Light Photons and Mass

[kilopi]

I'm going to concede your point, because this is an argument of terminology. You're arguing that if someone on earth gives their weight in kilograms, they are correct because the mass they give is commonly thought of as "equivalent" to a certain weight.

I may be going out on a limb here in defending JS, but I'm willing to bet dollars to donuts that he definitiely did not mean that!

In fact, because that is such a common well-understood error, and such a gross mischaracterization of what he said, that I think you should apologize.

[dlack]

I may be going out on a limb here in defending JS, but I'm willing to bet dollars to donuts that he definitiely did not mean that!

In fact, because that is such a common well-understood error, and such a gross mischaracterization of what he said, that I think you should apologize.

I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?

[kilopi]

I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?

Until he gets done with class, we'll talk. What do you say about it? Do you agree with all four of those? Or, is there any of them that you definitely do not agree with?

[dlack]

I'd rather hear what he has to say about it, but:

W = mg

E = mc^2

Force = mass

Energy = mass

What's the difference?

Until he gets done with class, we'll talk. What do you say about it? Do you agree with all four of those? Or, is there any of them that you definitely do not agree with?

I would say that the first two are true while I have issues with the second two.

I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances. Likewise we could speak of the weight of a particle instead of its mass, because for a given mass having a given acceleration due to gravity, we could calculate its weight.

However, to me, that doesn't say that weight is the same as mass or that energy is the same as mass. That says that the weight of a particle is intimately related to its mass, and that the rest energy of a particle is intimately related to its rest mass.

For me, the statement that mass and energy are so intimately related that a certain amount of energy can be created from a certain amount of mass is a strong enough statement without trying to say mass and energy are "the same thing." They are closely related, but not "the same thing."

I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.

[Bill Thmpson]

I'm going to concede your point, because this is an argument of terminology. You're arguing that if someone on earth gives their weight in kilograms, they are correct because the mass they give is commonly thought of as "equivalent" to a certain weight.

I still don't like the idea of saying mass and energy are "the same." A certain amount of mass can be converted to a certain amount of energy under certain circumstances, but to me, that doesn't mean they are "the same."

I think everyone should be able to agree that to get energy from mass, you have to do something to the mass to convert it into energy.

I just thought I would run that up the flag pole and see if anyone salutes.

I also like this weight VS mass argument. stated here as an analogy. Weight and Mass are not one and the same thing. THis is a good point. It also shows how everyone can think something is true and yet what they think is completely wrong.

Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.

Also, everyone use bathroom scales as a measurement of weight. Strangely, since it works on a spring mechanism, it really does measure weight... but the dial shows Pounds or Kilograms which is not correct. Kilograms is a measurement of mass. Take the bathroom scale up to the moon, and it would give you an incorrect measurement of mass.

And, everyone seems to overlook that the doctor's scale -- or the scale at a gym -- that uses a counter-balance system really does measure mass (while the bathroom scale measures weight because of the pull on a spring, the doctor's scale compares you against other object's graviataional pull, but the comparison would be the same on any gravity -- even on the moon, since it is a comparison of objects equally affected)-- take that mechanism (the doctor's scale) to the moon and it will give you the same result as on the Earth because the mass would not change. But the doctor says "step on the scale and let's see how much you weigh" not, "step on the scale and let's measure your mass" which would be more correct.

All this shows how people can get things all mixed up. The same is true for this discussion. And it is not clear who has the proper scientific credentials to back up what he or she is saying.

[kilopi]

I would say that the first two are true while I have issues with the second two.

Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances.

And how much would that be?

I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.

It's not as arbitrary as that, though. It's c and G that are set equal to one. Since g = GM/r^2, so g has units of Kg/m^2, but G/c^2 would also be one, so meters and kilograms cancel to leave meters in the denominator. So, mass is measured in meters, but weight is not.

It also shows how everyone can think something is true and yet what they think is completely wrong.

Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.

Well, clumsy as it is, pound is a measure of weight, so I guess that's a good example.

All this shows how people can get things all mixed up. The same is true for this discussion. And it is not clear who has the proper scientific credentials to back up what he or she is saying.

Credentials? We don't need no steenkin' credentials.

[daver]

Everyone uses Kilograms or Pounds as a measurement for weight. But the correct measurement should be in Newtons.

Oh well. Nit picking time. "Pounds" is a legitimate unit of force; its associated unit of mass is a "slug". A "pound" is also a legitimate unit of mass; its associated unit of force is a "poundal". One of the many reasons people prefer to use the metric system.

