Link between 'time' component of 4-momentum and energy

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1. Nov 28, 2016

bananabandana

1. The problem statement, all variables and given/known data

$$E = -\vec{v_{obs}} \cdot \vec{p}$$

Where $\vec{p}$ is the four momentum, and $\vec{v_{obs}}$ the velocity of the observer.

2. Relevant equations

3. The attempt at a solution

This was a stated result in a GR course. I look through my SR notes and find that I have actually never seen a proof that $p^{0}$ is exactly,definitely the energy.

I understand that the norm of the four momentum will be invariant etc...
But the phrase in the notes I have "For a free, non-interacting particle there is no physically meaningful property of "energy"" is making me uncomfortable - have I fundamentally misunderstood something?

Thanks!

2. Nov 28, 2016

Ray Vickson

Actually, if $c$ is the in vacuo speed of light, then energy is $c p^0$; you need that extra $c$ to make the units come out right.

Anyway, how you can take an inner product of a 3-vector $\vec{v_{obs}}$ and a 4-vector $p$?

3. Nov 28, 2016

bananabandana

Good question - I presume it's the four velocity of the observer (these are directly from my lecture notes).

I guess another part of my confusion is the fact that we have :

$$p^{\mu} = m \frac{dx^{\mu}}{d\tau} \\[2mm] p^{\mu} = \bigg(E,\vec{\dot{x}} \bigg) \\[2mm] \implies m\frac{dx^{0}}{d\tau} = E$$
Which I find a bit odd/ how is this the same quantity as my familiar 'classical' energy? Even if the units do work out (I have missed a factor of c somewhere).

Edit - The oddness to me being that we somehow have some kind of rate of change of a time co-ordinate associated with an energy.... so that somehow having higher energy, is associated, in some sense with a higher rate of motion in time?? But that sounds bananas to me...

4. Nov 28, 2016

Ray Vickson

No! For a particle of mass $m$ moving with 3-velocity $\vec{u}$ the 4-momentum is $p = (m \gamma c, m \gamma \vec{u})$, where $\gamma = (1 - u^2/c^2)^{-1/2}$, while the (total) energy is $E = m c^2 \gamma$. Note that $E = p^0 c$, NOT $p^0$. Why do you continue to ignore this distinction? If nothing else, look at the units: the units of momentum are $\text{mass}\times \text{velocity}$ while the units of energy are $\text{mass} \times \text{velocity}^2$. It has been that way ever since the time of Newton.

5. Nov 29, 2016

bananabandana

Sure, sure - the units are important and I entirely agree that $p^{0}c$ has the correct units of energy. What I find confusing is not the units, but the idea that somehow a translation in time is related to my classical picture of energy :)