Question to the mass and energy equivalence

[icarus2]

Hello,

I’m sorry, I can’t English well.

 

But, I have a question!

From the mass and energy equivalence

E=mc^2

 

I had some read the book( Gravitation and Spacetime, Hans C. Ohanian and Remo Ruffini )

 

In there, gravitational potential energy is gravity source also.

 

However, from the mass and energy equivalence

Gravitational potential energy is

[math]U = - \frac{{GMm}}{r} = + m_{gp}c^2 [/math]

 

or

[math]U = - \frac{{GMm}}{r} = - m_{gp}c^2 [/math]

 

([math]m_{gp} = \frac{{GMm}}

{{rc^2 }}[/math])

Which one is true?

 

In the chapter 2.7( Gravitation and Spacetime, Hans C. Ohanian and Remo Ruffini )

 

At the solar system, difference of action and reaction

 

[math]\frac{{GM_1 M}}

{{(r + \vartriangle r)^2 }} \simeq \frac{{GM_1 M}}

{{r^2 }} - \frac{1}

{2}\frac{{GM_1 }}

{{c^2 }}(\frac{{2GM_1 M}}

{{r^3 }})[/math]

 

M1=Sun mass, M=Earth mass, r=distance

 

From [2.195]

[math]\frac{{GM_1 M}}

{{r^2 }} - \frac{{GM_1 M}}

{{(r + \vartriangle r)^2 }} = \frac{1}

{2}\frac{{GM_1 }}

{{c^2 }}(\frac{{2GM_1 M}}

{{r^3 }}) = (\frac{{GM_1 M}}

{{rc^2 }})(\frac{{GM_1 }}

{{r^2 }})[/math]

 

However, this expression is same to the negative mass expression.

 

[math](\frac{{GM_1 M}}

{{rc^2 }})(\frac{{GM_1 }}

{{r^2 }}) = ( - \frac{{GM_1 M}}

{{rc^2 }})( - \frac{{GM_1 }}

{{r^2 }}) = - \frac{{GM_1 ( - \frac{{GM_1 M}}

{{rc^2 }})}}

{{r^2 }}[/math]

 

[math] = - \frac{{GM_1 ( - m_{gp} )}}

{{r^2 }} = - m_{gp} a[/math]

 

 

 

Therefore, I have a question.

 

------------------------------

From the mass and energy equivalence

Gravitational potential energy is

 

[math]U = - \frac{{GMm}}

{r} = + m_{gp}c^2 [/math]

or

 

[math]U = - \frac{{GMm}}

{r} = - m_{gp}c^2 [/math]

 

Which one is true?

Please explain to that!

 

Have a nice day!

-------------------------------

[icarus2]

Hmmm…

No answer…

 

May be, because that my question was wrong.

 

From the E=mc^2,

m = E/c^2,

 

Gravitational potential energy U is negative energy.

Also, Gravitational potential energy is gravity source.

It is energy, so it is mass.

 

m = U/c^2

 

So, it means that negative gravitational mass is being.

 

At the below sentence, which is right and wrong?

 

1. Mass and energy equivalence

This law can be applied only positive energy.

2. Gravitational potential energy(or gravitational self energy) is negative energy.

3. The law of mass and energy equivalence says that gravitational potential energy has a negative gravitational mass.

 

I want to your answer or opinion.

[ajb]

I am worried that you mixed Newtonian ideas with relativistic ones. I believe you can discuss some things without general relativity in this way, but it will inherently be ill-founded.

 

You may be interested in the positive mass theorem in general relativity.

[icarus2]

Thanks AJB.

I can’t use to the general relativity.

But, from your advice, I search to the positive energy(mass) theorem.

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

 

In Wikipedia,

“In general relativity, the positive energy theorem (more commonly known as the positive mass theorem in differential geometry) states that, assuming the dominant energy condition, the mass of an asymptotically flat spacetime is non-negative; furthermore, the mass is zero only for Minkowski spacetime.

