What is gravitational slowing of time in center of mass?

[DimaMazin]

How to define gravitational slowing of clock in center of mass?

[swansont]

What do you mean by "in center of mass"? In the center of mass frame of reference? In the physical center of a uniform mass distribution? Something else?

[DimaMazin]

7 hours ago, swansont said:

What do you mean by "in center of mass"? In the center of mass frame of reference? In the physical center of a uniform mass distribution? Something else?

In center of the Earth or in center of any planet or star.

[swansont]

3 minutes ago, DimaMazin said:

In center of the Earth or in center of any planet or star.

It would be the gravitational potential divided by c^2, just like at any other point.

[DimaMazin]

On 15.09.2018 at 8:16 PM, swansont said:

It would be the gravitational potential divided by c^2, just like at any other point.

Can we use the formula:

t_o= t_f(1- 2GM/rc²)^1/2

r is midle radius of all points of the massive body

R is radius of the body

(4Pi R³/3)/(4Pi r³/3)=2

r = R/2^1/3

t_o=t_f(1 - 2 * 2^1/3GM/Rc²)^1/2

                            so?

[swansont]

I don't know what "midle radius of all points of the massive body" means

If r is the coordinate of the observer, then that's time dilation using the Schwarzschild metric. But the center of a massive body is r=0, and as there is no mass inside of the observer but there is mass outside, it doesn't make sense to me to use this formulation.

[mistermack]

As I understand it, the gravitational potential is zero at the centre of a planet. (if you ignore external gravity sources like the star or galaxy).

If there was a hollow in the centre, a stone placed at the centre would not fall left, right, up or down, so there is no potential energy. 

[DimaMazin]

1 hour ago, mistermack said:

As I understand it, the gravitational potential is zero at the centre of a planet. (if you ignore external gravity sources like the star or galaxy).

If there was a hollow in the centre, a stone placed at the centre would not fall left, right, up or down, so there is no potential energy. 

Potential energy doesn't slow time. Did you mean "g" slows time? Do you think there is no gravitational slowing of time?

3 hours ago, swansont said:

I don't know what "midle radius of all points of the massive body" means

If r is the coordinate of the observer, then that's time dilation using the Schwarzschild metric. But the center of a massive body is r=0, and as there is no mass inside of the observer but there is mass outside, it doesn't make sense to me to use this formulation.

How do you define gravitational slowing of time in the center of a massive body mathematicaly?

[Janus]

2 hours ago, mistermack said:

As I understand it, the gravitational potential is zero at the centre of a planet. (if you ignore external gravity sources like the star or galaxy).

If there was a hollow in the centre, a stone placed at the centre would not fall left, right, up or down, so there is no potential energy. 

Incorrect.  g ( gravitational force) is zero at the center, but the Specific(per unit mass) Gravitational potential is  -3GM/2R

where M is the mass of the planet, and R is its radius. 

At the surface of the planet (or any point above it) the Specific Gravitational potential is -GM/r 

where r is the distance from the center of the planet ( on the surface r=R)

Specific Gravitational Potential Energy tells us how much energy it would take to move a unit mass from one point in the field to another.  It takes energy to lift a mass from the center of the Earth to the surface, just like it takes energy to lift the same unit mass from the surface of the Earth to a point above it.

 

48 minutes ago, DimaMazin said:

Potential energy doesn't slow time. Did you mean "g" slows time? Do you think there is no gravitational slowing of time?

It is the difference in potential gravitational energy is responsible for gravitational time dilation, not the difference in g.

An easy way to demonstrate this is to calculate the gravitational time dilation factor for the surface of the Earth vs the surface of a planet with twice the radius and 4 times the mass of the Earth.   You will get two different answers even though the value of g will be the same at the surface of both planets.

A clock at the center of the Earth is at a lower potential than one on the surface and thus will run slower than one on the surface, even though it feels 0g compared to the surface clock at 1g.

