Chapter 1
Euclidean relativity.
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Euclidean relativity is a view of time and space that matches
what is intuitive to all of us. It is easy to understand.
It is perhaps too simple but it is important to review
it to prepare us for the following special relativity discussion.
Yes it is very simplistic.
The units chosen are feet (ft) for distance. and
nanoseconds (ns) for time. The speed of light is
assumed to be 1 foot per nanosecond. These units
are chosen for simplicity. In these units the speed of
light is assumed to be one ft per ns. The error
due to this assumption is less than 2%.
We will start by considering the situation of a high speed train
passing a platform at a train station. Thus we
have two frames of reference. The platform frame where
distance and time are measured by X and T and the moving train frame
where distance and time are represented by X' and T'.
The velocity of the train is represented by V. In
this case V = 0.6 times the speed of light or 0.6 ft per
ns.
Figure 1 - 1
Above we see the train passing the platform in figure
1-1. The train consists of a
engine 4 passenger cars and a caboose. Each car is
100 feet long. The floors of the cars are inscribed with
feet marks starting with 0 at the center of the caboose up to 500 at
the center of the engine. The units in the train frame are X'
and T'
The platform platform is 600 feet long with markers every 100
feet. X is used to denote the distance measured on
the platform frame. and T is used to represent time in
the platform frame.
The train is shown at the instant when the zero on the caboose is
opposite the zero on the platform.
The picture above in figure 1 -2 shows what happens
after 500 ns have elapsed. The train is
shown at a new position relative to the platform. As we
can see the caboose of the train has progressed to the 300 feet
platform mark. This is predicted from D = V * T or
0.6 ft/ns * 500 ns = 300 ft. The
prediction that D = V x T is true both in
Euclidean Relativity and Special Relativity.
As we can see the 100 foot marker on the train is opposite the
400 foot platform mark and the 200 foot train mark is opposite the
500 foot platform mark.
Rather than drawing this last pictures we could simply compute
the answers we are looking for. Recall we already defined X
and T as distance and time in the platform frame and X' and T' as
distance and time in the train frame. The
equations we use are:
We should note that the first (equation 1-1) allows us
to find X' as a function of two variables, X
and T. The second (equation 1 - 2
) seems trivial at this point but it is
important.
So lets go ahead and compute X' and T' for the three locations
chosen.
Problem 1 - 1
Using the above equations to find X' and T' when X
= 300 ft and T = 500ns.
X' = X - VT = 300 - (500 * 0.6) = 0 or X' =
0.
T' = T or T' = 500 ns
Problem 1 - 2
Find X' and T' when X = 400 ft. and T = 500ns.
X' = X - V*T 400 -( 0.6 * 500 ) = 100
T' = T = 500 ns
Problem 1 - 3
Find X' and T' when X = 500 ft. and T = 500ns.
X' = X - VT = 500 - ( 0.6 * 500) = 200
T' = T = 500ns
All of the above seem simple but there is a problem.
The problem is that the answers given are wrong. In
order to get true answers we need to apply the Lorenz equations in
place of the equations that we used.
Chapter 2
The Lorenz equations
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The Special Theory of Relativity is used to solve problems like
those in the preceding chapter. It does this by replacing the
equations in Chapter 1 with a new set of equations
called the Lorenz equations. We will leave out the
derivation of the Lorenz equations as this can be found in other
sources.
The Lorenz equations that replace equation 1 - 1
and equation 1 -2 are:
Equation 1-1 is replaced by equation 2-1 .
Comparing these two equations we see that equation 1 -1 is
similar to equation 2-1 with the difference being the factor
G..
The second of these equations is where things get really
strange. If we compare equation 1-2 to
2-2 Not only does equation 2-2 have the factor G in it
but it has a additional term X*V/C^2.
G is a function of the velocity. If V = 0.6 then G
= 1/ square root ( 1 - 0.36) = 1/ square root (0.64)
= 1.25.
And now we will look at the clocks on the platform
positions as well as the train cars.
We will set all the clocks on the platform to 0 ns. Then
from the Lorenz equations we will compute the X' dimensions of the
train cars as well as their time T'.
X' = G ( X - V*T
)
equation 2 - 1
T' = G * ( T - X * V / C^2
)
equation 2 - 2
G = 1.25
Problem 2 - 1
X = 0 T = 0
X' = 1.25 ( 0 - 0.6 * 0 ) = 0
T' = 1.25 ( 0 - .6 * 0 ) = 0
Problem 2 - 2
X = 100 T = 0
X' = 1.25 ( 100 - 0.6
* 0 ) = 125
T' = 1.25 ( 0 - .6 * 100 ) = - 75
Problem 2 - 3
X = 200 T = 0
X' = 1.25 ( 200 - 0.6
* 0 ) = 250
T' = 1.25 ( 0 - .6 * 300 ) = - 150
Problem 2 - 4
X = 300 T = 0
X' = 1.25 ( 300 - 0.6
* 0 ) = 375
T' = 1.25 ( 0 - .6 * 300 ) = - 225
Problem 2 - 5
X = 400 T = 0
X' = 1.25 ( 400 - 0.6 * 0 ) =
500
T' = 1.25 ( 0 - .6 * 400 ) = - 300
Problem 2 - 6
X = 500 T = 0
X' = 1.25 ( 500 - 0.6 * 0 ) =
625
T' = 1.25 ( 0 - .6 * 500 ) = - 375
This is what we get.
