prove e=mc^2

haael

Rest mass is not conserved. Consider an electron and a position annihilating into two photons. Before the annihilation, the rest mass is two times rest mass of an electron. After the annihilation, the rest mass of the system is zero.

However, the total mass of the system is conserved and is the same before and after the annihilation. Total mass of the two photons (m = hv / c^2) will be exactly the total mass of the two electrons (m = m0 / √(1 - v^2/c^2)).

Also, the real measurement of inertia is total mass, not rest mass. This is one of the explanations why any particle can not be boosted to superluminal speed - the more you try, the greater the inertia.

Rest mass is not conserved. Any particular form of energy (rest, potential, kinetic) is not conserved. Only the total mass is conserved, which is also equivalent to total energy conservation.

I am amazed how poorly this topic is understood.

DrStupid

The rest mass of the system is not zero. It is unchanged during the annihilation.

What is your definition for the "total mass"?

haael

Roughly speaking: rest mass, plus kinetic mass, plus potential mass. Equivalent definition: rest energy, plus kinetic energy, plus potential energy, divided by c^2.

A free particle is characterized by 4 independent quantities: its total mass, and momentum. To convert from momentum to mass units, you have to divide it by c. These 4 numbers form a Lorentz vector called 4-momentum. Length of the 4-momentum is the rest mass, a Lorentz scalar.

The rest mass (scalar) is the same in any inertial reference frame. The total mass (and momentum) can change with reference frames. In a rest frame momentum of a particle is zero and the total mass equals the rest mass. In a moving frame momentum is nonzero and total mass increases by a factor "1 / √(1 - v^2/c^2)". This also has a beautiful geometrical explanation, as these are 4-momentum space and time coordinates in the new frame, respectively.

When you not care about Lorentz covariance, you can subtract total mass and rest mass and the result will be kinetic mass. When you multiply it by c^2, you get kinetic energy. In a small speed limit, it reduces to "m v^2 / 2". This is another way of looking at 4-momentum. It is not Lorentz-covariant, but it is preferred by some, possibly because of similarity to normal Newtonian physics.

To sum things up.
4D view:
- 4-momentum, a Lorentz vector.
- Rest mass, the length of the 4-momentum, a Lorentz scalar.

3D view:
- Momentum, space-like part of 4-momentum. It is a spatial vector.
- Total mass, time-like part of 4-momentum. It varies with speed.
- Rest mass, the same as in 4D view.
- Kinetic mass. It varies with speed. It is proportional to kinetic energy. When added to the rest mass, it gives the total mass.

Now, what it means to be conserved? Imagine a 4D spacetime with some process happening in it. Suppose we have an inertial frame. Take one moment (a space-like surface). Compute some quantity. Take some other moment. Is the quantity the same?

With this definition, 4-momentum is conserved. Total mass and momentum are also conserved, as they are components of the 4-momentu. No other quantity defined above is conserved.

Note that conservation is a different thing from Lorentz covariance. If we take some different inertial frame, momentum and total mass will be different. But they will still be conserved, that means they sum on each moment of the same reference frame will be the same.

Conservation of a quantity is its immunity to time translations. Lorentz invariance is about spacetime rotations.

This is extremely important to understand, as a common mistake is comparing total mass values from different inertial frames on different moments and "proving" its non-conservation. Namely, the rest mass is equal to the total mass in some reference frame. Some people first consider the time evolution (with some reference frame defining the time direction), then unconsiously move to the rest frame of the particle. This is not right.

To sum things up, once again:
- 4-momentum is a Lorentz 4-vector. It changes covariantly with boosts and rotations.
- Rest mass is a Lorentz scalar. It does not change with boosts and rotations. This is pure mathematics.
- Total mass is a time component of the 4-momentum.

- 4-momentum is conserved, that means for any closed system it does not change with time translations (and space-like translations too, for that matter).
- Rest mass is not conserved. It can change with time, as we can see with annihilation of two electrons.
- Total mass is conserved, since it is a time component of the 4-momentum.

I wanted to write about potential energy too, but this is too much for today. I hope I was clear enough.

DrStupid

That sounds like the classical inertial mass (also known as relativistic mass) but the potential energy is a problem because the potential energy between parts of the system is already part of its rest energy and the potential energy between the system and external systems is not part of the system only.

This is given for rest mass as well as for the relativistic mass.

