Why Is the Proportionality Constant in E=mc² Just c²?

[clinden]

It is pretty easy for a physics student to grasp the equivalence between mass and energy.
But, Units of length, mass, and time were established before Einstein came along with
E=MCsquared
My question is: "Why is the proportionality constant between M and E simply Csquared?"
Is this just a coincidence, or, more likely, what am I failing to see.

 

[PeterDonis]

First, compare the conventional units for mass and energy; you will see that the energy unit is the mass unit times a velocity squared. So any proportionality constant between mass and energy must have those units.

Now, since this proportionality constant must be universal, it must be the square of some universal velocity, some velocity that is the same everywhere and for all observers. The only such velocity is the speed of light, $c$. So the proportionality constant must be $c$ squared.

 

[Orodruin]

First, compare the conventional units for mass and energy; you will see that the energy unit is the mass unit times a velocity squared. So any proportionality constant between mass and energy must have those units.

Now, since this proportionality constant must be universal, it must be the square of some universal velocity, some velocity that is the same everywhere and for all observers. The only such velocity is the speed of light, cc. So the proportionality constant must be cc squared.

Technically, this argument does not exclude a dimensionless multiplicative constant such as c^2/2. The correct factor of one must be derived from the underlying theory.

 

[clinden]

Thanks peter. I understood the dimensional analysis logic you addressed. But might there not have been some other constant "K" in the equation?

 

[Nugatory]

Einstein originally derived $E=mc^2$ as a consequence of the rest of special relativity. Basically you start with the relativistic relationship between the energy and the momentum of an object moving relative to some observer, and see what it looks like when the momentum is zero (the observer and the object are moving at the same speed).

However, as PeterDonis points out above, Einstein's derivation pretty much had to come out the way it did. There's no other possible answer; anything else would inevitably lead to an internal contradiction and a logically inconsistent theory.

 

[PeterDonis]

might there not have been some other constant "K" in the equation?

On the basis of units alone, there could have been; but, as Orodruin pointed out, when you look at the full theory, it shows that this factor must be 1.

 

[Herman Trivilino]

It is pretty easy for a physics student to grasp the equivalence between mass and energy.

I don't agree with that. In fact, it's a rather tricky concept that many people, including some authors of physics books, misunderstand. It's actually an equivalence between mass and rest energy.

Even Einstein himself had serious reservations about its validity.

My question is: "Why is the proportionality constant between M and E simply Csquared?"
Is this just a coincidence, or, more likely, what am I failing to see.

Coincidence with what? It's possible to write it simply as $E_o=m$ by choosing a system of units where $c^2=1$.

 

[guywithdoubts]

Coincidence with what? It's possible to write it simply as $E_o=m$ by choosing a system of units where $c^2=1$.

But wouldn't there be missing units in this case?

 

[Demystifier]

Thanks peter. I understood the dimensional analysis logic you addressed. But might there not have been some other constant "K" in the equation?

This can be traced down to the fact that in non-relativistic mechanics the change of energy $dE$, change of momentum $dp$, and instantaneous velocity $v$ are related as
$$dE=vdp$$
Dimensionsionally you could write $dE=Kvdp$, but I hope you understand why $K=1$.

Of course, that's not a derivation. It's just a hint. The full derivation can be found in many textbooks. However, see also
http://lanl.arxiv.org/abs/0805.1400

 

[nikkkom]

But wouldn't there be missing units in this case?

No. The system of units where c=1 is a system of units where time and space have the same units. Say, meters. A meter of a spatial dimension is, well, the usual meter we all know and like. A meter in our only temporal dimension is approximately 3 nanoseconds (= "how much time light needs to propagate 1 meter").

 

[guywithdoubts]

No. The system of units where c=1 is a system of units where time and space have the same units. Say, meters. A meter of a spatial dimension is, well, the usual meter we all know and like. A meter in our only temporal dimension is approximately 3 nanoseconds (= "how much time light needs to propagate 1 meter").

I think I'm going a little offtopic, but I don't get it. A joule is 1 kg * (m/s)². Even if you decide c = 1, you still need the m²/s² for the equivalence to be valid.

 

[nikkkom]

I think I'm going a little offtopic, but I don't get it. A joule is 1 kg * (m/s)². Even if you decide c = 1, you still need the m²/s² for the equivalence to be valid.

In the "c=1" system of units, unit of energy will be just kg.

 

[nikkkom]

I think I'm going a little offtopic, but I don't get it. A joule is 1 kg *
(m/s)². Even if you decide c = 1, you still need the m²/s² for the equivalence to be valid.

"One joule is equal to the energy transferred to an object when a force of one Newton acts on that object in the direction of its motion through a distance of one metre".

"One Newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared".

Now, what are the analogous units in "c=1 and time and distance is measured in meters" system?

"c1_Newton is the force needed to accelerate one kilogram of mass at the rate of one metre per (3 nanoseconds) squared".
IOW: c1_Newton is a much, much, MUCH bigger unit of force than Newton: it's a force which accelerates one kg to near-lightspeed in 3 nanoseconds.

"c1_joule is equal to the energy transferred to an object when a force of one c1_Newton acts on that object in the direction of its motion through a distance of one metre".
No change, except Newton is replaced by c1_Newton. Since c1_Newton is a huge force, c1_joule is also a huge amount of energy. In fact, it's exactly the amount of energy in 1 kg of mass.

 

[nikkkom]

In the "c=1" system of units, unit of energy will be just kg.

And BTW, it's not the end. You can also decide to have h=1, and get rid of yet another unit :D
Then you'll have E=ν instead of E=hν, and thus energy (and mass) will have the same unit as frequency, i.e. 1/time_unit. I.e. 1/meter in our case.

Which makes more intuitive why high-energy physics is linked to very small spatial spaces, why you need high energy particle accelerators to probe internal structure of the proton. Because energy = 1/distance.

 

[Herman Trivilino]

I think I'm going a little offtopic, but I don't get it. A joule is 1 kg * (m/s)². Even if you decide c = 1, you still need the m²/s² for the equivalence to be valid.

In a system where c = 1 units of distance and time are the same. For example, light years and years, respectively. A system like SI, where distance and time are measured in different units, meters and seconds, is not such a system.

Another example is of a system where c = 1 is meters for distance and meters for time, where a meter of time is the amount of time it takes light to travel a distance of one meter. A unit of energy could have, for example, units of kg*(m/m)².

 

[clinden]

Thanks to all of you. I now understand why no additional non-dimensional constant is necessary and that it is not
just a coincidence.