 

In Wikipedia,

"assuming the dominant energy condition,"

So, I search to the dominant energy condition.

 

http://en.wikipedia.org/wiki/Energy_conditions#Mathematical_statement

 

However,

[ A counter example is exist ]

Dark energy or cosmological constant

 

I can’t English well. So my understanding was lacked.

 

In my think,

Positive energy theorem is seems to the assumption or not complete.

 

Also, in general situation,

Newton dynamics

 

[math]\frac{{GMm}}{r} < < mc^2 ;\frac{{GM}}{{rc^2 }} < < 1[/math]

 

In general relativity

(My statement is insufficient. so please think to the logical aspect, please forgive me!)

 

Even so in the strong gravitational potential,

The condition is

 

[math]| - \frac{{GMm}}{r}| \le mc^2 [/math]

 

[math]mc^2 + ( - \frac{{GMm}}{r}) \ge 0[/math]

 

 

Namely, in my guess,

 

Positive energy theorem seems to connect the total energy.

It is seems to not say that gravitational potential energy has a negative gravitational mass or positive gravitational mass. Total energy(namely, total mass) has a non-negative.

 

Again,

I’m very sorry for the English ability and understanding of the general relativity(energy-momentum tensor...).

 

 

[The law of mass and energy equivalence says that ]

1. gravitational potential energy has a negative gravitational mass.

 

2. gravitational potential energy has a positive gravitational mass.

 

3. gravitational potential energy has a zero mass.

 

4. nothing say the gravitational potential energy. This law can be applied only positive energy

 

5. others

 

Please one more explain to me!

Edited March 22, 2010 by icarus2

[ajb]

I can’t use to the general relativity.

 

If you really want to understand gravity you will have to.

 

The positive mass theorem can be proved under some reasonable physical assumptions. Like all theorems it is only as strong as the assumptions/conditions/axioms etc you put in.

[Royston]

I can’t use to the general relativity.

 

I suggest you do, you can derive the Newtonian limit via the Ricci tensor, it should reduce to Poisson's equation, from there you have gravitational potential.

 

So (unless I'm missing something) your derivation is somewhat back to front, you certainly can't derive any constants (e.g cosmological) via Newtonian methods. You need the assumptions of SR and GR.

 

EDIT: I'm a GR newb, so someone correct me if the above, isn't quite right.

[Farsight]

Icarus: the positive one is true. See The Foundation of the General Theory of Relativity and look at page 185 where Einstein says "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". This energy has a mass-equivalence.

 

Consider a region of space where the stress-energy "density" is uniform. Now add an immense concentration of energy tied up as matter, in the form of a planet. The surrounding space is "conditioned" by this, and its energy density is increased with proximity, and is no longer uniform. The result is a gravitational field. There is no negative energy at any location therein.

 

Gravitational energy is often considered to be a negative because "Newtonian" principles are applied, wherein a falling body is considered to have more energy than a stationary body at altitude. What's actually happening is that some of the energy within the stationary body is converted into the kinetic energy of the falling body. This energy within the stationary body is called gravitational potential energy. It's the difference between the internal energy of a body "at rest" and at some given temperature up in space, as compared to the internal energy of the same body at rest and at the same temperature in a region of lower "gravitational potential".

 

Gravitational time dilation is key to this, whereupon internal subatomic motion occurs at a reduced rate, and conservation of energy applies. The easiest way to grasp it is to say to yourself that the energy difference between a fast-spinning disk and a slow-spinning disk pays for the kinetic energy of the latter.

Edited March 28, 2010 by Farsight
doublepost merge correction

[swansont]

Gravitational energy is often considered to be a negative because "Newtonian" principles are applied, wherein a falling body is considered to have more energy than a stationary body at altitude.

 

This is either poorly worded or flat-out wrong. In Newtonian physics, F = -grad U, so any attractive force will have a negative potential energy, or will decrease potential energy as the objects move closer to each other. The falling body has more energy only because it actually has more energy — if it is falling, it has kinetic energy in addition to the potential energy.

