Chapter 26 Michelson-Morley Continued Relativistic Energy and Momentum

Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! Monday, May 10, :30am - 11:20am Chs. 20, 26, and 27 You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you. Same format!

Hint: Review notes from my review lectures! Try to do some of the old homework recommended homework, and exam problems. Monday, May 10, :30am - 12:30pm everything we’ve covered You may bring one 8.5”X11” sheet (hand-written on both sides), a pencil or pen, and a scientific calculator with you.

Monday, December 15, :30am - 12:30pm everything we’ve covered Format: 5 problems, pick 4. 1 problem on each of the following topics: ElectrostaticsCircuitsMagnetism Optics Modern

So, what happens to a clock when it moves? Well, let’s use the following clock: Here, the flashbulb goes off, light bounces off the mirror, is detected at the photocell, triggering a “tick” sound and another flash of the bulb. A simple clock... photocellflashbulb mirror 1.5 m

photocellflashbulb mirror 1.5 m This clock should produce a tick every how many seconds? t = sec So what happens when you put this clock on our boxcar (in a moving reference frame)?

GM R/R Physics Rules u Flashbulb first goes off here GM R/R Physics Rules u Photocell receives light at this point, causing another flash How did the photon get there?

How far did this photon travel? ut/2 1.5m d/2 How long does it take the photon to travel this distance?

GM R/R Physics Rules u Which means that the clock on the train (the moving clock) appears to be ticking more slowly than the stationary clock!!! Which means that the clock on the train (the moving clock) appears to be ticking more slowly than the stationary clock!!!

A little geometry tells us the relationship between the two clocks: c  t/2 u  t/2 1.5m Recall, however, that according to Einstein, the speed of light is the same in all frames.

Which means that inside the boxcar, 1.5m is the distance light travels in /2 s! Call the time between flashes as measured in the boxcar  t 0, then

What does this mean??? To a stationary observer, moving clocks appear to be ticking more slowly! And the faster the clock is moving, the more slowly it appears to tick! Although we used a light clock, the type of clock used in this experiment is immaterial. It really is TIME itself that is moving at different rates for the two observers! Although we used a light clock, the type of clock used in this experiment is immaterial. It really is TIME itself that is moving at different rates for the two observers!

So what happens when we put the Michelson- Morley experiment on our moving boxcar? Remember, the Michelson-Morley experiment provides a null result, which means that interference in never observed! GM R/R Physics Rules u

GM R/R Physics Rules u So, whether we sit on the boxcar or stand alongside the tracks, light from the two paths will appear to arrive at the detector simultaneously. But to the observer on the side of the tracks, we know that the time required for light to travel the two, equal-length paths should appear to be different.

Again, using our analogy of the boat in the stream, the time difference for light travelling the two different paths would be: But this time difference is NOT observed by the stationary observer! So our equation must not be correct... Which is to say, one of the assumptions we made in writing the equation is incorrect. WHICH ONE?

Well, what if we allowed the two distances to at least appear to be different to the observer alongside the track than to the observer on the boxcar? L || is the distance parallel to the motion L | is the distance perpendicular to motion What relation between L || and L | will result in  t = 0?

To the observer alongside the tracks, the distance parallel to the motion appears shorter than the distance perpendicular to the motion!

Of course, to the observer on the boxcar these two lengths are the same: L 0 is called the “proper length” and is defined to be the length of an object in the frame in which the object is at rest. An observer who sees the object in motion will describe its length in the direction of motion as SHORTER than its proper length.

L is the apparent length. L 0 is the “proper length.”

A train passes a railroad crossing at 100 km/hr. At rest, the train is 1 km long. How long does the train appear to a car waiting at the signal? L = m

m 0 rest mass L 0 proper length  t 0 proper time m 0 rest mass L 0 proper length  t 0 proper time In the rest frame of the object: In the rest frame of the object: In the lab (where the observer is): In the lab (where the observer is): m mass L length  t time m mass L length  t time

This one’s pretty easy. Our Newtonian definition told us that p = m v The relativistic formula has exactly the same form, but recall that the mass changes with velocity:

So you will often see the relativistic momentum defined as: As was the case in Newtonian mechanics, the relativistic momentum must also be conserved in all collisions. Furthermore, it reduces to the classical expression when the velocities are small compared to the speed of light.

I’m not going to try to derive this one (although you can achieve this results with some simple mathematical tricks)... Naturally, this expression reduces to our familiar Newtonian expression when v << c.

Here, m 0 c 2 is known as the “rest energy” (E 0 ) of the particle. Which raises the startling possibility that a particle at rest stores an enormous amount of energy. What is the rest energy of a 0.5 kg baseball? Enough energy for 100-W bulb for 10 million years!

The total energy of a particle consists of two parts... Total Energy = KE + Rest Energy

Remember that m is NOT the rest mass in this expression, and m changes with velocity. This is the familiar expression you’ve all probably seen related to Einstein (even if you had no idea what it meant)

It is useful to derive a relationship between relativistic momentum and energy.

Rewritten into its more usual form... (Total Energy) 2 = (momentum) 2 c 2 +(rest energy) 2