Deriving Einstein's most famous equation: Why does energy = mass x speed of light squared?

The

equation

e equals mz

squared

is perhaps the

most

famous

equation

in all of physics, but very few people really know what the equation means or where it comes from. In this video I would like to show a method to derive this equation and give some insight. In what the equation really means along the way we'll also touch on some of the

most

fascinating features of Einstein's theory of special relativity, so let's get started. The purpose of mechanics is to describe how bodies change their position in space with respect to time. Next I stand next to the window of a train that is traveling uniformly and I drop a stone without throwing it out the window, what do I see?

Without taking air resistance into account, you would see the stone descend in a straight line, while a person observing the stone from the side of the platform would see the stone fall to the earth in a parabolic curve, so a natural question arises. : Are the positions traversed by the stone actually on a straight line or on a parabolic curve in space? Well, first of all we will stop. using the vague word space and rather talking about motion relative to a coordinate system that is useful for a mathematical description, then we are in a position to say that the stone traverses a straight line relative to a coordinate system rigidly attached to the train but relative to a coordinate system rigidly attached to the ground describes a parabolic curve it is now clear that there is no such thing as an independently existing trajectory but only a trajectory relative to a particular reference system the fundamental law of mechanics of Galileo and Newton can be express as follows: a body removed sufficiently far from other bodies continues in a state of rest or uniform motion in a straight line unless acted on by an external force.

This is called the principle of inertia. A reference frame or coordinate system that is in a state of motion such that the principle of inertia holds relative to it is called an inertial reference frame. You can think of an inertial reference frame as a non-accelerated reference frame. This is why Einstein's theory of special relativity is known as special because it only deals with inertial reference frames and it is these special inertial reference frames that we will focus on in the rest of this video. The principle of relativity was actually introduced by the great Italian scientist Galileo. who used it in his dialogue about the two main world systems, the dialogue was published in 1632 and consists of an imaginary conversation between three characters, one of whom salviati tried to convince the other two of the validity of the Copernican theory that the Earth orbits around the sun.

Instead of the other way around, another Simplicio character presents several arguments against the Copernican view, all of which are cleverly demolished by salviality. One of Simplicio's arguments is that the Earth cannot move around the Sun because we do not seem to do so. Surely if the Earth were moving, we would know that Salviati dismisses this idea through a simple thought experiment. He asked the reader to imagine locking himself in the main cabin below deck of some large ship and performing a series of simple experiments while the ship proceeds in uniform motion in a straight line, observing that he will not discover the s

light

est change in all the experiments. nor will he be able to determine in any of those experiments whether the ship was moving or stopped, and concludes that because there is no way to perform an experiment to distinguish between a stationary reference frame and a reference frame that is moving at a

speed

constant in a straight line, there should be no difference in the laws of physics used to describe the motion of objects within those reference frames, therefore we will assume that the principle of relativity is valid and we assume that the laws of physics must be the same in all inertial reference frames, which are simply reference frames that move at a constant

speed

relative to each other.

Now suppose that a train travels along a railroad track with a constant speed zed with respect to the track and that a person on the train walks in the direction in which the train is moving with a speed w with respect to the train. , what is the speed of person v with respect to the railroad track? Well, we see that if the person stayed still, for one second he would travel a distance of z meters, which is the distance the train travels in one second with respect to the track. When walking, the person travels an additional distance of w meters in one second, therefore, in relation to the ground they travel with a speed of w plus z meters per second, this is known as the Galilean sum of speeds.

Let's now think about the consequences of applying this idea to the propagation of

light

, one of the first things you are taught in school is that the propagation of light in a vacuum is carried out in a straight line and with a fixed speed. 3 times 10 to 8 meters per second. This follows from Maxwell's wave equation, which contains information about the speed of waves in terms. of the constants of the theory but when we say that light propagates at three times ten to eight meters per second what frame of reference we are referring to, although Maxwell's equations predict the speed of light they do not tell us what frame of reference it is find this speed mentioned to emphasize this point let's take the train as our frame of reference once again we will assume that the train is moving with a speed of z meters per second relative to the track when a light pulse is emitted onto the train with speed c meters per second relative to the train, this situation is analogous to the situation where a person walks on the train, we simply replace the person with a beam of light, so the natural question arises what is the speed of the light in relation to the road, adopting the same logic.

