Evidence for a Young Earth (Not Billions of Years Old)

[Subduction Zone]

Well, the gravitational potential difference between a point and infinity.



I notice that you don't include gravitational potential. That is directly related to both escape velocity and the curvature. The force is a different thing.



OK, and where is that curvature largest? At the center. Why? Because that is where the potential is largest. The *force* is zero there *because* the potential is a maximum.

I do not fully understand this concept yet, but the equations that I can find on this use a factor for the gravitational potential, not the curvature of space. Gravitational potential is going to be the greatest at the center of a body, and so will the escape velocity which is why I used that tack. I was shocked out of my shoes by the claim it would be highest at the surface.

 

[Justatruthseeker]

You are confused. Relativity says the effect of curved space causes gravity. But it does not say that about time dilation. It says that is due to gravitational potential. Do you need the formula? I can provide it. That is why escape velocity matters. Perhaps you finally see your error now.

The only reason I am avoiding curvature is because you do not know how to apply it.

Because gravitational potential or energy increases with curvature and decreases as curvature lessens.

But you don’t understand that either.

“The gravitational potential (V) at a location is the gravitational potential energy (U) at that location per unit mass:
where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

It takes more energy to move something when the curvature is greatest. Linked directly to the gravitational field which decreases with depth. It’s not my fault you think potential energy isn’t directly linked to the gravitational field which is caused by curvature due to mass...

 

[Polymath257]

OK, the metric tensor of a gravitational field with potential function V is given by

ds^2 = -(c^2 +2V)dt^2 + (1-2V/c^2)dr^2 +r^2 d(Omega)^2

The potential is given by

V(r)=int_r^infty F(r) dr.

Want to go from there?

 

[Justatruthseeker]

OK, the metric tensor of a gravitational field with potential function V is given by

ds^2 = -(c^2 +2V)dt^2 + (1-2V/c^2)dr^2 +r^2 d(Omega)^2

The potential is given by

V(r)=int_r^infty F(r) dr.

Want to go from there?

I already know gravity is the same as acceleration...

Your point being what, that it is the same as acceleration?

 

[Justatruthseeker]

I do not fully understand this concept yet, but the equations that I can find on this use a factor for the gravitational potential, not the curvature of space. Gravitational potential is going to be the greatest at the center of a body, and so will the escape velocity which is why I used that tack. I was shocked out of my shoes by the claim it would be highest at the surface.

You can’t find anything because the gravitational potential is linked directly to the gravitational field.

They already expect you to understand that the gravitational field is the result of curvature due to mass.... they expect you to already have that understanding....

 

[Polymath257]

Because gravitational potential or energy increases with curvature and decreases as curvature lessens.

But you don’t understand that either.

“The gravitational potential (V) at a location is the gravitational potential energy (U) at that location per unit mass:
where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

It takes more energy to move something when the curvature is greatest. Linked directly to the gravitational field which decreases with depth. It’s not my fault you think potential energy isn’t directly linked to the gravitational field which is caused by curvature due to mass...

Yes, curvature is directly linked to the gravitational *potential*. That is different than the gravitational *force*.

Now, what is the connection between the force and the potential?

 

[Polymath257]

I already know gravity is the same as acceleration...

You have to be a bit careful what you mean by 'the same as' in this context. Gravity induces acceleration and acceleration is *equivalent* to a gravitational force.

Your point being what, that it is the same as acceleration?

Nope. my point is that the metric (and thereby the curvature) is determined by the potential. Also, the time dilation effect can be easily read from the metric.

Also, I wanted to point out the difference between the force and the potential. In essence, the force is the derivative of the potential (technically, the gradient), so the force is zero where the potential (and thereby the curvature) is a maximum or minimum.

 

[Justatruthseeker]

Yes, curvature is directly linked to the gravitational *potential*. That is different than the gravitational *force*.

Now, what is the connection between the force and the potential?

Energy.

Gravitational potential is the work required to move an object against the gravitational force.

“The gravitational potential (V) at a location is the gravitational potential energy (U) at that location per unit mass:

where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

“In physics, a gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body.”

Can we just say energy and stop all the idiotic roundabouts? Or do you all prefer to keep showing your lack of understanding?

 

[Polymath257]

Energy.

Gravitational potential is the work required to move an object against the gravitational force.

“The gravitational potential (V) at a location is the gravitational potential energy (U) at that location per unit mass:

where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

“In physics, a gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body.”

Can we just say energy and stop all the idiotic roundabouts? Or do you all prefer to keep showing your lack of understanding?

