# Special Relativity

## Introduction

This page strives to describe the core elements of the Special Theory of Relativity in easy to understand terms. An understanding of classical mechanics, algebra, and calculus is needed to fully understand what is presented.

The following topics are discussed:

• Science of the late 19th century that led to Special Relativity
• Fundamental principles that are the basis of Special Relativity
• Relativity of simultaneity.
• Speed of light limit
• Time dilation
• Length contraction
• Mass increase
• Mass-Energy equivalence

## Luminiferous Aether

in the late 19th century scientists believed that light waves must travel through a medium; similar to sound waves traveling through a medium (such as air).

The name given this medium was Luminiferous Aether, or Aether, or Ether.

The hypothesized Ether would be motionless in space, and the Sun and Earth would be moving through the Ether. A testable consequence of the Ether hypothesis is that the speed of a photon traveling in one direction should differ from the speed of a photon traveling in a perpendicular direction. This speed difference could be used to determine the direction of Earth's motion through the Ether.

## Michelson–Morley Experiment

In 1887, scientists Michelson and Morley performed an experiment to measure Earth's speed through the Ether, by comparing the speed of light in perpendicular directions.

Their experiment failed to detect the expected difference in the speed of light, which contradicts the Luminiferous Aether hypothesis. The results from this experiment, and similar experiments done by others, led to the theory that the speed of light is constant, regardless of the relative speed of the source of the light.

The Michelson–Morley Experiment has been called 'The Most Famous Failed Experiment in History".

## Lorentz Transformation

Lorentz and Poincare, derived coordinate system transformations to explain the negative result of the Michelson-Morley experiment. Their work was based on the existence of the Ether.

Interestingly, the Lorentz Transformations are identical to the equations found in Einstein's Special Theory of Relativity even though the founding principles used by Lorentz/Poincare were different than the founding principals of Einstein's Special Theory of Relativity.

## Einstein's Special Theory of Relativity

In 1905 Einstein published a paper titled "On the Electrodynamics of Moving Bodies", which later became known as Einstein's Special Theory of Relativity (STR).

Also published, later in 1905, "Does the Inertia of a Body Depend Upon Its Energy Content?", in which Einstein proposed mass-energy equivalence.

### Fundamental Principles

Einstein based his Special Theory of Relativity on the following fundamental principles:

• the equivalence of inertial frames, such that the laws of physics apply equally in all inertial coordinate systems;
• the constancy of the speed of light in empty space, independent of the relative motion of its source.

The principle of equivalence of inertial frames conflicts with the existence of the Luminiferous Aether because the Ether is a special reference frame. So, the STR does not allow the existence of the Ether.

### Relativity of Simultaneity

An interesting thought experiment can be performed based on a constant speed of light. This thought experiment shows that events that occur simultaneously in one reference frame are not simultaneous when viewed from another reference frame.

The two reference frames for this thought experiment are (1) a moving train car, and (2) the rail bed.

Imagine you are in the center of the moving train car, and cause baseballs to be simultaneously launched at equal speeds towards opposite ends of the car. Timekeepers at both ends of the car will measure the baseball's arrival time, which will be the same.

Next imagine you are standing near the rail bed, watching the train pass. You observe the baseballs being thrown, and timekeepers lined up along the rail bed record the time that the baseballs arrive at the two ends of the car. These timekeepers will also measure identical arrival times, however a measurement of the speed of the 2 baseballs will show they are moving at different speeds. For example, suppose the train car is moving at 30 MPH, and the 2 baseballs are thrown at 30 MPH. Then from the perspective of observers on the rail bed, the baseball thrown toward the rear of the train car will be stationary, and the baseball thrown toward the front will be moving at 60 MPH.

Now, lets run this experiment again, this time replacing the baseballs with photons. Remember photon move at a constant speed of light, regardless of the motion of the photon's source. As before, the 2 timekeepers riding the train car, will observe the photons arrive at each end of the train car at identical times. However, the situation for the observers on the rail bed has changed. From the perspective of the rail bed, the 2 photons are traveling at the constant speed of light. The photon traveling toward the rear of the train car arrives at the rear of the train car earlier than the photon traveling toward the front of the train car arrives at the front. This is because the motion of the train car is moving the rear of the car toward the launch point, and the front of the car away from the launch point.

