# In E=mc2, Why C?

#### Mileman10

Basic Question:

Of the infinite number of other values which could have been the multiplier in E=mc2, it surely cannot be a coincidence that the value of the speed of light squared was the number. So why c?

Ted

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#### Dale

Mentor
Hi Mileman10, welcome to PF!

What other combination of dimensionful universal constants could you use to get dimensions of L²/T²?

#### Mileman10

And, (big disclosure here), I'm just an interested novice, trying to understand things conceptually, if that's possible.

In my college years, I studied Greek at Oxford, and was privileged to play chess once with Paul Dirac. (He won.) He explained basic stellar evolution in an hour, and was so clear and simple even I could get it.

That's what I was hoping for here. So, please, if possible, keep it basic.

Thanks,

Ted

#### Doc Al

Mentor
Why do you say that? I think it was a reasonable answer.

Do you realize that the speed of light plays a special role in relativity? Assuming you do, is it so surprising that c appears in many of the equations of relativity, including this one?

#### Dale

Mentor
OK, let me try again.

We are looking for a universal factor which converts a mass into an energy. So just by looking at the units we know that the factor that we are looking for must have units of L²/T². So we look at the universal constants (e.g. c, G, h, etc.) and see how we can compose a factor with the right units of L²/T². There are only a small number of such constants so we quickly find that the only way to do so is to use c².

So the c² should be completely clear just from dimensional considerations. What is not obvious is why it is 1c² instead of 0.5c² or something else. But for some reason nobody ever asks about the subtle question and everyone focuses on the obvious question.

#### Mileman10

I appreciate your explanations, but on a very basic level, c is just a very large number, and c squared even more so.

Forgetting desired units for the moment, this large number is most often used in simple discussions of the power of atomic fusion to explain how a pea-sized amount of Uranium could level Hiroshima, for example. The potential energy in even tiny amounts of matter is enormous, by a factor of c2.

Please explain, why it is that particular factor, and not a google, for example.

Ted

#### Dale

Mentor
I appreciate your explanations, but on a very basic level, c is just a very large number, and c squared even more so.
No, c is not just a very large number. In fact, when doing relativity most of the time we use units where c=1. The actual numerical value of c is not very relevant and is completely arbitrary since we can set it to any number we wish simply by choosing our units appropriately.

The important thing about c is not its numerical value, but its units, and the fact that it is frame invariant, finite, and isotropic.

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#### Mileman10

OK, I think I'm beginning to follow, a little.

So is it still accurate to say that "Energy equals mass times the speed of light squared", or is this technically incorrect?

#### Dale

Mentor
Yes, that is accurate.

#### K^2

Again, one has to be very careful about rest mass vs inertial mass. E = mc² holds if m is the inertial mass, but the symbol m is rarely used for it in relativity. For object in motion, a better equation is E² = p²c² + (mc²)². Here, m is the rest mass of the object, and p is momentum of the object. Note that this equation gives correct energy even for massless particles, such as light.

#### ChrisJackson

Basic Question:

Of the infinite number of other values which could have been the multiplier in E=mc2, it surely cannot be a coincidence that the value of the speed of light squared was the number. So why c?

Ted

Piggybacking off Ted's question, why not just say that the equation is E=ML? [ Where L = kinetic energy in light MV^2 with no M]

The equation E= MC^2 is solving for total energy in a quantity of matter and since light is weightless, Kinetic energy in light is simply its velocity, no? What is the difference? Are we really just parsing terms?

The implication is that matter isn’t simply energy like Einstein said. Matter is a specific type of energy, its just light in another form. It is bound up and tied in loops or knots or whatever. I can see why Einstein might not want to make that claim. It sounds outrageous and maybe a little unscientific. Is that the reason we stick to C^2 or am I off base?

How much energy is in one Kilogram of matter?
E=MC^2 => E =C^2 => E= the amount of energy in light

Summary: The total amount of energy in a particular bit of matter is just a question of how much light is wrapped up in the matter.. The answer is the quantity of matter multiplied by energy of light (C^2)

Is there something I am missing here? If light were the basic building block of matter, it would come as no surprise that C^2 keeps popping up all over the place.

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#### Matterwave

Gold Member
What's L in that equation of yours?

You are getting a lot of the basics of physics wrong. Try reading some faqs or something. The equation E=mc^2 does not read like "energy is equal to matter times light squared". A statement such as that makes no sense.

#### Drakkith

Staff Emeritus
2018 Award
Piggybacking off Ted's question, why not just say that the equation is E=ML? The equation is solving for total energy in a quantity of matter and since light is weightless, its energy reduces to its velocity, no? What is the difference? Are we really just parsing terms?
We don't make a unit equal to C^2 because the value C is a constant and appears in more than one place. It makes it much easier to use it squared or whatnot instead of making a new term in a couple of equations.

Also, E=MC^2 is not useful for light as it has no mass. E² = p²c² + (mc²)² is actually the full equation. We simplify it to E=MC^2 to convert the invariant mass of an object to energy. In the full equation you would still have to square whatever you set C^2 equal to, so we can just use C and square it and it will be easier.

#### Agerhell

I think that Mileman10 comes up with a very good question that deserves a better answer. Perhaps someone has a link to the original paper where Einstein is supposed to have come up with this relation?

Regarding the energy containment in a kilogram of matter it is also dependent on the gravitational field. A one kilogram steel ball at rest placed very close to the Schwarzschild radius of a black hole should contain a lot less energy than the same steel ball placed on earth right?

#### ChrisJackson

What's L in that equation of yours?