As we did with the person walking on the train, we have concluded that light travels with a speed c plus z meters per second with respect to the track, in other words, the speed relative to the track is greater than three times 78 meters per second. Second, but this seems to conflict with the constancy of the speed of light implied by Maxwell's equations, to make this point in a slightly different way, consider another example, isn't it evident that if you are in a car traveling at 20 meters per second with respect to the road? and a car passes at 30 meters per second relative to the road, then the speed of the second car relative to the first is simply 10 meters per second, then surely it is also true that if a spaceship travels at 200,000 kilometers per second with respect to the road and a ray of light passes at 300,000 kilometers per second with respect to the road, then the speed of light with respect to the spacecraft is one hundred thousand kilometers per second, but if you really did the experiment, this is not what you would find that when you do the experiment it looks like the light is traveling at 300,000 kilometers per second relative to the spacecraft and it looks like the light is traveling at 300,000 kilometers per second relative to the road so, How can both statements be made?

The real Einstein realized that the only possible way a stationary person and a moving person could measure the same speed of light is if their sense of space and time were not the same; In other words, relative to the spacecraft on which light travels, three hundred thousand spacecraft kilometers per spacecraft second, and relative to the road, light travels 300,000 road kilometers per road second. How can it be that one person's measurement of time is different from that of another who is in constant relative motion? To answer this question we will have to think about what it means to measure the duration of time.

Let's consider a very simple type of clock: it consists of two perfectly parallel mirrors separated by one meter and if we send a light signal between the two ends, the light continues to rise and fall, reflecting back. outside the mirrors ticking every time it goes up and talking every time it goes down like a standard clock, we build two of those clocks with exactly the same length and synchronize them by running them together, then they always match afterwards because they are the same length and As we have seen, light always travels with the same speed c meters per second if we give one of these clocks to a person to put it on a train that moves at a constant speed with respect to the platform and mount the clock perpendicular to the direction of the train's motion, so what we want to do is consider how this moving clock compares to a stationary clock.

We can calculate the time it takes for the stationary clock to tick, which we call t zero, by observing that light traveling at speed c travels a distance of two meters during one tick of the clock and therefore the time it takes for the tick to tick. The tick of the stationary clock is two over You can see in the diagram we immediately see that due to the movement of the train the path taken by the light during the ticking of the moving clock is longer than the path taken by the light during the ticking of the stopped clock if we assume that the train is moving with a speed v meters per second with respect to the platform and if t large is the time it takes for light to travel from the lower mirror to the upper mirror, then we see that the train will move a distance v t during the time it takes the clock in motion.

We also see from the diagram that the distance traveled by the light beam during this time is ct because the train is moving at a constant speed. The same goes for the top of the moving clock. Now, if we look at this diagram carefully, we see that there is a right triangle involving vt ct and the distance between the two. mirrors which, if you remember, are one meter long, so we should be able to use the Pythagorean theorem to find a relationship between the three sides of this triangle. If we focus on this right triangle, then we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two legs.

Applying this to our example we see that c t al

squared

is equal to one squared plus v t squared and if we rearrange for t then we find that the time taken for the ticking of the moving clock is one over the square root of c squared minus v squared, but we want the time for the ticking and the moving clock talking so we need to multiply this by two if we use lowercase t for the time it takes for the moving clock to tick then we see that t is equal to two over the square root of c squared minus v squared then if we take a factor of c squared into the square root, then we can write t as 2 over c divided by the square root of 1 minus v squared over c squared the reason we've done this It is because 2 over c is none other than the time it takes for the stationary clock to tick which we label t zero and so we see that I can write t is equal to t zero divided by the square root of one minus v squared over c squared This expression is often written in the compact form t equals gamma t zero where gamma is the factor that tells us how significant the difference between t and t is.

