Why does my apple not weigh 500 tons?

Question

Related to this question: What is potential energy truly?

$E=mc^2$ - energy equals mass. So, if an object has gravitational potential energy relative to another object, it should have additional mass to represent that extra energy.

So, if I have an apple in my hand, it should have quite a large amount of gravitational potential energy relative to a black hole at the edge of the observable universe. I did a quick calculation of how much this gravitational potential energy would be (using Newton's Law of Gravitation) and it came out to be the equivalent of around 500 tons (and that's just for one black hole!).

So, why does my apple not weigh a lot more than it appears to?

EDIT: Here's my calculation of the GPE:

Assumptions:

1. The apple has mass $m_a=0.3 kg$
2. The black hole has a mass equal to 1000 solar masses, $m_b=2\times10^{33} kg$
3. Distance between the apple and black hole is 60 billion light years

Obviously $GPE=mgh$ can't be used, because the gravitational acceleration is not constant and varies with distance, so it has to be integrated. I used Newton's Law of Gravitation to provide the acceleration:

$$ F=\frac{G{m_1}{m_2}}{h^2} $$

A small change in height produces a small change in energy, which can be used to integrate:

$$ \begin{eqnarray*} \delta E&=&F \delta h\\ \int_0^EdE&=&G{m_a}{m_b}\int_1^R\frac{1}{h^2}dh \end{eqnarray*} $$

I'm integrating the force starting at 1m from the center of the black hole, as the force goes to infinity at $h=0$. It's a little crude, but it should provide a conservative estimate.

$$ \begin{eqnarray*} E&=&G{m_a}{m_b}(1-\frac{1}{R})\\ E&\approx& G{m_a}{m_b}\;\;\;\;\;\;(R>>1)\\ G&=&6.673\times 10^{-11}\;\;N(\frac{m}{kg})^2\\ G{m_a}{m_b}&=&4\times 10^{22}\;\;Nm\\ E&=&mc^2\\ \Delta m&=&\frac{4\times 10^{22}}{(3\times 10^8)^2}\\ &\approx&444,000\;\;kg \end{eqnarray*} $$

So, a little short of 500 tons, but still pretty big!

Answer 1

The short answer, as Kyle notes in a comment, is that your apple is larger than its Schwarzschild radius.

The longer answer though, is that gravitational potential energy is a property of the system as a whole, not of an individual object in the system. Thus it doesn't matter that your apple has a lot of gravitational potential energy relative to some object (earth, the sun, a black hole, etc.) unless you are asking why the whole system collapses into a black hole (i.e. why doesn't the earth/sun system collapse into a black hole). So if you have a system with a large amount of gravitational potential energy, you still have to think of the scale of the system. If two heavy objects together have a large amount of gravitational potential energy, but are very far apart, then the system as a whole (the two objects) will be much larger than its Schwarzschild radius.

Edit: I answered this question when the title was "Why does my apple not collapse into a black hole?" However the answer is still relevant. Gravitational potential energy is a property of the system as a whole, not merely one constituent particle in the system.

Answer 2

Potential energy is the amount of energy stored in an object with respect to a particular reference point. For instance, look at the image of a sphere above a table below:

We could define the potential of the sphere with respect to the table, $P=mgh$, or with respect to the ground, $P=mgy$. Clearly the potential of the two positions are different, so does that mean the mass is different? The mass should not be different simply by changing your reference point, so the mass is the same. Potential energy only matters due to the change from the reference point.

Consider also the proper definition of Einstein's mass-energy relation: $$ p^\mu p_\mu=E^2-p^2=m^2 $$ (using $c=1$ units). However, this is for a free particle; one that has no potential. When encountering potentials, the energy relation becomes $$ \left(E-U\right)^2=\left(\mathbf p-\mathbf Q\right)^2+m^2 $$ (also in $c=1$ units) where $U$ is the potential energy and $\mathbf Q$ is the potential momentum such that the four-potential is $Q^\mu=(U,\mathbf Q)$. I have not done the math, but I would presume that this fixes your error.

Answer 3

For a bound system which is not itself massive enough to significantly curve spacetime, I believe it's true that the system's resistance to acceleration (i.e. its inertial mass) in the rest frame of its center of mass is proportional to the total internal energy of the system. This would include the mass energy of all the constituent particles, the kinetic energy of all the particles in the center-of-mass frame, and the internal potential energy, defined so that zero potential energy would correspond to all the constituent particles being moved far away from each other (keep in mind that only differences in potential energy have real physical meaning, like the difference in potential between particles far apart and bound together--it is an arbitrary matter what configuration we choose to define as "zero" potential). That's why, as noted on this page, a Helium nucleus consisting of two protons and two neutrons actually weighs a bit less than the sum of weights of two isolated protons and two isolated neutrons, because if the potential of these nucleons is defined as zero when they are far apart, then their potential energy when bound together into a nucleus is negative. In general, if there is an attractive force between two objects then the objects will have lower potential energy when close together than far apart.

Similarly, the difference in potential between a collection of particles bound into an apple and the same collection if all the particles were scattered far apart does contribute to the apple's inertial mass (it contributes negatively to the inertial mass, making the apple weigh a little less than the sum of weights of all its individual particles if they were measured when far apart, since the forces binding the particles of the apple together are attractive). But the apple's potential energy relative to something external like the Earth does not, although it would contribute to the inertial mass of the combined apple/Earth system.

