Download Links between the Einstein`s Special Relativity DS and

Links between the Einstein’s Special Relativity DS and the Synchrotron DS
The obvious link between these two studies is that the electrons in a synchrotron are travelling at speeds over
99.99999% of the speed of light which makes them well and truly ‘relativistic’. Because of this the ‘relativistic
mass’ of the electrons is far higher than their rest mass – and so as they travel through the bending magnets of
the synchrotron the radius of their circular path is far higher than it would be for ‘classical’ electrons. Indeed the
whole synchrotron would fit on a table top if Newton’s physics applied! Furthermore, the ‘synchrotron light’
would be mere radio waves. This gives us a number of opportunities to link these two new exciting detailed
studies. While the theory of relativity was one of the great triumphs of twentieth century theoretical physics, the
synchrotron was a great achievement of applied physics.
The relationship between accelerating voltage and speed of an electron.
If we simply use the non-relativistic equations to determine the speed of an electron which has been accelerated
through a given potential we find that the speed simply increases with the square root of the potential. Let’s see
why.
In classical physics the kinetic energy (½mv²) of the
electron is equal to the work done on it by the
accelerating potential (qV). Taking m as the mass of
the electron, and e as the charge (q) on an electron we
can write:
and so v
= √(2eV/m)
A graph of the speed v as a fraction of the speed of
light against the kinetic energy of the electrons is
shown at right. The horizontal line at 100% represents
the speed of light (c = 3x108 m/s). We can see that
with an accelerating potential of about 255 kV,
classical physics tells us that the speed of the electron
would reach c. Even at the accelerating potential in an
ordinary TV set, about 25 kV, we find that the electron
is doing almost one third of c.
200%
180%
160%
140%
Speed (v/c)
½mv² = eV
Non-Relativistic speed (% of c)
120%
100%
80%
60%
40%
20%
0%
0
200
400
600
800
1000
Energy of electrons (keV)
However, Einstein tells us that as the speed of the electron approaches ‘relativistic’ speeds, say over 10% of c
(where the Lorentz factor γ is about 1.005) we cannot use the simple expression for kinetic energy we used
above. Einstein’s expression for kinetic energy is:
Ek = (γ – 1)moc²
where γ is the Lorentz factor γ = 1/√(1 – v²/c²)
We will see shortly where this expression comes from, but it is not too difficult (by using the binomial theorem
to expand the denominator of γ) to show that this rather strange expression does indeed reduce to Ek = ½mv² if v
<< c. In a rough way we can see that if v is zero, γ is exactly 1 and so the (γ – 1) part of the expression is zero,
that is zero kinetic energy as we expect. If v is an ‘ordinary’ value, γ is very close to, but a little greater than 1.
This means (γ – 1) will have a very small positive value. However, this small value is to be multiplied by a very
large number, c², and this leads to the normal values we expect. Let’s try an example:
An example
For an electron in a TV set, accelerated by a potential of 25 kV, the kinetic energy of the electron would be 25
keV, or 4 x 10-15 J. We can see from the graph above that this electron, according to classical physics, has a
speed of close to ⅓c. Now let’s use Einstein’s expression to find the energy of an electron travelling at this
speed:
Ek = (γ – 1)moc² where γ = 1/√(1 – v²/c²) = 1/√(1 – ⅓²) = 1.061
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And so Ek = (γ – 1)moc² = 0.061moc² = 0.061 x 9.1x10-31 x 9x1016 = 5.0 x 10-15 J.
This is a little more than the 4 x 10-15 J we calculated above for the obvious reason: In order for the electron in
the TV tube to reach ⅓c it will require more than 4 x 10-15 J because, as we know from relativity, its mass will
have increased somewhat and so it will have needed more energy to bring it up to ⅓c. This calculation does,
however, show that Einstein’s expression is reasonably consistent with Newtonian calculations at speeds not too
close to c. If you repeat this type of calculation for slower speeds you will find very close agreement.
Speed and energy of relativistic electrons
We should be able to use Einstein’s equation to find a new graph of the actual speed of electrons accelerated by
a certain potential. The expression for the kinetic energy Ek = qV is still perfectly valid at relativistic speeds.
The qV is simply the energy given to a charge q going through a potential V. This energy will appear as kinetic
energy (provided it is not being lost in some other way).
The algebra is a little messy, but it is not difficult to rearrange the expression Ek = (γ – 1)moc² into one giving v
(which is hidden in the γ) in terms of Ek (where Ek = eV for the electron) and moc². For convenience we will
replace v/c simply with v (that is, we are measuring v as a fraction of c) and moc² with Eo (the ‘rest energy’ of
the electron, the so called ‘mass energy’). With this in mind, the equation above can be rearranged to become:
γ = Ek/Eo – 1. If γ is now replaced by 1/√(1 – v²), both sides squared and inverted, a little more rearranging leads
to:
v² = 1 – Eo²/(Ek + Eo)²
where Ek = eV, and Eo = moc² (both can be measured in electronvolts as the expression is a ratio). This may
appear to be a rather strange expression, but it is worth spending a moment looking at it and thinking about it:
Note that the second term is to be subtracted from 1. The second term is the ratio of the rest energy to the total
energy. At zero or slow speeds this ratio is effectively 1 and so v is very close to zero (relatively speaking!). If
the kinetic energy increases, the second term becomes smaller, and so v increases. As the kinetic energy gets
larger and larger, the second term gets smaller and smaller, and so v approaches 1, that is, the speed of light.
