# Clarification on 1st law of thermodynamics and e=mc^2

1. May 21, 2010

### Null_

I'm sure each of you is well-versed in the first law of thermodynamics: energy cannot be created or destroyed. Thus, there is the law conservation of energy.

E=mc^2 states that energy and mass are functions of each other, basically one and the same less a conversion factor. I take mass to be rest energy.

Matter can be created from photon collisions, and particles can be anhillated by collision as well, thus establishing that there is no law of conservation of mass. I have heard that photons do have energy but do not have mass. I don't see how this is possible based on E=mc^2. E would =0 if m=0, and m=0 if E=0.

I would greatly appreciate it if someone would clarify why there is a law of conservation of energy but not of mass if the two are one and the same. (Pardon my rudimentary knowledge..I'm in high school and have yet to take a physics class.) I have taken calculus, so mathematical explanations of a light degree will be understood.

2. May 21, 2010

### Feldoh

Have you ever seen a photon at rest?

There are both laws of conservation of mass and energy, it just depends on the context to when they are applicable to a given system.

Last edited: May 21, 2010
3. May 21, 2010

### Null_

That's what I'm confused about. A photon is continually moving at a constant speed in a vacuum, so it has kinetic energy but no rest energy (m). So a positive#E= 0c^2.

4. May 21, 2010

### Kynnath

Photons don't have rest mass, because there is no valid inertial reference frame where they are at rest, which is required to calculate the rest mass. Their energy is not 0 because the equation is not applicable to them.

You can't use the photon itself as a reference frame because you can't translate to and from a reference frame moving at c, and photons move at c in every other reference frame.

They do have relativistic mass. Their energy divided by c^2. Rest mass is constant across all inertial reference frames, while relativistic mass is not, however.

5. May 21, 2010

### Pengwuino

Mass is not conserved, it's as simple as that. You don't even have to look at light, simply look at nuclear reactions; mass isnt conserved. What you will find is that mass-energy is conserved. If you have a particle that decays into such and such other particles, you'll see that if you add up all the masses, they won't add up to the original particle. However, if you add up all the masses and their kinetic energies, you'll find the energy was conserved. Consider the same idea except now a photon has flown out of the reaction; again, energy is still conserved using a different method of finding the energy of the photon. As stated before, the energy of a photon does not follow E = mc^2 as it is massless.

6. May 22, 2010

### netheril96

Your opinion is a common mistake.The so-called mass defect is the defect of rest mass,while the total mass,as well as energy,is always conserved.
As for the photon thing,it doesn't have rest mass,but it DO have mass and momentum

The key is to understand what is rest mass and what is mass

7. May 22, 2010

Arnt you just defining total mass and energy to be the same there?

8. May 22, 2010

### diazona

The real formula is
$$E^2 = m^2c^4 + p^2c^2$$
where p is momentum. Photons have zero mass, but they do have momentum, and that's where their energy comes from.

$E=mc^2$ is merely a special case of the above formula, which applies only to objects at rest.

9. May 22, 2010

### netheril96

Another one who doesn't understand what is the difference between rest mass and mass

10. May 22, 2010

### diazona

There is no difference. The terms are synonymous in modern usage.

11. May 22, 2010

### netheril96

Well,you may have a different definiton for mass.Then I emphasize that there is a difference between rest mass and so-called relativistic mass.

12. May 22, 2010

### diazona

Agreed... rest mass = m, relativistic mass = E/c2 (that's what you mean, right?)

The modern convention (at least, among a large majority of the references I look at and the people I talk to) is that mass = m, and the terms "rest mass" and "relativistic mass" are deprecated.

13. May 22, 2010

### netheril96

In my book,m0 denotes the rest mass while m denotes mass or relativistic mass.

So the two formula should be
$$\begin{array}{l} {E^2} = m_0^2{c^4} + {p^2}{c^2} \\ E = m{c^2} \\ \end{array}$$

14. May 22, 2010

### DrGreg

There are two schools of thought

Old school

E/c2 is called "mass"
m0 is called "rest mass"

Preferred modern view

E/c2 is called "energy divided by c2"
m0 is called "mass"

The majority of professional physicists seem to subscribe to the modern view but you will find the old-school view in some older text books or coffee-table books aimed at the general public. When you read any book or website, you'll need to work out which school the author is in, because unfortunately there is no Physics Police to enforce one terminology over the other.

15. May 22, 2010

### Null_

Ah, I have never seen the real formula before LaTeX Code: E^2 = m^2c^4 + p^2c^2 .

So, basically I had a misunderstanding due to the commonly used formula...is acceleration only applicable when referring to protons? And now that I think about it, why do photons have acceleration at all? If they are always in continuous motion, isn't the acceleration a constant 0?

16. May 22, 2010

### Feldoh

Why are you talking about acceleration?

17. May 22, 2010

### Null_

Haha, I saw the p=momentum post then came back after doing a problem dealing with acceleration. My mistake.

18. May 22, 2010

### Mute

To elaborate on DrGreg's post, in the first view people interpret the mass of an object to change with speed, according to the formula $m = \gamma m_0$. For a photon the RHS is an indeterminate form which turns out to be p/c.

The modern interpretation is that mass is mass and it's momentum itself that has this strange behaviour as speed increasing.

i.e., the first view interprets the non-relativistic formula $p = mv$ to be the correct relativistic formula, except that now mass changes with speed. The second view interprets the momentum formula to be incorrect, which the correct formula being $p = \gamma m v$, where m is the rest mass, which does not change with speed. So, really it's mathematically all the same and up to one's interpretational preference, grouping the gamma with the mass or the momentum, but the first interpretation sure seems to give people lots of conceptual difficulties.