1. Lecture 4: PHYS 344
Homework #1 Due in class Wednesday, Sept 9th Read Chapters 1 and 2 of Krane, Modern Physics Problems: Chapter 2: 3, 5, 7, 8, 10, 14, 16, 17, 19, 20

2. The Galilean Transformations
Time (t) for all observers is a Fundamental invariant, i.e., it’s the same for all inertial observers.

3. Einstein’s Two Postulates
With the belief that Maxwell’s equations must be valid in all inertial frames, Einstein proposed the following postulates:
The principle of relativity: All the laws of physics (not just the laws of motion) are the same in all inertial systems. There is no way to detect absolute motion, and no preferred inertial system exists.
The constancy of the speed of light: Observers in all inertial systems measure the same value for the speed of light in a vacuum.

4. The constancy of the speed of light
Consider the fixed system K and the moving system K’.
At t = 0, the origins and axes of both systems are coincident with system K’ moving to the right along the x axis.
A flashbulb goes off at both origins when t = 0.
According to postulate 2, the speed of light will be c in both systems and the wavefronts observed in both systems must be spherical. K’ K

5. The constancy of the speed of light is not compatible with Galilean transformations.
Spherical wavefronts in K:
Spherical wavefronts in K’:
Note that this cannot occur in Galilean transformations:
There are a couple of extra terms (-2xvt + v2t2) in the primed frame.

6. Finding the correct transformation
What transformation will preserve spherical wave-fronts in both frames?
Try x’ = g (x – vt) so that x = g’ (x’ + vt’) , where g could be anything.
By Einstein’s first postulate: g’ = g
The wave-front along the x’- and x-axes must satisfy: x’ = ct’ and x = ct
Thus: ct’ = g (ct – vt) or t’ = g t (1– v/c)
and: ct = g’ (ct’ + vt’) or t = g’t’(1 + v/c)
Substituting for t in t’ = g t (1– v/c) :
which yields:

7. Finding the transformation for t “prime”
Now substitute x’ = g ( x – v t ) into x = g ( x’ + v t’ ):
x = g [g (x – v t) + v t’ ]
Solving for t’ we obtain: x - g2 (x – v t) = g v t’
or: t’ = x / g v - g ( x / v – t )
or: t’ = g t + x / g v - g x / v
or: t’ = g t + (g x / v) ( 1 / g2 - 1 )
or:
1 / g2 - 1 = -v2/c2

8. Lorentz Transformation Equations

9. Lorentz Transformation Equations
A more symmetrical form:

10. Properties of g Recall that b = v / c < 1 for all observers.
g equals 1 only when v = 0.
In general:
Graph of g vs. b:
(note v < c)

11. The Binomial Approximation
(1+x)a≈1+ax iff x<<1 We want to approximate: g=(1-v2/c2)-1/2 So we have: g≈1+1/2v2/c2 This is just one example of the Taylor expansion of a function f(x)

12. The complete Lorentz Transformation
If v << c, i.e., β ≈ 0 and g ≈ 1, yielding the familiar Galilean transformation.
Space and time are now linked, and the frame velocity cannot exceed c.

13. Let’s be precise in our definitions!
Events and Spacetime Coordinates: An event is any physical occurrence that happens at a definite place in space and at a definite instant in time. The event’s spacetime coordinates are the 4 numbers that specify where and when an event took place: 3 spatial coordinates (x, y, z) and 1 for time. A reference frame’s fundamental task is to provide a means of measuring the spacetime coordinates of events.

14. Timing events occurring in different places can be tricky
Timing events occurring in different places can be tricky. Depending on how they’re measured, different events will be perceived in different orders by different observers.
Simultaneity
Frank
Fil
Fred -L L
Due to the finite speed of light, the order in which these two events will be seen will depend on the observer’s position. The time intervals will be: Fred: -2L/c; Frank: 0; Fil: +2L/c
But this obvious position-related simultaneity problem disappears if Fred and Fil have synchronized watches.

15. Synchronized clocks in a frame
It’s possible to synchronize clocks throughout space in each frame.
This will prevent the position-dependent simultaneity problem in the previous slide.
But there will still be simultaneity problems due to velocity.

16. Simultaneity
So all stationary observers in the explosions’ frame measure these events as simultaneous.
What about moving ones?
Compute the interval as seen by Mary using the Lorentz time transformation. K’
Mary -L L
Mary experiences the explosion in front of her before the one behind her. And note that Dt’ is independent of Mary’s position!

17. Time Dilation and Length Contraction
More very interesting consequences of the Lorentz Transformation:
Time Dilation:
Clocks in K’ run slowly with respect to stationary clocks in K.
Length Contraction:
Lengths in K’ contract with respect to the same lengths in stationary K.

18. We must think about how we measure space and time.
In order to measure an object’s length in space, we must measure its leftmost and rightmost points at the same time if it’s not at rest.
If it’s not at rest, we must ask someone else to stop by and be there to help out.
In order to measure an event’s duration in time, the start and stop measurements can occur at different positions, as long as the clocks are synchronized.
If the positions are different, we must ask someone else to stop by and be there to help out.
Ruler:
Clock:

19. Proper Time
To measure a duration, it’s best to use what’s called Proper Time.
The Proper Time, T0, is the time between two events (here two explosions) occurring at the same position (i.e., at rest) in a system as measured by a clock at that position.
Same location
Proper time measurements are in some sense the most fundamental measurements of a duration. But observers in moving systems, where the explosions’ positions differ, will also make such measurements.
What will they measure?

20. Time Dilation and Proper Time
Frank’s clock is stationary in K where two explosions occur.
Mary, in moving K’, is there for the first, but not the second. Fortunately, Melinda, also in K’, is there for the second. K’
Mary
Melinda
Mary and Melinda are doing the best measurement that can be done. Each is at the right place at the right time.
If Mary and Melinda are careful to time and compare their measurements, what duration will they observe? K
Frank

21. Time Dilation
Mary and Melinda measure the times for the two explosions in system K’ as and By the Lorentz transformation:
This is the time interval as measured in the frame K’. This is not proper time due to the motion of K’:
Frank, on the other hand, records x2 – x1 = 0 in K with a (proper) time: T0 = t2 – t1, so we have:

22. Time Dilation
1)  T ’ > T0: the time measured between two events at different positions is greater than the time between the same events at one position: this is time dilation.
2) The events do not occur at the same space and time coordinates in the two systems.
3) System K requires 1 clock and K’ requires 2 clocks for the measurement.
4) Because the Lorentz transformation is symmetrical, time dilation is reciprocal: observers in K see time travel faster than for those in K’. And vice versa!