We will see that gamma appears in many of the equations of modified relativity, so, What

does

this expression tell us? The first thing we notice is that if v is equal to 0, then gamma is simply equal to 1 and we see that t is equal to t nothing. This simply tells us that if two clocks are stationary with respect to each other, they will run at the same rate; If, on the other hand, v is greater than zero, then the square root appearing in the expression for gamma will be less than one. and therefore gamma will be greater than 1 and therefore the time required for the ticking of the moving clock will be greater than the time required for the ticking of the stationary clock.

Now it is important to emphasize that not only

does

this particular moving clock run slower, but if the principle of relativity is correct, then any other type of clock would also have to run by exactly the same amount. Good because? Because otherwise the person on the train could use the mismatch between the clocks to determine the speed of the train, which would violate the principle of relativity which we have already seen states that there is no way to perform any experiment to distinguish between a stationary reference system and a reference system that moves at a constant speed in a straight line.

Now if all moving clocks go slower by the same amount then we should say in a sense that time itself appears to be slower in the moving train all the phenomena in the moving train the person's pulse his processes Of thinking how long it takes to grow and age, all of these things must slow down in the same proportion to preserve the principle of relativity. This difference inTime elapsed between stationary and moving reference frames is known as time dilation and is one of the great predictions of Einstein's theory of special relativity. Now you may have been wondering what happens to the ticking of the moving clock when v It approaches the speed of light.

Well we see in the gamma equation that as v approaches c, gamma approaches infinity and therefore tends to infinity, in other words the ticking of the moving clock would become slower and slower until eventually would freeze and time would literally stop, but we will come back to very fast moving objects later, for now we want to answer a simpler question: if all this is true, why don't we notice the slowing down of time day by day? To answer this question, let's consider an example. Let's imagine that Usain Bolt runs the hundred meters and travels at an average speed. speed of 10 meters per second if I am standing on the side of the track and watching Usain Bolt pass by, what factor do I watch the time in?

Let's go slower, we can calculate the answer, we know that t is equal to t zero divided by the square root of one minus v squared over c squared and in this example v over c is equal to 10 over 300 million, but we need v squared over c squared, which is one over 900 billion, which is an incredibly small number and, therefore one minus this incredibly small number gives a number that is very close to being one, so we see that t is essentially the same as t zero and this is true for the vast majority. Many of the moving objects we encounter in our lives simply do not move fast enough for us to notice the time dilation effect.

So how do we know if time dilation is really good? Turns out it's a very interesting example of time slowing down with movement. It is provided by a certain class of fundamental particles known as muons. Muons are created when cosmic radiation bombards Earth's upper atmosphere, and muons created in the upper atmosphere spontaneously decay after an incredibly short average lifetime of 2.2 microseconds. It should now be clear that in such a short lifetime, muons should not be able to travel much further than about 600 meters, even if they travel close to the speed of light and yet even though muons are created in the upper atmosphere about 10 kilometers above the earth's surface, they are actually detected. in the laboratories here on land, so how can that be?

The answer is time dilation. If, for example, a muon travels at 99.8 percent of the speed of light, then we can substitute v for 0.998 c in the time dilation equation and see that mu1's lifetime increases to about 35 microseconds, time during which is capable of traveling more than 10,000 meters and reaching a laboratory on the surface of the earth. Well, suppose we are willing to accept the validity of the time dilation equation and the notion that time runs slowly for moving clocks. Our next task is to understand how this links to Einstein's most

famous

equation e equals a m c squared now we move on to our strategy for determining the relativistic

energy

of an object is to first think about how we define

energy

in classical In physics we know from classical mechanics that if a force acts on an object causing it to accelerate from one place x1 to another place x2, then we can determine the kinetic energy gain by calculating the work done, which is equal to the integral of f dx where dx represents a small displacement along the path connecting x1 and x2.