Most people probably use "weight" when they're really concerned with "mass". And in most people's lives, the difference is insignificant. If the force of gravity in Europe were 10% higher than in the US there might be some incentive to disambiguate. However, for rather obvious reasons, the people who concocted the language weren't aware of the difference. I don't think you'll have much luck changing common parlance.

Sometimes people who know should know better get the two confused. Thrust being measured in kg is one example. Isp in 1/sec is another.

[dlack]

I would say that the first two are true while I have issues with the second two.

Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.

I understand how someone could speak of the energy of a particle instead of its mass, because for a certain mass, we can calculate the energy that can be gotten from it in the proper circumstances.

And how much would that be?

If you have two particles with given mass and momentum colliding to give you photons, the energy is given by E^2 = p^2c^2 + m^2c^4. (Notice their momentum matters).

However, you couldn't convert a single particle of given mass into a photon for the same reason you can't convert a single photon into a particle of given mass. (I show why below).


The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.

The equation E^2 = p^2c^2 + m^2c^4 must be satisfied. Conservation of momentum must be satisfied. Conservation of Energy must be satisfied. Conservation of charge must be satisfied (start with a neutral photon, you better end up with a net charge of zero). Other conservations probably must be satisfied that I haven't studied yet.

A photon is never stationary (it moves at speed, um, c), so it never has zero momentum, so E = mc^2 does not apply.

For a similar reason you couldn't take one photon and convert it into one particle of a certain mass, because momentum will never be conserved.

The initial energy of the system is (for a photon):
E1 = p1c
The final energy (for a single massive particle):
(E2)^2 = (p2c)^2 + (mc^2)^2
Which means:
(E1)^2 = (E2)^2 = (p1c)^2 = (p2c)^2 + (mc^2)^2

But, remember momentum has to be conserved in the collision. The only possible way that p1 can equal p2 here (for a single particle) is if the mass of the new particle is zero! So a photon can't be changed into a single particle of given mass.

Now I think it would be possible to convert a single photon into two particles, but then the mass of the particles would depend on how much momentum each particle had, so you can't say a photon can give you a certain amount of mass.

I do understand how the argument could be made that you could set c = 1 and have mass = energy, but then you have to admit that I can set g = 1 and have mass = weight.

It's not as arbitrary as that, though. It's c and G that are set equal to one. Since g = GM/r^2, so g has units of Kg/m^2, but G/c^2 would also be one, so meters and kilograms cancel to leave meters in the denominator. So, mass is measured in meters, but weight is not.

I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently). And what's the significance of the fact that G/c = 1 here? They don't ever appear in the same equation (of the ones you've used, anyways). Your argument looks like an argument to show that Force can be measured in (kg^2)/(m^2).

W = mg = (kg)(kg/m^2)

Which goes back to what I said before. If you want to think of it that way, you can, and apparently it's useful sometimes, but what you are doing is playing with units. You aren't proving that mass is the same thing as energy or that force is the same thing as mass or whatever is measured in (kg^2/m^2).

[kilopi]

I would say that the first two are true while I have issues with the second two.

Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.

No, that's using your reasoning. Of course, you think you're duplicating his reasoning, but I don't think you are. That's what makes it a straw man.

The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.

Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently).

No, I wasn't claiming that you said it was. I was merely following the scheme laid out in physics textbooks. For instance, the sidebar on page 29 of Misner, Thorne, and Wheeler's huge Gravitation.

[dlack]

I would say that the first two are true while I have issues with the second two.

Kinda what I figured. But he didn't say the third one, so it's unfair to include it. That's called a straw man.

My point is that, using his reasoning, if the fourth follows from the second, the third must follow from the first.

No, that's using your reasoning. Of course, you think you're duplicating his reasoning, but I don't think you are. That's what makes it a straw man.

I'd still like to know if he agrees with my use of his reasoning.

The energy of a massless photon can be converted to mass, but it's not a simple matter of E=mc^2.

Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I already addressed this:

If you have two particles with given mass and momentum colliding to give you photons, the energy is given by E^2 = p^2c^2 + m^2c^4. (Notice their momentum matters).

However, you couldn't convert a single particle of given mass into a photon for the same reason you can't convert a single photon into a particle of given mass. (I show why below).

A single particle can't be converted into a single photon of energy because momentum won't be conserved. If a couple particles are used, or however you do it (provided you conserve energy, momentum, etc.), the energy is given by the formula.