[Amr Morsi]

What is meant by "gravitational potential energy is gravity source also" is that it is another manifistation (face) of the gravitational energy. The mass of the potential energy is still "M", icarus2.

 

The only relation connecting it with mc^2 is the following relation, which can be deduced from SR (Special Relativity), neglecting GR effects:

 

-GMm/r+gamma*mc^2=E

 

Where E is the total energy of motion of m.

This is equivalent to what swansont is saying.

[Farsight]

Swanson, let me rephrase: hold an object 100m above the ground. It has no discernible kinetic energy, but it does have gravitational potential energy. Now drop it. Ignoring air resistance, at the instant just before it touches the ground, it now has kinetic energy, and it also has less potential energy. However the total energy of that object is unchanged. See http://www.physicsclassroom.com/class/energy/U5l2a.cfm

 

"When the only type of force doing net work upon an object is an internal force (for example, gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant. In such cases, the object's energy changes form."

 

 

Icarus: if I can reiterate: Like Einstein said, "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". The energy is positive. It only looks negative if you invert the energy density to plot gravitational potential like the depiction at http://en.wikipedia.org/wiki/Gravitational_potential

 

We know from conservation of energy and from relativity that the kinetic energy of a falling body comes from the potential energy of that body. Throw a body up into the air and the kinetic energy is converted to potential energy. Throw it hard enough to give it escape velocity and the potential energy is gone forever - because it's in the body, not in the gravitational field. Now play it backwards: as a body falls to earth, its potential energy is converted into kinetic energy, which is dissipated and radiated away on impact. The potential energy is no longer present, and the mass of the body has been reduced in line with Does the Inertia of a Body depend upon its energy content?. Hence that potential energy is [math]+ m_{gp}c^2 [/math].

[icarus2]

Thanks! Snail, Farsight, swansont, Amr Morsi

From Amr Morsi’s explain,

In SR, total energy is

[math] E = mc^2 = \gamma m_0 c^2 + ( - \frac{{GMm_0 }}{r})[/math]

 

 

For the consistency (refer to the first posting),

[math]E = M'c^2 = \gamma Mc^2 + ( - \frac{{GM_1 M}}{r})[/math]

 

[math]( - \frac{{GM_1 M}}{r}) = - m_ - c^2 [/math]

 

([math] - \frac{{GM_1 M}}{{rc^2 }} = - m_ - [/math] : Negative mass define (m_≥0))

 

So,

[math]E = M'c^2 = (\gamma Mc^2) + ( - m_ - c^2 ) = (\gamma M + ( - m_ - ))c^2 [/math]

 

If, c>>v, γ~1,

[math]E = M^{'} c^2 = (M + ( - m_ - ))c^2 [/math]

 

Mass deficit is happen, because of gravitational potential energy(It is a kind of binding energy)

 

Semi-Classically, or SR

Total Energy = ((γX rest mass) + (mass of the gravitational potential energy))c^2

 

===================================

In the chapter 2.7( Gravitation and Spacetime, Hans C. Ohanian and Remo Ruffini ))

 

P116~118

 

Our Gedanken-experiment is less farfetched than it seems. The Earth is in continuous acceleration as it orbit the Sun, and action and reaction are in fact out of balance all the time. We can give a rough estimate of the difference between action and reaction. The force of the Sun on the Earth is GM1M/r^2, where M1 and M are the masses of Sun and Earth, respectively. The force of the Earth on the Sun differs from this by a term depending on the acceleration of the Earth. In the time r/c, the radial displacement of the Earth caused by the acceleration is

 

[math]\Delta r \simeq \frac{1}{2}\frac{{GM1}}{{r^2 }}(\frac{r}{c})^2 = \frac{1}{2}\frac{{GM1}}{{c^2 }} [/math] [194]

 

Hence, without the acceleration the Earth would have moved off tangentially and increased its distance by this amount. Fig. 2.8 shows the present position of the Earth and retarded position, that is, the position at a time r/c before the present position. The figure also shows the “extrapolated” position, or the position that the Earth would have reached if it had moved at constant velocity along a straight line beginning at the retarded position. This extrapolated position is nearly on the same radius as the present position, but the radial distance is large by the amount given by Eq. [194].

