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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