Putting all this into a spreadsheet and letting the program do the arithmetic we can plot the results on a chart
similar to the earlier one. Here we have two charts covering a wider range of accelerating potentials. This time,
as we might expect, the speed gets closer and closer to c but never actually reaches it.
Speed of accelerated electrons (relativistic)
100%
100%
90%
90%
80%
80%
70%
Speed (fraction of c)
Speed (fraction of c)
Speed of accelerated electrons (relativistic)
60%
50%
40%
30%
70%
60%
50%
40%
30%
20%
20%
10%
10%
0%
0%
0
100
200
300
400
500
0
Energy of electrons (keV)
1
2
3
4
5
Energy of electrons (MeV))
Comparing these graphs with the non-relativistic one above we can see that they are similar up to around 50 keV
(v is a little over 40% of c) but by 100 keV the relativistic velocity is about 55% of c whereas in the non-
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relativistic case it was 63% of c.
At the energies the electrons in the Australian synchrotron have it is hard to show the
relationship on a graph as the speeds are so close to the speed of light. The table at
right shows some speeds in the range of GeV energies. Our synchrotron will have a
maximum energy of 3 GeV giving the electrons speeds of 99.9999985% of c.
GeV
0.2
1
1.5
2
3
4
5
v/c
0.999 996 741
0.999 999 869
0.999 999 942
0.999 999 967
0.999 999 985
0.999 999 992
0.999 999 995
The mass and momentum of the accelerated electrons
We said above that the reason the electrons don’t exceed the speed of light is that their mass increases with
increased total energy. In fact, Einstein did not like to talk of an increase in the mass of a relativistic object, only
the increasing energy and momentum. Indeed modern particle physicists don’t usually speak of increasing mass
either. However, as we will see, it is a reasonable simplification, based on Einstein’s momentum and energy
equations, to imagine that the mass of a particle increases toward infinity as it nears the speed of light, and that
this is why it can never actually reach the speed of light – because the work required to increase the energy also
approaches infinity.
Einstein’s expression for relativistic momentum is:
p = γmov
Clearly for normal speeds, where γ = 1, this is equivalent to the usual expression for momentum. For high speeds
however, the expression looks like the normal expression for momentum with the mass replaced by m = γmo.
Hence the notion that as gamma increases, the effective, or ‘relativistic mass’ also increases.
In order to find the relationship between the accelerating voltage, or energy of the electron, and its relativistic
mass we could find γ in the way we did above and then simply multiply the rest mass by gamma. However, it is
probably more instructive to go back to Einstein’s basic relationship between energy and momentum. He showed
that the fundamental relativistic relationship between the two is not that which we use in classical physics (that
is, Ek = p²/2m) but something rather different:
E² = mo²c4 + p²c²
The first difference to note here is that when p = 0 (object at rest) the energy is not zero but: E = moc². This of
course is the famous ‘rest mass energy’, energy associated with the mass of the object itself and not
with its motion. The ‘E’ in the above equation is then interpreted as the ‘total energy’, not the kinetic
energy. The total energy is the sum of the rest mass energy and the kinetic energy. If the expression for
momentum, p = γmov, is now substituted into this equation it becomes:
E² = mo²c4 + γ²mo²v²c² = mo²c²(c² + γ²v²)
Perhaps a little surprisingly, the term in brackets can then be simplified by a little algebraic manipulation to
become just c²γ². When substituted back into the expression above this gives us E² = γ²mo²c4 which becomes:
E = γmoc²
If we adopt the idea that relativistic mass can be written as γmo this then becomes the most famous equation of
the twentieth century:
E = mc²
We need to be very conscious of the meaning of the letters in this equation however! The ‘E’ is the total energy
which includes both the energy due to mass as well as the kinetic energy. The ‘m’ here is the ‘relativistic mass’,
that is, gamma times the rest mass. And c² is of course the square of the velocity of light. For an object at rest,
this equation reduces to E = moc² as we saw before. This is the context in which the famous equation is normally
used, that is, the equivalence between mass and energy. Notably, in any reaction (chemical or nuclear) where
energy is released, there is a small loss of mass given by Δm = E/c² as this mass is, in effect, the mass of the
energy that has been released. Of course in normal chemical reactions this mass is totally undetectable.
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Kinetic energy
The relativistic equation for kinetic energy follows easily from what we have seen above. Given that E = γmoc²,
the kinetic energy must be the difference between the total energy of a moving object and the total energy of the
stationary object. As the rest energy is moc², the kinetic energy must therefore be Ek = γmoc² – moc², or, as we
saw earlier:
Ek = (γ – 1)moc²
As noted earlier, although this looks most unlike our normal expression for kinetic energy, it does indeed reduce
to ½mv² for non-relativistic velocities.