 

 

As a rough estimate for the force felt at the Sun, we take that force which the Earth would exert if stationary at the extrapolated position. This estimate for the force gives

 

[math] \frac{{GM1M}}{{(r + \Delta r)^2 }} \simeq \frac{{GM1M}}{{r^2 }} - \frac{1}{2}\frac{{GM1}}{{c^2 }}(\frac{{2GM1M}}{{r^3 }}) [/math] [195]

 

Of course, Eq. [195] is no more than an educated guess. Essentially, we are assuming that the force depends mainly on the velocity of the Earth at the retarded time, and that to this the acceleration contributes only some insignificant corrections (which have to do with the emission of gravitational radiation). If only the velocity is important, then the force exerted by the Earth is the same as that for a mass that moves with uniform velocity from the retarded position and reaches the extrapolated position at the present time. (For a mass moving with uniform velocity, the center of force will coincide with the position of the mass, as becomes immediately obvious if one asks for the force as seen from the rest frame of that mass. However, the transformation to the rest frame of that mass introduces relativistic corrections which are typically of the same order of magnitude as the extra term in [195]).

 

The difference between action and reaction is then of the order

 

[math](\frac{{GM1M}}{{rc^2 }})(\frac{{GM1}}{{r^2 }})[/math] [196]

 

 

The inequality of action and reaction implies a violation of momentum conservation. The question is now; Where does the momentum go? There is only one place it can go—it must be stored in the gravitational field. What is conserved is not the momentum of the Earth-Sun system, but rather the momentum of the system of the Earth-Sun system, but rather the momentum of the system consisting of Earth, Sun, and gravitational fields. This interpretation is in accord with our estimate [196].

 

To see this, note that GM1M /rc^2 may be interpreted as the “mass” associate with the interaction energy (potential energy) between the Sun and the Earth. This energy is stored in the gravitational fields, and as the Earth orbit the Sun, so will this extra “mass.” To obtain the rate of momentum transfer to the fields, we must take the product of this “mass” by a characteristic acceleration. A reasonable estimate for the acceleration is GM1 /r^2, and the product then has the form [196]

 

========================================

From [195]

 

[math]\frac{{GM_1 M}}{{(r + \Delta r)^2 }} \simeq \frac{{GM_1 M}}{{r^2 }} - (\frac{{GM_1 M}}{{rc^2 }})(\frac{{GM_1 }}{{r^2 }})[/math] [195]

 

 

[math]\frac{{GM_1 M}}{{r^2 }} - \frac{{GM_1 M}}{{(r + \Delta r)^2 }} = (\frac{{GM_1 M}}{{rc^2 }})(\frac{{GM_1 }}{{r^2 }})[/math] [A]

 

In the eq. [A]

Below term is eq [196]

 

[math](\frac{{GM_1 M}}{{rc^2 }})(\frac{{GM_1 }}{{r^2 }})[/math] = (mass of gravitational potential energy) X (acceleration) term

 

From the gravitational potential energy

 

[math]| - \frac{{GM_1 M}}{r}| = mc^2 [/math] --> [math]\frac{{GM_1 M}}{{rc^2 }} = m[/math]

 

Generally gravitational acceleration term a = - GM1/r^2

([math]\vec F = m\vec a = - \frac{{GM_1 m}}{{r^2 }}\hat r = = > \vec a = - \frac{{GM_1 }}{{r^2 }}\hat r[/math])

 

If [math]\vec a = \frac{{GM1}}{{r^2 }}\hat r [/math],

Namely, if acceleration term is positive, it means antigravity. So, gravitational acceleration term [math](\frac{{GM_1 }}{{r^2 }})[/math] is absolute value of acceleration.

