The Mystery of c^2 in Einstein's Famous Formula E=mc^2 Explained

[MarsWTF]

Why c^2 in "e=mc^2"?

Ok, i'v got that e=mc^2 meens some mass at max possible speed will equal max possible energy.
But max possible speed is c, not c^2. Nothing can have speed greater than speed of light, isn't it?

 

[cjl]

You're kind of misinterpreting the equation. It doesn't mean that the mass has the maximum possible energy at maximum speed. It means that the mass itself has an inherent energy, even when it isn't moving, that is equal to m*c². The full equation is actually E² = (m*c²)² + (p*c)², which accounts for any motion that the object has in addition to the rest energy.

Oh, and part of the reason that the c needs to be squared is simply for the units to work out - mass times velocity gives units of kg*m/s, which is momentum. Energy has units of kg*m²/s² (J).

 

[MarsWTF]

the reason that the c needs to be squared is simply for the units to work out

Exactly. In my view this is like do some precise mechanism with duct tape.
There is no objectivity reasons for c^2 in this formula, only for the units to work :/
Imho, the formula must be
E=mc
what do you think? :)

 

[cjl]

Exactly. In my view this is like do some precise mechanism with duct tape.

You'll notice I said that was part of the reason, not the whole reason. In order for the equation to be valid, the units must work out, but that alone isn't sufficient.

I remember that there's a fairly good explanation for why E=mc² is correct, but I don't remember it off the top of my head at the moment. I might have to glance through one of my textbooks to refresh my memory.

 

[Academic]

Exactly. In my view this is like do some precise mechanism with duct tape.
There is no objectivity reasons for c^2 in this formula, only for the units to work :/
Imho, the formula must be
E=mc
what do you think? :)

I think your units don't work.

You could give mass the units of energy and have E=m.

 

[MarsWTF]

I think your units don't work.

Yes it would not work. But squaring с isn't correct cause light cannot exist on the speed of c^2.

 

[Theorem.]

Yes it would not work. But squaring с isn't correct cause light cannot exist on the speed of c^2.

The equation is not suggesting light can reach speeds of c^2 at all.

 

[maxverywell]

The rest mass of a photon is zero, so its energy is E=pc.

 

[MarsWTF]

light can reach speeds of c^2 at all

how? when? where? 0_o

 

[maxverywell]

What are you talking about?

Take for example the classical kinetic energy which is K=1/2 m u^2. This doesn't mean that the speed of particle is u^2.

 

[russ_watters]

Yes it would not work. But squaring с isn't correct cause light cannot exist on the speed of c^2.

Again, C^2 is not a speed so what you are saying is meaningless. Check the units!

...and consider that this equation is very similar to the kinetic energy equation: e=.5mv^2

 

[Dale]

light cannot exist on the speed of c^2.

I agree with Russ, c² isn't a speed. The units are different. I think you need to learn to be more careful with units. That should have been the first thing you were taught in physics.

Let's say that we begin with the supposition that mass has some intrinsic amount of energy, even when it is at rest, and we want to find the formula that determines how much energy is in a given amount of mass. First, we know that the units of mass and energy differ by a speed² so we can immediately write E=kmv².

So, what v can we use? The mass is at rest (v=0). So we cannot use its speed; it must be some sort of universal constant with units of speed. The only such constant is the speed of light, so we know E=kmc² just from consideration of units.

Determining that k=1 requires some actual derivation, so if anything you should ask why k=1 instead of k=1/2. But c² should be obvious: what else could it possibly be?

 

[Jimmy Snyder]

In Einstein's original paper, it said E = mc with a reference to footnote 2. He intended to write in the footnote: E = ma and E = mb are rejected for orthographical reasons. However, he forgot to add the footnote and thus his fame was assured.

 

[DrGreg]

In Einstein's original paper, it said E = mc with a reference to footnote 2. He intended to write in the footnote: E = ma and E = mb are rejected for orthographical reasons. However, he forgot to add the footnote and thus his fame was assured.

Nice joke, but of course totally untrue.

 

[Theorem.]

how? when? where? 0_o

You ignored the rest of the sentence entirely...

 

[MarsWTF]

and consider that this equation is very similar to the kinetic energy equation: e=.5mv^2

i like kinetic energy equation. because some speed v' may be grater than v (if only v<<<c), so its allowed to be squared

 

[Doc Al]

i like kinetic energy equation. because some speed v' may be grater than v (if only v<<<c), so its allowed to be squared

Huh?