I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/reso...photonmass.htm

I'm not sure what you're doing here. I never said weight was measured in meters (which is what you're trying to disprove, apparently).

No, I wasn't claiming that you said it was. I was merely following the scheme laid out in physics textbooks. For instance, the sidebar on page 29 of Misner, Thorne, and Wheeler's huge Gravitation.

You said:

So, mass is measured in meters, but weight is not.

I took this to mean that you were trying to show that weight isn't measured in meters.

I still don't understand what you were trying to do.

I don't have your particular textbook, so I don't know what exactly they are doing on that page, but I'm looking at p. 324 of Physics by Tipler, and Newton's Law of Gravity is laid out fairly simply. Nowhere does it say anything about measuring mass in meters. In no formula does G/c appear.

(edit: particluar. blah.)

[kilopi]

I'd still like to know if he agrees with my use of his reasoning.

Well, if he did, I wouldn't, so we can continue.

Well, try the other way then. What if you had a body of mass M in space that was gravitationally affecting other bodies, and you converted it entirely to energy. (1) How much energy would you get? (2) What would the effect be on the other bodies, gravitationally?

I already addressed this:

OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?

I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/reso...photonmass.htm

That link discusses rest mass, and no one in this thread disputes that the photon has zero rest mass. That doesn't seem to be of any contention. Since you're unaware that photons contribute to gravitational attraction, I guess your answer to (2) would be zero?

I still don't understand what you were trying to do.

::snip::

Nowhere does it say anything about measuring mass in meters. In no formula does G/c appear.

They're called geometrized units, and I apologize, they appear on page 36, not page 29, of MTW.

Basically, to address the first comment at the top of this post, I was illustrating why the reasoning could not be extended to include the formula that you thought it could be, using your reasoning. That is, speaking for myself and not necessarily JS, the reason for our conclusions is not just a simple matter of re-interpreting any old equal sign.

[dlack]

OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?

NO!

If the momentum of the initial mass is zero, it can't be converted into a single photon of given energy!

Conservation of momentum must be obeyed, and a photon never has zero momentum. So if you try to convert a mass M with momentum 0 into a photon of energy, you get . . . a mass M with momentum 0.

I think you could possibly have a particle change into several photons moving in opposite directions. Then momentum is conserved, and the total energy would be mc^2. So the energy of each photon depends on its individual wavelength.

I'd like to see some sources about the idea that photons have gravitational attraction. I don't think gravity is part of this discussion, but you keep bringing it up. Photons don't have mass. A quick google search gives:
http://www.usatoday.com/weather/reso...photonmass.htm

That link discusses rest mass, and no one in this thread disputes that the photon has zero rest mass. That doesn't seem to be of any contention. Since you're unaware that photons contribute to gravitational attraction, I guess your answer to (2) would be zero?

I'm saying that I haven't studied the effects on nearby particles of converting massive particles (with gravitational attraction) into photons of energy (which don't have gravitational attraction). My guess would be that the effects of gravity are pretty small for this case since we're dealing with quantum particles with extremely small mass. No one's talking about converting entire planets into energy, right?

[dlack]

They're called geometrized units, and I apologize, they appear on page 36, not page 29, of MTW.
Basically, to address the first comment at the top of this post, I was illustrating why the reasoning could not be extended to include the formula that you thought it could be, using your reasoning. That is, speaking for myself and not necessarily JS, the reason for our conclusions is not just a simple matter of re-interpreting any old equal sign.

According to your link:

As a result of converting to geometrized units, all quantities are expressed in terms of a unit of distance

So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

As I said, I'm not saying this isn't a legitimate way of playing with units to make calculations easier, I'm saying this doesn't show that energy is the same thing as mass. They're intimately related and can be converted into one another, but they aren't the same thing.

[kilopi]

OK, since the momentum of the mass M is relatively zero, your equation reduces to E=mc^2, as you'd think. So your answer to (1) would be Mc^2?

NO!

I'm surprised at the vehemence of this answer.

If the momentum of the initial mass is zero, it can't be converted into a single photon of given energy!

Especially if that mass M is even as large as a baseball maybe. But why must the energy be just a single photon?

Conservation of momentum must be obeyed, and a photon never has zero momentum. So if you try to convert a mass M with momentum 0 into a photon of energy, you get . . . a mass M with momentum 0.

I think you could possibly have a particle change into several photons moving in opposite directions. Then momentum is conserved, and the total energy would be mc^2.

