# Potential Energy and energy conservation in General Relativity

1. May 4, 2009

### yuiop

This is a spin off from another thread ( https://www.physicsforums.com/showthread.php?t=310455&page=2 ) to avoid hijacking the original thread.

DrGreg offered this expression for Potential Energy from the text book "General Relativity" by Woodhouse:

$$PE = m_o c^2*\sqrt{1 - 2GM/(rc^2)}$$

where $m_o$ is rest mass of the test particle, G is the gravitational constant, M is the mass of the large gravitational body, c is the speed of light and r is the radius of the test particle from the centre of M.

Setting units of c=G=1 (geometrised units) and $\gamma = 1/ \sqrt{1-2GM/(rc^2)}$ for the gravitational gamma factor, the equation simplifies to:

$$PE = m_o/\gamma}$$

To find out how potential energy is converted into Kinetic Energy (KE) we need to find the velocity of the falling particle at a given height.

The equation for the local velocity of a free falling object is:

$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$

where R is the release height or apogee of the free falling particle. See http://www.mathpages.com/rr/s6-07/6-07.htm for the source of this equation.

When R is set to infinity and G=c=1 the equation simplifies to:

$$v = \sqrt{ 1 - {1}/{\gamma^2}}$$

The Total Energy (TE) of a particle is given by:

$$TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$$

where Rest mass Energy $(RE) = m_o c^2$, Momentum Energy (ME) = pc and momentum $(p) = \gamma m_o v$.

The Kinetic Energy (KE) of a falling particle is that part of the Total Energy that is not rest energy so KE = TE-RE.

It should be noted that the gravitational gamma factor is numerically equal to the velocity gamma factor of special relativity for the velocity of an object that has fallen from infinity at any given height.

We are now in a position to do a numerical example.

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0
RE = 1.0
PE = 1,0
TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzchild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: $$\gamma = 4.582576$$
The terminal velocity is $$v = 0.975900$$ (97% of the speed of light)
RE = 1.0
PE = 0.218218
ME = 4.472136
KE = 3.582576
TE = 4.582576

A problem arises here. By assumed Rest mass Energy is invariant it turns out that energy is not conserved as promised by Baez. (See http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html "The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls.")

The Total Energy of the particle after falling is greater than the combined total of the initial Rest Energy of the particle and the Potential Energy of the particle at infinity.

What if Potential Energy and Rest mass Energy are the same thing?

Here is the numerical example reworked with the new assumption PE=RE :

Initial quanties at infinity per unit mass of the test particle.

Velocity = 0.0
RE = 1.0
PE = 1,0
TE = 1.0

After the particle has fallen to a height (r) of 2.1*(Schwarzchild radius) and assuming a mass of 10,000 units for the black hole

The gravitational gamma factor is: $$\gamma = 4.582576$$
The terminal velocity is $$v = 0.975900$$ (97% of the speed of light)

RE = 0.218218
PE = 0.218218
ME = 0.975900
KE = 0.781782
TE = 1.000000

Now, by assuming the Rest Mass Energy and gravitational Potential Energy of a falling particle are the same thing, energy is conserved, and the gain in Kinetic Energy is a result of loss of Potential Energy which is the same thing as the loss of Rest mass.

The implication, if correct, is that the Rest mass Energy of a particle falling from infinity is completely converted to Kinetic Energy by the time it gets to the event horizon. In effect the falling massive particle becomes a photon at the event horizon which is consistent with the particle attaining the speed of light by the time it gets to the event horizon. If it is true that a particle has no rest mass as it reaches the event horizon then that would be something very difficult to transform away by use of some alternative metric. For example it would not be possible to have a physical observer (with rest mass) fall smoothly and continuously through the event horizon. If it is not true, the equation for PE in the Woodhouse textbook is wrong OR Baez's claim that energy is conserved in Schwarzschild coordinates is wrong, OR the equation for the terminal velocity of a falling particle from the mathpages website is wrong OR my calculations are wrong.

It should be noted that the Potential Energy of a particle goes to zero at the event horizon and the final velocity goes to c at the event horizon for ANY initial height that the particle is dropped from, not just infinity.

Also, as mentioned in previous thread, a photon has no Rest mass Energy and as such it would seem that it is not possible for a photon to have gravitational Potential Energy, so if energy is conserved in Schwarzschild coordinates the Total Energy of a photon does not change as it rises or falls.

Those are my observations on the subject. I look forward to the members of this forum telling me where I have gone wrong.

Last edited: May 4, 2009
2. May 4, 2009

### nutgeb

Last edited by a moderator: Apr 24, 2017
3. May 4, 2009

### yuiop

I had a quick look. This point you made:

... is exactly right.