[DimaMazin]

43 minutes ago, Janus said:

Incorrect.  g ( gravitational force) is zero at the center, but the Specific(per unit mass) Gravitational potential is  -3GM/2R

where M is the mass of the planet, and R is its radius. 

At the surface of the planet (or any point above it) the Specific Gravitational potential is -GM/r 

where r is the distance from the center of the planet ( on the surface r=R)

Specific Gravitational Potential Energy tells us how much energy it would take to move a unit mass from one point in the field to another.  It takes energy to lift a mass from the center of the Earth to the surface, just like it takes energy to lift the same unit mass from the surface of the Earth to a point above it.

 

Swansont said that gravitational potential should be divided by c²

-3GM/2Rc²    it has no unit therefore it would be a factor of slowing of time, but haw should I use negative factor for time dilation definition?

what is 3/2 ?

[mistermack]

2 hours ago, Janus said:

5 hours ago, mistermack said:

As I understand it, the gravitational potential is zero at the centre of a planet. (if you ignore external gravity sources like the star or galaxy).

If there was a hollow in the centre, a stone placed at the centre would not fall left, right, up or down, so there is no potential energy. 

Incorrect.  g ( gravitational force) is zero at the center, but the Specific(per unit mass) Gravitational potential is  -3GM/2R

where M is the mass of the planet, and R is its radius. 

At the surface of the planet (or any point above it) the Specific Gravitational potential is -GM/r 

where r is the distance from the center of the planet ( on the surface r=R)

Yes, agreed. Got it now, thanks. 

[swansont]

2 hours ago, DimaMazin said:

Swansont said that gravitational potential should be divided by c²

-3GM/2Rc²    it has no unit therefore it would be a factor of slowing of time, but haw should I use negative factor for time dilation definition?

what is 3/2 ?

The - sign tells you that the potential is smaller at the center (you need to add energy to move a particle to larger r) and that the frequency  is lower.

[DimaMazin]

On 10.11.2018 at 11:33 PM, swansont said:

The - sign tells you that the potential is smaller at the center (you need to add energy to move a particle to larger r) and that the frequency  is lower.

What about slowing of time? Is it there :

t₀=t_f( 1- 3GM/2Rc²)^1/2 ?

[Janus]

2 hours ago, DimaMazin said:

What about slowing of time? Is it there :

t₀=t_f( 1- 3GM/2Rc²)^1/2 ?

No. If you look at the standard time dilation equation:

t₀ =t_f(1-2GM/Rc²)^1/2

One thing stands out. It is that 2GM/R also appears in the escape velocity equation:

V_e= (2GM/R)^1/2

and so, 2GM/R = V_e²

Ergo, we could rewrite the gravitational time dilation equation as:

t₀ =t_f(1-V_e²/c²)^1/2

So what we need to do is substitute the escape velocity from the center of the Earth for V_e

Escape velocity is reached when the sum of the kinetic energy and gravitational potential energy equal zero.

Using specific energies:

v²/2-3GM/2R = 0

v²/2 = 3GM/2R

v² = 3GM/R

Thus:

t₀ =t_f(1-3GM/Rc²)^1/2

 

[DimaMazin]

13 hours ago, Janus said:

No. If you look at the standard time dilation equation:

t₀ =t_f(1-2GM/Rc²)^1/2

One thing stands out. It is that 2GM/R also appears in the escape velocity equation:

V_e= (2GM/R)^1/2

and so, 2GM/R = V_e²

Ergo, we could rewrite the gravitational time dilation equation as:

t₀ =t_f(1-V_e²/c²)^1/2

So what we need to do is substitute the escape velocity from the center of the Earth for V_e

Escape velocity is reached when the sum of the kinetic energy and gravitational potential energy equal zero.

Using specific energies:

v²/2-3GM/2R = 0

v²/2 = 3GM/2R

v² = 3GM/R

Thus:

t₀ =t_f(1-3GM/Rc²)^1/2

 

Thank you very much. Now it is clear.