E=mc^2 in a gravitational field

[Agerhell]

Let us say that we have small object of mass $m$ at some location far away from the Earth (with zero velocity compared to the earth). The energy of this object is according to relativity $E=mc^2$.

Now we drop this object and it starts falling towards the earth, transforming potential energy to kinetic. Finally the objects hits the Earth. It may heat up from the impact but finally it cools down again.

What is the energy of the object now? It still contains the same number of atoms as initially but energy has been converted into heat that has radiated away. It can not really be $E=mc^2$ again if the mass has remained contant and by $c$ we mean velocity of light as measured locally which is invariant. Is $E=mc^2$ only supposed to hold in a non-gravitational setting?

If we assume something like:

$E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$

Then at least the energy of an object at rest infinitelly close to the Schwarzschild radius of a black hole (using Schwarzschild coordinates) have zero energy, which seems logical? What is the correct interpretation of the energy formula $E=mc^2$ when the objects whose energy we are interested in is sunk into the gravitational field of (for instance) a planet?

 

[PAllen]

See Komar Mass definition. It has just this type of factor effectively saying that after radiation thermal equilibrium are reached, and you can treat the situation as static again, there is redhshift factor applied to mass contribution. I haven't checked through the wiki entry in detail, but the essentials look ok on quick inspection:

http://en.wikipedia.org/wiki/Komar_mass

 

[phinds]

I think you are confusing forms of energy. Potential energy and kinetic energy have varying effect in gravitational fields (e.g. higher up = greater potential energy). E=mc*2 is just a conversion equivalence between mass and energy, having nothing to do with gravity.

 

[ZapperZ]

Let us say that we have small object of mass $m$ at some location far away from the Earth (with zero velocity compared to the earth). The energy of this object is according to relativity $E=mc^2$.

Now we drop this object and it starts falling towards the earth, transforming potential energy to kinetic. Finally the objects hits the Earth. It may heat up from the impact but finally it cools down again.

What is the energy of the object now? It still contains the same number of atoms as initially but energy has been converted into heat that has radiated away. It can not really be $E=mc^2$ again if the mass has remained contant and by $c$ we mean velocity of light as measured locally which is invariant. Is $E=mc^2$ only supposed to hold in a non-gravitational setting?

If we assume something like:

$E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$

Then at least the energy of an object at rest infinitelly close to the Schwarzschild radius of a black hole (using Schwarzschild coordinates) have zero energy, which seems logical? What is the correct interpretation of the energy formula $E=mc^2$ when the objects whose energy we are interested in is sunk into the gravitational field of (for instance) a planet?

I'm always puzzled when I see questions similar to this.

When you are asking this, are you not aware of the FULL relativistic equation (see the FAQ in the Relativity forum). If the object is moving, at what point to you include the relativistic KE? You do know that that is the main reason why we accelerate particles to such high energy at these particle colliders, don't you?

Zz.

 

[Agerhell]

See Komar Mass definition. It has just this type of factor effectively saying that after radiation thermal equilibrium are reached, and you can treat the situation as static again, there is redhshift factor applied to mass contribution. I haven't checked through the wiki entry in detail, but the essentials look ok on quick inspection:

http://en.wikipedia.org/wiki/Komar_mass

Hmm... My interpretation of the Komar mass as described in wikipedia is that, for a stationary object in a spherically symmetric gravitational field, to get the gravitational acceleration right, one must replace:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}$

with:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2(1-\frac{2GM}{rc^2})}\hat{r}$.

I do not immediately see how this relates to the relation $E=mc^2$ for an object in a gravitational field...

It seems like you get the Komar mass by replacing the central mass $M$ by $\frac{M}{1-\frac{2GM}{rc^2}}$, but the central mass does not appear in $E=mc^2$.

Should both the central mass "M" and the small test-mass "m" be multiplicated by the factor $\frac{1}{1-\frac{2GM}{rc^2}} \frac{1}{1-\frac{2Gm}{rc^2}}$ ?

This can not really be combined with $E=mc^2$ because then you get more energy the closer the the two masses are to each other...

 

[PAllen]

Hmm... My interpretation of the Komar mass as described in wikipedia is that, for a stationary object in a spherically symmetric gravitational field, to get the gravitational acceleration right, one must replace:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}$

with:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2(1-\frac{2GM}{rc^2})}\hat{r}$.

I do not immediately see how this relates to the relation $E=mc^2$ for an object in a gravitational field...

It seems like you get the Komar mass by replacing the central mass $M$ by $\frac{M}{1-\frac{2GM}{rc^2}}$, but the central mass does not appear in $E=mc^2$.

Should both the central mass "M" and the small test-mass "m" be multiplicated by the factor $\frac{1}{1-\frac{2GM}{rc^2}} \frac{1}{1-\frac{2Gm}{rc^2}}$ ?

This can not really be combined with $E=mc^2$ because then you get more energy the closer the the two masses are to each other...

I meant that it has the general idea that the 10 hydrogen molecules at local temparature T contribute less to overall mass of of a body than expected from either a local measurement or if they were part a much small body. And the difference is the same as loss of potential energy using the standard potential for the SC metric. E=Mc^2 is never going to recovered except locally, in GR. See the math leading up to this phrase: "Furthermore, the contribution of local energy and mass to the system mass is multiplied by the local "red shift" factor ".

I think this is the closest you can come to your idea within GR.

 

[Agerhell]

I meant that it has the general idea that the 10 hydrogen molecules at local temparature T contribute less to overall mass of of a body than expected from either a local measurement or if they were part a much small body. And the difference is the same as loss of potential energy using the standard potential for the SC metric. E=Mc^2 is never going to recovered except locally, in GR. See the math leading up to this phrase: "Furthermore, the contribution of local energy and mass to the system mass is multiplied by the local "red shift" factor ".

I think this is the closest you can come to your idea within GR.

I am sure you are right, yet it is disturbing that the fact that $E=mc^2$ does not hold up when gravity is involved is, as far as I can see, not mentioned at all at for instance the wikipedia page on the relation:

http://en.wikipedia.org/wiki/Mass–energy_equivalence

It is a bit disturbing that $E=mc^2$ is, at least to laymen, taken to be a very general formula when in a weak field spherical symmetric gravitational field (which is basically what we have on earth) a more accurate formula reads:

$E=mc^2(1-\frac{GM}{rc^2})$

 

[PAllen]

I am sure you are right, yet it is disturbing that the fact that $E=mc^2$ does not hold up when gravity is involved is, as far as I can see, not mentioned at all at for instance the wikipedia page on the relation:

http://en.wikipedia.org/wiki/Mass–energy_equivalence

It is a bit disturbing that $E=mc^2$ is, at least to laymen, taken to be a very general formula when in a weak field spherical symmetric gravitational field (which is basically what we have on earth) a more accurate formula reads:

$E=mc^2(1-\frac{GM}{rc^2})$

No, that's not right either. Measured locally, you still have E=mc^2, and locally is how all SR formulas go over into GR. It is just that globally, from far away, if you add up all the locally measured mass you will not get the mass corresponding the the orbits produced. The Komar integral, discounting each locally measured value (not just mass) will produce the right result (for its domain of applicability).