Solving the Mystery of E=mc2: Exploring c2 and Potential Energy

[Hallan]

Okay, I started thinking about relativity earlier, and got stuck on how to square speed.
I went with some basics to try and figure it out, this is some of what i came up with;

Main Question: Does (10mph)^2 equal?
A. 10 m^2 per h^2
B. 100 m^2 per h^2
C. 100 mph

Logically "B." makes sense, except for the fact that it seems to be measuring the acceleration of size?

oh, on a slightly different note, how do you calculate the potential energy of an object?

 

[Redbelly98]

B. is indeed correct.

c² is not physically meaningful all by itself. It could however be thought of as a conversion factor between mass and energy.

 

[maverick_starstrider]

m*c^2 = M*L^2/T^2 (where M is a unit of mass, L a unit of length, T a unit of time). This is the unit of energy (for comparison look at the kinetic energy equation (1/2)*m*v^2, unit are again M*L^2/T^2)

 

[Hallan]

Are the two so easily comparable?

the kinetic energy equation (1/2)*m*v^2 takes mass times speed.
whereas M*L^2/T^2 takes mass times the changing speed of growth.

 

[Doc Al]

the kinetic energy equation (1/2)*m*v^2 takes mass times speed.

No, that's mass times speed squared.

whereas M*L^2/T^2 takes mass times the changing speed of growth.

I don't know what "changing speed of growth" means, but M*L^2/T^2 shows the proper dimensions for energy.

 

[Redbelly98]

Are the two so easily comparable?

Yes.

the kinetic energy equation (1/2)*m*v^2 takes mass times speed.
whereas M*L^2/T^2 takes mass times the changing speed of growth.

I think you're attributing more meaning to simple units than is merited.

EDIT:
Welcome to Physics Forums, by the way

 

[Hallan]

@Doc AI: what i meant by "changing speed of growth" is acceleration of size (L^2/T^2) (IE. square feet per second per second) you go from making something move to making it bigger.
Thank you for pointing out that it is speed squared as well. I hadn't considered the implications that it wasn't mass times speed but mass times speed squared which is exactly what e=mc^2 is doing.

@Redbelly98: Thanks, I like that you all are so willing to help.

 

[russ_watters]

@Doc AI: what i meant by "changing speed of growth" is acceleration of size (L^2/T^2) (IE. square feet per second per second) you go from making something move to making it bigger.

But you just made that up! That's your interpretation of what those units mean and as was said before, it isn't correct. Those units have no useful meaning on their own. The C^2 or V^2 is part of an expression and the end result of that expression is the units of energy.

 

[Hallan]

Am I correct in saying that "Miles per hours per hour" is a measurement of acceleration, and that Square miles is a measurement of size?
If so where am I going wrong to say that square miles (a measurement of size) per hour per hour(a measurement of Time^2) (mph)^2 is the "acceleration of size"?

 

[Hootenanny]

Am I correct in saying that "Miles per hours per hour" is a measurement of acceleration,

Yes. L/T² is indeed the units of acceleration.

and that Square miles is a measurement of size?

More accurately L² is a measure of area.

If so where am I going wrong to say that square miles (a measurement of size) per hour per hour(a measurement of Time^2) (mph)^2 is the "acceleration of size"?

So you have L²/T², which is L*L/T², which is a length times acceleration. Note that 1/T² in itself is not a unit of acceleration.

 

[Vanadium 50]

Hootenanny is correct.

Also, one needs to keep straight the distinction between units and dimensions. Units of torque and energy have the same dimensions, but torque and energy are not the same thing nor are they measured in the same units.

 

[Redbelly98]

I think the OP's point is that L²/T could represent the rate of change of an area. And L²/T² could represent the rate of change of L²/T, i.e. the 2nd derivative of a changing area w.r.t. time.

Hallan, while it's true that the 2nd derivative of area has units of L²/T², it is not true that those units must always represent the 2nd derivative of an area. Just as L can represent many things: distance traveled, the radius of a circle, or a circle's circumference.

The units alone do not imply what type of quantity is being represented.

 

[Hallan]

So when you throw in the OP's topic, we know the quantities (299,792,458m/s)^2 what then does it represent?

 

[Redbelly98]

I repeat:

c² is not physically meaningful all by itself. It could however be thought of as a conversion factor between mass and energy.

 

[Hallan]

Okay, On an very similar but slightly different note, does anyone know why the kinetic energy equation ((1/2)*m*v^2) takes into account the speed of the mass, but E=MC² is the speed of light?
Is it because E=MC² isn't just kinetic but all energy?

 

[dx]

Okay, On an very similar but slightly different note, does anyone know why the kinetic energy equation ((1/2)*m*v^2) takes into account the speed of the mass, but E=MC² is the speed of light?
Is it because E=MC² isn't just kinetic but all energy?

E = mc² also has the speed of the mass in it, but it is hidden in the definition of m:

m = m₀/√(1 - v²/c²)

where m₀ is actually what we call 'mass' these days, and m is what was previously called the 'relativistic mass' but is no longer considered to be a useful concept. So the correct equation in terms of the modern definition of mass is

E = mc²/√(1 - v²/c²)

 

[russ_watters]

While both discuss energy, they are different equations, describing differen things. One is kinetic energy, the other describes the equivalence of matter an energy. Note that there are several other equations that deal with energy. You should really try not to get too hung up on this.

 

[dx]

Just to tie russ' answer with mine, the formula I gave for the energy:

E = mc²/√(1 - v²/c²)

gives the total energy of the particle. By expanding the expression in a series, we get

E = mc² + mv²/2 + ...

The first term can be considered to be the 'mass energy', which doesn't depend on velocity, and the rest of the terms which do contain the velocity is the relativistic generalization of kinetic energy.

 

[Superstring]

think of it like simple algebra. c represents length and time, so if you had for example (5x)^2, the result is 25x^2 and not just 25x.

 

[Klockan3]

E = mc² also has the speed of the mass in it, but it is hidden in the definition of m:

m = m₀/√(1 - v²/c²)

where m₀ is actually what we call 'mass' these days, and m is what was previously called the 'relativistic mass' but is no longer considered to be a useful concept. So the correct equation in terms of the modern definition of mass is

E = mc²/√(1 - v²/c²)

Um, E=m₀c^2 is still a correct equation though, it is just that now it isn't the energy of a moving particle but instead how energy and mass are directly related with each other.

This is also the most intriguing way to read the formula, if you teach it like you show then it is just an altered formula for calculating kinetic energy which is dull.

As for C, C is a conversion factor between seconds and meters to make every axis on a time-space diagram have the same dimensions, it is the natural speed to compare everything else to. If you use natural units with C=1 then energy, mass and momentum all have the same units which is just natural since they are all parts of the same tensor.