Printer-friendly version

09 May 2003

More on Lengths and Times

by George Hrabovsky, President of MAST

News from MAST

I will be giving details of the school program we have just started this week. It is a very exciting program. We are offering Associate and Bachelor degrees in Archaeology, Astronomy, Atmospheric Science, Botany, Cell Biology, Chemistry, Computer Science, Ecology, Electronics, Engineering Science, Geology, Hydrology, Mathematics, Microbiology, Molecular Biology, Oceanography, Physics, and Zoology. We are emphasizing guided self-study, where the student is given tasks to complete by the next meeting with their instructor (either in person or on the Web). The student gets credit for doing things, not for attending classes (an obsolete technology that was necessary before textbooks were affordable). There is no time limit in getting a degree, and we are charging only $15 per meeting. For more details, please contact me. We will also be offering the same program for distance learning over the web (using Net Meeting or Yahoo Chat). If you wish to explore this method you will need to acquire the free MathReader program from Wolfram Research.

Lorentz Contraction

Recall from last time that we established the invariance of the spacetime interval

FormBox[RowBox[{Δ Overscript[s, ~]^2, =, RowBox[{- Δ Overscript[t, ~]^2 + Δ Ove ... 6; t^2,   , +,   , Δ x^2}], , =, , Δ s^2 .}]}]}], TraditionalForm]

What does this, along with the Lorentz transformations, tell us about what one observer will experience while watching another? Let us consider the length of an object in a moving frame. We begin by stating that in the Overscript[÷~, ~] frame, Overscript[t, ~] = 0 and Overscript[x, ~] = l. By the Lorentz transformation we have,

l = γ (x - v t)

    =    x γ

x =    l/γ

This tells us that, in the ÷~ frame, the moving object's length seems shortened. This is called the Fitzgerald-Lorentz contraction, or just the Lorentz contraction. For example, let us say that v = 0.5, half the speed of light, then γ = 1/(1 - 0.5^2 )^(1/2) = 1.1547, then x = 0.8660 l , in other words the length seems around 13% shorter than it really is.
    We can look at a spacetime diagram and see this effect graphically.

[Graphics:Images/index_12.gif]

Let's say that we have a rod of length l on the Overscript[x, ~]-axis.

[Graphics:Images/index_14.gif]

Here this length is along the Overscript[x, ~]-axis from the origin to the blue arrow. The blue arrow is the Overscript[t, ~]-coordinate for the head of the rod. Note the distance between the Overscript[t, ~]-axis and the blue arrow is shorter on the x-axis than on the Overscript[x, ~]-axis. The object moving in the Overscript[÷~, ~] frame has its length contracted in the ÷~ frame.

Time Dilation

We can apply a similar argument when thinking about the relative passage of time. Let us consider the spacetime diagram again.

[Graphics:Images/index_22.gif]

This time we observe a time interval in the Overscript[÷~, ~] frame.

[Graphics:Images/index_24.gif]

Here we see that Δ Overscript[t, ~] seems physically longer along the Overscript[t, ~]-axis than it is along the t-axis. This implies that in the rest frame the time interval is shorter than in the moving frame, thus the clock in the moving frame seems to run slower than it does in the rest frame.

We can see this symbolically by considering the Lorentz transformation for the time interval, where the interval is defined

Δ Overscript[t, ~] = Overscript[t, ~] _ end - Overscript[t, ~] _ begin .

We then have

Overscript[t, ~] _ end = γ (t _ end - v x),

and

Overscript[t, ~] _ begin = γ (t _ begin - v x) .

We the rewrite the interval,

Δ Overscript[t, ~] = Overscript[t, ~] _ end - Overscript[t, ~] _ begin

         = γ (t _ end - v x) - γ (t _ begin - v x)

         = γ [(t _ end - v x) - (t _ begin - v x)]

         = γ (t _ end - t _ begin)

         = γ Δ t .

This gives us the same result that we had from the spacetime diagram, showing that the time in the moving frame passes more slowly than time in the rest frame.