Relativistic mass
If we return to this equation in the form Ek = γmoc² – moc², and then replace γmoc² with simply mc² where m is
the relativistic mass (γmo) we have Ek = mc² – moc². A little manipulation of this expression leads to:
m = Ek/c² + mo
The Ek/c² part of this expression can be thought of as the ‘mass equivalent’ of the kinetic energy. For an
accelerated electron the kinetic energy is equal to eV, the electron charge times the accelerating voltage. We see
then, that the relativistic mass of the electron increases (from mo) linearly with the voltage. This is shown in the
graphs below. Notice that the relativistic mass of an electron is about twice its rest mass at 0.5 MeV. Another
way of putting that is that the kinetic energy of a 0.5 MeV electron is equal to its rest energy. Or as particle
physicists would put it, the rest mass of an electron is about 0.5 MeV (more closely 512 keV). As mass and
energy are equivalent, they tend to talk of mass in what we normally refer to as energy units.
Relativistic Mass
12
12000
10
10000
Relative mass (m o )
Relative mass (m o )
Relativistic Mass
8
6
4
2
8000
6000
4000
2000
0
0
0
1
2
3
4
5
0
Energy of electrons (MeV)
1
2
3
4
5
Energy of electrons (GeV)
At the sort of energies we find in accelerators such as the synchrotron (in the GeV range) the relativistic mass is
thousands of times the rest mass. This explains why the synchrotron has to be so large. To bend the path of these
‘heavy’ electrons into a circle very large forces are required. These forces are supplied by the ‘bending magnets’.
The size of the synchrotron
The path of the electrons in the synchrotron is not actually circular, but a series of circular segments with straight
sections between them. The circular segments are the result of the electrons moving through strong bending
magnets. The force on the electrons as they move through a magnetic field is given by F = qvB. As this force is
at right angles to their velocity, they move in a circular path. The radius of this path is found from Newton’s 2 nd
law with the expression for circular acceleration: F = mv²/R. Equating this to the magnetic force we find R =
mv/qB. However, we must remember that the mass of these electrons is thousands of times greater than their rest
mass. In fact m = γmo and so for an electron of mass γmo and charge e:
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R = γmov/eB
For example, the radius of the path for an electron travelling in a magnetic field of 1 tesla at 99.999994% of c,
when γ = 2887, is found from this equation to be 4.9 m. This is a much smaller radius than that of the
synchrotron, but of course the magnets are not continuous. In fact there are a series of bending magnets around
the Australian synchrotron each 1.7 m long and separated by straight sections in which the ‘wigglers’ and
‘undulators’ are placed. (More details to come.)
Synchrotron Radiation
The synchrotron radiation is produced when the electrons move through the three types of magnets: bending
magnets, wigglers and undulators. The bending magnets are uniform in field and produce a circular path to keep
the electrons in the ring. The wigglers and undulators have alternating vertical fields which cause the electrons to
oscillate in a horizontal plane.
A obvious question arises: As the periodic spacing of the wigglers and undulators is of the order of centimetres,
why does the radiation have a wavelengths in the range of micrometres down to nanometers? We need to turn to
relativity for a satisfactory answer to this question. In fact there are two parts to the answer:
1. The Doppler effect
Although the electrons are moving at almost the speed of light, the radiation they emit still reaches us at the
speed of light. This is Einstein’s second postulate at work – the speed of light is constant for all observers.
However, because of the electron speed the frequency and wavelength of the radiation is changed as a result of
the Doppler effect. Now we should not be surprised that at relativistic speeds the usual equations for the Doppler
effect do not work. However, the relativistic equation for a situation where radiation is emitted by a source
moving toward us at a speed v (expressed as a fraction of c) is f = fo√[(1+v)/(1-v)]. Here f is the frequency we
see and fo is the frequency as seen in the frame of reference of the source. Given that v is about 0.99999994, this
leads to f = 5773 fo. We can assume that the wavelength will decrease by this same factor. But the question is
what is the wavelength in the frame of reference of the electrons? This brings us to the other part of our answer:
2. Length contraction
Although the periodic variation in the magnets of the wigglers is of the order of 10 cm as we see it, to the
electrons moving at high speed, this distance is contracted according to the Lorentz contraction, l = l o/γ. As we
have seen, gamma is of the order of 3000 and so the electrons ‘see’ a wiggle in their path of about 0.1/3000 =
3x10-5 m. We can assume that this is about the wavelength of the radiation in the frame of reference of the
electrons. Now the Doppler effect is responsible for a decrease of around a 6000 times in the wavelength and so
putting these two factors together we will have a wavelength of around 3x10-5/6000 = 5x10-9 m which is of
course of the correct order of magnitude.
In fact the synchrotron radiation spreads over a wide range of wavelengths with the figures above being roughly
the highest energy.
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