# Energy in Schwarzschild geometry

1. May 12, 2010

### stevebd1

From In Wheeler and Taylor's 'Exploring Black Holes', on pages 3-12, the equation for energy in Schwarzschild geometry for an object in free fall is-

$$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}=1$$

where $\tau$ is proper time conventionally expressed in Schwarzschild geometry as $d\tau=dt\sqrt{(1-2M/r)}$. This would appear to give an answer greater than 1.

I'm guessing that the Lorentz factor come into play as the object is in free-fall so that $d\tau$ is the sum of gravitational and velocity time dilation-

$$d\tau=dt\sqrt{1-\frac{2M}{r}} \cdot \sqrt{1-\frac{v^2}{c^2}}$$

where $v=\sqrt{(2M/r)}\ c$ for an object in freefall. This would maintain e/m=1. Would this be correct?

2. May 12, 2010

### George Jones

Staff Emeritus
This is true for an observer who falls freely from rest at infinity.
This relation is not true for an observer who falls freely from rest at infinity.

3. May 12, 2010

### stevebd1

Thanks for the reply George. There's some derivation on pages 3-8 & 3-9 of EBH which implies that for an object in free fall from infinity, $d\tau=dt(1-2M/r)$ as apposed to being square rooted as would be the case for an object hovering at a specific r, and because E/m must equal 1 for an object in free fall, the equation for energy in Schwarzschild geometry is derived using this time dilation equation, hence the introduction of the term $(1-2M/r)$ to maintain E/m=1.

4. May 12, 2010

### yuiop

$$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}=1$$

is only true when when an initially stationary particle falls from infinity and has the value e/m=1 for the reason you give, i.e. the total time dilation (dt/dtau) is the product of its gravitational and velocity time dilation factors.

For now, lets define k = e/m = "conserved energy per unit rest mass" = constant. If a projectile is launched upwards, it eventually comes to rest briefly (at its apogee) before falling back down again. At its apogee the value of dt/dtau is $1/\sqrt{(1-2M/r)}$ and the value of k is:

$$k = \left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}= \frac{(1-2M/r)}{\sqrt{(1-2M/r)}}= \sqrt{(1-2M/r)}$$

This is the value of k for a particle at its apogee at r, or for a particle at rest at r in the gravitational field, because dt/dtau has the same value for both particles.

Now because k is a conserved quantity, it remains unchanged as the particle falls from its apogee, so this fact can be used to compute dt/tau for a particle falling from any initial height.

Lets say a particle is initially at rest at r = 4M, then $k = \sqrt{(1-2M/r)}$ = 0.7071 and as it falls the time dilation factor of the particle at any given r is given by:

$$\frac{dt}{d\tau}= \frac{k}{(1-2M/r)} = \frac{0.7071}{(1-2M/r)}$$

That much I think I understand. What I am not sure about, is exactly what the "conserved energy per unit mass" represents. It is obviously not e/m = mc^2/m = c^2. Is e/m or k the sum of the potential energy and the kinetic energy of the falling particle? I think that would make some sort of sense because as it gains kinetic energy it loses potential energy, but poential energy seems to be more of a Newtonian concept that a GR concept (not sure about that either). k does not seem to represent the total energy of the particle in the normal sense that includes the rest mass energy. On the other hand, if the value of k is unity for a particle at infinity this might suggest that k is the ratio of its total energy to its rest energy and since it is initially at rest all of its energy is the form of rest mass, but that would imply that as the particle is falling it is gaining kinetic energy at the cost of losing rest mass energy. I would be very interested in any opinions on the exact physical nature of the "conserved energy per unit mass".

Certainly we can note that the energy per unit rest mass of a hydrogen atom at r=4M is 0.7071 and the energy per unit rest of a hydrogen atom at infinity is 1. What is the significance of that?

Last edited: May 12, 2010
5. May 13, 2010

### stevebd1

Hi Kev. Here are a couple of other equations from EBH-

E/m relationship for a particle dropped from a specific radius (ro)-

$$\frac{E}{m}=\left(1-\frac{2M}{r_o}\right)^{1/2}\,<\ 1$$

Energy/mass ratio relative to a shell frame (r) for a particle dropped from a specific radius-

$$\frac{E_{shell}}{m}= \left(1-\frac{2M}{r_o}\right)^{1/2}\left(1-\frac{2M}{r}\right)^{-1/2}$$

compared to that of an object in free fall from infinity-

$$\frac{E_{shell}}{m}=\left(1-\frac{2M}{r}\right)^{-1/2}$$

6. May 13, 2010

### yuiop

Hi Steve,

So:

$$\frac{E}{E_{shell}} = \left(1-\frac{2M}{r}\right)^{1/2}$$

That all seems to agree with what I said in #4 and in case I did not make it clear, I basically agree with everything you concluded in your original posts. (You have to carefully define how v is measured but you are probably aware of that.) Just the tiny quible that when you said greater than 1 in #1, you perhaps meant less than 1.

P.S. I am still interested in any ideas on the nature of this conserved energy.