To claim that clocks lower down run slower than clocks higher up, is the same as claiming the frequency of a falling photon does not change...

... and to claim that the frequency of a photon increases as it falls, is the same as saying clocks higher up and lower down run at the same rate.

Last edited by a moderator: Apr 24, 2017
4. May 5, 2009

### Ich

That's just another coordinate singularity. It's measuring kinetic energy with lightspeed observers (thus very large) and then rescaling to a finite value.
Always bear in mind that at the EH, many Schwarzschild quantities are defined via an impossible set of observers. Simply expect impossible observers to measure impossible things.
That has nothing to do with the properties of the massive falling body.

5. May 5, 2009

### quZz

I'd like to mention that $$E = m_0 c^2 \sqrt{1-2GM/rc^2}/\sqrt{1-v^2/c^2}$$ is conserved. Anything else is the work of the evil one

6. May 5, 2009

### atyy

Does this hold in Schwarzschild space? I was thinking something like TE=KE+PE, like Eq 7.47 of http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html.

7. May 5, 2009

### yuiop

I should have made it clear that I had substituted $m_o*\sqrt{1-2GM/(rc^2)}$ for $m_o$ in the original $TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$ equation when I proposed that PE is substituted for RE to make the energy balance in Schwarzschild coordinates.

The link you posted is very helpful but a little difficult to follow as it seems to assume the reader is already an expert in General Relativity. That reference defines the effective potential as:

$$\frac{E^2}{2} = \frac{1}{2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + V(r)$$

By substituting equation 7.48 given in your link for V(r) and removing the terms that contain L as I am not concerned with angular momentum here, the effective potential is:

$$\frac{E^2}{2} = \frac{1}{2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + \frac{1}{2}\epsilon-\epsilon \frac{GM}{r}$$

To obtain an equation that is in the correct units for energy, I have reintroduced the not explicitly stated $m_o$ and $c^2$ variables and then multiplied by 2 and taken the square root on both sides to obtain an expression for total energy :

$$TE = m_o c^2 * \sqrt{\frac{1}{c^2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + \epsilon-\epsilon \frac{2GM}{r c^2}$$

Earlier in the link given by atyy just after equation 7.39 it states that for a massive particle we should use $\lambda = \tau$ (proper time of the falling particle) and $\epsilon = 1$ so I will use those parameters to give:

$$TE = m_o c^2 * \sqrt{\frac{1}{c^2} \left(\frac{\delta r}{\delta \tau} \right)^2 + \left(1-\frac{2GM}{r c^2} \right)$$ (Eq 1a)

where ${\delta r}/{\delta \tau}$ is the instantaneous radial falling velocity measured in proper time and $(1-2GM/(r c^2))$ is the familiar gravitational gamma factor squared and inverted.

The equation for ${\delta r}/{\delta \tau}$ is conveniently given by mathpages here http://www.mathpages.com/rr/s6-07/6-07.htm as

$$\frac{\delta r}{\delta \tau} = - \sqrt{\frac{2GM}{r} - \frac{2GM}{R}}$$

where R is the release height of the particle and r is the height the particle has fallen to.

By substituting this new expression for the radial proper falling velocity into (Eq 1a) the equation for Total Energy becomes:

$$TE = m_o c^2 * \sqrt{ \left( \frac{2GM}{r c^2} - \frac{2GM}{R c^2} \right) + \left(1-\frac{2GM}{r c^2}\right)$$ (Eq 1b)

For any initial release height (R) the TE given by that equation is constant for any r. This is especially obvious when R is set to infinity so that term goes to zero and the equation simplifies to:

$$TE = m_o c^2 * \sqrt{1}$$

If we use the symbol 'u' to represent the instantaneous radial falling velocity measured in proper time ${\delta r}/{\delta \tau} = - \sqrt{\left({2GM}/{r} - {2GM}/{R}\right)}$ then the Carroll TE equation (Eq 1a) can be expressed as :

$$TE = m_o c^2 * \sqrt{\frac{u^2}{c^2} + \left( 1-\frac{2GM}{r c^2} \right)$$ (Eq 1c)

It turns out that the equation given by Carroll (Equation 1a, 1b or 1c) is numerically equal to the equation by quZx which I will call (Eq 2a) :

,when v is calculated using this mathpages equation given in my earlier post:

