# Tag Archives: Jim Parsons

## The Big Bang Theory of the Doppler Effect

Season 01, Episode 06: “The Middle-Earth Paradigm”

Sheldon Cooper dresses as the Doppler Effect on the Season 1, Episode 6 “The Middle-Earth Paradigm” of the Big Bang Theory

Yes. It’s the apparent change in the frequency of a wave caused by relative motion between the source of the wave and the observer.

-Sheldon Cooper

In the “Middle-Earth Paradigm” episode, Sheldon Cooper dresses as the “Doppler Effect” for Penny’s Halloween party. The Doppler Effect (or Doppler Shift) describes the change in pitch or frequency that results as a source of sound moves relative to an observer; moving relative can mean either the source is moving while the observer is stationary or vice versa. It is commonly heard when a siren approaches and recedes from an observer. As the siren approaches, the pitch sounds higher and lowers as it moves away. This effect was first proposed by Austrian physicist Christian Doppler in 1842 to explain the color of binary stars.

In 1845, Christophorus Henricus Diedericus (C. H. D.) Buys-Ballot, a Dutch chemist and meteorologist conducted the famous experiment to prove this effect. He assembled a group of horn players on an open cart attached to a locomotive. Ballot then instructed the engineer to rush past him as fast as he could while the musicians played and held a constant note. As the train approached and receded, Ballot noted that the pitch changed as he stood and listened on the stationary platform.

# Physics of the Doppler Effect

A stationary sound source has sound waves radiating outward and can be viewed as concentric circles.

As a stationary sound source produces sound waves, its wave-fronts propagate away from the source at a constant speed, the speed of sound. This can be seen as concentric circles moving away from the center. All observers will hear the same frequency, the frequency of the source of the sound.

When either the source or the observer moves relative to each other, the frequency of the sound that the source emits does not change but rather the observer hears a change in pitch. We can think of the following way. If a pitcher throws balls to someone across a field at a constant rate of one ball a second, the person will catch those balls at the same rate (one ball a second). Now if the pitcher runs towards the catcher, the catcher will catch the balls faster than one ball per second. This happens because as the catcher moves forward, he closes in the distance between himself and the catcher. When the pitcher tosses the next ball it has to travel a shorter distance and thus travels a shorter time. The opposite is true if the pitcher was to move away from the catcher.

If instead of the pitcher moving towards the catcher, the pitcher stayed stationary and the catcher ran forward. As the catcher runs forward, he closes in the distance between him and the pitcher so the time it takes from the ball to leave the pitcher’s hand to the catcher’s mitt is decreased. In this case, it also means that the catcher will catch the balls at a faster rate than the pitcher tosses them.

## Sub Sonic Speeds

The source radiates sound waves outward. As it moves, the center of each new wavefront is slightly displaced to the right and the wavefronts bunch up on the right side (front) and spread out further apart on the left side (behind) the source.

We can apply the same idea of the pitcher and catcher to a moving source of sound and an observer. As the source moves, it emits sounds waves which spread out radially around the source. As it moves forward, the wave-fronts in front of the source bunch up and the observer hears an increase in pitch. Behind the source, the wave-fronts spread apart and the observer standing behind hears a decrease in pitch.

### The Doppler Equation

When the speeds of source and the receiver relative to the medium (air) are lower than the velocity of sound in the medium, i.e. moves at sub-sonic speeds, we can define a relationship between the observed frequency, $$f$$, and the frequency emitted by the source, $$f_0$$.
$f = f_{0}\left(\frac{v + v_{o}}{v + v_{s}}\right)$
where $$v$$ is the speed of sound, $$v_{o}$$ is the velocity of the observer (this is positive if the observer is moving towards the source of sound) and $$v_{s}$$ is the velocity of the source (this is positive if the source is moving away from the observer).

#### Source Moving, Observer Stationary

We can now use the above equation to determine how the pitch changes as the source of sound moves towards the observer. i.e. $$v_{o} = 0$$.
$f = f_{0}\left(\frac{v}{v – v_{s}}\right)$
$$v_{s}$$ is negative because it is moving towards the observer and $$v – v_{s} < v$$. This makes $$v/(v - v_{s})$$ larger than 1 which means the pitch increases.