Conclusion, if you accept the constant speed of light hypothesis then the definition of simultaneous events needs to be modified. Events that are simultaneous in one reference frame may not be simultaneous in another reference frame.

Time travel to the past has been called the Grandfather Paradox, because the time traveler could kill his grandfather before his grandfather meets his grandmother.

Continuing with the Relativity of Simultaneity thought experiment, it is possible to describe a procedure to travel back in time.

We will again be using the train car and rail bed, but this time the train car will be traveling at 0.5 C, and the length of the car will be 1 light minute. Time zero is when an observer on the rail bed sees the photons released. Other observers on the rail bed will see the photon arrive at the rear of the car at time 20 sec, and arrive the front of the car at time 60 sec.

Now we will perform some steps to send a time traveler back in time. Marty will be the time traveler. Initially Marty is the observer, on the rail bed, who sees the photon arrive at the rear of the train car.

• At time 20 sec Marty sees the photon arrive the rear of the train car.
• Marty instantly transports himself to the location along the rail bed of the front of the train car. The time is still 20 sec, because the transportation was instantaneous. The photon has not yet arrived at the front of the train car, it will arrive there at time 60 sec.
• Marty jumps on the front of train car. Marty is now in the train car reference frame, where the photons arrive simultaneously at the front and rear of the car. The photon has not arrived at the front of the car, so the other photon must not yet have arrived at the rear of the car.
• Marty instantly transports to the rear of the train car. The photon has still not arrived at the rear of the train car.
• Marty watches the rail bed from the rear of the train car. When he sees his original self standing on the rail bed, he jumps off the train car and prevents his original self from making this journey.
• Now there are 2 Martys.

Now, do the same thought experiment, but instead of Marty being able to travel instantly from one end of the train to the other, limit Marty's speed to slightly less than the speed of light. By limiting Marty's speed he is not able to arrive at the rail bed at the front of the train car prior to the photon arriving at the front of the train car. Thus Marty can not travel back in time.

In the early 20th century scientists must have started to have similar concerns. In 1904, Poincare said "Perhaps we must construct a new mechanics ... in which the speed of light would become an impassable limit."

### Implication of the STR Principles

#### Speed of Light Limit

The principal of the constancy of the speed of light implies that it is not possible for an object to move from point A to point B in less time than it takes for a photon to move from point A to B. Refer to the prior section, Relativity of Simultaneity.

#### Time Dilation

The principle of equivalence of inertial frames implies that an observer on a spaceship traveling from point A to point B can make the journey in an arbitrarily short amount of time by moving at an arbitrarily large velocity. The velocity of the spaceship as viewed from point A or point B can not exceed the speed of light. However the spaceship's reference frame is not constrained by the existence of the A/B reference frame, and the observer on the spaceship can make the journey in an arbitrarily short amount of time. This implies that clocks run at a slower rate in the traveling spaceship's reference frame as compared to the A/B reference frame.

#### Length Contraction

The observer on the spaceship going from point A to point B must not measure speeds exceeding the speed of light. Suppose in the A/B reference frame, the points A and B are 100 light minutes apart. And suppose in the spaceship's reference frame, that the trip from points A to B is completed in 10 minutes. At first glance it seems that the travelers would believe they have traversed a 100 light minute distance in just 10 light minutes, at a speed 10 times C. This contradicts not exceeding the speed of light. STR deals with this with Length Contraction. The observer on the spaceship will not measure the distance between A and B to be 100 light minutes, but instead the distance between A and B will appear to be about 9 light minutes.

#### Mass Increase

According to Newton's F=ma, the velocity increase of a rigid object being acted on by a constant force F, for time t is given by:

$v_{2}=v_{1}+t*{\frac {F}{m}}$

Thus it seems that by applying a force to an object which is traveling at speed just less than the speed of light (c), that it would be possible to increase the speed of the object to greater than c. The STR deals with this problem by having the mass of the object increase as it's speed approaches the speed of light (the mass approaching infinity as speed approaches c). Thus, we can still use Newton's law, but the increasing value of mass prevents velocity from exceeding c.

#### Mass / Energy Equivalence

In classical mechanics, when a force is applied over a distance to a rigid object then the kinetic energy of the object is increased by the force times the distance. At small velocities of the object we observe a change in velocity of an object when a force is applied over a distance.