 

[MarsWTF]

Huh?

my English is bad, i don't know how to say my thought more clearly, sorry :(

 

[Mute]

i like kinetic energy equation. because some speed v' may be grater than v (if only v<<<c), so its allowed to be squared

You are seriously confused about what these formulas mean. As has already been said, $E = mc^2$ is the amount of energy in an object of mass m in its rest frame, it is NOT "the energy of a mass m moving at maximum speed". i.e., it is the object's internal energy due to the fact that it has mass. The $c^2$ is the conversion factor between mass and rest energy. It's no different that how the Boltzmann constant is a conversion factor between energy and temperature, e.g. $E = k_B T/2$.

 

[russ_watters]

i like kinetic energy equation. because some speed v' may be grater than v (if only v<<<c), so its allowed to be squared

There is no v' in any of the equations we've discussed so far...

...you're not trying to say that v'=v^2, are you? If you are, you need to reread the entire thread (actually read it this time!) and then repeat the following over and over again until it sinks in:

v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.

 

[Jimmy Snyder]

v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.
v^2 is not a speed.

Neither is v for that matter.

 

[brainstorm]

Energy has units of kg*m²/s² (J).

So why does this squaring occur in the energy equation but not in the momentum equation? I always think of kinetic energy as equivalent to particle momentum in qualitative terms.

 

[novop]

Why should it? Momentum is defined as the product of mass and velocity. What other units would you expect it to have?

Of course, the units for energy are a direct consequence of it's definition as well.

Since $$E = \int{F dx}$$ then $$E = \int{v dp} = \frac{mv^2}{2}$$

which means that energy has units of $$\frac{kg*m^2}{s^2}$$

 

[brainstorm]

Of course, the units for energy are a direct consequence of it's definition as well.

Since $$E = \int{F dx}$$ then $$E = \int{v dp} = \frac{mv^2}{2}$$

which means that energy has units of $$\frac{kg*m^2}{s^2}$$

I googled "energy" to find out why it is defined in this way mathematically, but there were too many different types of energy and corresponding equations for me to figure out why this one you mention is general and what it refers to, specifically - let alone why the units are squared in this way.

 

[novop]

My choice of deriving kinetic potential was arbritary, but you can be assured that the units of energy are the same for all types. In all cases, energy is a quantity with units of force times distance. Why?

The definition of energy is $$E = \int{F dx}$$

Since the force "F" has units of Newtons, and $$1~N = 1~\frac{kg\cdot m}{s^2}}$$, and $$dx$$ has units of metres, then $$F dx$$ = $$\frac{kg*m^2}{s^2}$$.

The "squared units" are a direct consequence of how energy is defined. We could have called the quantity $$\int{F dx}$$ "spaghetti tuesday", but then you'd be asking why the units of "spaghetti tuesday" are what they are.

 

[HallsofIvy]

No, no, no! It is the numbers that are squared. The fact that they are speeds is irrelevant to the arithmetic.

 

[brainstorm]

The "squared units" are a direct consequence of how energy is defined. We could have called the quantity $$\int{F dx}$$ "spaghetti tuesday", but then you'd be asking why the units of "spaghetti tuesday" are what they are.

What is your point with your "spaghetti tuesday" comment? That physics terms are arbitrary names for relationships between arbitrary empirical data? I don't see how you can defend that. Force over a distance is "work," I thought. Energy is defined as the potential to do work but I don't think it is the same thing as work itself because potential energy doesn't occur over a distance, although the work it causes occurs over a distance. Radiation does seem to be a form of potential energy as long as it is not interacting with matter, though, doesn't it?

So, if I understand the logic of definitions, force is an intensity that only becomes practically applied when it is exerted in the form of motion as work, unless it is stored in a relatively static system as potential. I suppose the tricky thing with light, and maybe you could also say this about objects orbiting or otherwise moving in a frictionless state, is that it is moving without exerting work - so those would be forms of potential energy with velocity.

So for objects with mass, velocity results in momentum. But for radiation without mass, energy exists as potential for momentum once matter is contacted and interacted with. So is all E=MC^2 really says that the potential work done by radiation is equal to the momentum of a corresponding amount of mass moving at the speed of light? That doesn't seem to make sense since mass can't move at the speed of light and if it would, it would require infinite energy/momentum, right? But it does make sense insofar as radiation is a conduit for momentum/work across a distance without a material medium. In other words, is it basically saying that radiation teleports momentum/work at the speed of massless waves/particles?

Now I'm still wondering about the relationship between the actual speed of propagation and the momentum transmitted, but that's a different thread.

 

[novop]

My point was just to use a little hyperbole to attempt to get my point across. The bottom line is that it doesn't make sense to expect momentum and energy to have the same units, they are fundamentally different quantities. Besides this, you've really confused yourself. Work is in fact defined as the capacity to change energy. Any change in energy is done by work. I don't really know how to respond to the rest of your post, I'll let someone else.