That famous equation - a better understanding

[DarkMatterHol]

OK Speed = Distance / time . For example Miles / Second .

I find this entirely intuitive and easy to understand.

Now with E = MC^2 i still have some problems. For example what units are we dealing with?

Is it joules = grams * speed of light in miles per second? Or is this equation not similar to the speed equation above? Does this have something to with dimensioned vs. dimensionless equations? For example, can i determine using this equation, how long a gram of material will run my household if i could convert it all into energy?

I am not studying physics formally but i do follow what i can and not fully understanding this equation and why it is so has been bothering me. Any help would be much appreciated. Thanks.

 

[JesseM]

OK Speed = Distance / time . For example Miles / Second .

I find this entirely intuitive and easy to understand.

Now with E = MC^2 i still have some problems. For example what units are we dealing with?

Energy has units of distance * force, and force has units of mass * acceleration or mass * distance / time^2, so that means energy has units of mass * distance^2 / time^2. For example, 1 joule = 1 kilogram * meter^2 / second^2. So you can use any units you like as long as they're consistent on both sides, for example if you use joules for energy on the left side then you should use kilograms for mass and meters/second for speed on the right side.

For example, can i determine using this equation, how long a gram of material will run my household if i could convert it all into energy?

Yes, it would tell you that. It also works the other way in the sense that energy contributes to inertia (resistance to acceleration) which in Earth's gravity can be measured as weight...for example, if you take a brick and weigh it, then add a little heat energy to the brick (adding to the average kinetic energy of the molecules that make it up), in theory the brick would weigh slightly more by an amount that could be calculated from E=mc^2, though I think it would be so tiny as to be difficult to measure in practice. But as another example, it can be observed that a hydrogen atom, made up of one proton and one electron, actually has slightly less inertial mass than the sum of the masses of the proton and electron that make it up, because for a proton and electron the potential energy is lower when they are brought together into an atom than when they are far apart.

 

[HallsofIvy]

OK Speed = Distance / time . For example Miles / Second .

I find this entirely intuitive and easy to understand.

Now with E = MC^2 i still have some problems. For example what units are we dealing with?

Is it joules = grams * speed of light in miles per second?

You could but why? It is not a good idea to mix unit systems like that. In the English system (which the English do not use any more) M would be in "slugs" or "poundals" (the mass that one pound of force will accelrate at one foot per second), C in feet per second, and E in poundals*feet^2/sec^2, for which there is no specific name.

In the MKS (meter-kilogram-second) system, M would be in kilograms, C in meters/second, and E in Joules.

In the CGS (centimeter-gram-second) system, M would be in grams, C in cm/second, and E in ergs.

Or is this equation not similar to the speed equation above? Does this have something to with dimensioned vs. dimensionless equations? For example, can i determine using this equation, how long a gram of material will run my household if i could convert it all into energy?

Yes, theoretically, of course, there is no way to "convert it all into energy". But I don't understand your reference to "dimensioned vs. dimensionless equations". Both of these equations have dimensions. There are no dimensionless equations here.

I am not studying physics formally but i do follow what i can and not fully understanding this equation and why it is so has been bothering me. Any help would be much appreciated. Thanks.

 

[vilas]

OK Speed = Distance / time . For example Miles / Second .

I find this entirely intuitive and easy to understand.

Now with E = MC^2 i still have some problems. For example what units are we dealing with?

Is it joules = grams * speed of light in miles per second? Or is this equation not similar to the speed equation above? Does this have something to with dimensioned vs. dimensionless equations? For example, can i determine using this equation, how long a gram of material will run my household if i could convert it all into energy?

I am not studying physics formally but i do follow what i can and not fully understanding this equation and why it is so has been bothering me. Any help would be much appreciated. Thanks.

I think maths can never be wrong. But its application can be. Possibly you are uncomfortable with the equation itself and not the units. This equation is derived (wrongly) from kinetic energy of mass. It is therefore expected that there is no leftover energy associated with so called ‘rest mass’.
In the derivation, mass and velocity are treated as independent variables, whereas mass is dependent on velocity and if you proceed with the conversion and then integrate then you don’t get this term.
But this equation is found to be true. Mass does get converted into energy. So answer should be found in the relationship between mass and charge.

 

[DarkMatterHol]

Thank you all for your explanations. I'll think about it some more. I think it will help to unpack all the variables.