 

So, strictly speaking eq. [A] is

[math](\frac{{GM_1 M}}{{rc^2 }})(\frac{{GM_1 }}{{r^2 }}) = ( - \frac{{GM_1 M}}{{rc^2 }})( - \frac{{GM_1 }}{{r^2 }}) = - \frac{{GM_1 ( - \frac{{GM_1 M}}{{rc^2 }})}}{{r^2 }} = - \frac{{GM_1 ( - m_ - )}}{{r^2 }}[/math]

 

---------------------------

[ Einstein said, "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". ]

means that below shape

[math] ma = - \frac{{GMm}}{{r^2 }} [/math]

 

namely, a= - GM/r^2

(In the book "Gravitation and Spacetime")

mass is not concerned it.

---------------------------

 

From the eq[A]

[math]\frac{{GM_1 M}}{{r^2 }} - \frac{{GM_1 M}}{{(r + \Delta r)^2 }} = - \frac{{GM_1 ( - m_ - )}}{{r^2 }}[/math] [C]

 

[math] - \frac{{GM_1 M}}{{(r + \Delta r)^2 }} = ( - \frac{{GM_1 M}}{{r^2 }}) + ( - \frac{{GM_1 ( - m_ - )}}{{r^2 }})[/math]

 

[math] - \frac{{GM_1 M}}{{(r + \Delta r)^2 }} = - \frac{{GM_1 (M + ( - m_ - ))}}{{r^2 }} = - \frac{{GM_1 M^{'} }}{{r^2 }}[/math] [D]

==============

 

From the above

in SR,

[math]E = M^{'} c^2 = (M + ( - m_ - ))c^2 [/math] or

[math]E = M'c^2 = (\gamma M + ( - m_ - ))c^2 [/math]

 

So, we can explain that gravitational potential energy(or negative binding energy) has a negative mass.

However, Negative mass is stable at the maximum point. Therefore the catastrophe to energy level of minus infinity never happens

http://www.scienceforums.net/forum/showpost.php?p=529741&postcount=5

 

 

 

Therfore, In my think

From the “mass and energy equvalance”

1) We can treat that gravitational potential energy has a negative mass.

(Standard expression is "mass deficit")

 

2) Newton’s eq. of the negative mass is

[math] - m_ - a = - \frac{{GM( - m_ - )}}{r}[/math]

([math]m_ - \ge 0[/math])

or

 

[math]m_ - a = - \frac{{GMm_ - }}{r}[/math]

([math]m_ - \le 0[/math])

 

Two notational systems are same.

Edited March 31, 2010 by icarus2

[Farsight]

Apologies, I couldn't follow that, icarus. If you raise an object you give it gravitational potential energy. You add energy to it. This energy has a mass-equivalence, so you increase the mass. Gravitational potential energy has positive mass.

 

There's some issue here regarding standpoint. See Gravitational binding energy and note where it says:

 

"The gravitational binding energy of an object consisting of loose material, held together by gravity alone, is the amount of energy required to pull all of the material apart, to infinity. It is also the amount of energy that is liberated (usually in the form of heat) during the accretion of such an object from material falling from infinity."

 

The important thing to note, is that this object hasn't gotthis energy. That's why it's a negative as per the next sentence:

 

"The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy, considering the system as a set of small particles."

 

There isn't actually any negative energy, and there isn't actually any negative mass, just less positive energy and so less positive mass in the bound system.

[swansont]

Discussing the sign of the potential energy is a distraction; it depends on what your reference value is. We're interested in changes in energy, so the reference is purely a matter of convenience. Because of the math in Newtonian gravity, the choice is that infinite separation corresponds to zero potential energy, and the potential energy of a bound system is negative. Regardless of the choice of a reference, the energy (and therefore mass) of a bound system is less than that of a free system, and the difference (also used in nuclear physics) is known as the mass defect.

