# Potential energy mass equivalency

1. Jan 11, 2004

### carl fischbach

I have a question.
If you have a closed system that contains a lot
of stored potential energy and you release that
potential energy and convert it to kinetic energy
is the total mass or energy of the system, the
same before and after the release of the potential
energy?

2. Jan 12, 2004

### Staff: Mentor

Yes. The total mass-energy does not change.

3. Jan 12, 2004

### carl fischbach

potential energy mass equivaleny

If there the same it brings up an interesting
possibility.
Lets say you take an isolated system of an
electron positron pair and seperate them by
1 meter and allow them to accelerate towards
each other and since the mass of the pair has to
remain constant,there will be no increase of
mass dectected as the pair accelerates towards
each other.That means no increase of energy or
mass with increased velocity,I can't see that
happening.

4. Jan 13, 2004

### Staff: Mentor

Re: potential energy mass equivaleny

I'm not sure what you mean by "mass" of the pair, but the equivalent mass-energy must include all contributions, including potential energy. Yes, the KE of the particles increases as they accelerate towards each other, but that energy must come from somewhere. Since the system is isolated, the total mass-energy does not change.

5. Jan 13, 2004

### Kurdt

Staff Emeritus
The Kinetic energy of the electron-positron pair would come from their electric potential and therefore no difference in the mass-energy of the system is detected.

6. Jan 13, 2004

### Arcon

Yes. The mass of a closed system is a conserved quantity. In fact what you described is exactly what happens in nuclear fission. For a concrete example please see
http://www.geocities.com/physics_world/sr/nuclear_energy.htm

Who said that there is no increase of mass with velocity? you've probably been listening to the folks who keep saying that mass does not change with speed. They're speaking of proper mass. Not mass-energy (aka relativistic mass). However you are speaking of mass-energy.

Suppose you held the electron one meter from the positron in the inertial frame S. The total energy of the system is the sum of the electron's proper mass, the positron's proper mass and U/c2 where U is electric potential energy of this system. Now releast them at the same time as measured in S. The particles will start to accelerate towards each other and pick up speed. As the potential energy increases U decreases. The sum remains constant. Thus as the mass of the electron's increase the mass associated with the potential energy decreases. Thus the mass of the system remains constant and the mass of each particle increases.
It sounds to me that he is refering to the sum of the proper masses of the electron and positron.

Arcon

7. Jan 14, 2004

### james fisher

This is actually Carl Fischbach, there was a
glitch in the system I had to reregister to
Maybe the point I'm trying to make is that
potential energy isn't detectable as mass.
If you looked at electron and positron
seperately as they accelerated towards each other
you would see an electron with an increasing
mass as measured by a magnetic field and you would
also see a positron with increasing mass as
measured by a magnetic field.The work would
come from the electric field and this
energy cannot be detected as mass,otherwise the
detectable mass of the system would remain constant.

8. Jan 14, 2004

### Arcon

Considere instead a (massless) bar with two charges attached to each end. Weigh this two charge system by placing it on a scale as one might place a barbell on a scale. Define the quantities

$$U = \frac {1}{4 \pi \epsilon_{0}} \frac {kq_{1}q_{2}}{r}$$

$$m_{U} = \frac {U}{c^{2}}$$

$$M_{0} = m_{01} + m_{02} + m_{U}}$$

where
r = distance between charges = length of bar
m01 = proper mass (aka rest mass) of particle 1
m02 = proper mass of particle 2
q1 = charge of particle 1
q2 = charge of particle 2

Then it can be shown that

W = M0g

where

g = the acceleration due to gravity (If the barbell having no mass bothers you simply add the rest mass of the bad into the first expression).

This shows that the dumbell has a proper mass of M0. This is actually the passive gravitational mass. But the same thing holds if you try to accelerate the charge and then measure the weight in the rest frame of the particle.

The above relation was proved in

Electrostatic potential energy leading to an inertial mass change for a system of two point charges,, Timothy H. Boyer, Am. J. Phys., 46(4), 383 (1978)

Electrostatic potential energy leading to a gravitational mass for a system of two point charges, Timothy H. Boyer, Am. J. Phys., 47(2), Feb. 1979

9. Jan 14, 2004

### james fisher

I have one last point to make on this issue.
Lets say you have a huge hydrogen cloud 10 billion
Kilometers across and the mass of our sun.
Then you test the mass of 1 hydrogen atom and find
to be the same mass as a hydrogen atom found on
earth.Then cloud collaspses under it's own
gavity to a diameter of 1 million Kilometers.
The work of gravity done by the collaspse has
given the hydrogen atoms an extreme velocity,
since the energy content of the cloud must
remain constant, the mass of a hydrogen atom moving at extreme velocity within the collasped
cloud must have the same mass as before the collaspse of the the cloud. I can't understand
this but I'll take your word for it that thats the way it is.

10. Jan 15, 2004

### Arcon

I don't see why the mass of the hydrogen atom should have the same mass before the collapse as after it. When a force is exerted on a particle it's mass changes. Its only the total mass which is a constant. In this case the mass associated with the gravitational field must be taken account when the mass of the entire cloud is calculated.

11. Jan 15, 2004

### james fisher

mass of cloud

You cannot detect the gravitational potential
energy in terms of mass before the cloud
collaspses.Lets put this way if you were to
accelerate the entire cloud before it collaspses
to velocity x then,
(momentum of cloud)=(mass of cloud)*x
(momentum of cloud after collaspse)=
(new mass of cloud)*x
You can change the momentum of system when you
release or store potential energy.

12. Jan 15, 2004

### Arcon

Re: mass of cloud

I don't understand what you mean by detect the gravitational potential. Can you clarify that for me? For example: Consider the Newtonian situation of a particle falling in a uniform gravitational field. The total energy, E, is given by

$$E = K + V = \frac {1}{2}m v^{2} + mgz$$

where z = height of particle above zero potential level, and g = acceleration due to gravity. Total energy is useful since it is often a constant of motion (E = constant when the field has a time independant potential energy function). That is what energy means. As the particle falls its speed increases. As such the K in that equations increases. However since the ball is falling the height, z, is decreasing and thus V is decreasing. The sum remains the same.

With this example in mind please describe what you mean by detect the gravitational potential.

IF we were to accelerate the entire cloud then we are no longer considering a closed system. The mass of the cloud is then no longer conserved. Conservation of both energy and momentum does not apply to systems which are not closed. They apply to closed systems. However that does not mean that the energy of a non-closed system is not conseved either. Keep in mind that mass is much like kinetic energy - kinetic energy is not a conserved quantity since it is total energy which is conserved.

Thank you

13. Jan 15, 2004

### Kurdt

Staff Emeritus
Re: potential energy mass equivaleny

I believe the answer to the question here is really that the total mass of the isolated system as you view it is jut the rest mass. Following from Einsteins equation this implies that isolated systems have the following property

rest mass of particles + potential energy = rest mass of particle + max. kinetic energy

So any system that is in transition from potential energy to kinetic energy will have mass equal to

rest mass of particle + (kinetic + potential energy)

Last edited: Jan 15, 2004