#### Source Stationary, Observer Moving

Now if the source of sound is still and the observer moves towards the sound, we get:
$f = f_{0}\left( \frac{v + v_{o}}{v} \right)$
$$v_{o}$$ is positive as it moves towards the source. The numerator is larger than the denominator which means that $$v + v_{o}/v$$ is greater than 1. The pitch increases.

## Speed of Sound

As the source of sound moves at the speed of sound the wave fronts in front of the source all bunch up at the same point.

As the source of sound moves at the speed of sound, the wave fronts in front become bunched up at the same point. The observer in front won’t hear anything until the source arrives. When the source arrives, the pressure front will be very intense and won’t be heard as a change in pitch but as a large “thump”.

The observer behind will hear a lower pitch as the source passes by.
$f = f_{0}\left( \frac{v – 0}{v + v} \right) = 0.5 f_{0}$

Early jet pilots flying at the speed of sound (Mach 1) reported a noticeable “wall” or “barrier” had to be penetrated before achieving supersonic speeds. This “wall” is due to the intense pressure front, and flying within this pressure front produced a very turbulent and bouncy ride. Chuck Yeager was the first person to break the sound barrier when he flew faster than the speed of sound in the Bell X-1 rocket-powered aircraft on October 14, 1947.

Bell X-1 rocket plane of the United States Air Force (NASA photography)

As the science of super-sonic flight became better understood, engineers made a number changes to aircraft design that led the the disappearance of the “sound barrier”. Aircraft wings were swept back and engine performance increased. By the 1950s combat aircraft could routinely break the sound barrier.

## Super-Sonic

As the source moves faster than the speed of sound, i.e. faster than the sound waves it creates, it leads its own advancing wavefront. The sound source will pass by the stationary observer before the observer hears the sound.

As the sound source breaks and moves past the “sound barrier”, the source now moves faster than the sound waves it creates and leads the advancing wavefront. The source will pass the observer before the observer hears the sound it creates. As the source moves forward, it creates a Mach cone. The intense preseure front on the Mach cone creates a shock wave known as a “sonic boom”.

### Twice the Speed of Sound

Something interesting happens when the source moves towards the observer at twice the speed of sound — the tone becomes time reversed. If music was being played, the observer will hear the piece with the correct tone but played backwards. This was first predicted by Lord Rayleigh in 1896 .

We can see this by using the Doppler Equation.
$f = f_{0}\left(\frac{v}{v-2v}\right)$
This reduces to
$f=-f_{0}$
which is negative because the sound is time reversed or is heard backwards.

# Applications

U.S. Army soldier uses a radar speed gun to catch speeding violators at Tallil Air Base, Iraq.

The Doppler Effect is used in radar guns to measure the speed of motorists. A radar beam is fired at a moving target as it approaches or recedes from the radar source. The moving target then reflects the Doppler-shifted radar wave back to the detector and the frequency shift measured and the motorist’s speed calculated.

We can combine both cases of the Doppler equation to give us the relationship between the reflected frequency ($$f_{r}$$) and the source frequency ($$f$$):
$f_{r} = f \left(\frac{c+v}{c-v}\right)$
where $$c$$ is the speed of light and $$v$$ is the speed of the moving vehicle. The difference between the reflected frequency and the source frequency is too small to be measured accurately so the radar gun uses a special trick that is familiar to musicians – interference beats.

To tune a piano, the pitch can be adjusted by changing the tension on the strings. By using a tuning instrument (such as a tuning fork) which can produce a sustained tone over time, a beat frequency can be heard when placed next to the vibrating piano wire. The beat frequency is an interference between two sounds with slightly different frequencies and can be herd as a periodic change in volume over time. This frequency tells us how far off the piano strings are compared to the reference (tuning fork).

To detect this change in a radar gun does something similar. The returning wave is “mixed” with the transmitted signal to create a beat note. This beat signal or “heterodyne” is then measured and the speed of the vehicle calculated. The change in frequency or the difference between $$f_{r}$$ and $$f$$ or $$\Delta f$$ is
$f_{r} – f = f\frac{2v}{c-v}$
as the difference between the speed of light, $$c$$, and the speed of the vehicle, $$v$$, is small, we can approximate this to
$\Delta f \approx f\frac{2v}{c}$
By measuring this frequency shift or beat frequency, the radar gun can calculate and display a vehicle’s speed.

