# Energy and reference frames

1. Oct 8, 2005

### trichop

After a dispute on the nature of kinetic energy on a previous topic,
a new question has arised.
If kinetic energy is different for each reference frame,
then what happens with mass, considering the mass-energy
equation?

2. Oct 8, 2005

### Staff: Mentor

Depends on which kind of mass you're asking about. The "invariant mass" a.k.a. "proper mass" a.k.a. "rest mass" doesn't change. The "relativistic mass" depends on the kinetic energy:

$$m_{rel} c^2 = m_0 c^2 + K$$

Last edited: Oct 8, 2005
3. Oct 8, 2005

### trichop

So for different reference frames we have different relativistic mass.
Can this be true?

4. Oct 8, 2005

### pervect

Staff Emeritus
This is definitely true. Only the invariant mass is the same for different frames.

5. Oct 8, 2005

### pmb_phy

Yep. That is 100% true.

Pete

6. Oct 8, 2005

### mezarashi

In relativistic terms, you add mass to the variant variables. It is no longer frame-independent. Don't let incredulity make you skeptical. The way physics works is sometimes not "common sense" to everyday experience.

Also, in a swing from classical mechanics, mass is also no longer conserved. Instead, the total mass-energy of the system is conserved. That's how we get nuclear fusion for example.

7. Oct 8, 2005

### pmb_phy

Mass is always conserved, whether you call "mass" relativistic mass or you call it the magnitude of the total 4-momentum.

Pete

8. Oct 8, 2005

### mezarashi

How do you explain the annihilation of an electron-positron pair?

9. Oct 8, 2005

### Staff: Mentor

People (like Pete) who use "mass" to refer to relativistic mass say that the photons produced in the annihilation have mass, and the sum of the masses of the electron and positron before the annihilation equals the sum of the masses of the photons afterwards.

People (like me) who use "mass" to refer to invariant mass, defined by

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

say that the mass of the system of electron and positron before the annihilation equals the mass of the system of two photons afterwards. In calculating the mass of a system, you use the total energy and the total (vector) momentum in the equation above.

In this view, the mass of a system of particles does not equal the sum of the masses of the individual particles, unless the particles are at rest with respect to each other.

[added] Also in this view, the mass of a photon is zero, because $E = pc$ for a photon.

Last edited: Oct 9, 2005
10. Oct 9, 2005

### mezarashi

I prefer using the conservation of mass-energy. As in the case of the electron-positron annihilation, it is rather odd to assign a mass to the photon. It would then be difficult then to explain photon absorption using that model. There seems to be too many inconsistencies as far as I can see.

11. Oct 9, 2005

### pmb_phy

Assumption: To make the derivationn easy we assum that all observations will be made from the zero momentum frame.

Since each of the two particles have the same proper mass then for momentum to be conserved the two photons produced in the resulting annihilation must have the same frequenecy, f and therefore the same energy E = hf and hence same momentum |p| = E/c.

According to the mass = relativistic mass people: The mass of each photon is defined as the ratio of the magnitude of the photon's momentum to the speed of light. This equals the ratio of the energy over c2, i.e. m = hf/c2. Since energy is conserved then so too is the total (relativistic) mass. Actually the conservation of mass can be proven to be a consequence of two facts (1) that (mc, p) is a 4-vector and (2) that p is conserved.

If you question whether a photon has a non-zero mass recall the definition of mass; m = p/v (E.g. Special Relativity, A.P. French, page 16 - See footnote #2 as well). Almost any text that uses rel-mass will explain that fact

According to the mass = magnitude of 4-momentum 4-vector - The magnitude of the 4-vector is simply proportional to the energy of the system. For the energy to remain constant then so too does the system mass.

For all the glorious details please see -
http://www.geocities.com/physics_world/sr/invariant_mass.htm

Ohanian has a nice discussion in his GR text.

Pete

Last edited: Oct 9, 2005