Answer to the Theory Challenge from Last Time

The task was to solve the differential equation,

FormBox[RowBox[{m d^2 x/d t^2 + b d x/d t + k x, , =, , RowBox[{RowBox[{Re[A _ 0 e^(i ω ... p;           ]}], (1)}]}], TraditionalForm]

From previous experience with oscillators we know that this kind of equation will have a solution of the form,

x = A _ 0 cos(ω t + θ),

or, in complex form

FormBox[RowBox[{x, , =, , RowBox[{x _ 0, RowBox[{e^(i ω t), ., Cell[   & ... p;           ]}], (2)}]}], TraditionalForm]

This gives us the velocity

Overscript[x, ·] = d/d t (x)

     = d/d t (x _ 0 e^(i ω t))

     = x _ 0 d/d t e^(i ω t)

     = i ω x _ 0 e^(i ω t) .

We also have

Overscript[x, · ·] = d^2/d t^2 (x)

     = d/d t (i ω x _ 0 e^(i ω t))

     = i ω x _ 0 d/d t e^(i ω t)

     = (i ω) i ω x _ 0 d/d t e^(i ω t)

     = i^2    ω ^2 x _ 0 d/d t e^(i ω t)

since i^2 = -1

Overscript[x, · ·] = - ω ^2 A _ 0    e^(i ω t) .

(1) then becomes

-m    ω ^2 x _ 0    e^(i ω t) + b i ω x _ 0 e^(i ω t) + k x _ 0 e^(i ω t) = Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]

x _ 0 e^(i ω t) (-m    ω ^2    + b i ω    + k ) = Re[A _ 0 e^(i ω t) e^(i    θ _ 0)] .

x _ 0 e^(i ω t) = Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]/(-m    ω ^2    + b i ω    + k ) .

We can then multiply the right hand side by m/m to remove that factor

x _ 0 e^(i ω t) = (Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]/m)/(-ω ^2    + b/m i ω    + k/m ) .

Recall that

γ = b/(2 m)

so

2 γ = b/m,

therefore

x _ 0 e^(i ω t) = (Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]/m)/(-ω ^2    + 2 γ i ω    + k/m ) .

Recall also that

ω _ 0 = k/m^(1/2),

so

x _ 0 e^(i ω t) = (Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]/m)/(-ω ^2    + 2 γ i ω    + ω _ 0^2 ) .

We can rearrange terms to make it look nicer

x _ 0 e^(i ω t) = (Re[A _ 0 e^(i ω t) e^(i    θ _ 0)]/m)/(ω _ 0^2 - ω ^2    + 2 γ i ω ) .

We can also cancel e^(i ω t),

x _ 0 = (Re[A _ 0    e^(i    θ _ 0)]/m)/(ω _ 0^2 - ω ^2    + 2 γ i ω ) .

We can then solve the differential equation by applying this result to (2),

x = x _ 0 e^(i ω t) = (Re[A _ 0    e^(i    θ _ 0)] e^(i ω t)/m)/(ω _ 0^2 - ω ^2    + 2 γ i ω ) .

Theory Challenge

Can we simplify the solution to the differential equation?

Books That I Like

Bernard F. Schutz (1990), A First Course in General Relativity, Cambridge University Press. This has a nice introduction to spacetime diagrams and derives how to view other coordinate systems.

Ray D'Inverno (1992), Introducing Einstein's Relativity, Oxford University Press. This is my favorite introduction to relativity.

Online Resources
Lorentz Contraction:

For a brief description go to:

http://scienceworld.wolfram.com/physics/LorentzContraction.html

For a slightly different approach,

http://musr.physics.ubc.ca/~jess/p200/str/str11.html

Time Dilation:

This site also has information about other topics in relativity and astrophysics.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

Here is a graphically oriented tutorial on relativity:

http://www.astro.ucla.edu/~wright/relatvty.htm

Forced Oscillators:

Here is a fairly elementary site that presents the topic in general terms.

http://www.pinkmonkey.com/studyguides/subjects/physics/chap10/p1010601.asp

Here is a more challenging site that gets similar results to those above.

http://colos1.fri.uni-lj.si/~colos/COLOS/TUTORIALS/JAVA/JAVAXYZET/RESONANCE/HTML/resonance_6.html

Converted by Mathematica  (May 9, 2003)