# “I am the Doppler Effect”

The Doppler Effect is an important principle in physics and is used in astronomy to measure the speeds at which galaxies and stars are approaching or receding from us. It is also used in plasma physics to estimate the temperature of plasmas. Plasmas are one of the four fundamental states of matter (the others being solid, liquid, and gas) and is made up of very hot, ionized gases. Their composition can be determined by the spectral lines they emit. As each particle jostles about, the light emitted by each particle is Doppler shifted and is seen as a broadened spectral line. This line shape is called a Doppler profile and the width of the line is proportional to the square root of the temperature of plasma gas. By measuring the width, scientists can infer the gas’ temperature.

We can now understand Sheldon’s fascination with the Doppler Effect as he aptly explains and demonstrates its effects. As an emergency vehicle approaches an observer, its siren will start out with a higher pitch and slide down as as it passes and moves away from the observer. This can be heard as the (confusing) sound he demonstrates to Penny’s confused guests.

# References

## The Big Bang Theory of the Inclined Plane

Season 01, Episode 02: “The Big Bran Hypothesis”

Sheldon and Leonard contemplate using the stairs as an inclined plane to move Penny’s furniture to her apartment.

In Season 1 Episode 2 of The Big Bang Theory, “The Big Bran Hypothesis”, Penny (Kaley Cuoco) asks Leonard (Johnny Galecki) to sign for a furniture delivery if she isn’t home. Unfortunately for Leonard and Sheldon, they are left with the task of getting a huge (and heavy) box up to Penny’s apartment.

To solve this problem, Leonard suggest using the stairs as an inclined plane, one of the six classical simple machines defined by Renaissance scientists. Both Leonard and Sheldon have the right idea here. Not only are inclined planes used to raise heavy loads but they require less effort to do so. Though this may make moving a heavy load easier the tradeoff is that the load must now be moved over a greater distance. So while, as Leonard correctly calculates, the effort required to move Penny’s furniture is reduced by half, the distance he and Sheldon must move Penny’s furniture twice the distance to raise it directly.

# Mathematics of the Inclined Plane

## Effort to lift block on Inclined Plane

Now we got an inclined plane. Force required to lift is reduced by the sine of the angle of the stairs… call it 30 degrees, so about half.

Free-body Diagram of a block on an inclined plane. It shows the forces acting on a block and the force needed to keep it stationary and not let it slip down.

To analyze the forces acting on a body, physicists and engineers use rough sketches or free body diagrams. This diagram can help physicists model a problem on paper and to determine how forces act on an object. We can resolve the forces to see the effort needed to move the block up the stairs.

If the weight of Penny’s furniture is $$W$$ and the angle of the stairs is $$\theta$$ then
$\angle_{\mathrm{stairs}}\equiv\theta \approx 30^\circ$
and
$\Rightarrow\sin 30^\circ = \frac{1}{2}$
So the effort needed to keep the box in place is about half the weight of the furniture box or $$\frac{1}{2}W$$, just as Leonard says.

## Distance moved along Inclined Plane

The relationship between the height $$h$$ the block is raised and the distance it moves $$d$$ is the sine of the angle $$\theta$$.

While the inclined plane allows Leonard and Sheldon to push the box with less effort, the tradeoff is that the distance they move along the incline is twice the height to raise the box vertically. Geometry shows us that
$\sin \theta = \frac{h}{d}$
We again assume that the angle of the stairs is approximately $$30^\circ$$ and $$\sin 30^{\circ} = 1/2$$ then we have $$d=2h$$.

# Uses of the Inclined Plane

We see inclined planes daily without realizing it. They are used as loading ramps to load and unload goods. Wheelchair ramps also allow wheelchair users, as well as users of strollers and carts, to access buildings easily. Roads sometimes have inclined planes to form a gradual slope to allow vehicles to move over hills without losing traction. Inclined planes have also played an important part in history and were used to build the Egyptian pyramids and possibly used to move the heavy stones to build Stonehenge.

## Lombard Street (San Francisco)

Lombard Street is one of the most visited street in San Francisco as seen from Coit Tower. It is best known for the one-way section on Russian Hill between Hyde and Leavenworth Streets, in which the roadway has eight sharp turns (or switchbacks) that have earned the street the distinction of being the crookedest “most winding “street in the world, though this title is contested. (Photo by David Yu).

