**Table of Contents**

*.......The Elegant Universe*

**THE ELEGANT UNIVERSE,****Brian Greene,**1999, 2003

```(annotated and with added

**bold highlights by Epsilon=One**)

**Chapter 2: Notes**

**1**. The presence of massive bodies like the earth does complicate matters by introducing gravitational forces. Since we are now focusing on motion in the horizontal direction—not the vertical direction—we can and will ignore the earth's presence. In the next chapter we will undertake a thorough discussion of gravity.

*Return to Text***2**. More precisely, the speed of light through the

*vacuum of empty space*is 670 million miles per hour. When light travels through a substance such as air or glass its speed is decreased in roughly the same way that a rock dropped from a cliff is dragged to a slower speed when it enters a body of water. This slowing of light relative to its speed through a vacuum is of no consequence for our discussion of relativity and is justifiably ignored throughout the text.

*Return to Text***3**. For the mathematically inclined reader, we note that these observations can be turned into quantitative statements. For instance, if the moving light clock has speed

*v*and it takes

*t*seconds for its photon to complete one round-trip journey (as measured by our stationary light clock), then the light clock will have traveled a distance

*vt*when its photon has returned to the lower mirror. We can now use the Pythagorean theorem to calculate that the length of each of the diagonal paths in Figure 2.3 is V(vt/2)² + h², where h is the distance between the two mirrors of a light clock (taken to be six inches in the text). The two diagonal paths, taken together, therefore have length 2V(vt/2)² + h2 . Since the speed of light has a constant value, conventionally called c, it takes light 2V'(vt/2)² + h²/c seconds to complete the double diagonal journey. And so, we have the equality t = 2V(vt/2)² + h²/c, which can be solved for t, yielding t = 2h/\/c² – v². To avoid confusion, let's write this as t

**moving**= 2h/Vc² – v², where the subscript indicates that this is the time we measure for one tick to occur on the moving clock. On the other hand, the time for one tick on our stationary clock is t

**stationary**= 2h/c and as a little algebra reveals, t

**moving**= t

**stationary**/V1 – v²/c², directly showing that one tick on the moving clock takes longer than one tick on the stationary clock. This means that between chosen events, fewer total ticks will take place on the moving clock than on the stationary, ensuring that less time has elapsed for the observer in motion.

*Return to Text***4**. In case you would be more convinced by an experiment carried out in a less esoteric setting than a particle accelerator, consider the following. During October 1971, J. C. Hafele, then of Washington University in St. Louis, and Richard Keating of the United States Naval Observatory flew cesium-beam atomic clocks on commercial airliners for some 40 hours. After taking into account a number of subtle features having to do with gravitational effects (to be discussed in the next chapter), special relativity claims that the total elapsed time on the moving atomic clocks should be less than the elapsed time on stationary earthbound counterparts by a few hundred billionths of a second. This is just what Hafele and Keating found: Time

*really does slow down*for a clock in motion.

*Return to Text***5**. Although Figure 2.4 correctly illustrates the shrinking of an object along its direction of motion, the image does not illustrate what we would actually see if an object were somehow to blaze by at nearly light speed (assuming our eyesight or photographic equipment were sharp enough to see anything at all!). To sec something, our eyes—or our camera—must receive light that has reflected off the object's surface. But since the reflected light travels to us from various locations on the object, the light we see at any moment traveled to us along paths of different lengths. This results in a kind of relativistic visual illusion in which the object will appear both foreshortened and rotated.

*Return to Text***6**. For the mathematically inclined reader, we note that from the spacetime position 4-vector x = (ct, x

**1**, x

**2**, x

**3**) (ct, x) we can produce the velocity 4-vector u = dx/d

**tau**, where

**tau**is the proper time defined by d

**tau**²= dt² – c-²(dx1²+ dx2² + dx3²). Then, the "speed through spacetime" is the magnitude of the 4-vector u, V((c²dt²– dx²)/(dt²– c-²dx²)), which is identically the speed of light,

*c.*Now, we can rearrange the equation c²(dt/d

**tau**)² – (dx/d

**tau**)² = c², to be c²(d

**tau**/dt)² + (dx/dt)² = c². This shows that an increase in an object's speed through space, V(dx/dt)² must be accompanied by a decrease in dt/dt, the latter being the object's speed through time (the rate at which time elapses on its own clock, d

**tau**, as compared with that on our stationary clock, dt).

*Return to Text*