When an object's velocity is near the speed of light, and a force is applied over a distance, the object's velocity will increase to become slightly closer to the speed of light, and the mass of the object will also increase. In this case it appears that the kinetic energy increase of the object has gone into increasing both the velocity and the mass of the object.

From this we can hypothesize that energy and mass are not distinct. Instead energy can be converted to mass and mass can be converted back to energy.

### STR Equations Derived

The following sections present derivations of some of the equations found in the Special Theory of Relativity.

An example scenario will be used, in which a spaceship travels from Earth to Jupiter at speed 0.99 C. Earth/Jupiter are assumed to be in the same inertial reference frame, and at a distance of 100 light minutes from each other. Observers on Earth and Jupiter have synchronized clocks. The spaceship will leave Earth at time zero. The spaceship also has a clock, which is also set to time zero when the spaceship leaves the Earth.

There is no acceleration of the spaceship during this example scenario. When the spaceship leaves Earth it is already at 0.99 C, and when it arrives at Jupiter the spaceship doesn't stop.

#### Time Dilation

To derive Time Dilation we need to calculate the time (t) on the spaceship's clock when it arrives at Jupiter. This can be done using STR principle of equivalence of reference frames.

• An observer on Jupiter with a telescope trained on the spaceship will observe the spaceship's clock ticking at a fast rate
• An observer on the spaceship with a telescope trained on Jupiter will observe Jupiter's clock ticking at a fast rate

These 2 rates must be equivalent.

Let's start with what will be seen by the observer on Jupiter.

• Jupiter clock will read 100 minutes when the spaceship is seen leaving Earth.
• Jupiter clock will read 101.01 minutes when the spaceship is seen arriving at Jupiter.
• During this 1.01 minute interval, the observer on Jupiter will see the clock on the spaceship advance from 0 to t.

Next what will be seen by the observer on the spaceship.

• Spaceship clock will read 0 minutes when Jupiter is first seen approaching the spaceship.
• Spaceship clock will read t minutes when the Jupiter is seen arriving at the spaceship.
• During this t minute interval, the observer on the spaceship will see the clock on Jupiter advance from -100 to 101.01 minutes.

Make an equation, setting the 2 rates equal to each other, and solve for t.

${\frac {t}{1.01}}={\frac {201.01}{t}}\qquad t=14.25\;minutes$

When the spaceship arrives at Jupiter, the spaceship clock will read 14.25 minutes. The observer on the spaceship will have aged just 14.25 minutes during the trip from Earth to Jupiter, while an observer on either Earth or Jupiter have observed the duration of the spaceship's trip to be 101.01 minutes.

Next, let's replace the numbers with variables, and hopefully arrive at a familiar equation from STR.

${\frac {t}{{\frac {d}{v}}-{\frac {d}{c}}}}={\frac {{\frac {d}{v}}+{\frac {d}{c}}}{t}}$
$t^{2}=({\frac {d}{v}}-{\frac {d}{c}})({\frac {d}{v}}+{\frac {d}{c}})$
$t^{2}={\frac {d^{2}}{v^{2}}}-{\frac {d^{2}}{c^{2}}}$
$t^{2}={\frac {d^{2}}{v^{2}}}(1-{\frac {v^{2}}{c^{2}}})$
$t={\frac {d}{v}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}$

#### Length Contraction

We can build on what we now understand about Time Dilation to derive Length Contraction.

To derive Length Contraction we will measure the distance between Earth and Jupiter in the spaceship's reference frame. To do this we will have the spaceship emit a photon at the moment at leaves Earth, and determine the time (ts) at which the photon arrives at Jupiter (ts is in the spaceship's reference frame).

First let's collect what we know:

• the photon arrives at Jupiter at time 100, in the Jupiter reference frame
• the spaceship arrives at Jupiter at time 101.01, in the Jupiter reference frame
• the spaceship arrives at Jupiter at time 14.25, in the spaceship's reference frame

So, the photon arrives at Jupiter 1.01 minutes prior to the spaceship, and the spaceship's clock is ticking at a rate of 0.141 (14.25 divided by 101.01) of the Jupiter clock. Therefore the photon's arrival time at Jupiter, in the spaceship's reference frame is:

$t_{s}=14.25-1.01*{\frac {14.25}{101.01}}$
$t_{s}=14.25*(1-{\frac {1.01}{101.01}})$
$t_{s}=14.11\;minutes$

Thus the distance (ds) from Earth to Jupiter, in the spaceship's reference frame is:

$d_{s}=c*t_{s}$
$d_{s}=1*14.25*(1-{\frac {1.01}{101.01}})$
$d_{s}=14.11\;light\;minutes$

Solve for ds, by Replacing the numbers with variables ...