Lombard Street in San Francisco is famous for its eight tight hairpin turns (or switchbacks) that have earned it the distinction of being the crookedest street in the world (though this title is contested). These eight switchbacks are crucial to the street’s design as the reduce the hills natural 27° grade which is too steep for most vehicles. It is also a hazard to pedestrians, who are more accustomed to a more reasonable 4.86° incline due to wheel chair navigability concerns.

Technically speaking, the “zigzag” path doesn’t make climbing or coming down the hill any easier. As we have seen, all it does is change how various forces are applied. It just requires less effort to move up or down but the tradeoff is that you travel a longer distance. This has several advantages. Car engines have to be less powerful to climb the hill and in the case of descent, less force needs to be applied on the brakes. There are also safety considerations. A car will not accelerate down the switch back path as fast than if it was driven straight down, making speeds safer and more manageable for motorists.

This idea of using zigzagging paths to climb steep hills and mountains is also used by hikers and rock climbers for very much the same reason Lombard Street zigszags. The tradeoff is that the distance traveled along the path is greater than if a climber goes straight up.

# The Descendants of Archimedes

We don’t need strength, we’re physicists. We are the intellectual descendants of Archimedes. Give me a fulcrum and a lever and I can move the Earth. It’s just a matter of… I don’t have this, I don’t have this!

We see that Leonard had the right idea. If we were to assume are to assume — based on the size of the box — that the furniture is approximately 150 lbs (65kg) and the effort is reduced by half, then they need to push with at least 75 lbs of force. This is equivalent to moving a 34kg mass. If they both push equally, they are each left pushing a very manageable 37.5 lbs, the equivalent of pushing a 17kg mass.

Penny’s apartment is on the fourth floor and we if we assume a standard US building design of ten feet per floor, this means a 30 foot vertical rise. The boys are left with the choice of lifting 150 lbs vertically 30 feet or moving 75lbs a distance of 60 feet. The latter is more manageable but then again, neither of our heroes have any upper body strength.

## The Big Bang Theory of the Rolling Ball Problem

Big Bang Theory Season 2, Episode 5: “The Euclid Alternative”

In the opening scene of the “The Euclid Alternative”, we see Sheldon (Jim Parsons) demanding that Leonard (Johnny Galecki)needs to drive him around to run various errands. Leonard, after spending a night in the lab using the new Free Electron Laser to perform X-ray diffraction experiments. In the background, we can see equations that describe a rolling ball problem on the whiteboard in the background.

Scene with Sheldon Cooper. The background shows a whiteboard with the physics equations of a rolling ball.

Rolling motion plays an important role in many familiar situations so this type of motion is paid considerable attention in many introductory mechanics courses in physics and engineering. One of the more challenging aspects to grasp is that rolling (without slipping) is a combination of both translation and rotation where the point of contact is instantaneously at rest.The equations on the white board describe the velocity at the point of contact on the ground, the center of the object and at the top of the object.

# Pure Translational Motion

An object undergoing pure translational motion

When an object undergoes pure translational motion, all of its points move with the same velocity as the center of mass– it moves in the same speed and direction or $$v_{\textrm{cm}}$$.

# Pure Rotational Motion

An object undergoing pure rotational motion. The velocity of each point of the object depends on the distance away from the center. The further away the faster it moves as can be seen by the length of the arrows.

In the case of a rotating body, the speed of any point on the object depends on how far away it is from the axis of rotation; in this case, the center. We know that the body’s speed is $$v_{\textrm{cm}}$$ and that the speed at the edge must be the same. We may think that all these points moving at different speeds poses a problem but we know something else — the object’s angular velocity.
The angular speed tells us how fast an object rotates. In this case, we know that all points along the object’s surface completes a revolution in the same time. In physics, we define this by the equation:

\omega=\frac{v}{r}

where $$\omega$$ is the angular speed. We can use this to rewrite this equation to tell us the speed of any point from the center:

v(r)=\omega r

If we look at the center, where $$r=0$$, we expect the speed to be zero. When we plug zero into the above equation that is exactly what we get:

v(0)= \omega \times 0 = 0
\label{eq:zero}

If we know the object’s speed, $$v_{\textrm{cm}}$$ and the object’s radius, $$R$$, using a little algebra we can define $$\omega$$ as:
$\omega=\frac{v_{\textrm{cm}}}{R}$
or the speed at the edge, $$v_{\textrm{cm}}$$ to be $$v(R)$$ to be:

v_{\textrm{cm}}=v(R) = \omega R
\label{eq:R}

# Putting it all Together

An object that rolls undergoes both translational and rotational motion. To determine the speed at any point we must add both of these speeds. We see the translational speed show in red and the rotational speed show in in blue.