$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$

Returning to the original equation I gave in OP of this thread:

$$TE = \sqrt{(ME)^2+(RE)^2}$$

where RE is Rest Energy and ME is Momentum Energy and using the assumption that Potential Energy (PE) is synonomous with Rest Energy in the context of a particle in a gravitational field, then:

$$TE = \sqrt{(ME)^2+(PE)^2}$$ (Eq 3a)

which can be expanded to the more explicit equation:

$$TE = \sqrt{\frac{(1-2GM/(r c^2)) m_o^2 v^2 c^2}{(1-v^2/c^2)} +(1-2GM/(r c^2)) m_o^2 c^4}$$

Using the symbols $\gamma$ for the velocity gamma factor and g for the gravitational gamma factor the equation can be stated as:

$$TE = \sqrt{\left(\frac{\gamma m_o v c}{g} \right)^2 +\left (\frac{m_o c^2}{g}\right)^2}$$ (Eq 3b)

Similarly, the equation given by quZz can be expressed as:

$$TE = \frac{\gamma m_o c^2}{g}$$ (Eq 2b)

The equations 1a, 1b, 1c, 2a, 2b, 3a and 3b are all numerically equal and all agree that the total energy of a falling particle is constant and Kinetic Energy is gained at the expense of Rest Energy/Potential energy of the particle. At the event horizon the Rest Energy/Potential Energy is zero and all the energy of the particle is purely Kinetic.

It is also correct that:

TE = KE + PE as pointed out by atyy which is just another way of expressing the more familiar:

TE = KE + RE of special relativity.

The Caroll equations (1a, 1b and 1c) are superior only in that they do not suffer from division by zero errors for values of r <= 2GM/c^2 but the other equations may be more convenient in certain situations.

It is also worth noting that effective potential is not the same thing as potential energy and the two are often confused.

Last edited: May 6, 2009
8. May 5, 2009

### atyy

Nice algebra! I'm pretty sure from other considerations that a massive particle does not turn into a photon at the event horizon, but am not sure what a correct way to answer your question is. How about considering that what one calls RE, PE etc is arbitrary - I suspect that only TE has a coordinate independent meaning as a conserved quantity related to a Killing vector. So how about trying some renaming (using ?) to think of sqrt(1-2GM/r) as representing RE?+PE?, so that RE? is constant, and PE? varies with r, and it just so happens that RE?+PE? is zero at the event horizon, but RE? is still non-zero, and so there is no massive particle turning into a photon?

9. May 6, 2009

### yuiop

I think I understand what you are getting at and I think you are on the right track. It occured to me that even though the RE and/or PE of a massive particle goes to zero at the EH, the Rest Mass $m_o$ of a massive particle is always non-zero even at the event horizon, while the Rest mass of a photon is always exactly zero, so perhaps it would be unsafe to consider that a massive particle becomes a photon at the EH. It also occurs to me that the Potential Energy is by definition the stored energy that has potential to be converted to kinetic energy by nuclear processes or by virtue of the particles location in the gravitational field. When a particle arrives at the EH its coordinate velocity goes to zero as does the PE. This suggests there are no potential energy reserves for the particle to continue anywhere unless we are talking about going into negative potential energy and where is this energy borrowed from?

There are still some paradoxical things about the EH. For example a particle with zero rest mass can never be stationary and yet a photon can be stationary (with respect to the centre of the gravitational field) at the EH. A particle with non-zero Rest Mass can never travel at the speed of light and yet it can at the EH but it just so happens that the speed of light is zero there. At the EH all the energy of of a massive particle is kinetic and yet it is not moving in coordinate terms.

Anyway, it seems that even when the PE or RE of a massive particle is zero at the EH, it still retains the property of non-zero Rest Mass and that the information of being a massive particle is retained and recovered when the particle is eventually released as Hawking radiation when the black hole evaporates... so no information loss... maybe...

Last edited: May 6, 2009
10. May 6, 2009

### nutgeb

I'm referring to Taylor & Wheeler's textbook Exploring Black Holes, which specifically says that a Newtonian analysis (potential energy goes down as kinetic energy increases) cannot be used to accurately assess a particle's relativistic energy in freefall in a gravitational field. It says that a "shell" observer (located on a theoretical mechanical shell surrounding the BH outside the EH) can perform a locally valid measurement of a falling particle's relativistic energy. This measurement is quite different than the energy calculated for the same falling particle by a very distant observer, which is always equal to the free falling particle's rest mass regardless of velocity. As you know, it is not valid for a distant observer to separate a particle's total energy into potential energy and kinetic energy components, nor into rest energy and kinetic energy components. On the other hand, it is valid for a shell observer to divide the particle's total energy into kinetic energy and rest energy components (but not a potential energy component). The shell observer will locally measure the particle's kinetic energy, and therefore its total energy, to increase as the particle falls farther toward the EH. The difference between the total energy measured locally by the shell observer and globally by an observer at infinity is:

$$\frac{E_{shell}}{E_{infinity}} = \frac{1}{\sqrt{1 - \frac{2GM}{c^{2}r}}}$$

Every shell observer, no matter how close to the EH, will locally measure light to be travelling at exactly the speed of light; not coming slowly to a rest at the EH. The shell observer will measure an infalling massive particle to approach the speed of light as it approaches the EH, not slow down.