For example, expressing E as Joules = kilograms * meters^2 / seconds^2, and C as meters/second should help.

I suppose i need to study up on the Joule and how it was derived, and also on the speed of light and how it was determined. For example did the speed of light come out of Maxwell's equation or did the equations come out from the measured speed of light.

I'm starting to realize that it is the 'cleanness' of the equation that has me intrigued.

At first glance there appeared to be no arbitrary values except perhaps C, but grams, meters, C, and seconds are all arbitrary, aren't they? I mean they are all fundamental except C and C is derived from meters and seconds.

So, if grams, meters, seconds, and the speed of light are all arbitrary, then i still find it a bit of a mystery why E=MC2 is so 'clean'. Does my confusion make any sense? I'll keep working on it...

I guess what i really want to know is: Why does the energy [in Joules] contained in matter equal exactly MC^2?

 

[DarkMatterHol]

You could but why? It is not a good idea to mix unit systems like that. In the English system (which the English do not use any more) M would be in "slugs" or "poundals" (the mass that one pound of force will accelrate at one foot per second), C in feet per second, and E in poundals*feet^2/sec^2, for which there is no specific name.

In the MKS (meter-kilogram-second) system, M would be in kilograms, C in meters/second, and E in Joules.

In the CGS (centimeter-gram-second) system, M would be in grams, C in cm/second, and E in ergs.


Yes, theoretically, of course, there is no way to "convert it all into energy". But I don't understand your reference to "dimensioned vs. dimensionless equations". Both of these equations have dimensions. There are no dimensionless equations here.

Ok. I think i will be able to form my question a bit better thanks to your replys.

First of all, as a preliminary question let me ask you if I'm thinking about this right.
Using the MKS system, the energy [in Joules], contained in 1 KG of matter is equal to the speed of light squared, or 299,792,458 Joules.
Is that precisely correct?

 

[JesseM]

Ok. I think i will be able to form my question a bit better thanks to your replys.

First of all, as a preliminary question let me ask you if I'm thinking about this right.
Using the MKS system, the energy [in Joules], contained in 1 KG of matter is equal to the speed of light squared, or 299,792,458 Joules.
Is that precisely correct?

It's equal to the speed of light squared multiplied by 1 kg, which would be the square of 299,792,458, or 89,875,517,873,681,760 joules.

 

[DarkMatterHol]

It's equal to the speed of light squared multiplied by 1 kg, which would be the square of 299,792,458, or 89,875,517,873,681,760 joules.

Oops! my mistake. Thanks JesseM. That's what i meant to say.

I'm still trying to figure out why the quantity of energy in a mass is directly dependant on the speed of energy propagation as measured in any frame. If there's no easy intuitive explanation, can you put me on the right path to understanding this relationship?

It appears to me that there must be a logical reason for this relationship, as opposed to the mass of an electron vs. a proton, which i understand has no real explanation so far.

 

[JesseM]

Oops! my mistake. Thanks JesseM. That's what i meant to say.

I'm still trying to figure out why the quantity of energy in a mass is directly dependant on the speed of energy propagation as measured in any frame. If there's no easy intuitive explanation, can you put me on the right path to understanding this relationship?

c isn't exclusively the "speed of energy propagation", even if no particles or waves traveled at c it would still appear as a constant in equations for time dilation for example, it's built into the way space and time are related to one another (also, even just in terms of energy propagation, obviously the energy in massive particles or sound waves propagates slower than c!) If you're looking for a derivation, I think the most intuitive one may be a thought-experiment involving light being shot from one end of a box and absorbed by the other, then conservation of momentum and energy must apply throughout the whole process (before the photon has been emitted, while it's in flight, and after it's been absorbed) and from that you can get E=mc^2. See here for example:

http://www.adamauton.com/warp/emc2.html

 

[DarkMatterHol]

Thanks JesseM - i guess instead of 'energy propagation' i should have simply said 'light' or radiant energy, but it looks like you knew what i meant anyway.

I'll take a look at the information in the link you provided.

Thanks again!
Bob

 

[DarkMatterHol]

OK, i have gone over the explanation in your link, JesseM. It looks good.

I have a question on the first equation:

Pphoton = E/c

This is the opposite of matter that has momentum Pbox = Mv

So momentum for a photon is its Energy divided by its speed while the momentum of matter is its mass times its speed.

My first thought is that this seems weird.

If the speed of light happened to be half what it is, then a photon with energy E would have twice the momentum.