You are getting a lot of the basics of physics wrong. Try reading some faqs or something. The equation E=mc^2 does not read like "energy is equal to matter times light squared". A statement such as that makes no sense.
Thanks for the response. I didn't say "energy is equal to matter times light squared". I said that it seems like E=MC^2 is similar to Energy = (the amount of matter) ( the amount of energy in light) and that the amount of energy in light is equal to the kinetic energy of light. I guess it would be fair to say that the kinetic energy of light is 0 since Ke = mv^2 would be a large number multiplied by 0.

#### Matterwave

Gold Member
c^2 is by no means some measure of "amount of energy in light". The energy of a single photon is given by E=hf where f is the frequency and h is the planck's constant. It is also E=pc. c^2 doesn't even have units of energy.

All of light's energy is kinetic. It has no rest energy.

#### questionpost

Why does it use c? Is it because at that speed matter would have to be equivalent to energy in order to travel that speed?

#### Matterwave

Gold Member
Why does it use c? Is it because at that speed matter would have to be equivalent to energy in order to travel that speed?
Matter (with mass) can never move at c. The speed of light, c, is simply a given constant in the theory of relativity. We assume a "maximum speed of information transmission", and call that c. This constant pops up obviously because we have used it as our main axiom. What value you give c depends purely on the choice of units. We just have to posit that such a maximum exists.

Why does such a maximum exist? Because the experiments show that it does. As far as special relativity is concerned, this is a postulate that requires experimental evidence, but is not proved by the theory itself. No theory can prove it's own axioms. That would be tautological.

#### ChrisJackson

We don't make a unit equal to C^2 because the value C is a constant and appears in more than one place. It makes it much easier to use it squared or whatnot instead of making a new term in a couple of equations.

Also, E=MC^2 is not useful for light as it has no mass. E² = p²c² + (mc²)² is actually the full equation. We simplify it to E=MC^2 to convert the invariant mass of an object to energy. In the full equation you would still have to square whatever you set C^2 equal to, so we can just use C and square it and it will be easier.
Understood.. E² = p²c² + (mc²)² adjusts for relative inertia of a mass. For the initial intent E=MC^2 was fine since the momentum of the matter in question wasn't significant.

#### ChrisJackson

We don't make a unit equal to C^2 because the value C is a constant and appears in more than one place. It makes it much easier to use it squared or whatnot instead of making a new term in a couple of equations.

Also, E=MC^2 is not useful for light as it has no mass. E² = p²c² + (mc²)² is actually the full equation. We simplify it to E=MC^2 to convert the invariant mass of an object to energy. In the full equation you would still have to square whatever you set C^2 equal to, so we can just use C and square it and it will be easier.
c^2 is by no means some measure of "amount of energy in light". The energy of a single photon is given by E=hf where f is the frequency and h is the planck's constant. It is also E=pc. c^2 doesn't even have units of energy.

All of light's energy is kinetic. It has no rest energy.
E=hf, this is the answer I was looking for.. thanks

#### Agerhell

Matter (with mass) can never move at c. The speed of light, c, is simply a given constant in the theory of relativity. We assume a "maximum speed of information transmission", and call that c. This constant pops up obviously because we have used it as our main axiom. What value you give c depends purely on the choice of units. We just have to posit that such a maximum exists.

Why does such a maximum exist? Because the experiments show that it does. As far as special relativity is concerned, this is a postulate that requires experimental evidence, but is not proved by the theory itself. No theory can prove it's own axioms. That would be tautological.
Now do we really take E=mc2 as an axiom or is it possible to derive that relation from the assumption that the energy of a moving object varies with the Lorentz factor? Or from some other assumption?

#### Drakkith

Staff Emeritus
2018 Award
Now do we really take E=mc2 as an axiom or is it possible to derive that relation from the assumption that the energy of a moving object varies with the Lorentz factor? Or from some other assumption?
Take an electron and annihilate it with a positron and measure the energy of the gamma rays produced. The energy of the photons will equal the rest mass of the two particles + any kinetic energy they had. It can be experimentally verified and has been. The Large Hadron Collider has measured trillions of particle collisions in the last few years, each one is 100% dependent on the validity of the equation. If that isn't verification, then I don't know what is.

#### Drakkith

Staff Emeritus
2018 Award
Regarding the energy containment in a kilogram of matter it is also dependent on the gravitational field. A one kilogram steel ball at rest placed very close to the Schwarzschild radius of a black hole should contain a lot less energy than the same steel ball placed on earth right?
Yes and no. It depends on where you measure each one from. Measuring the energy of each from a local frame near each ball will give you the same energy for each. However if you measure the energy of the ball on Earth from the frame of the black hole, it has far more energy. For example, take the ball and annihilate it with another ball made of antimatter. If you measure the energy of the emitted photons from a frame near the black hole the photons will be extremely blue shifted and have far more energy than they would if you were measuring the event on Earth. The opposite would happen if you did the same thing near the black hole and measured it on Earth.

In it's own frame an objects mass (and therefore the energy you can convert the mass into) does not change no matter the gravitational field it's in.

#### Agerhell

Take an electron and annihilate it with a positron and measure the energy of the gamma rays produced. The energy of the photons will equal the rest mass of the two particles + any kinetic energy they had. It can be experimentally verified and has been. The Large Hadron Collider has measured trillions of particle collisions in the last few years, each one is 100% dependent on the validity of the equation. If that isn't verification, then I don't know what is.
Yes absolutely, it is an empirically well proven fact. That is not the question here though, the question is if this fact can be derived from some underlying assumption.