$d_{s}=c*({\frac {d}{v}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}})*(1-{\frac {d/v-d/c}{d/v}})$
$d_{s}=c*({\frac {d}{v}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}})*({\frac {v}{c}})$
$d_{s}=d*{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}$

#### Mass Increase

The following rational for deriving Mass Increase may involve assumptions that go beyond the founding principles of STR.

In classical mechanics an Impulse (a fast acting force or impact), applied to an object will change the momentum of the object. If the magnitude of the impulse is increased by a factor of two then the momentum is increased by a factor of two, and the time it takes the object to traverse a distance is cut in half.

The same should hold true for our relativistic spaceship. Suppose the first trip from Earth to Jupiter is initiated by an impulse of 1 unit, and the duration of the trip is 1 time unit (in the spaceship's reference frame). Then, a second trip from Earth to Jupiter, initiated by an impulse of 2 units would have a duration of 1/2 time units, also in the spaceship's reference frame.

So, for the two trips, the ratio of the momentum is equal to the inverse ratio of the duration. This can be expressed in the following equation.

${\frac {m_{2}v_{2}}{m_{1}v_{1}}}={\frac {t_{1}}{t_{2}}}$
${\frac {m_{2}v_{2}}{m_{1}v_{1}}}={\frac {{\frac {d}{v_{1}}}{\sqrt {1-{\frac {v_{1}^{2}}{c^{2}}}}}}{{\frac {d}{v_{2}}}{\sqrt {1-{\frac {v_{2}^{2}}{c^{2}}}}}}}$
${\frac {m_{2}}{m_{1}}}={\frac {{\sqrt {1-{\frac {v_{1}^{2}}{c^{2}}}}}}{{\sqrt {1-{\frac {v_{2}^{2}}{c^{2}}}}}}}$
$m_{2}=m_{1}*{\frac {{\sqrt {1-{\frac {v_{1}^{2}}{c^{2}}}}}}{{\sqrt {1-{\frac {v_{2}^{2}}{c^{2}}}}}}}$

Now, by letting v1 approach 0 the above equation simplifies to the following, where m0 is the rest mass. Note that v1 cannot be set to zero because v1 was used in the denominator in one of the above equations.

$m_{2}=m_{0}*{\frac {1}{{\sqrt {1-{\frac {v_{2}^{2}}{c^{2}}}}}}}$

Or simply the following, where v is the velocity of the object, and m is the relativistic mass of the object.

$m={\frac {m_{0}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$

Note that in STR momentum is m*v, same as in classical mechanics. However, in STR, the definition of mass differs from the constant mass of classical mechanics. In STR the mass of an object approaches infinity as the velocity of the object approaches the speed of light.

#### Mass / Energy Equivalence

The kinetic energy gained by a rigid object when a force is applied to that object is equal to the force times the distance over which the force is applied.

For example, if the object is initially at rest, and a constant force of 5 newtons is applied to the object over a distance of 10 meters then the kinetic energy of the object will become 50 joules.

##### Kinetic Energy in Classical Mechanics

So far we have discussed the kinetic energy in terms of force and distance; however our goal is to determine the kinetic energy of an object as a function of the velocity of the object. Let's see how this can be done in classical mechanics when a constant force of 1 Newton is applied to an object whose mass is 1 Kg. The acceleration is F/m, equals 1 meter/sec2. Now we'll examine how much the kinetic energy of the object increases as the object's velocity increases.