[Farsight]

No problem with that Swanson, but icarus seems to be trying to justify "a negative mass", and I don't buy it.

[icarus2]

I’m sorry, I can’t English well.

 

There's some issue here regarding standpoint. See Gravitational binding energy and note where it says:

 

 

The important thing to note, is that this object hasn't gotthis energy. That's why it's a negative as per the next sentence:

 

"The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy, considering the system as a set of small particles."

 

 

Above wiki explanation is

Gravitational binding energy=| ∑ gravitational potential energy |= [math]|\sum\limits_{i < j}^n {U_{ij} |} = - \sum\limits_{i < j}^n {( - \frac{{Gm_i m_j }}{{r_{ij} }}} )[/math]

 

Namely, sign of the gravitational binding energy has an absolute sign of the gravitational potential energy.

 

At wiki, gravitational binding energy is

 

[math]U = \frac{{3GM^2 }}{{5r}}[/math]

 

"The gravitational binding energy of a system is equal to the negative of the total gravitational potential energy, considering the system as a set of small particles."

 

Above sentence means

 

[math]U_{bindingenergy} = - U_{gravitationalpotentialenergy} = - ( - \frac{{3GM^2 }}{{5r}})[/math]

 

In other words,

They say that gravitational potential energy has a negative value.

==========

 

To swansont,

In SR, gravitational potential energy is gravity source.

In my think

That means, gravitational potential energy has a fixed value such as rest mass.(at proper frame)

 

[math]E_T = M'c^2 = \gamma Mc^2 + ( - \frac{{GM_1 M}}{r})[/math]

 

 

So, the reference is not purely a matter of convenience.

Also, for the mass defect

 

[math]U = - \frac{{GM_1 M}}{r}[/math] is correct.

 

Therefore,

Gravitational potential energy is negative energy.

However, mass-energy equivalence says that negative energy has a negative mass.

 

[math]E_T = M'c^2 = \gamma Mc^2 + ( - \frac{{GM_1 M}}{r})[/math] is E=A+B structure

So, this equation can be explain by A+B or E

 

Although total mass of the system has a positive mass, we can say that gravitational potential energy corresponds to the negative mass.

[michel123456]

there is a mention on this point in Dr Allan Guth's lecture at MIT

[icarus2]

Thanks! Michel

Good Lectures!

 

At fourth files!

 

 

 

The energy of gravitational field is negative!

 

The positive energy of the false vacuum was compensated by the negative energy of gravity. The TOTAL ENERGY of the universe may very well be zero.

 

 

At third files ( 3:55s ~ 4:23s )

 

 

Gravitational Repulsion!

 

The combination of general relativity and modern particle theories predicts that at very high energy, there exists forms of matters that create a gravitational repulsion!

 

 

Gravitational potential energy is negative energy.

However, mass-energy equivalence says that negative energy has a negative mass.

 

Therefore, we can say that gravitational potential energy has a negative mass.

 

========

http://www.scienceforums.net/forum/showthread.php?t=46362

Edited April 13, 2010 by icarus2

[pioneer]

I don't think gravitational potential is negative energy/mass. One way to test this is to give a steel ball some gravitational potential, by lifting it into the air. Next, we will let it fall attached to a string connected to a magnetic that will rotate as the ball falls; making electricity. We are using the conservation of energy to test for positive or negative energy.

 

We will connect our electrical output to a special circuit with a dim light bulb. If it is negative energy we should suck electricity out of the system and make the bulb even dimmer. If it is positive energy we will make electricity and bulb will get brighter. We will make positive energy.

[michel123456]

Icarus, don't forget this part of the lecture is called by Dr. Guth "Miracle Of Physics"...

Gravitational repulsion is "Miracle #1"

Negative energy of gravitational field is "Miracle #2"