Maybe I'm misreading your posts, but it seems to me you tend to shift your reference frame at various points by using distant observer attributes sometimes and shell observer attributes at other times, in the course of your calculations. If you are actually doing that (I'm not entirely sure), you will obtain seemingly paradoxical results. There is no doubt that relativistic analysis is abstract because the calculated and measured results are so different depending on the observer's situation.

I believe it also is fundamentally invalid to introduce the concept of potential energy into relativistic equations such as the Schwarzschild metric. At best, PE can be treated as a Newtonian analogy for only limited parts of the relativistic calculations.

11. May 6, 2009

### yuiop

Agree, but ratio going to infinity at the EH is a nuisance.

This is the equation I gave for the velocity used in the equations:

$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$

You will notice that if r is set to 2GM/c^2 that v=1 and so this equation is the local measurement of velocity. I modified the original coordinate velocity of a falling particle given in mathpages http://www.mathpages.com/rr/s6-07/6-07.htm as:

$$v = \left(1-\frac{2GM}{r c^2} \right) \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$

by removing the $\left(1-{2GM}/{r c^2} \right)$ factor on the left of the square root term precisely for the purpose of expressing velocity in local terms rather than coordinate terms.

I am aware of that possibility and I am exploring it further, but if I am guilty of doing that, then so is Sean M. Carroll here http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html and N.M.J. Woodhouse on page 100 of his text book "General Relativity". Woodhouse gives the Total Energy of a falling particle as :

$$E = \frac{\sqrt{1-2M/r}}{\sqrt{1-v^2}}$$

which is essentially the same as the equation given by quZz in his earlier post.

I agree that a local observer is unable to determine the gravitational gamma factor without reference to the measurements of other observers not located at his locality. I also agree that if the coordinate velocity is used then it is difficult to make an energy balance that conserves energy because the coordinate velocity goes to zero as the EH is approached.

I also admit I am not sure of the exact nature of potential energy in GR and that is the reason for starting this thread, so that I might learn more.

I would also be interested if anyone could post an equation and or derivation ( a link will do) for gravitational potential in GR as I can not seem to find much information on that subject.

[EDIT] It occurred to me after posting this, that the falling velocity could be expressed in coordinate terms as long as the speed of light is also expressed in coordinate terms, i.e. $c_{(COORD)} = c_{(LOCAL)}(1-2GM/(r c^2))$. That would probably work out fine and the Total Energy expression would then be completely in coordinate terms rather than the mixed coordinate and local terms, which is a valid concern.

Last edited: May 6, 2009
12. May 6, 2009

### yuiop

Sure, a local observer can measure the local falling velocity, but I am curious how he would determine the rest mass of the falling object? A mass of 2kgs at infinity and a mass of 1kg at infinity fall at exactly the same rate so the local falling velocity tells you nothing about the rest mass of the falling object.

Maybe the falling objects can be arranged to collide with lightly suspended stationary test masses at a given radius and the mass of the falling object estimated from momentum considerations?

To be sure that the total energy is increasing, the local observer would have to have complete knowledge of the all the energy components of the particle.

Last edited: May 6, 2009
13. May 6, 2009

### nutgeb

Please be more specific about what you think would be wrong in Carroll or Woodhouse. The link to Carroll isn't specific.

Is there a place where they mix the perspectives of an observer at infinity (usual for the Scwharzschild metric) and a shell observer?

14. May 6, 2009

### yuiop

No, I am not saying that there is anything wrong with the Carroll or Woodhouse equations. I am saying their equations numerically agree with mine, so if mine is wrong, so are theirs. However, you are right that I was guilty of introducing a local measurement of velocity, but that did not affect the results, as I was using a local measurement for the speed of light also.

$$\frac{v^2_{(COORD)}}{c^2_{(COORD)}} = \frac{ v^2_{(LOCAL)}(1-2M/r)^2} { c^2_{(LOCAL)}(1-2M/r)^2} = \frac{v^2_{(LOCAL)}}{c^2_{(LOCAL)}}$$

Last edited: May 6, 2009
15. May 6, 2009

### quZz

gravitational "force" is not potential in GR in general =) even in Schwarzschild case it depends on the velocity

16. May 6, 2009

### atyy

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