4-momentum question

1. May 25, 2010

Jerbearrrrrr

Basic 4 momentum questions. Trying to understand SR a bit better.

Suppose m is non zero for a particle (m being the rest mass of the particle). Then the 4-momentum is related to the 4-velocity by p=mu (in 4 coordinates).
The zeroth coordinate of p is therefore m*gamma(v) where v is the 3-velocity of the particle.
If we approximate this in the non-relativistic limit, we get mc²+½mv²+...
We call this quantity "E".
Why do we interpret (mc²+½mv²) as something's total energy?
Sure, the ½mv² term is some form of energy (and by dimensional analysis, the mc² term is also measured in energy units)...but why should this total thing be a useful quantity?

eg I can add two random things that happen to be measured in joules, and get a not-very-useful quantity.

I was going to ask why we call the zero'th component of 4-momentum "E" but I guess it's just because if we evaluate it in the rest frame, it turns out to be E (or E/c) and that lets us deal with the m=0 case.

Relativistically, mc² is a conserved quantity. Does this derive conservation of energy (for simple systems at least), or is it just consistent with it?

Thanks

2. May 26, 2010

haushofer

Well, if you have a particle which is not in some sort of field, the ONLY energy it has is the rest energy and kinetic energy. That's why we add the two and call it TOTAL energy.

That's not so strange, right?

3. May 26, 2010

Jerbearrrrrr

Why is mc^2 the rest energy?
-because it's the 0th component of the 4-momentum in the rest frame
Why is the 0th component in the 4-momentum the total energy?

4. May 26, 2010

dx

It has to do with the definition of the four-momentum itself. To really understand it, we have to think in terms of the relationship between symmetries and conserved quantities. The basic reason is that the relationship between energy and time is the same the one between x-momentum and x and so on.

From this viewpoint, there are four equally important symmetries: translation in the t direction, translation in the x direction, translation in the y direction, translation in the z direction. The conserved quantities associated with these are -E, Px, Py and Pz

The symmetries listed above can be summarized more generally as arbitrary spacetime translations. This is a vector ξa. The components of the four-momentum arise as the gradient of the action dS acting on the respective basis vectors.

Last edited: May 26, 2010
5. May 26, 2010

Jerbearrrrrr

I'm still seeing E as an arbitrary first integral/constant of integration.
what action do we use?$$m\sqrt{\eta_{ab} \dot{x}^a \dot{x}^b}$$ ?

6. May 26, 2010

Staff: Mentor

We do not interpret (mc²+½mv²) as the total energy. We interpret m*gamma(v) as the total energy. Doing a series expansion about v=0 we get that the total energy is mc²+½mv²+... which is good because we get a term that looks like the Newtonian definition of kinetic energy in our approximation, but we also get the mc² term, which we interpret as rest energy, and all of the higher order terms which are relativistic corrections to the Newtonian formula for kinetic energy.

7. May 26, 2010

starthaus

We don't. $$E$$ is by definition $$\frac{m_0c^2}{\sqrt{1-(v/c)^2}}$$ .
A Taylor expansion of the above , for $$v/c<<1$$ produces:

$$E=m_0c^2(1+\frac{1}{2}\frac{v^2}{c^2}+....).=m_0c^2+\frac{1}{2}m_0v^2+...$$

8. Jun 10, 2010

Jerbearrrrrr

Sorry to resurrect the thread, but it'd probably be worse to make a new one.
Thanks for the replies.

But I'm still not seeing the significance of the quantity 'E', other than in an expansion, one term just happens to be the kinetic energy.
Not sure how to explain what I don't understand but...
How do we start from E:=m.gamma, and end up with "mass and energy are the same"? Where energy is...well, what is energy? The capacity to do work?

dx - can you elaborate? I'm guessing you're talking about a (the) theorem due to Noether.

You hear everywhere (in popular science) that relativity establishes (some sort of) equivalence between mass and energy. But I'm having trouble going from special relativity to atomic bombs. Where does binding energy even enter the discussion (from a theoretical physics point of view, rather than a coincidence in observed quantities).