Next, we can make use of Newton's second law and rewrite the force as the rate of change of momentum dp by dt. We also know that momentum is defined as the

mass

of an object multiplied by its velocity and therefore if we consider that we are in the rest frame of the moving object then we can write p equals m dx times dt zero where m is known as the rest

mass

of the object, then we can use the chain rule to rewrite dx times dt zero as dx times dt multiplied by dt times dt zero and if we then differentiate t with respect to t zero then we simply get a general factor of gamma and therefore our expression for momentum becomes gamma multiplied by mv or if we write the full expression get mv divided by the square root of one minus v squared over c squared now that we have an expression for relativistic momentum we can differentiate this with respect to t to find the relativistic force acting on our object to do this we will use the product rule which simply states that if we have two functions a of t and b of t multiplied together then the derivative of the product a b is simply a times db times dt plus gives times dt times b, applying this to our example we see that v is playing the role of a and one minus v squared over c squared to the power of minus half is playing the role of b applying the rule of product we find the following scary green expression, next we want to calculate the highlighted derivative of one minus v squared over c squared to the power of negative half if we do this then we find the following blue expression which simplifies to v over c squared dv times dt times 1 minus v squared over c squared to the power of negative 3 over 2.

So if we put all this together, we find the following yellow expression for the derivative of momentum. Note that both terms in this expression contain m times dv times dt, so we can factor this out to get to the next green. expression the next trick is to write this as a single fraction with denominator 1 minus v squared over c squared to the power 3 over 2 as you can see in the blue expression and we notice that the terms v squared over c squared in the numerator cancel and so when the dust settles we finally find this simplified pink expression which is simply equal to gamma cubed m dv times dt now that we have a concise expression for the derivative of momentum we can substitute this back into our integral and finally determine an equation for the kinetic energy of our moving object, so if we combine everything we have calculated so far we find the following expression for the kinetic energy, the trick to solving this integral is to change the variables so that we are integrating with respect a v instead of x if we do this then we can rewrite dv times dt times dx as dx times dt times dv this change of variables will also change the limits of the integral and if we imagine that our object starts at rest then we can write v is equals zero at Now let's take a closer look at what this means, after a bit of a long calculation we have arrived at the following equation which tells us the kinetic energy of an object with rest mass m moving at a speed of v meters per second, what we now want.

The first thing we must do is understand each of the terms in this equation. The first thing to note about this equation is that if we want, we can write this equation in a little more compact form as m c gamma squared minus 1. Next, let's see what happens when we set v. is equal to 0 in other words when the object is stationary in that case we see that gamma is equal to 1 and therefore we see that the kinetic energy is equal to 0 as expected our equation gives the correct answer for a stationary object but does what else is this equation? telling us to reveal the true depth and power of this equation, we can rewrite it as mz squared over the square root of 1 minus v squared over c squared equals kinetic energy plus mz squared and this expression is simply equal to e where e is the total energy of the object, so we see that the total energy of an object of rest mass m moving at speed v is equal to gamma times mz squared.

Now we want to understand well what this equation means to have a better idea. First write the total energy as m c squared times one minus v squared over c squared to the power of negative one-half. If we assume that the speed v is much less than the speed of light, then we can expand this parenthesis by making use of the binomial theorem. and arrive at the following blue expression because v is much smaller than c, we can ignore all terms of order higher than v squared over c squared in the expansion, since they will be insignificant if we then simplify this expression that we find in the limit of low speed that the total energy is equal to m c squared plus half mv squared we immediately recognize the second term of this expression as the classical equation for the kinetic energy of an object of mass m moving at a speed v so, what happens to the first term?