Last edited: May 13, 2010
7. May 13, 2010

### stevebd1

While it's easy to understand why an object in free fall which already had velocity would have a relationship of E/m>1 (the equation being-

$$E/m=\left(1-v_{far}^2\right)^{-1/2}\,>\,1$$

for an object hurled inward at speed $v_{far}$ from a great distance) the idea of an object having a relationship of E/m<1 when dropped at a specific radius is a little less easier to comprehend. It's as if the object isn't at it's full kinetic energy potential, hence E/m<1, and won't achieve it unless it has a boost of energy bringing it to E/m=1. For an object in free fall from infinity, as r reduces, potential energy reduces but kinetic energy increases maintaining a balance but for an object dropped at a specific r, potential energy still reduces with r but it would appear that kinetic energy isn't increasing proportionally enough to maintain a balance.

Last edited: May 13, 2010
8. May 13, 2010

### yuiop

The balance is still being maintained even in the case when the object is dropped from a radial coordinate of less than infinity. I gave the example of an object dropped from r=4M having an initial energy of E/m=0.7071 and as it falls its energy stays at E/m=0.7071 with gain in kinetic energy being balanced by loss of potential energy just as in the case of an object dropped from infinity.

What is interesting is that if an object is dropped from infinity with an initial energy of 1.0, to a surface at r=4M, it will will have a remaining energy of 0.7071 when it comes to rest at the surface, after its kinetic energy is dispersed as heat after the impact. Since this energy is described as conserved energy per unit rest mass, I guess we have to conclude that the rest mass of the object is unchanged.

What is a little puzzleing is that in relativity (in SR at least) the total energy of an object can be defined in terms of its rest mass and linear and angular momentum and usually there is no mention of potential energy. In Newtonian physics, this potential energy is thought of as being stored in the gravitational field, but I thought in GR it was defined differently. Not really sure about that and that is why I am asking.

9. May 13, 2010

### stevebd1

In EBH, an example of an object falling from infinity to a neutron star is given, an aspirin tablet weighting 0.5 of a gram, the neutron star has a mass of 1.4 sol and has a radius of 10 km. Using the Eshell/m, the tablet arrives at the star surface with a total energy of 1.3, it then states that the KE is the shell energy minus the rest energy (m) which produces the equivalent of 0.15 grams of energy. The remaining energy of the object is 1.

In your case, still for an object falling from infinity, you've established a radius of 4M and used the E/m equation for a particle dropped from a specific radius (ro) (or k) to imply what energy the object would have remaining once it hit at 4M, in this case 0.7071. Basically, for a free fall object hitting a specific radius, EBH states KE=Eshell/m - m while you state that KE=Eshell/m - k and maybe I see why you've done this because if the object was to strike at 4M, disperse it's kinetic energy and then continue to fall from 4M, then the E/m equation for a particle dropped from a specific radius (ro) would apply. There just seems to be a discrepancy here.

Last edited: May 13, 2010
10. May 13, 2010

### yuiop

Hi steve,
I can not exclude the possibilty that I may be wrong because as I have already said I am not certain of the form of the enrgy being used here. This is how I read the above example.

It starts at infinity with rest mass m =0.5 gram and energy per unit rest mass E/m of 1.0

It arrives at the shell with shell energy Eshell/m = 1.3 which means the total energy is 1.3*0.5 = 0.65 grams. Of this the KE is 06.5-0.5 = 0.15 grams.

If this KE is dispersed the remaining energy of the aspirin is 0.5grams which is its rest energy. The final energy per unit rest mass of the aspirin is Eshell/m = 1.0

I think my example differs from the book because I am considering the energy as viewed from infinty (the coordinate energy) while the book is considering the proper energy as measured locally. In the book example the aspirin starts with rest energy =0.5 grams and ends up with rest energy+KE =0.65 grams and because potential energy has not been shown it appears to have made a profit in energy. Anyway, GR does not claim to always conserve energy and is what Emmy Noether describes as an "improper energy theorem". In some ways potential energy is an accounting device and can not be directly measured. When you have missing energy you preserve conservation of energy by calling the missing energy potential energy :tongue:.

The quantity I call k is the total energy as measured from infinity and is conserved. I probably have to give all this a lot more thought.

[EDIT] OK, I think I have figured out the discrepancy now.

The initial E/m of the aspirin at infinity is 1. When it falls to the neutron star its final E/m (as measured from infinity is k = 1/1.3 = 0.7692 and the final total energy is (E/m)*m = 0.7692*0.5 = 0.38462 grams which represents the coordinate KE as measured at infinty. The local (proper) KE is 0.38462*1.3 = 0.15 grams.

So I conclude that (coord KE + coord PE)/(rest E) = constant = (coord E)/$(m_0c^2)$ = $(1-2M/(rc^2))dt/d\tau$ for a free falling particle and (shell E) = (proper E) = $E_0$ = (coord E)$(1-2M/r)^{(-1/2)}$.

Last edited: May 13, 2010
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