Can you explain why this is so? Or will i need to read and understand Maxwell's equations to understand this relationship? If so, maybe you could point me in the right direction since there are a very many Maxwell equations and topics.

Thanks,
bob

 

[GrayGhost]

OK, i have gone over the explanation in your link, JesseM. It looks good.

I have a question on the first equation:

Pphoton = E/c

This is the opposite of matter that has momentum Pbox = Mv

If the speed of light happened to be half what it is, then a photon with energy E would have twice the momentum.

Well, let's see ...

E = mc² ...

P_photon = E/c
mc = E/c​

If c magically became half its current value, then ...

m(c/2) = E/(c/2)
mc/2 = 2E/c
mass*velocity = 2E/c​

So at first glance it would seem you are correct. However ...

E = mc²​

so if c became c/2, then what happens to the energy? ...

E = m(c/2)²
E = mc²/4​

Thus, the energy must drop by the factor (1/2)² when light's speed drops by the factor of 1/2. So it seems to me that we really have this ...

mc=E/c
m(c/2) = E/(c/2)
mc/2 = 2E/c ... but energy drops by (1/2)², so ...
mc/2=2(E/4)/c
mc/2 = E/2c
mc = E/c
mc² = E​

So if the speed of light dropped by 1/2, the energy would not remain as it was. The energy E would drop by the factor (1/2)² = 1/4. Hence, the momentum drops by 1/2 as well. Yes?

The momentum is ... p=mc. So if c dropped to c/2, then the momentum becomes p=m(c/2) = 1/2 x mc, a 50% drop.

GrayGhost

 

[DarkMatterHol]

Well, let's see ...

E = mc² ...

P_photon = E/c
mc = E/c​

If c magically became half its current value, then ...

m(c/2) = E/(c/2)
mc/2 = 2E/c
mass*velocity = 2E/c​

So at first glance it would seem you are correct. However ...

E = mc²​

so if c became c/2, then what happens to the energy? ...

E = m(c/2)²
E = mc²/4​

Thus, the energy must drop by the factor (1/2)² when light's speed drops by the factor of 1/2. So it seems to me that we really have this ...

mc=E/c
m(c/2) = E/(c/2)
mc/2 = 2E/c ... but energy drops by (1/2)², so ...
mc/2=2(E/4)/c
mc/2 = E/2c
mc = E/c
mc² = E​

So if the speed of light dropped by 1/2, the energy would not remain as it was. The energy E would drop by the factor (1/2)² = 1/4. Hence, the momentum drops by 1/2 as well. Yes?

The momentum is ... p=mc. So if c dropped to c/2, then the momentum becomes p=m(c/2) = 1/2 x mc, a 50% drop.

GrayGhost

Oh wow. OK. Thanks GreyGhost.

Let me ask, because it will take me quite some time to go through and follow the math - my first reaction to this is to say that the momentum of a photon is directly proportional to it's speed. Is that true? If so this part is already becoming a little more intuitive Of coarse light speed is constant so it hardly matters, but it's nice to know to understand the first equation.

 

[GrayGhost]

Oh wow. OK. Thanks GrayGhost.

Let me ask, because it will take me quite some time to go through and follow the math - my first reaction to this is to say that the momentum of a photon is directly proportional to it's speed. Is that true?

Yes. In fact, the magnitude of the momentum of any entity (be it light or matter) is directly proportional to its speed ... momentum = mass x velocity.

If so this part is already becoming a little more intuitive Of coarse light speed is constant so it hardly matters, but it's nice to know to understand the first equation.

Glad I could help! You had me wondering there for a minute too :)

GrayGhost

 

[DarkMatterHol]

Yes. In fact, the magnitude of the momentum of any entity (be it light or matter) is directly proportional to its speed ... momentum = mass x velocity.
Glad I could help! You had me wondering there for a minute too :)

GrayGhost

OK so if that's true then the mass of a photon remains the same when the speed of light changes, but the photon's energy changes by the square of the velocity. right? OK, that sounds pretty basic then.

So as far as the E=MC2 goes then, it looks like it also applies to a massive particle in a similar way to a photon - an electron's energy would = it's rest mass times the square of it's velocity too would it not?
[sorry to ask such fundamental questions]

 

[cristo]

Yes. In fact, the magnitude of the momentum of any entity (be it light or matter) is directly proportional to its speed ... momentum = mass x velocity.t

This isn't true.

