General relativity -- Proof of energy measured by observer

[Jonsson]

I want to prove that $E = -g_{\mu \nu}u^\mu p^\nu$ is the energy measured by an observer with velocity $u^\mu$ of an object with momentum $p^\mu$. My reasoning is that in special relativity we know that $\gamma m = E$. We can transform to coordinates where $u'^\mu = (1,\vec{0})$. Since the measurement is happening locally to the observer, the space appears flat, and $g = \eta$ and hence $dt/d\tau = \gamma$.
$$
-g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =
-(1,\vec{0})\cdot(-\gamma m, \gamma \vec{v}) = \gamma m = E
$$
My question. Is this kind of reasoning legal? Is it enough to argue that since the measurement happens locally, we can take the space to be flat? The reason I ask is because it would surprise me since pretty much any kind of measurement I can think of is a local concept.

 

[PeroK]

I want to prove that $E = -g_{\mu \nu}u^\mu p^\nu$ is the energy measured by an observer with velocity $u^\mu$ of an object with momentum $p^\mu$. My reasoning is that in special relativity we know that $\gamma m = E$. We can transform to coordinates where $u'^\mu = (1,\vec{0})$. Since the measurement is happening locally to the observer, the space appears flat, and $g = \eta$ and hence $dt/d\tau = \gamma$.
$$
-g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =
-(1,\vec{0})\cdot(-\gamma m, \gamma \vec{v}) = \gamma m = E
$$
My question. Is this kind of reasoning legal? Is it enough to argue that since the measurement happens locally, we can take the space to be flat? The reason I ask is because it would surprise me since pretty much any kind of measurement I can think of is a local concept.

I think that's reasonable, although I don't think you need to introduce $\gamma$ into the equation. In the observer's LIRF, the energy of a particle is the first component of its four-velocity. In this LIRF, the observer's four-velocity is, as you say, $(1, \vec{0})$. Therefore, the measured energy is the inner product of these two four velocities.

The inner product is coordinate-independent (assuming that it is defined using the metric appropriate to the coordinates in each case), hence the result follows.

 

[Jonsson]

Thanks!

 

[PeterDonis]

I don't think you need to introduce $\gamma$ into the equation.

You do if you want to get the right answer.

In the observer's LIRF, the energy of a particle is the first component of its four-velocity

Not quite; it's the first component of its 4-momentum. That's where the factor of $m$ in the correct answer $\gamma m$ comes from. If it were the first component of the object's 4-velocity, that would just be $\gamma$, with no factor of $m$.

 

[PeterDonis]

Is it enough to argue that since the measurement happens locally, we can take the space to be flat?

Yes. (With the proviso that it's spacetime that we take to be flat, not space.)

The reason I ask is because it would surprise me since pretty much any kind of measurement I can think of is a local concept.

Plenty of quantities of interest do not involve purely local measurements. For example, the redshift of distant objects due to the expansion of the universe.

 

[Jonsson]

You do if you want to get the right answer.

What do you mean? Is it wrong to use the gamma? I was taught that we obtain GR locally. Thanks

 

[PeterDonis]

Is it wrong to use the gamma?

No, it's right. I was responding to PeroK who appeared to be implying that you didn't need to include $\gamma$ in the equation; you do.

 

[PeroK]

You do if you want to get the right answer.

Not quite; it's the first component of its 4-momentum. That's where the factor of $m$ in the correct answer $\gamma m$ comes from. If it were the first component of the object's 4-velocity, that would just be $\gamma$, with no factor of $m$.

Yes, of course, four-momentum rather than just the four-velocity.

The point I was making is that:

$-\mathbf{p}.\mathbf{u_{obs}}$

In invariant in any coordinate system, including in the LIRF, where we know that it equals the particle's observed energy. It's irrelevant to the argument that $E= \gamma m$ in this reference frame.

In other words, you could write more simply:

$$
-g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =
-(1,\vec{0})\cdot(E, \vec{p}) = E
$$

 

[PeterDonis]

you could write more simply:
$$
- g_{\mu \nu} u^\mu p^\nu = - \eta_{\mu \nu} u'^\mu p'^{\nu} = - (1, \vec{0}) \cdot (E, \vec{p}) = E
$$

Ah, I see. Yes, this is correct.