Answer 4

You are reading the formula the other way round. With $$\tag{1}E = mc^2$$ we mean that the energy associated to the mass $m$ of a particle at rest is $E=mc^2$. In oher words, (1) strictly holds for a particle at rest, not interacting with anything else.

What about all the potential energy that a particle has? That must be included on the right hand side of (1). So the version of (1) including potential energy would be $$ \tag{2} E = mc^2 + E_g,$$ with $E_g$ denoting all the gravitational potential energy contributions that you want. So you can now replace $E_g$ with your calculated gravitational potential energy (and then of course all the gravitational energy contributions coming from the interaction of the apple with the rest of the universe... good luck with that).

Now it is crucial to understand that (as also pointed out by Phoenix in the comments) energy is always defined up to a constant, and what physically matters in the dynamic of a particle (or an apple) is the change in the energy between two states (say the apple at point A and the apple at point B). So $E_g$ can be as big as you want (it can be infinite for that matter), and that would not change anything in the dynamic of the apple. This is way it is not generally included, unless the gravitational energy plays a role in the calculations. You could think of a more general version of (1) as being something like $$ \tag{3} E = \sqrt{m^2c^4 + c^2p^2} + E_{gravitational} + E_{em} + E_{whatever} + C, $$ where $E_{em}$ keeps count of the electromagnetic interaction of the object with the rest of the universe (if there is one), $E_{whatever}$ keeps count of all the other possible interaction you can think of, and $C$ is a constant that you can always choose arbitrarily. Obviously, writing (1) like (3) would be in most cases useless, and hence only the relevant contributions are included.

So to conclude, the mass in $F=ma$ keeping count of relativistic corrections is just $$ \tag{4} m = \gamma m_0 \equiv \frac{m_0}{\sqrt{1-v^2/c^2}},$$ where $\gamma$ is the Lorentz factor, $m_0$ the rest mass used in (1), and $v$ the velocity of the particle. Einstein's formula is meant to tell you the amount of energy that you would obtain by completely disintegrating the mass of a particle, not as a way to calculate the inertial mass of an object.

Answer 5

I think the fundamental issue has nothing to do with black holes, but has to do with distinguishing between gravitational force (weight) and gravitational potential energy.

The change in potential energy for the apple and the distant black hole is (if one carries out the appropriate path integral) indeed quite large. But the gravitational force itself is small. This is possible because the gravitational force is the (negative) spatial rate of change of gravitational potential energy; that is, $F=-dU/dx$. There can be a large $\Delta U$, but $dU/dx$ still be quite small it some point.

Answer 6

There must be some error in your calculation. My calculation using Newtons law of gravity suggests that half of objects energy is potential energy. The thing below is a simulation of lowering a 1 kg object hanging on an non-stretching string to event horizon of a black hole (online version).

# A Python script:
G= 6.673*10**-11
M=10**27
m=1.0
c=299792458.0

def swartrad(m):
    return (2*G*m) / (c**2)

print "schwartschildradius" , swartrad(M)   

def force(m,M,d):
    return (G*m*M) /(d**2)

pos=200*swartrad(M) 
endpos= swartrad(M)

step=0.001
energy=0
while pos>endpos:
    energy += force(m,M, pos)*step
    pos -= step


print "potential energy of one kilo" , energy
print "total energy of one kilo    " , c**2
print "ratio" , energy / c**2

Answer 7

You have essentially calculated the difference in potential energy at two points and equated that to the mc2 formulea a valid thing from the point of view of energy conservation. However when your apple was in the black hole you assumed it to be at rest and it therefore had a total negative potential energy equal to your mc2 value. Then when you take it out from there to 60 billion ly away by giving it the energy equal to your mc2 value it comes to rest and you simply cancel the negative PE it had . So it would weigh 0.3kg.

Answer 8

I see several things wrong with your calculation (for instance, units of $1-1/R$?). But mostly, you aren't calculating what you think you are.

1) $\int_0^E dE$ is the change in energy from where you set $E=0$ to the location of the apple. Since the black hole is extremely far away, you would expect this change in energy to be very large, as you found. Several people have made similar comments - only the change in energy is relevant here, since you can set the zero of potential to be anywhere you choose.

2) You cannot equate this change in potential energy to the rest energy of the apple through $E=mc^2$. This energy is only related to the apple itself and not at all to the environment or physical situation it is in (since it is the first component of the apple's four-momentum in a frame at rest with respect to the apple).

So, your calculation is something like "what is the equivalent mass associated with the total energy of the apple-black hole system?" It makes total sense that this is ridiculously huge. Imagine dropping your apple - it would fall a nearly infinite distance and be moving incredibly fast when it hits the black hole, when it has converted all its potential into kinetic energy.

Answer 9

In Newton's mechanics cooling ideal gas releases an infinite amount of energy as the gas shrinks into a point. That's not a problem, because energy is massless and weightless in Newton's mechanics.

In general relativity you can cool ideal gas, and release energy by doing that. But at some point the gas cloud becomes a black hole, which when cooled eventually disappears.

(If a massive, large, low density black hole could become a massive, small, high density black hole, then it's possible to imagine that energy would be released when the black hole collapses under its own gravity, but such thing does not happen in general relativity.)

There's also the fact that in general relativity energy has mass, so if you take away the released energy of a gas cloud being cooled, you end up with a black hole with extra low mass.