This is where Special Relativity comes into play.
In the platform frame both distance and time behave just as we
expect with no surprises but when we look across to the frame of a
moving train passing the platform we see a entirely different
scenario. We see distances changed so they no
longer match those on the platform and when we look at the time on
the moving train we see times that do not agree with the platform
time. Also they do not even agree with each
other. Each car has a different time when observed
from different points on the platform.
From here lets wait 500 ns and then redraw the platform and
train. After 500 ns the caboose will have traveled
to the 300 foot platform mark. Now we will compute the
distances and times seen from the platform as the train is passing
the platform.
Using the Lorenz equations to compute
T' and X'
X' = G( X - V*T/C^2)
(2-1 )
T' = G ( T - X*V )
( 2-2)
We have determined that G = 1.25 '
V = 0.6
Problem 2 - 7
X = 300 T =
500
X' = 1.25 ( 300 - 0.6 * 500) = 0
T' = 1.25 ( 500 - 300 * 0.6) = 400
Problem 2 - 8
X = 400 T =
500
X' = 1.25 ( 400 - 0.6 * 500) =
125
T' = 1.25 * ( 500 - 400 * 0.6 ) = 325
Problem 2 - 9
X = 500 T =
500
X' = 1.25 ( 500 - 500 * 0.6 )
= 200
T' = 1.25 ( 500 - 500 * 0.6 ) =
200
The results are shown on figure 2 - 2
Once again the distances and times do not match those on the train
platform.
One very interesting fact is that if C is set equal to
infinity then the Lorenz equations all become identical to the
euclidean relativity equations.
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CHAPTER 3 The concept of the interval and
proper time
The next question is, is there a pattern to the
numbers above. It turns out there
is. It is called the interval. The
platform interval is defined as a function of two events the
zero event or X = 0 T = 0 and the end event
or X' and T'. Similarity the zero event of
the train is X' = 0 and T' = 0 and the end event
is a function of X' and T'. The
following equation gives the interval given T and X.
So lets look at the positions of the caboose in figure 2 - 1 and
figure 2 - 2. We get:
Start on the platform from Figure 2 -1
X = 0 T = 0
End from figure 2 - 2 we get
X = 300ft T = 500ns or I =
400.
Now looking at the train interval we get
start X' = 0
T' = 0
end from figure 2 is
X' = 0
T' = 400
and computing the train interval we get I = 400.
Thus the singular event, the caboose passing the 300
foot mark on the platform , has the same interval whether
measured on the platform frame or the train frame.
This is very important
Again lets look at the car next to the caboose and measure the
interval between zero and it's passing the 400 foot platform
mark.
Start X = 0
T = 0
end X = 400
T = 500
so the interval is 300.
And looking at the same car and measuring the interval between the
zero mark in the train frame and the X' and T' in
the train frame we get
Start X' = 0
T' = 0
end X' = 125 ft
T' = 325 ns
which results in I = 300.
All of the above reminds me of a situation of a X Y grid on the
floor and a dozen 12 inch long straws distributed randomly on the
floor. Each straw has a delta x and a delta y
between it's ends but if we compute the square root of delta x
squared added to the square root of delta y squared we always
arrive at 12 inches.
Proper time.
The proper time of an event is used when the frame looked at
has a delta X of zero. Thus in the problem
above the moving train with X' equal to zero at the start of the
event and the X' = 0 at the end of the interval then 400 ns is the
proper time of this event. If we switch to
the platform frame X distance is 300 feet and the time found
is 500 ns. This is not proper time.
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Chapter 4
The Haskill Method.
Richard Haskill has a PDF available that describes a graphic
way of solving Lorenz Equations. It is well worth
looking at and may be found at,
http://www.cse.secs.oakland.edu/haskell/Special%20Relativity%20and%20Maxwells%20Equations.pdf
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Chapter 5
One of the interesting results of special relativity is the fact
that given a current carrying conductor by applying the Lorenz
equations we can deduce the existence of the magnetic
field. That is the existence of the magnetic
field is entirely due to special relativity.
For more on this subject visit Chapter 13 section 6 of the Feynman
lectures in Physics.
http://www.feynmanlectures.caltech.edu/II_13.html
This approach was also taken by Purcell It is available here
courtesy of Dan Schroeder.
http://physics.weber.edu/schroeder/mrr/MRRhandout.pdf
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