                              Distance Travel    Kinetic Energy Gain
Time        Velocity       In this Interval      In this Interval
----        --------       ---------------    ------------------
1 sec         1 m/s          0.5 m                1 * 0.5 joules
2             2              1.5                  1 * 1.5
3             3              2.5                  1 * 2.5
4             4              3.5                  1 * 3.5
5             5              4.5                  1 * 4.5
6             6              5.5                  1 * 5.5


What the above table is showing is that the object

• gained KE of 0.5J as it's velocity increased from 0 to 1
• gained KE of 1.5J as it's velocity increased from 1 to 2
• gained KE of 2.5J as it's velocity increased from 2 to 3
• gained KE of 3.5J as it's velocity increased from 3 to 4
• gained KE of 4.5J as it's velocity increased from 4 to 5
• gained KE of 5.5J as it's velocity increased from 5 to 6

Summing all these KE gains, we see that the object gained KE of 18 Joules due to it's velocity increase from 0 to 6 meters/sec. This answer is confirmed using $E_{k}={\tfrac {1}{2}}mv^{2}={\tfrac {1}{2}}*1*6^{2}=18\;joules$

In the example above we see that the distance traveled in each velocity interval varies. To arrive at the total kinetic energy gain we had to sum the kinetic energy gains for each velocity interval.

To do this mathematically we need to work with the following Physics relationships, using Calculus.

• Kinetic Energy Gained by a rigid object when a force is applied equals the force times the distance over which the force is applied.
$\textstyle E_{k}=\int Fds$
• Newton's Second Law: force on an object is equal to the rate of change of its momentum.
$F={\tfrac {dp}{dt}}$
• Definition of velocity.
$v={\tfrac {ds}{dt}}$

We need a form of the first equation that is expressed in terms of velocity or momentum, and not in terms of distance. This is easily done by substituting he 2nd and 3rd of the above equations into the first equation.

$E_{k}=\int Fds$
$E_{k}=\int {\frac {dp}{dt}}ds$
$E_{k}=\int vdp$

Let's check this new equation using classical mechanics, where dp = m dv.

$E_{k}=\int v*(mdv)=\int mvdv={\frac {1}{2}}mv^{2}$

Great, this works!

FYI Integral Calculator can be used to evaluate integrals. To evaluate the above integral, first enter "mv", then using the Options tab set variable of integration to v, and click Go.

##### Kinetic Energy in Special Relativity

To derive the kinetic energy of an object in Special Relativity, we will again use

$E_{k}=\int vdp$

However dp is not simply mdv, as it was in classical mechanics, because m is no longer constant. In STR, dp is:

$dp=(m+dm)*(v+dv)-mv$
$dp=mv+vdm+mdv+dmdv-mv$
$dp=vdm+mdv$

Next we need to substitute dm with a function involving dv. Take the derivative of the following to find dm as a function of dv. The derivative is taken using Derivative Calculator.

$m={\frac {m_{0}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$
${\frac {dm}{dv}}={\frac {m_{0}v}{c^{2}*(1-{\frac {v^{2}}{c^{2}}})^{{1.5}}}}$
${dm}={\frac {m_{0}v}{c^{2}*(1-{\frac {v^{2}}{c^{2}}})^{{1.5}}}}dv$

We can now determine dp as a function of v and dv.

$dp=vdm+mdv$
$dp=v{\frac {m_{0}v}{c^{2}*(1-{\frac {v^{2}}{c^{2}}})^{{1.5}}}}dv+{\frac {m_{0}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}dv$
$dp=\left({\frac {m_{0}v^{2}}{c^{2}*(1-{\frac {v^{2}}{c^{2}}})^{{1.5}}}}+{\frac {m_{0}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}\right)dv$

And the kinetic energy is

$E_{k}=\int vdp$
$E_{k}=\int v\left({\frac {m_{0}v^{2}}{c^{2}*(1-{\frac {v^{2}}{c^{2}}})^{{1.5}}}}+{\frac {m_{0}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}\right)dv$

Evaluating with Integral Calculator, and simplify.

$E_{k}={\frac {c^{2}\vert c\vert m_{0}}{{\sqrt {c^{2}-v^{2}}}}}+C$
$E_{k}={\frac {m_{0}c^{2}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}+C$

Now let's consider the Constant of Integration, 'C'. When the object's velocity is zero, the kinetic energy of the object is zero; therefore C=-m0c2.

Finally , the Kinetic Energy can be expressed in either of these 2 ways.