To determine the absolute speed of any point of a rolling object we must add both the translational and rotational speeds together. We see that some of the rotational velocities point in the opposite direction from the translational velocity and must be subtracted. As horrifying as this looks to do, we can reduce the problem somewhat to what we see on the whiteboard. Here we see the boys reduce the problem and look at three key areas, the point of contact with the ground ($$P)$$, the center of the object, ($$C$$) and the top of the object ($$Q$$).

Rendering copy of the whiteboard showing the rolling ball problem as seen on the Big Bang Theory.

We have done most of the legwork at this point and now the rolling ball problem is easier to solve.

## At point $$Q$$

At point $$Q$$, we know the translational speed to be $$v_{\textrm{cm}}$$ and the rotational speed to be $$v(R)$$. So the total speed at that point is

v = v_{\textrm{cm}} + v(R)
\label{eq:Q1}

Looking at equation \eqref{eq:R}, we can write $$v(R)$$ as

v(R) = \omega R

Putting this into \eqref{eq:Q1} and we get,
\begin{aligned}
v & = v_{\textrm{cm}} + v(R) \\
& = v_{\textrm{cm}} + \omega R \\
& = v_{\textrm{cm}} + \frac{v_{\textrm{cm}}}{R}\cdot R \\
& = v_{\textrm{cm}} + v_{\textrm{cm}} = 2v_{\textrm{cm}}
\end{aligned}
which looks almost exactly like Leonard’s board, so we must be doing something right.

## At point $$C$$

At point $$C$$ we know the rotational speed to be zero (see equation \eqref{eq:zero}).
Putting this back into equation \eqref{eq:Q1}, we get
\begin{aligned}
v & = v_{\textrm{cm}} + v(r) \\
& = v_{\textrm{cm}} + v(0) \\
& = v_{\textrm{cm}} + \omega \cdot 0 \\
& = v_{\textrm{cm}} + 0 \\
& = v_{\textrm{cm}}
\end{aligned}
Again we get the same result as the board.

## At point $$P$$

At the point of contact with the ground, $$P$$, we don’t expect a wheel to be moving (unless it skids or slips). If we look at our diagrams, we see that the rotational speed is in the opposite direction to the translational speed and its magnitude is
\begin{aligned}
v(R) & = -\omega R \\
& = -\frac{v_{\textrm{cm}}}{R}\cdot R \\
& = -v_{\textrm{cm}}
\end{aligned}
It is negative because the speed is in the opposite direction. Equation \eqref{eq:Q1}, becomes
\begin{aligned}
v & = v_{\textrm{cm}} + v(r) \\
& = v_{\textrm{cm}} – \omega R \\
& = v_{\textrm{cm}} – \frac{v_{\textrm{cm}}}{R}\cdot R \\
& = v_{\textrm{cm}} – v_{\textrm{cm}} = 0
\end{aligned}
Not only do we get the same result for the rolling ball problem we see on the whiteboard but it is what we expect. When a rolling ball, wheel or object doesn’t slip or skid, the point of contact is stationary.

# Cycloid and the Rolling ball

The path a point on a rolling ball draws is called a cycloid.

If we were to trace the path drawn by a point on the ball we get something known as a cycloid. The rolling ball problem is an interesting one and the reason it is studied is because the body undergoes two types of motion at the same time — pure translation and pure rotation. This means that the point that touches the ground, the contact point, is stationary while the top of the ball moves twice as fast as the center. It seems somewhat counter-intuitive which is why we don’t often think about it but imagine if at the point of contact our car’s tires wasn’t stationary but moved. We’d slip and slide and not go anywhere fast. But that is another problem entirely.