \alpha }{ R } \text{ (3)}$

The angular acceleration $\alpha$ is given by $\alpha = a/R$ where $a$ is the linear acceleration. So, we can re-write Eq. (3) as

$ma = 0.5mg - \frac{ Ia }{ R^{2} } \text{ (4)}$

Now we will put in the moments of inertia for the solid cylinder and the hollow cylinder. For the solid cylinder, we can write

$ma = 0.5mg - \frac{ 0.5mR^{2}a }{ R^{2} } = 0.5mg - 0.5ma$

The mass $m$ can be cancelled out, and assuming $g=9.8 \text{ m/s/s}$, we have

$a = 0.5g - 0.5a \rightarrow 1.5a = 4.9 \rightarrow \boxed{ a = 3.27 \text{ m/s/s (5)} }$

Notice that Equation (5) does not have the mass $m$ in it, as this cancels out. It also does not have the radius $R$ of the cylinder in it; the acceleration of the cylinder as it rolls down the slope is independent of both the mass and the radius of the cylinder.

For the hollow cylinder, again using Eq. (4), we have

$ma = 0.5mg - \frac{ 0.82maR^{2} }{ R^{2} } = 0.5mg - 0.82ma$

This simplifies to

$a = 4.9 - 0.82a \rightarrow 1.82a = 4.9 \rightarrow \boxed{ a = 2.69 \text{ m/s/s} (6)}$

As with Equation (5), Equation (6) is independent of both mass and radius.

So, as we can see, the linear acceleration $a$ for the hollow cylinder is 2.69 m/s/s, less than the linear acceleration for the solid cylinder, which was 3.27 m/s/s. This is why the solid cylinder rolls down the slope quicker than the hollow cylinder! And, the result is independent of both the mass and the radius of either cylinder. Therefore, a less massive solid cylinder will roll down a slope faster than a more massive hollow one, which may seem contradictory.

## Summary

All objects falling vertically fall at the same rate, but this is not true for objects which roll down a slope. We have shown above that a solid cylinder will roll down a slope quicker than a hollow one. This is because their moments of inertia are different, it requires a greater force to get the hollow cylinder turning than it does the solid cylinder. Remember, the meaning of the word ‘inertia’ is a reluctance to change velocity, so in this case a reluctance to start rolling from being stationary. A larger moment of inertia means a greater reluctance to start rolling.

The solid cylinder will start turning more quickly from being stationary than the hollow cylinder, and this means that it will roll down the slope quicker. This result is independent of the masses (and radii) of the two cylinders; even a less massive solid cylinder will roll down a slope quicker than a more massive hollow one, which may be counter-intuitive.

## Beagle 2 “close to Mars success”

New images of the European Space Agency’s Beagle 2 have emerged recently, suggesting that it came closer to success than has long been thought. These new images have been analysed more thoroughly and carefully than previous images of Beagle 2, and with the help of a computer simulation it is being suggested that Beagle 2 did not crash land. Instead, this team led by Professor Mark Sims of Leicester University are arguing that Beagle 2 deployed, but not completely correctly. They suggest that, due to not deploying correctly, that it may well have done science for a period of about 100 days, before shutting down due to lack of power. They even suggest that there is a very slim possibility that it is still working.

I do have to take issue, however, with the way this story is worded on the BBC website. It implies that we now know, with certainty, that Beagle 2 operated for some period on the surface of Mars. This is not true. One study has argued that it did. One swallow does not make a summer. This particular team’s analysis and study will need to be looked at by others before we can say with any reasonable certainty that Beagle 2 survived its landing.

New images of Beagle 2 taken by NASA’s Mars Reconnaissance Orbiter have been analysed by a computer model, suggesting it may have actually worked for a short period of time.

As with any suggestion which flies in the face of conventional wisdom, this claim will need to be checked and followed up by others. But, if the evidence is sufficiently strong that Beagle 2 did not crash, then it will come as a relief to those who worked on it and have long felt that it failed in a crash. Sadly, even if it did work, we have not received any data back from it; and that is not going to change.