But, that's what I said, Mc^2. Now I'm even more suprised at your first answer.

I'm saying that I haven't studied the effects on nearby particles of converting massive particles (with gravitational attraction) into photons of energy (which don't have gravitational attraction). My guess would be that the effects of gravity are pretty small for this case since we're dealing with quantum particles with extremely small mass. No one's talking about converting entire planets into energy, right?

Sure, why not?

OK, let's start with a baseball. So, your answer to (2) would be zero?

[kilopi]

So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

No, energy can be expressed in kilograms, it's just not the standard way of it. If ergs can be expressed in centimeters, and grams expressed in centimeters, it should be easy to see how the transition from ergs to grams could be made.

[Ring]

Mass is equal to the energy of a system that cannot be transformed away. In other words if the system has a zero momentum frame then the system has mass. So a single photon cannot have mass, but a system of photons can.

This is also why mass cannot be converted to energy. If a closed system initially has zero momentum then it must always have zero momentum and therefore the mass cannot change. A vault containing a nuclear weapon weighs the same both before and after detonation.

Even though a single photon doesn’t have mass it does have energy and energy effects the geometry of spacetime. However if you travel along with a beam of light you can reduce its energy to any value you please and this is why curvature is covariant not invariant.

[kilopi]

Mass is equal to the energy of a system that cannot be transformed away. In other words if the system has a zero momentum frame then the system has mass. So a single photon cannot have mass, but a system of photons can.

That's still "rest mass." Everybody seems to agree that a photon has zero rest mass.

This is also why mass cannot be converted to energy. If a closed system initially has zero momentum then it must always have zero momentum and therefore the mass cannot change. A vault containing a nuclear weapon weighs the same both before and after detonation.

There is no energy released in a nuclear explosion?

Actually, that's probably an illustration of the point. The gravitational effect doesn't change, even after the mass has been converted to energy.

Even though a single photon doesn’t have mass it does have energy and energy effects the geometry of spacetime.

I think that's what we're talking about.

However if you travel along with a beam of light you can reduce its energy to any value you please and this is why curvature is covariant not invariant.

You can't travel along with a beam of light. You can set a reference frame to do so. Or am I wrong?

[Ring]

]
There is no energy released in a nuclear explosion?

Fusion/fision isn't about converting mass to energy its about converting potential energy to kinetic and electromagnetic energy. In other words it's about changing one type of energy into another. The remnant of the nucleus has less internal energy and as a result of this decrease in internal energy it has less mass, not vice versa. Mass and energy are properties of a system, and not things in their own right

You can't travel along with a beam of light. You can set a reference frame to do so. Or am I wrong?

Well, if you move towards a light source it will appear blueshifted and vice versa, I think that's what I was actually trying to say.

[dlack]

Kilopi, my insistence that the energy of a mass can't be released in a single photon is due to the fact that I thought the discussion was about assigning a "mass" to a photon. You can't assign any kind of mass to a single photon. That's where my vehemence comes from. Apparently the discussion has shifted from there, but it seemed like you were trying to get back to the "a photon has some kind of mass associated with it" issue.

My other point has been that when you transform energy into mass, the mass does not come from the (partial) equation E = mc^2. The whole equation I've typed several times must be used.

Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.

As for the effect this has on gravity, I don't know the answer. I'm sure there would be some dramatic results if you took an entire planet and converted it into energy. I don't think that loss of gravity would be the biggest concern in such a scenario, however. I think the resulting enormous amount of energy would be the biggest concern.

As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

[daver]

As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

c is a universal constant, it's the same on the earth, the moon, or omicron ceti 5. g varies, even on earth. I memorized 9.80665 m/sec/sec a while back, i don't know if that is currently accepted as the standard or not.

[Ring]

Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.

Use your own equation:

m^2 = E^2 - p^2 (c=1)

If the system has zero momentum before detonation then it must have zero momentum after detonation. So m = E = no conversion of mass to energy.

Nucleons have orbitals similar to electrons in an atom. And the nucleons in these orbitals have potential energy. When a reaction occurs the nucleus rearanges itself and the potential energy decreases (binding energy increases) and this decrease in potential energy results in kinetic and electromagnetic energy. (This is a gross simplification). There's a local mass defect but the system energy and mass remains the same.

[dlack]

Ring is absolutely wrong. Mass can be converted into energy, and energy into mass. This occurs in the explosion of a nuclear weapon, and it occurs in the formation of molecules.

