# Why are mass and energy related by the speed of light?

1. Oct 12, 2010

### Kyxha

Im a 3rd year physics major. I understand the equation E=mc$$^{2}$$.
But i don't understand why mass and energy are related by the number c.

How are mass, energy, and the speed of light so inextricably related that
we can describe their relationship with one simple equation?

2. Oct 12, 2010

### Sleuth

The point should be that they are not simply related... they are EXACTLY the very same thing....
Mass IS energy, and energy IS mass...

3. Oct 12, 2010

### Kyxha

Okay, they are each other.
Why are the forms of pure energy and pure mass related by c^2?

4. Oct 12, 2010

### tom.stoer

Einstein introduced the idea that the speed of light - c - is identical in all reference frames. From this idea one can derive the relation between space and time, i.e. "spacetime"; one can show that the four-vector (ct,x) has an "invariant length c²t² - x² which is identical in all reference frames. In addition one can show that structurally the same applies to energy and momentum: the four-vector (E,cp) has an "invariant length E² -c²p² = (mc²)² which is identical in all reference frames; it's value (mc²)² is the invariant mass (squared) i.e.the rest mass (squared).

So the relation via rest mass m and energy E = mc² is basically derived from the requirement of universal c.

5. Oct 12, 2010

### diazona

EDIT: looks like tom.stoer just said the same thing I did but more concisely.

It has to do with the energy-momentum four-vector.
$$p^\mu = \begin{pmatrix}E/c \\ p_x \\ p_y \\ p_z\end{pmatrix}$$
(The c there is just a unit conversion thing) You know, of course, that if you change from one inertial reference frame to another, the values of E and p that you get for any particular particle (assuming the particle is not accelerating) will change; for example, you might have both px and E decrease.

One of the insights of special relativity is that those changes in E and p are analogous to a rotation. Think about rotating a vector in 3D space. As an example, suppose the original vector is (1,0,0), then when you rotate it, its x component will decrease and its y and/or z components will increase. The energy-momentum four-vector acts similarly, except that instead of one component increasing when another decreases, both the energy and momentum increase or decrease together. Mathematically, though, both cases look very similar.

Anyway, Einstein realized that if you just plug in the usual expressions for energy and momentum,
$$\begin{pmatrix}\frac{1}{2}mv^2/c \\ mv_x \\ mv_y \\ mv_z\end{pmatrix}$$
and look at how that four-vector changes as you move from one reference frame to another, it doesn't correspond to a rotation. For example, if you go from a reference frame in which a particle is moving to another one in which it is at rest, the vector goes from having nonzero elements to being all zeros, which means its norm goes from nonzero to zero - but a rotation can't change the norm.

In order to make a vector that transforms according to the rules of a generalized rotation (i.e. invariant norm), you need to ensure that not all of the components of the vector go to zero at the same time. Now, if you have a particle at rest, clearly its momentum will be zero. That means its energy has to be nonzero. If you were to calculate what the norm of the vector has to be to keep it invariant, you'd find out that you need to add $mc$ to the energy component, which means that every particle has a "rest energy" of $mc^2$ in addition to whatever kinetic energy it may have. That's where the formula $E = mc^2$ comes from: it's the inherent energy that a particle has just by virtue of being a massive particle.

Now, you may wonder, why c specifically? That also has to do with the idea of shifts in reference frame being like rotations. When you perform one of these "rotations" by speeding up or slowing down, you shift time into space or vice-versa. Since time and space are measured in different units, we need a conversion factor to convert between them. The conversion factor will obviously have to have units of speed. Now, one of the postulates of relativity is that the there is a particular speed which is invariant across all reference frames, i.e. no matter how much you speed up or slow down, something traveling at this invariant speed will never appear to go any faster or slower relative to you. It's that speed that appears in $E = mc^2$, or in the full version, $E^2 = p^2c^2 + m^2c^4$. It just happens that the first reason anyone had to suspect that that speed was special was that it was the speed of light, and that's why we call it "the speed of light".

6. Oct 12, 2010

### enotstrebor

Though the question may have some ambiquity do to interpretation of E=mc^2 equation (today it is a invariant/rest mass only relationship), it is clearly applicable to the invariant rest mass and thus any answer to the question must explicitly apply to the invariant/rest mass to energy relationship.

For the invariant mass energy relationship (E=mc^2), why is energy and mass related by c? As the question applies to the invariant mass energy, the answer can not have anything to do with the energy momentum four-vector.

For the invariant mass energy relationship (E=mc^2), why is energy and mass related by c? As the question applies to the invariant mass energy, the answer can not have anything to do with reference frames.

Yes that is true, but why for every particle is the rest mass and rest energy related by the square of a value that is the same as the value of the speed of light.

