# Is E=mγc2 correct?

1. Nov 6, 2012

### Antuanne

I've just saw this other equation E2=m2c4+p2c2 but what's going on with that? I thought E=mγc2 was the equation including relativistic mass. What is going on?

2. Nov 6, 2012

### Ibix

Just algebra. If you substitute the expression for relativistic 3-momentum (p=γmv) into E2=m2c4+p2c2, you'll end up at E=γmc2.

3. Nov 6, 2012

### Antuanne

I get it now, but, if you we're traveling near the speed of light and needed the Lorentz factor, would you make it E2=(mγ)2c4+(mvγ)c2 since you need to put in for relativistic mass in both places?

4. Nov 6, 2012

### PAllen

No, in the formula E2=m^2c^4+p^2c^2, m is rest mass (or invariant mass). In substituting for momentum, it is better to just treat p=γmv as a definition of momentum, not as 'relativistic mass' times velocity.

5. Nov 6, 2012

### Staff: Mentor

No. The m in E2=(mc2)2 + p2c2 is the rest mass.

One of the advantages of this formulation is that it works for photons and other massless particles as well.

6. Nov 6, 2012

### Antuanne

That makes sense, but in the simplified thing E=mγc2, where does the γ come from then?

7. Nov 6, 2012

### PAllen

mc^2 is rest energy. mγc^2 is total energy, including kinetic energy. Not quite sure what you are asking. In newtonian mechanics, do you ask: where does the 1/2 come from in:

KE = (1/2) mv^2 ?

8. Nov 6, 2012

### Antuanne

What I'm asking is what is the equation E2=m2c4+p2c2 used for if E=mγc2 is used for total energy? And is the γ in the just there for relativistic mass or what is that there for?

9. Nov 6, 2012

### Ibix

It is always there. But if you are at rest, the gamma factor is 1 and it reduces to the famous form. If you are travelling at lightspeed then gamma is infinite, but m is zero and you need to use the other form and a different definition of momentum - p=2pi hk.

10. Nov 6, 2012

### PAllen

Normally, for massless particles traveling at c, you simply use p=E/c.

11. Nov 6, 2012

### Ibix

The two equations are equivalent for particles with non-zero rest mass. Use whichever you like the look of better. However, the longer form is also useful for massless particles, albeit with a different expression for momentum.

The gamma isn't "for" anything. Some people do like to define a "relativistic mass" and make the relativistic equations look like the Newtonian ones. Consensus around here is that this is not helpful - for example you need two different relativistic masses to deal with force, and the acceleration is not parallel to the force (in general) anyway. Better to acknowledge that from the beginning.

The gamma is a consequence of being in a universe with a Minkowski space-time. It is not a modification to Newton. If you wish to recover the low velocity limit, do a binomial expansion of gamma and neglect terms of order v4 and higher. But you can't get from Newton to Einstein that way.

Does that make some kind of sense?

12. Nov 6, 2012

### Antuanne

Do E2=m2c4+p2c2 and E=mγc2 give you the same answer, because, it doesn't seem like they do?

13. Nov 6, 2012

### PAllen

Of course they do. Just plug p=mγv in and you get E=mγc2 after algebra.

14. Nov 6, 2012

### Staff: Mentor

They do. For p2, just substitute m2v2γ2 = m2c4(v/c)2γ2

15. Nov 6, 2012

### Antuanne

But E2=m2c4-p2c2 has v velocity in p so where does that go?

Last edited: Nov 6, 2012
16. Nov 6, 2012

### PAllen

Did you try the algebra?

Here are a few stages along the way. Try to fill in the gaps.

E^2 = (mc^2)^2 + m^2γ^2v^2c^2

= (mc^2)^2 ( 1 + v^2/(c^2 - v^2) )

= (mc^2)^2 (c^2 / (c^2 - v^2) )

then E = mγc^2

17. Nov 7, 2012

### Antuanne

I still can't figure out how to factor or whatever your doing to get to that!

18. Nov 7, 2012

### PAllen

Don't take this wrong, but while SR needs no higher math, it needs basic facility with algebra. It would be much better for you to review algebra and work it out for yourself than for me fill in one or two more steps between each posted intermediate result.

19. Nov 7, 2012

### DrGreg

Antuanne, don't forget that$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$You will need to use that at some point.

20. Nov 7, 2012

### micromass

The OP won't be coming back.