Well, let's see ...

E = mc² ...

P_photon = E/c
mc = E/c​

This is also incorrect, since a photon is massless, so m=0. In order to obtain the momentum of a photon you use
$$E^2=m^2c^4+p^2c^2$$
which, for a massless photon, reduces to
$$E=pc$$

 

[DarkMatterHol]

This isn't true.
This is also incorrect, since a photon is massless, so m=0. In order to obtain the momentum of a photon you use
$$E^2=m^2c^4+p^2c^2$$
which, for a massless photon, reduces to
$$E=pc$$

So back to my question then:

If Pphoton = E/c then...

What happens to the momentum of a photon if the speed of light were to be half it's value?
It looks like it would have twice the momentum at half the speed... That doesn't seem right, so what is the right answer?

Oh wait... by E = pc we get half the Energy at half the velocity, so does that mean that the momentum of a photon is unchanged at any speed for c? [if so, then the equation Pphoton = E/c would be kind of useless if Pphoton never changes, so i'll just wait for the answer before i get ahead of myself]

 

[DrGreg]

So back to my question then:

If Pphoton = E/c then...

What happens to the momentum of a photon if the speed of light were to be half it's value?
It looks like it would have twice the momentum at half the speed... That doesn't seem right, so what is the right answer?

Oh wait... by E = pc we get half the Energy at half the velocity, so does that mean that the momentum of a photon is unchanged at any speed for c? [if so, then the equation Pphoton = E/c would be kind of useless if Pphoton never changes, so i'll just wait for the answer before i get ahead of myself]

This is really pointless speculation. The value of c is fixed, so asking what if it had a different value is a question about a different universe than the one we live in. How do you compare two different universes?

Another way of looking at this is via the equation

$$p = \frac {Ev} {c^2}$$​

which is true for all particles, whether they be photons or not.

 

[GrayGhost]

This is also incorrect, since a photon is massless, so m=0. In order to obtain the momentum of a photon you use
$$E^2=m^2c^4+p^2c^2$$
which, for a massless photon, reduces to
$$E=pc$$

Indeed. Thanx much for the correction there cristo. I began to ponder that just after shutting down and hitting the sack very late, but did not feel like getting back up to mess with it, and I couldn't quite recall the relativistic formula precisely either. Never good to post at O-dark-30.

GrayGhost

 

[GrayGhost]

So back to my question then:

If Pphoton = E/c then...

What happens to the momentum of a photon if the speed of light were to be half it's value?
It looks like it would have twice the momentum at half the speed... That doesn't seem right, so what is the right answer?

Oh wait... by E = pc we get half the Energy at half the velocity, so does that mean that the momentum of a photon is unchanged at any speed for c? [if so, then the equation Pphoton = E/c would be kind of useless if Pphoton never changes, so i'll just wait for the answer before i get ahead of myself]

OK, so ... p_photon = E/c
p = hf/c ... h being Planck's constant and f being the photon frequency.

So the big question is ... what all changes in this equation if c changed? Does only p or f or h also change? Do some combination of these 3 change? If only the momentum changes with a change in c, then in the limit as c approaches zero, nothing could move. Relativity suggests that the speed-of-light is equivalent to the speed-of-proper-time (ie one's speed thru 4-space), and so c=0 means the rate-of-proper-time = 0, and so nothing as we know it would exist let alone move. If c dropped by 50%, then the rate-of-proper-time would drop by 50%. Could we ever know that time's proper rate changed (with a change in c), since it should change to the same tune for all? Hmmm. Most would hope light's speed does not change, because the math would get fairly messy right quick :)

My gut feeling, and that's all it is mind you, would be that only the momentum would change if light's speed changed. The photon's energy would not. Just a guess. The reason I figure it as such, is because one should not know time's proper rate changed with a change in c. Therefore, > if < atomic structure remains unchanged with a change in c, then it would stand to reason that only the momentum would change, and not the photonic energy, because it would seem that Planck's constant and the frequency of generated light should not change. I'm probably mistaken though :)

GrayGhost

 

[DarkMatterHol]

OK, so ... p_photon = E/c
p = hf/c ... h being Planck's constant and f being the photon frequency.