A good physics theory should not only mathematically relate values, but explain why the values are related. As far as I know, the SM does not indicate any reason why rest mass and rest energy are related this way.

7. Oct 12, 2010

### diazona

What?! That doesn't make sense at all. Just because the equation deals with rest mass does not mean a derivation based on four-momentum or Lorentz transformations is invalid.

8. Oct 12, 2010

### enotstrebor

If mass is energy and energy is mass then this statement requires the photon to have mass.

Clearly the photon is energy but it has no mass!

The particle mass energy form is an energy form where (by todays view) the measured energy value is dependent on the reference frame but whose mass (characterized by resistance to change in velocity within a given reference frame) never goes to zero in any reference frame.

Photon (particle) energy form is an energy form whose energy value is frame dependent and if the reference frame is moving in the same direction as the photon, but ahead of it, as this reference frame approaches the speed of light, the measured value goes toward zero.

So it is clear that the mass form of energy does not have the properties of the photon form of energy. Thus the above statement is incorrect.

9. Oct 12, 2010

### enotstrebor

It is not a derivation question, it is a fundamental question. One that goes below the mathematics, and asks what is the fundamental nature of particle mass energy.

As the E in E=mc^2 can be photon energy, let me rephrase the question using the particle mass energy and the equivalent photon energy. Note experimentally 2 equivalent rest mass particles, the electron and the anti-electron (positron) can annihilate to produce two photons. I will reference this specific case (the answer to the question must be universally true so it must be true it this specific case also) with the condition that both electron and anti-electron are at rest at time of annihilation (even if not possible to physically do this experiment, the present theory allows us to consider this possibility).

As the two particles are of equal mass and as conservation of momentum requires the two resultant photons must be of equal energy and momentum and must head out in opposite directions (net momentum 0), the rest mass relationship is 2 m c^2 = 2 E_ph

Thus for, lets say the anti-electron (or electron) the single particle's mass is converted to a single photon where the relationship between the single particle's rest mass has the equivalent energy of a photon given by the relationship m c^2 = E_ph.

What is the physics that allows the mass energy form to be at rest and resist a change to velocity and what is the substantive physics difference (not the mathematical difference) of the anti-particle that interacts with the particle nature that causes the transform of the mass form of energy to the photon energy form and why does this energy form always and immediately accelerate to the velocity c and what is the source nature of the photon acceleration which unlike mass energy occurs without requiring any energy input and without any energy loss?

Once you have answered this fundamental question, which requires a deeper understanding of the source of nature (why nature is this way) not just a mathematical form that indicates how nature behaves, then one can answer the question of why the relationship E=mc^2 exists in the first place rather than how relationships are known to behave.

For example string theory attempts to give an underlying foundation (source of behavior) to the present theory, the SM, that models the behavior of particles.

Many/most believe the model of the behavior of the particle is a model of the particle itself.

Some, including some famous physicists do not. If you are interested, read Roger Penrose's book, The Road to Reality. It is both a mathematically informative and a philosophically informative.

10. Oct 13, 2010

### tom.stoer

The italic sentences are not correct. We tried to explain above what happens (perhaps the explanation is not clear for you).

Lorentz transformations are relations for general reference frames. The name "rest mass" is missleading; it should read "invariant mass" b/c it has the same value in all reference frames. The Lorentz transformations automatically generate a relation like E = mc² were c is a numerical constant (speed) used to construct the Lorentz transformations (in SR: speed of light). That's why these relations clearly have something to do with reference frames and Lorentz transformations.

No.

A good physical theory makes some basic assumptions (a small set of axioms and principles) and derives falsifiable predictions. So the "why" is encoded in the axioms and principles. This is what SR does very efficiently. One basic principle is the universal speed of light.

The SM is based on relativistic QFT and therefore automatically has this property E = mc². So the SM can't say anything about the "why", it must simply reproduce this relation as a consistency check.

This is wrong

As I said, the relation E² = mc² applies to the rest mass or invariant mass only. Therefore this relation says that the photon is massless (zero rest mass) which is correct. In general the relation E² - (cp)² = (mc²)² which is modified for photons: E² - (cp)² = 0

Last edited: Oct 13, 2010
11. Oct 13, 2010

### tom.stoer

No.

Both SM and ST do neither question theory of relativity, nor do they add any additional structure which could explain these relations.

They may provide mechanism to generate rest mass like spontaneous symmetry breaking etc., but they do not explain E = mc² in a more fundamental sense. The SM uses SR for its construction is therefore by construction not able to explain its own basics. The same applies to string theory: String theory uses a target space with certain space-time symmetries for its construction is therefore not able to explain the symmetry structure of spacetime.