OK, so *mathematically*, what does that say about the link between the force and the potential?

In particular, what can you say about the potential if the force is zero?

 

[Justatruthseeker]

You have to be a bit careful what you mean by 'the same as' in this context. Gravity induces acceleration and acceleration is *equivalent* to a gravitational force.

A rocket engine induces acceleration and acceleration is the same as gravity......

Equivalent: “equal in value, amount, function, meaning, etc.”

Bandying words won’t change reality.

Nope. my point is that the metric (and thereby the curvature) is determined by the potential. Also, the time dilation effect can be easily read from the metric.

Which is determined by the amount of curvature due to mass... which induces acceleration. So of course the time dilation effect can be easily read from metric.

Also, I wanted to point out the difference between the force and the potential. In essence, the force is the derivative of the potential (technically, the gradient), so the force is zero where the potential (and thereby the curvature) is a maximum or minimum.

The force is zero where the curvature is at a minimum. No maximum belongs in that sentence.

 

[Polymath257]

The force is zero where the curvature is at a minimum. No maximum belongs in that sentence.

No. Any ciritcal point (max or min) will give a zero force. Now we agreed that the force is zero at the center of the Earth. So where is the curvature a maximum?

 

[Subduction Zone]

Because gravitational potential or energy increases with curvature and decreases as curvature lessens.

But you don’t understand that either.

“The gravitational potential (V) at a location is the gravitational potential energy (U) at that location per unit mass:
where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

It takes more energy to move something when the curvature is greatest. Linked directly to the gravitational field which decreases with depth. It’s not my fault you think potential energy isn’t directly linked to the gravitational field which is caused by curvature due to mass...

Please, since you do not understand this at all do not say that someone else does not understand it. I asked if you had an equation that supports your claim. You obviously can't find one.

Let's go over your error of escape velocity and then I will provide the equation that you need.

 

[Subduction Zone]

A rocket engine induces acceleration and acceleration is the same as gravity......

Equivalent: “equal in value, amount, function, meaning, etc.”

Bandying words won’t change reality.



Which is determined by the amount of curvature due to mass... which induces acceleration. So of course the time dilation effect can be easily read from metric.


The force is zero where the curvature is at a minimum. No maximum belongs in that sentence.

You don't like the term potential but it is more correct. To have "energy" one needs a specific mass. For example the gravitational potential of a kilogram is a thousand times that of a gram at any point.

By the way, understanding escape velocity, something that you appear to know that you are wrong about, will also let you understand gravitational potential since they are directly related.

 

[Justatruthseeker]

No. Any ciritcal point (max or min) will give a zero force. Now we agreed that the force is zero at the center of the Earth. So where is the curvature a maximum?

Between the outer core and the mantle......

We are not discussing theoretical black holes where the mass is all concentrated at a single point.....

And where therefore the greatest curvature would exist at that point, hence the greatest force.

So then you would agree that the greatest curvature on a black hole could be zero force? Or do you concur that due to a single point source of mass it would be the greatest curvature and the greatest force?

I don’t think you can help but conflict with your own views at this point without even realizing you are doing so.

Just as if we leave the earth and head into space, the curvature decreases as does the force. Never to a non-zero force because the curvature is never non-zero.

But curvature is due to mass, and the mass inside any radius decreases as one goes below the surface, hence the curvature also decreases. The mass of the earth is not concentrated at a single point in space as are the singularities you are confusing it with......

And go tell your fable to engineers who understand the maximum critical point can never have a non-zero force... they’ll laugh at you, but hey..... maybe you can convince them that all their engineering is flawed.....

 

[Polymath257]

Between the outer core and the mantle......

We are not discussing theoretical black holes where the mass is all concentrated at a single point.....

And where therefore the greatest curvature would exist at that point, hence the greatest force.

This is your mistake. The force isn't greatest where the curvature is greatest. The *potential* is greatest when the curvature is greatest.

But curvature is due to mass, and the mass inside any radius decreases as one goes below the surface, hence the curvature also decreases. The mass of the earth is not concentrated at a single point in space as are the singularities you are confusing it with......

Nope. The curvature is due to mass, yes. But the mass density, not the mass inside of some radius. The curvature *increases* as we go into the Earth, with the maximum at the center (where the force is zero since the potential is a maximum).

I never claimed that the mass of the Earth is concentrated at a single point. I'm also not using the black hole solutions. I am simply using the fact that curvature is related to the potential and not the force. Both are produced by the presence of mass, yes.

