E=m*c 2 or

1. Jul 15, 2011

solarblast

E=m*c**2 or ...

I'm reading the book The Ring of Truth, written in the 80s as a companion for the PBS series of the same name.

The author is discussing energy and the Subject equation. He remarks that another form is E=M, and c**2 is considered a "conversion factor". Is he using E=M as a concept or an "actual" equation? Is the conversion factor malleable depending upon circumstances? I would think a concept.

2. Jul 15, 2011

Pythagorean

Re: E=m*c**2 or ...

Not any circumstances I've ever heard of...

Yes, it can be considered a "conversion factor", but it's a physically meaningful conversion factor.

The full equation, by the way, is:

$$E^2 = (mc^2)^2 + (pc)^2$$

for P = 0 (a motionless particles), it simplifies to what you know n' love, but light always has a non-zero P (though it has M = 0).

So in the case of photons,

$$E = pc$$

clasically, p would have been mv, (mass times velocity), but since photons have 0 m and nonzero p, p (momentum) is not always dependent on mass... HRRRMMM....

3. Jul 15, 2011

the_house

Re: E=m*c**2 or ...

It's just a choice of units, and indeed it is quite common to use a set of units (http://en.wikipedia.org/wiki/Natural_units" [Broken]) where c=1. In that case, for a zero momentum particle, one indeed has the relation E=m.

Last edited by a moderator: May 5, 2017
4. Jul 15, 2011

BruceW

Re: E=m*c**2 or ...

$$E = m c^2$$
can mean one of two things:
a) m is the rest mass of a stationary particle.
b) m is the relativistic mass of a particle.
In either case, 'the house' is right that the choice of natural units is just as legitimate as any other set of units.
And in fact, natural units are the most natural set of units to use (hence their name).
So yes, E=m is an actual equation.

5. Jul 15, 2011

Pengwuino

Re: E=m*c**2 or ...

If the idea of setting a value arbitrarily equal to 1 seems odd, there is good reason to do so. There are a few constants in nature which you typically see set to 1: the speed of light, c, Planck's constant, $\hbar$, the Gravitational constant, G, and Boltzmann's constant, $k_B$. So something normally very complicated such as the Hawking temperature, which is given by $T = {{\hbar c^3}\over{8 \pi G M k_B}}$ reduces down to simply $T = {{1}\over{8 \pi M}}$ where M is the mass of the black hole that's radiating with that temperature.

Of course, by changing the numerical values of all those constants, every other quantity needs to be changed when you actually sit down and do numeric computations. So you'll see at the back of some texts a small guide that basically says "Okay, your temperature is in units of $1/mass$, to get it back to degrees Kelvin, multiply by this number and you'll have a temperature". It's way simpler than having to keep those constants around in every single line in your calculation.

Last edited: Jul 15, 2011
6. Jul 15, 2011

RK1992

Re: E=m*c**2 or ...

edit: oops, ignore this

7. Jul 15, 2011

Pythagorean

Re: E=m*c**2 or ...

Well, sure, as long as you manipulate E and m to account for the change. You're basically just hiding the conversion factor in the units of E and m.

8. Jul 15, 2011

BruceW

Re: E=m*c**2 or ...

Yeah, the idea is to change to a system of units that make equations easier to write down.

9. Jul 16, 2011

solarblast

Re: E=m*c**2 or ...

Thanks to the above for their responses. Now let ask a similar question on a different manner. In the final chapter or The Grand Design by Hawking and Mlodinow, they mention on page 181 "...the positive energy of the matter [in the universe] can be balanced by the negative gravitational energy ..." How is gravity energy? We have something like M+G = 0. M=total energy, and G total negative gravitational energy.

10. Jul 16, 2011

Antiphon

Re: E=m*c**2 or ...

Gravity isn't energy but it can store and release it like a spring. When you lift a box and put it on a shelf, the chemical energy in your muscles is transformed into gravitational potential energy. The additional energy in the gravitational field shows up as a slight reduction of the net gravity of the earth+box. Since it's a reduction and not an increase of the gravity, it's considered that the gravitational potential is negative. To understand what this means, try lifting a charged box against an electric field. In that case the field gets stronger not weaker when it stores potential energy.

11. Jul 16, 2011

BruceW

Re: E=m*c**2 or ...

In classical physics, the gravitational potential energy that exists between all masses is negative.
Think about it this way: When you have two masses, far apart, the gravitational field pulls them closer together, while increasing their kinetic energy (they speed up). Energy is conserved, so we say the increase in kinetic energy accompanies a decrease in gravitational potential energy.
So the gravitational potential energy is less when two masses get close to each other. The gravitational potential energy is defined to be zero when the two masses are infinitely far apart from each other. Therefore, the gravitational potential energy is always some negative value.

A true explanation of cosmological theory would require general relativity, instead of classical physics, but unfortunately I don't know much about general relativity.

12. Jul 16, 2011

solarblast

Re: E=m*c**2 or ...

The above sounds reasonable, so let me try this. M-!G|=0. We are now talking about the entire universe. Does this allow me to say there is "nothing" to the universe? I quote Feynman with The physicist uses ordinary words in a peculiar manner." What I'm moving toward is this a way of describing the early universe when it essentially nothing. If not the above equation, some such equation that is zero, so a physicist can say it is nothing, but in fact it is in equilibrium between two entities.

13. Jul 16, 2011

Pengwuino

Re: E=m*c**2 or ...

Equilibrium and "nothingness" are two different things. It is not true to say there is nothing to the universe, you can only say the universe is in equilibrium (which is not true either).

Now, to clarify what is meant by energy being gravity, you need to look at general relativity. The Einstein Field Equations equate energy with gravity; any form of energy creates a gravitational field. This means suns, planets, atoms, photons, everything. Of course, since this includes rest energies, suns and planets have gravitational fields so strong that individual atoms and photons have gravitational interactions so tiny in comparison that you need some rather complex experiments to verify this.

If you're reading a book written in the 80s, you're going to be misled on the cosmology in it. Since then, we have discovered the universe is actually acceleration - space-time is expanding at an increasing rate. The expansion of the universe is not balanced by the gravitational pull of the actual stuff inside the universe trying to contract it back in.