So the big question is ... what all changes in this equation if c changed? Does only p or f or h also change? Do some combination of these 3 change? If only the momentum changes with a change in c, then in the limit as c approaches zero, nothing could move. Relativity suggests that the speed-of-light is equivalent to the speed-of-proper-time (ie one's speed thru 4-space), and so c=0 means the rate-of-proper-time = 0, and so nothing as we know it would exist let alone move. If c dropped by 50%, then the rate-of-proper-time would drop by 50%. Could we ever know that time's proper rate changed (with a change in c), since it should change to the same tune for all? Hmmm. Most would hope light's speed does not change, because the math would get fairly messy right quick :)

My gut feeling, and that's all it is mind you, would be that only the momentum would change if light's speed changed. The photon's energy would not. Just a guess. The reason I figure it as such, is because one should not know time's proper rate changed with a change in c. Therefore, > if < atomic structure remains unchanged with a change in c, then it would stand to reason that only the momentum would change, and not the photonic energy, because it would seem that Planck's constant and the frequency of generated light should not change. I'm probably mistaken though :)

GrayGhost

Great! Thanks for the help GreyGhost!

"So the big question is ... what all changes in this equation if c changed?"

Yes. That's exactly what i was wondering, as a way to understand it all better. Perhaps I'm pulling on the wrong thread though. - As the good doctor said, we live in a universe where the speed of light in vacuum [4-space] is constant and presumably cannot be changed. Although i don't think we yet know why it has the value it has, do we? That's why i thought there might be a point to trying out different values for c.

I'll try simply accepting the first equation Pphoton = E/c for now, and see how far i can get through the explanation as provided by JesseM above http://www.adamauton.com/warp/emc2.html There may also be other things i need to know before tackling why the relationship between the speed limit of the universe and the energy contained in matter is what it is...
Bob

 

[GrayGhost]

Great! Thanks for the help GreyGhost!

"So the big question is ... what all changes in this equation if c changed?"

Yes. That's exactly what i was wondering, as a way to understand it all better. Perhaps I'm pulling on the wrong thread though. - As the good doctor said, we live in a universe where the speed of light in vacuum [4-space] is constant and presumably cannot be changed. Although i don't think we yet know why it has the value it has, do we? That's why i thought there might be a point to trying out different values for c.

Here's a hyperlink, which I hope you are able to see ...

http://discovermagazine.com/2003/apr/cover"

João Magueijo has challenged the notion as to whether light's speed has always been the value as defined today. However, it seems he challenges Alan Guth's rapid inflation theory more than GR, as even João Magueijo assumes light's speed to be 300k km/s everywhere locally "in our epoch of time". There's little doubt that he has asked the same question you just did here.

There may also be other things i need to know before tackling why the relationship between the speed limit of the universe and the energy contained in matter is what it is

Indeed, such questions tend to lead to many more yet unanswered questions. Many nobel prizes sit begging :)

GrayGhost

 

[SMRN]

Here's a hyperlink, which I hope you are able to see ...

http://discovermagazine.com/2003/apr/cover"

João Magueijo has challenged the notion as to whether light's speed has always been the value as defined today. However, it seems he challenges Alan Guth's rapid inflation theory more than GR, as even João Magueijo assumes light's speed to be 300k km/s everywhere locally "in our epoch of time". There's little doubt that he has asked the same question you just did here.



Indeed, such questions tend to lead to many more yet unanswered questions. Many nobel prizes sit begging :)

GrayGhost

Hello all

I was searching for a thread in connection to the famous equation, and stumbled upon this thread.

I have a relatively simple question - Why did Einstein include the speed of light, why not any other speed ? and how did he arrive on the decision to include the speed of light, how did he know it has to be the speed of light ?

I am looking for an answer other than "This is how the universe is built, everything is based on the speed of light"

Thanks

 

[JesseM]

Hello all

I was searching for a thread in connection to the famous equation, and stumbled upon this thread.

I have a relatively simple question - Why did Einstein include the speed of light, why not any other speed ? and how did he arrive on the decision to include the speed of light, how did he know it has to be the speed of light ?

I am looking for an answer other than "This is how the universe is built, everything is based on the speed of light"

Thanks

See my earlier comment, you can derive it in a scenario with light being emitted and absorbed in a box if you use conservation of energy and momentum:

If you're looking for a derivation, I think the most intuitive one may be a thought-experiment involving light being shot from one end of a box and absorbed by the other, then conservation of momentum and energy must apply throughout the whole process (before the photon has been emitted, while it's in flight, and after it's been absorbed) and from that you can get E=mc^2. See here for example:

http://www.adamauton.com/warp/emc2.html

 

[SMRN]

See my earlier comment, you can derive it in a scenario with light being emitted and absorbed in a box if you use conservation of energy and momentum:

I understood this, the only problem I encountered is you are considering something that moves at the speed of light (photon) ... what if we consider something, that does not move at 'c', is emitted from the system ? Is it that the relation for energy will change ?