What does m c squared represent? Well first notice that this term appears whether the object is moving or not, if we set v to zero then we see that the total energy of the object is equal to m c squared as this energy exists even when an object is at rest , this energy is Known as rest mass energy, how should we correctly interpret this new rest mass energy? Even before the advent of relativity, it was understood that the energy of an object is more than just the kinetic energy of the object. Kinetic energy is the energy due to motion. of an object, but even when the object is at rest it can still contain energy; that energy was considered the energy necessary to assemble the system.

Now what is particularly interesting about this energy of assembly or creation is that it does not depend on the speed at which it moves. is a completely different form of energy, you can think of the rest energy of an object as the energy required to create the mass of the object, the more massive an object is, the more energy is required to create that mass, once it is created this mass, you can think of energy at rest as a form of stored energy energy is stored in the form of mass now it is possible to release this mass energy in certain circumstances this is, for example, how stars produce their energy through conversion of mass into energy during the process of nuclear fusion We also know that if a matter particle meets its antimatter partner, for example, an electron meets a positron, then the mass of both particles will be converted into energy in a process known as annihilation and The amount of energy released can be calculated.

Using e equals m c squared we see that in the case of an electron and a positron annihilating each other at rest, the energy released is 1.6 times 10 to the negative 13 joules, which may seem like a small amount of energy , but it's about 75,000 times the ionization energy of a hydrogen atom if that doesn't mean anything to you don't worry, let's think about it another way if we annihilated 1.2 kilograms of matter with 1.2 kilograms of antimatter then that would release 2.2 times 10 to 17 joules of energy, which is approximately the annual energy consumption of New York and all we would need is 1.2 kilograms of antimatter.

Unfortunately, antimatter is not as easy to produce in large quantities as it is quickly annihilated by matter, so it is not currently a feasible option for mass-scale energy production on Earth, so let's pause for a moment. and let's take stock of what we have seen so far. We have seen that for an object of mass m moving freely with speed v, the total energy of the object is given by e equals gamma. multiplied by mz squared and looking at the low speed limit we see that this total energy is essentially made up of two parts: the kinetic energy of the object which depends on the speed and the rest energy of the object which is the energy required to create mass. of the object, this rest energy equation is perhaps the most famous equation in all of physics and now we see where it comes from and what it means for any object, whether it is an elementary particle, a star or a black hole. in a reference frame in which the object is at rest, then the energy of that object is its mass multiplied by the speed of light squared.

This is perhaps one of the most profound realizations in the history of physics. Okay, but let's focus now on the situation where the object is moving well, in that case we have seen that the appropriate equation to use for the energy of the object is e equals gamma mz squared, so if you think about it , this should be the famous equation that everyone knows, since it is the most general expression we can write to express the energy of an object moving with any speed below the speed of light. Why do we say below the speed of light?

What happens if we try to increase the speed of a massive object all the way? the speed oflight, we see in our definition of gamma that if v tends to c then gamma tends to infinity and therefore energy tends to infinity; In other words, an infinite amount of energy would be required to accelerate a massive object to the speed of light and this is simply not possible and therefore we see that the speed of light represents an upper limit to the speed that an object can possess. in our universe. We can visualize this universal speed limit by plotting the kinetic energy function for a relativistic object and comparing this. with classical kinetic energy according to classical physics an infinite kinetic energy would require that the speed of an object tend to infinity while we see that according to Einstein's theory of relativity the kinetic energy tends to infinity when the speed tends to a finite number, that is, the speed of light and therefore we see once again that the speed of light represents a fundamental limit for the speed of any object until now we have only looked at the properties of massive objects, these are objects that when left at rest have a value other than zero energy at rest but not all objects have mass, for example, we know that the fundamental particles of light known as photons are massless particles that travel at the speed of light, so, do What do we get from the energy of a photon if we substitute in m is equal to zero and v is equal to c in the equation e is equal to gamma mz squared, well, we see that in this case both the numerator and the denominator are equal to zero, which is not particularly insightful or perhaps this is telling us that we should not think of the energy of a massless particle in terms of speed because in that case we would expect all massless particles to have the same energy since they all travel at the same speed, so this leads to a natural question: can massless particles have different energies if they all travel at the same speed? the same speed Well, it turns out that the answer is yes, but to see this we will need to look at the energy of a particle in a slightly different way.