DrGreg

Actually the real problem here is that there's no universally agreed language for explaining this. "Conservation of (rest) mass" could be interpreted in one of two ways:

1. Conservation of the sum of the rest masses of the particles of a system. This is not true.
2. Conservation of the "invariant mass", or "system mass", or "rest mass", of the whole system. Invariant mass means E/c² where E is the total energy of the system relative to the frame in which the total momentum is zero, or equivalently [tex]\frac{\sqrt{\left( \Sigma E \right)^2 - \left| \Sigma \textbf{p} \right|^2 c^2}}{c^2}[/tex] in any frame. This type of conservation is true.

Similarly "total mass" could mean "total of the masses" or "mass of the total".

Much of the recent disagreement in this thread has been due to different contributors interpreting the language differently. I don't think they are disagreeing over the actual physics, just the terminology.

DrStupid

This makes no sense because rest mass is not additive. The sum of the rest masses of the particles of a system is not the rest mass of the system. The rest mass of the system is the mass of the system at rest.

SMarioKingdom

http://www.youtube.com/watch?v=hW7DW9NIO9M

This is an Youtube video made by a physicist. It is made with simple drawing and it is easily understandable.

Neandethal00

Here is a dumb thought.
Say, I have a ball in my hand of mass 1 Kg.
I throw it with speed 20m/sec.
Its kinetic energy is 200J. Mass equivalent of this energy is
2.22x10^-15 Kg.
Can we say mass of the ball is increased by 2.22x10^-15 Kg.
In another words, mass of the moving ball is 1+2.22x10^-15Kg, and it will act like a ball of mass 1+2.22x10^-15Kg.
Is this the mass increase we see in special relativity caused by speed?
In such case, it is not a physical increase of mass at all.
Also, then measurements of masses of planets etc are not rest masses.

There must be a big hole in my thinking.

DrStupid

The classical inertial mass is increased but the rest mass remains unchanged.

Define "act like"!

jtbell

How do you define "classical inertial mass"? If you mean the "m" in F = ma, its value (in relativistic physics) depends on whether F is parallel or perpendicular to the motion of the object, or somewhere in between (at an angle).

DrGreg

Yes to all of the above*. However, there is more than one definition of mass in relativity, and this version ("relativistic mass") is a version most physicists don't like to use any more. They prefer to work in terms of rest mass only (so they usually call rest mass just "mass") and they use equations involving rest mass and not relativistic mass.


*To be precise, the kinetic energy calculated using relativity rather than Newtonian physics would be a tiny, tiny amount larger than 200 J.

DrStupid

It was implicit defined by Newton as the "m" in p=m·v

Neandethal00

Then we have an interpretation problem with special relativity.
Usually we say, 'one reason speed faster than light is not possible is mass of the
object will become so high it would require all energy of the universe to move'.
This statement then is not correct. The actual mass the energy would work on
is always the rest mass, increased mass is just its energy.

For example, mass of the moon we use for gravitational pull may not be its rest mass,
but rest mass + energy from its motion.

arindamsinha

There does not seem to be any real 'exact' proof of E = mc². This was a very good and logical approximation made by Einstein in his 1905 paper "Does the Inertia of a Body depend on its Energy-Content?"

No substantial experimental proof has yet appeared to disprove this, so we must assume this is close to correct (perhaps similar to accepting that Newtonian gravitational theory is close to correct, till we learnt better). There is the possibility that there may be additional correction factors with small influence which have not yet been discovered.

I have seen some claims that this has been proved 'exactly' later on, using 4-vectors etc. in GR, but from what I can make out, those depend ultimately on the assumption of correctness of this equation itself.

I am no real expert in relativity theory, so the above is based on my limited knowledge and understanding.

DrGreg

You just need to argue in terms of energy instead of mass. As I hinted in the tiny footnote to my last post, the correct relativistic formula for kinetic energy isn't [itex]\tfrac{1}{2}mv^2[/itex], it is really[tex]
mc^2 \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right)
[/tex]where m is rest mass, which approximates to [itex]\tfrac{1}{2}mv^2[/itex] when (v/c) is very small. So an object with non-zero rest mass would have infinite kinetic energy if it could travel at the speed of light -- impossible.

DrStupid

Gravitational mass of a moving body does not necessarily be equal to its inertial mass. The weak equivalence principle is limited to bodies at rest.

krash661

http://en.wikipedia.org/wiki/Cosmological_constant

The cosmological constant

is equivalent to an energy density in otherwise empty space.