$E_{k}={\frac {m_{0}c^{2}}{{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}-m_{0}c^{2}$
$E_{k}=mc^{2}-m_{0}c^{2}$

This equation for kinetic energy looks quite different from the $E_{k}={\tfrac {1}{2}}mv^{2}$ of classical mechanics. It is important to show that STR is approximately equivalent to classical mechanics for velocities that are small relative to the speed of light. The Taylor series expansion of the STR kinetic energy is shown below, which does indeed approximate $E_{k}={\tfrac {1}{2}}mv^{2}$ for small values of v.

$E_{k}=(m_{0}c^{2}+{\frac {1}{2}}m_{0}v^{2}+{\frac {3}{8}}m_{0}{\frac {v^{4}}{c^{2}}}+...)-m_{0}c^{2}$
##### Energy Contained in Rest Mass

The previous section did not derive equivalence between rest mass and rest energy.

In the previous sections the mass-energy equivalence was derived as it relates to the Kinetic Energy of an object. We saw that the Kinetic Energy of an object can be expressed as a function of the object's relativistic mass and the object's rest mass.

$E_{k}=mc^{2}-m_{0}c^{2}$

This equation can be re-arranged to show that the kinetic energy of an object is proportional to a change of mass of the object. This change in mass is referred to as mk below.

$E_{k}=(m-m_{0})c^{2}$
$E_{k}=m_{k}c^{2}$

Which leads to the hypothesis that, similarly, the Rest Energy of an object is proportional to the rest mass of an object.

$E_{0}=m_{0}c^{2}$

The Total Energy of the object is the sum of the object's Rest Energy and Kinetic Energy.

$E=E_{0}+E_{k}$
$E=(m_{0}c^{2})+(mc^{2}-m_{0}c^{2})$
$E=mc^{2}$

Fast forward to modern physics, and let's consider what the rest mass of a drop of water is comprised of. Note that many of the items listed are energy; the equivalent rest mass can be calculated by dividing the quantity of energy by the speed of light squared. Also, note that some of the items listed below contribute only slightly to the rest mass of the drop of water. And finally, this is not a complete list.

• Thermal Energy: The molecules that make up the drop of water are in constant motion. The average kinetic energy of the molecules is proportional to the absolute temperature of the water drop. The sum of the kinetic energy of all the water molecules is the Thermal Energy.
• Chemical Binding Energy: This is the energy required to disassemble the water molecule into its constituent hydrogen and oxygen atoms. This results in the mass of the water molecule to be slightly less than the mass of the constituent hydrogen and oxygen atoms.
• Electron Binding Energy: This is the energy required to remove the electrons from the nucleus of an atom. Similar to chemical binding energy, the electron binding energy causes the mass of the atoms to be slightly less than the mass of the constituent electrons, and atomic nucleus.
• Nuclear Binding Energy: This is the energy required to remove all of the protons and neutrons that comprise the nucleus from the nucleus. Similar to chemical and electron binding energy, the nuclear binding energy causes the mass of the atomic nucleus to be slightly less than the mass of the constituent protons and neutrons.
• Gravitational Binding Energy: For a water drop, the gravitational binding energy is is very small. If the object was much larger than a water drop, such as our sun, then the gravitational binding energy is significant. The gravitational binding energy causes the mass of the gravitationally bound object to be less than the mass of the constituent parts, were the parts to be separated by a great distance.
• The protons and neutrons are comprised of quarks and gluons. The quarks are currently considered elementary particles (that is they can't be further divided). The quarks have mass, but the total mass of the quarks accounts for only about one percent the mass of the proton or neutron. Much of the remainder of the mass of a proton or neutron comes from the energy associated with the massless gluons that bind the quarks together.
• The electrons are currently considered to be an elementary particle. The mass of an electron is about 1/1800 the mass of a proton.

In summary, the components that add to the rest mass of the water drop include:

• the energy associated with the gluon field that binds the quarks together in a proton or neutron; this is the most significant contributor to the mass of the water drop
• the mass of the quarks themselves
• the mass of the electron
• the thermal kinetic energy of the molecules

And the components that subtract from rest mass of the water drop include:

• chemical binding energy
• electron binding energy
• nuclear binding energy
• gravitational binding energy

String Theory attempts to unify theories of General Relativity and Quantum Mechanics. String theory is still very speculative, and no part of string theory has been experimentally confirmed. However, if string theory does prove to be viable, then the mass of the electron and the mass of the quark will be thought of, not as mass, but instead as vibrating bands of energy.