Instead of expressing the total energy of an object in terms of its speed, we can Rewrite the energy in terms of the particle's momentum. This is something we already know from classical physics, where we can take the kinetic energy expression half mv squared and then use the definition of momentum p equals mv and write this as ke equals p squared. about 2m so let's try to do something similar for relativistic energy if we take the equation e is equal to gamma mz squared and we square both sides we see that e squared is equal to m squared c a four divided by one minus v squared about c squared if we are a little clever and add the following green expression which is simply equivalent to adding zero, then we see that we can combine the first and third terms circled in blue and write these two terms as m squared c to the power of four minus m squared c squared v squared all over one minus v squared over c In other words, we can write e squared in the following way and we notice that the term inside the parentheses is none other than the relativistic momentum that we derived earlier In addition, we can take a factor of m squared c squared into the numerator in the yellow expression which allows us to then simplify our equation as can be seen in the green expression when the dust finally settles we see that the total energy can be represented Using the famous momentum relation e squared equals p squared c squared plus m squared c a 4, which should be familiar to anyone who has studied particle physics, we see that this expression gives the energy in terms of moment and mass.

The advantage of this is that it describes all particles, whether their masses are zero or non-zero. The added advantage of this expression is that we can immediately see what happens if we set the mass equal to zero; In that case, we see that the second term in the energy momentum relation becomes zero and therefore the energy of a massless particle can be written as e equals pc where p is the magnitude of the momentum of the particle. Now you might be concerned that by the classical definition of momentum where p equals mv, if a particle has no mass then its momentum must be zero, but this is no longer true according to relativity a massless particle has momentum and the The magnitude of the impulse is simply equal to the energy of the particle divided by the speed of light.

As an interesting note, we see that if we combine the relativistic expression for the energy of a massless particle with the famous equation for the energy of a photon proposed by Einstein To explain the photoelectric effect, we then see that the momentum of a massless photon can be expressed as Planck's constant divided by the lambda wavelength and it is fascinating to note that in 1923 it was like this. The equation that Louis de Bruy suggested applies to all particles of matter and which later inspired Schrödinger to develop his famous wave equation of quantum mechanics. In closing, I would like to make a final observation about the energy moment relationship.

This equation lies at the heart of modern particle physics and is interesting to understand why, as I'm sure you've heard before, both the energy and momentum of an isolated physical system are conserved, which is simply another way to say that the total energy and total momentum of an isolated system do not change with time. and this is true from the point of view of each observer, however, according to Einstein's theory of special relativity, different observers moving relative to each other will assign a different amount of energy and momentum to the system, what is notable is that if those Different observers substitute their own particular values ​​for the energy and momentum of a moving object into the energy momentum relation and then calculate the mass, they will all measure the same value regardless of their frame of reference, so we see that the mass of an object is a fundamental invariant of the theory of relativity is something that all observers can agree on and it is this fact that helps particle physicists determine the masses of fundamental particles because, by experimentally measuring energy and momentum, they can use the energy momentum relationship to determine the invariant mass.

It was through the use of this method that the Higgs boson was experimentally discovered and its mass determined in 2012, a topic I will return to in a future video, for now I think it is appropriate to leave you with the wise words of Albert Einstein, the important thing . It is not stopping to question oneself, curiosity has its own reason for existing, one cannot help but be amazed when contemplating the mysteries of the eternity of life, the wonderful structure of reality, it is enough for one to try to understand a little of this mystery every day. I hope May this video have helped you understand a little of the mystery of relativity.

Thanks for seeing it.