 

[Polymath257]

A rocket engine induces acceleration and acceleration is the same as gravity......

Equivalent: “equal in value, amount, function, meaning, etc.”

Bandying words won’t change reality.

More precisely, a gravitational field is *locally* equivalent to an acceleration. Once you go global, this is no longer the case.

 

[gnostic]

"The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 kn) at the surface."

No other velocity needs be considered, because it is greatest at the surface...

COMPREHEND: If starting at the center of the earth, you would initially only need to overcome the half of the mass beneath you, while the mass above you would aid in your acceleration. Only when reaching the surface would the entire mass of the earth be beneath you and need to be overcome. It would gradually increase to the maximum of 11.186 km/s.

We can calculate for the center of mass only when on or above the surface. Once below the surface all the mass is no longer beneath you..... It is for convenience only because one never launches a ship from below the surface. And regardless of where you launched the ship from, it would only take 11.186 km/s to reach escape velocity.....

No, it is not a matter of convenience that they don’t launch deep underground, like launching at the core or the centre of the Earth.

It is unrealistic, because it is never going to happen.

What you are talking about, isn’t science. It is just pseudoscience conjecture.

Science is about WHAT it is, or WHAT is feasible, or HOW does it work, or HOW would you do it, etc, and once you have the explanation, you would test each and every ones of those questions and explanations.

What you are asking is doing the impossible, which amounts to being highly improbable to achieve.

Seriously, Justatruthseeker, how do you excavate all the way to the centre of Earth and at the same time construct the launch tube?

Do you magically make the tube appear? Or make a wish? Do you pray for the launch tube?

A matter of “convenience only” is an understatement. Your conjectures can never be achieve, so essentially they are pointless with the “what-if” sophistry.

You are forgetting that science have to be something that feasible. Wasting time on something that feasible, is nothing but waste of everyone’s time.

 

[Polymath257]

And go tell your fable to engineers who understand the maximum critical point can never have a non-zero force... they’ll laugh at you, but hey..... maybe you can convince them that all their engineering is flawed.....

Yes, when the potential energy is a maximum, the force is zero. I'm sure any competent engineer knows this.

 

[Subduction Zone]

No, it is not a matter of convenience that they don’t launch deep underground, like launching at the core or the centre of the Earth.

It is unrealistic, because it is never going to happen.

What you are talking about, isn’t science. It is just pseudoscience conjecture.

Science is about WHAT it is, or WHAT is feasible, or HOW does it work, or HOW would you do it, etc, and once you have the explanation, you would test each and every ones of those questions and explanations.

What you are asking is doing the impossible, which amounts to being highly improbable to achieve.

Seriously, Justatruthseeker, how do you excavate all the way to the centre of Earth and at the same time construct the launch tube?

Do you magically make the tube appear? Or make a wish? Do you pray for the launch tube?

A matter of “convenience only” is an understatement. Your conjectures can never be achieve, so essentially they are pointless with the “what-if” sophistry.

You are forgetting that science have to be something that feasible. Wasting time on something that feasible, is nothing but waste of everyone’s time.

He may have realized his error in that post. He surely will not discuss it. Part of being a creationist is that one can never own up to one's mistakes since that leads to an acceptance of reality.

 

[Justatruthseeker]

This is your mistake. The force isn't greatest where the curvature is greatest. The *potential* is greatest when the curvature is greatest.

The potential is greatest where the curvature is the least. That’s why a ball on top of the hill has more potential than a ball at the bottom of the hill.

Your reasoning is unsound and against all of physics.

Nope. The curvature is due to mass, yes. But the mass density, not the mass inside of some radius. The curvature *increases* as we go into the Earth, with the maximum at the center (where the force is zero since the potential is a maximum).

We have already seen your reasoning is flawed. A ball at the top of a hill possesses more potential than a ball at the bottom. A ball at the bottom possesses none.....

I never claimed that the mass of the Earth is concentrated at a single point. I'm also not using the black hole solutions. I am simply using the fact that curvature is related to the potential and not the force. Both are produced by the presence of mass, yes.

NO. Potential is greatest at the top of a hill, not the bottom.

You are confusing gravitational potential energy with Kinetic energy.

Gravitational Potential Energy – College Physics

“The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy.

Don’t try to explain physics to me when you don’t yet understand it yourself.

I’ll repeat what every science book in existence will tell you. Any object looses gravitational potential energy as it goes into the well or downhill, not gains it. It gains Kinetic energy.