Use your own equation:

m^2 = E^2 - p^2 (c=1)

If the system has zero momentum before detonation then it must have zero momentum after detonation. So m = E = no conversion of mass to energy.

Suppose a system of mass m has momentum p. The energy of this system is found using the equation.

If this mass is converted into energy, the total momentum of the (massless) photons of energy will be equal to p, the initial momentum. The total energy of the photons will be equal to the initial energy.

Thus, momentum is conserved and energy is conserved. MASS is NOT conserved.

I'm taking a break from this thread for the day. If someone could help Ring out, that would be great.

[dlack]

As for the issue with the units, I've conceded that you can play with units in order to make calculations simpler. I don't understand why you won't concede that if you do it with c, I can just as easily and legitimately do it with g.

c is a universal constant, it's the same on the earth, the moon, or omicron ceti 5. g varies, even on earth. I memorized 9.80665 m/sec/sec a while back, i don't know if that is currently accepted as the standard or not.

Alright, I see how that difference can be argued. I concede this point.

[Ring]

Suppose a system of mass m has momentum p. The energy of this system is found using the equation.

If this mass is converted into energy, the total momentum of the (massless) photons of energy will be equal to p, the initial momentum. The total energy of the photons will be equal to the initial energy.

Thus, momentum is conserved and energy is conserved. MASS is NOT conserved.

Question: What physical system are we talking about?
Answer: A stationary nuclear weapon

Question: What’s the momentum of a stationary nuclear weapon?
Answer: Zero.

Question: What’s the momentum of this system after it detonates?
Answer: Zero. It’s a closed system so momentum must be conserved.

Question: Given m^2 = E^2 – p^2 what’s the mass of the detonated system?
Answer: The same as the pre detonated system, since p =0, m = E!

SYSTEM MASS IS a universally conserved quantity.

[dlack]

Ok, I couldn't stay away for a whole day.

MOMENTUM IS A VECTOR!

The magnitudes of the individual momenta of the photons doesn't have to be zero, the vector sum of the momenta must be zero.

Suppose we have a massive particle at rest at the origin.

Initial momentum is zero.

Now suppose this particle is converted into massless photons of energy.

As long as the momentum in the +x direction is the same as the momentum in the -x direction, momentum is conserved.

The sum of the momenta of the photons is zero.

Mass has been converted into energy.

The algebra may be a little more complicated for a 3-D space, but the idea is the same. The vector sum of the momenta must be zero.

[Ring]

Ok, I couldn't stay away for a whole day.

MOMENTUM IS A VECTOR!

So is the energy monentum four-vector and its magnitude is the system mass, its time component is energy and its space component is momentum.

m^2 = e^2 - p^2

The magnitudes of the individual momenta of the photons doesn't have to be zero, the vector sum of the momenta must be zero.

That's correct and since the sum of the momenta equals zero: m = E

Suppose we have a massive particle at rest at the origin.

Initial momentum is zero.

And therefore the final momentum of the system must also equal zero and: m =E

As long as the momentum in the +x direction is the same as the momentum in the -x direction, momentum is conserved.

The sum of the momenta of the photons is zero.

That's correct and since p = 0: m = E.

Mass has been converted into energy.

Incorrect. m = E

[dlack]

When I say mass, I mean rest mass.

We started out with a particle of mass m.

We got photons with mass zero.

There was less mass after the "event" than there was before the "event."

Mass was not conserved.

[Ring]

When I say mass, I mean rest mass.

So do I. That's what the magnitude of the energy momentum four-vector is.

We started out with a particle of mass m.

We got photons with mass zero.

The individual photons have zero mass but the system comprised of two photons with anti parallel momentum has a mass of m = E.

There was less mass after the "event" than there was before the "event."

No, there is exactly the same amount of mass both before and after the event because p of the system = 0 and therefore m = E.

Mass was not conserved.

Not true. m = E.

[dlack]

The individual photons have zero mass but the system comprised of two photons with anti parallel momentum has a mass of m = E.

I have real issues with this statement. Do you have some sources as to how a system comprised of massless particles can be said to have mass?

[Bill Thmpson]

So if you're using geometrized units, all quantities are in centimeters, so energy can't be expressed as kilograms, as was stated initially to start this particular aspect of the debate.

Is this a thowback to my comment about radius and area of a circle?

If so, then it is wrong.

Area would be centimeters sqared, not centimeters.