12. Oct 13, 2010

### Sleuth

It may seem stupid to you, but I think that special relativity does not explain WHY.
The only possible WHY is that for dimensional reasons, if you want to obtain an energy from a mass you ha to multiply it for a velocity squared, and the ONLY universal constant in general relativity with the right dimensions is just c...

13. Oct 13, 2010

### DrDu

I would argue like this: For a massive particle, the energy in the rest frame is M.
Then $$E=\sqrt{M^2+p^2c^2}$$. At low speeds, we get $$E\approx M \left(1+\frac{1}{2} \frac{p^2c^2}{M^2}\right)=M+\frac{1}{2} \frac{ p^2c^2}{M}$$. If we want the second term to have the non-relativistic form $$\frac{1}{2} \frac{p^2}{m}$$, then the constant M has to be chosen as $$mc^2$$.

Last edited: Oct 13, 2010
14. Oct 13, 2010

### enotstrebor

I have never seen a derivation of E=mc^2 from the Lorentz transformation which results in the invariance of E - p^2c^2. Can you show it or give a reference.

I do know Einstein derived the relationship without using the Lorentz transform and one can go to Wiki Mass-energy equivalence and see E=mc^2 derived independent of a Lorentz transform.

Thus, as the relationship E=mc^2 can be derived independently and without using the Lorentz transform, I maintain my stand and statement that the answer to the original question is also independent of the Lorentz transform, (i.e. has nothing to do with Lorentzian transforms or four-vectors).

15. Oct 14, 2010

### tom.stoer

Einstein's derivation was not directly based on the Lorentz trf., instead he found the Lorentz transformation properties; so we know that his derivation and the Lorentz transformation are deeply related. Length contraction and time dilation are nothing else but special cases cases of the Lorentz transformation.

The Lorentz transformation itself follows from various reasons; a physical reason is the Lorentz-covariance of Maxwell’s equations which can be checked both theoretically and experimentally; in addition one finds from Maxwell’s theory a constant speed c of el.-mag. waves.

In the following we need the Lorentz transformation in momentum space. It is structurally identical to the trf. for space-time itself. Einstein gave a derivation for E and p, but the Lorentz symmetry can be seen independently via transforming Maxwell’s equation into momentum space.

The idea is rather simple:

Take the energy-momentum four-vector (in the following c=1)

$$p^\mu = (E, \vec{p})$$

It transforms under a Lorentz-transformation as

$$p^\mu \to (p^\prime)^\mu = \Lambda ^\mu_\nu p^\nu$$

A Lorentz transformation is exactly the transformation that leaves the four-vector "length" invariant:

$$p^\mu p_\mu = (p^\prime)^\mu (p^\prime)_\mu$$

Calculating the invariant length one finds

$$p^\mu p_\mu = E^2 - p^2$$

The reason why the same trf. applies in position and in momentum space is that the expression

$$p^\mu x_\mu$$

must be invariant as well.

As you are familiar with Einstein's derivation of the expressions for E(v) and p(v) you can simply check that

$$p^\mu p_\mu = E^2 - p^2 = m^2$$

holds for all v, that means of all reference frames. Use

$$\gamma(v) = \frac{1}{\sqrt{1-v^2}}$$

$$E(v) = \gamma(v) m$$

$$p(v) = \gamma(v) mv$$

and do the calculation explicitly.

This is all you need. You have the relation between E, p, v and m and you have an invariant "length" m. All what remains is to find out what this m means physically. The simplest way is to make a Taylor expansion for small velocities. You get

$$E(v) = \gamma(v) m = m + \frac{m}{2}v^2 + \ldots$$

That means that you must identify m with the rest mass of the particle. So rest energy and rest mass are related via

$$E = m$$

Introducing the omitted factors of c again into the equations results in

$$E = mc^2$$

So rest mass and rest energy are related via this famous equation.

Your statement that the derivation can be done w/o using the Lorentz transformation is not really true. The Lorentz transformation is somehow hidden in all these expressions, but in principle it is the underlying mathematical structure.

Last edited: Oct 14, 2010
16. Oct 14, 2010

### DrDu

Nowadays, rest mass is defined as a Casimir invariant of "massive representations" of the Poicare algebra. So Lorentz invariance lies at the very heart of the definition of mass.

17. Oct 14, 2010

### tom.stoer

Exactly

18. Oct 14, 2010

### Staff: Mentor

Since mass has units of kg and energy has units of kg m²/s² it is clear that the conversion factor must have units of m²/s². The only combination of universal constants with those units is c².

19. Oct 15, 2010

### enotstrebor

If you would actually go and look on wiki at the evidence you are given, rather than just insisting you are right, you might find there are other avenues you haven't considered.

20. Oct 15, 2010

### tom.stoer

could you give an explanation of the ideas from on wiki?
do you have a link? there are many sites on wiki