Or what if I fill the box with some medium that changes the speed of the photon, will the relation E=mc^2 change ?

 

[JesseM]

I understood this, the only problem I encountered is you are considering something that moves at the speed of light (photon) ... what if we consider something, that does not move at 'c', is emitted from the system ?

In this case the momentum of the object would no longer be given by the simple relation p=E/c. From relativity we know the for an object with nonzero rest mass M, the energy is given by E²=M²c⁴ + p²c², so we would have $$p = \sqrt{\frac{E^2}{c^2} - M^2 c^2}$$, but obviously this is a relativistic equation already so it would be silly to start out by assuming this and then "derive" E=mc² from it (here m is the "relativistic mass", distinct from the rest mass M which is zero for light). In contrast p=E/c for light was already known from Maxwell's electromagnetism so it's reasonable to use it in a derivation.

Or what if I fill the box with some medium that changes the speed of the photon, will the relation E=mc^2 change ?

No, in this case you would presumably have to take into account the change in momentum of the medium when it periodically absorbs and re-emits photons, which would be complicated. But there are more abstract and general derivations of E=mc² which show that the resistance to inertia (m) of any contained system (like any collection of particles in a box) will always be proportional to the total energy of the system divided by c².

 

[GrayGhost]

Hello all

I was searching for a thread in connection to the famous equation, and stumbled upon this thread.

I have a relatively simple question - Why did Einstein include the speed of light, why not any other speed ? and how did he arrive on the decision to include the speed of light, how did he know it has to be the speed of light? I am looking for an answer other than "This is how the universe is built, everything is based on the speed of light". Thanks

Einstein derived E=mc² from a thought experiment whereby a light source emits light. He showed that the mass of the light source must decrease as light emits, and the lost mass is carried by the photons in the form of momentum and is transferrable to other bodies it strikes later. Einstein relates a POV at rest with the light source, and a POV in motion wrt the light source, using SR. Since light is involved in the scenario, it is not surprising that c arises in the result. Here's the paper ...

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/​

He summarizes at the end as follows ...

If a body gives off the energy L in the form of radiation, its mass diminishes by L/c². The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are led to the more general conclusion that

The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense (by L/9 × 1020, the energy being measured in ergs, and the mass in grammes.)​

At the end of the derivation, it states ...

KineticEnergy-before-light-emission - KineticEnergy-after-light-emission = 1/2(delta-mass)v² = 1/2(L/c²)v²​

the change in mass (delta-mass) is equal to the energy of the emitted light divided by c².

delta-mass=L/c². L is energy light energy E, and so m=E/c² ... or E=mc²​

GrayGhost

 

[SMRN]

Einstein derived E=mc² from a thought experiment whereby a light source emits light. ... L is energy light energy E, and so m=E/c² ... or E=mc²[/INDENT]

GrayGhost

Got you but again the same question to you as well (just want to know your view on this). Why do we always consider that a Photon is emitted, radiating energy ? why not any other thing ?
Something which does not travel with the speed of light ?

I got Jesse's view on this

Thanks

 

[GrayGhost]

SMRN,

Einstein could have modeled an electron from an emitter instead of light from an emitter, however nothing new would have been gained. He did eventually show the conservation of momentum between material entity using the LTs. However, in Einstein's E=mc² paper here, his goal was specific. If light carries momentum, and light is generated from the light source, then mass must be lost from the source during photonic emission. It's all about conservation of momentum, but not between 2 material entities, but rather between some material entity and light. Einstein merely needed to show the mass/energy relation for 2 differing POVs was consistent.

Remember, Einstein was applying his new model of space and time to other various aspects of classical mechanics. In this particular case, it was the relation of the change in kinetic energy during photonic emission. Make no mistake, Einstein's model also applies for the firing of an electron from an emitter, and the value of c surely arises in that derivation as well (as it always does). Only when v<<c does c drop out, because relativistic effects become negligible and relativity reduces to classical mechanics.

GrayGhost