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

[Justatruthseeker]

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

I’m sure none of science agrees with you. See above post.

Learn science first, then come back to this discussion.

I’ll repeat.....

“The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy.”

There isn’t more gravitational potential energy at the bottom, but less. Stop confusing it with Kinetic energy.

 

[Polymath257]

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.

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.....

And what is the net force on the ball at either the top or the bottom?

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

For potential wells like gravity, the potential is defined to be the energy per mass to get to infinity. That reverses a sign.

You are confusing gravitational potential energy with Kinetic energy.

Gravitational Potential Energy – College Physics

No, I am not. Potential energy is only determined up to a constant (because it is the integral of a force). For potential wells, we set that constant so the potential at infinity is zero.

“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.

Yes, the potential is negative everywhere for gravity. Where is it the largest absolute value? So where is the curvature largest?

 

[Polymath257]

I’m sure none of science agrees with you. See above post.

Learn science first, then come back to this discussion.

I’ll repeat.....

“The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy.”

There isn’t more gravitational potential energy at the bottom, but less. Stop confusing it with Kinetic energy.

OK, so the potential is negative everywhere for gravity. Where does it have the largest negative value?

 

[Polymath257]

I’m sure none of science agrees with you. See above post.

Learn science first, then come back to this discussion.

I’ll repeat.....

“The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy.”

There isn’t more gravitational potential energy at the bottom, but less. Stop confusing it with Kinetic energy.

Yes, and curvature can also be both positive and negative. In our discussion, we are talking about the largest absolute value of both.

 

[Subduction Zone]

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.

You appear to be conflating slope with curvature. The slope at the top of a hill is zero, not the curvature. The slope would be the first derivative of the line, the curvature is the second derivative.

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.....


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.

Gravitational potential is the amount of energy per unit mass that it would take to move a body from a known point out to infinity. At least check out the meaning of the terms that you do not understand before accusing others of not getting it.

Gravitational potential - Wikipedia

 

[Justatruthseeker]

And what is the net force on the ball at either the top or the bottom?

We aren’t discussing force, but potential energy.

For potential wells like gravity, the potential is defined to be the energy per mass to get to infinity. That reverses a sign.

And yet science just told you that it looses potential energy as it goes downwards into the gravitational well, not gains it. If the potential energy was greater at the bottom then it would gain potential energy as it descended.

No, I am not. Potential energy is only determined up to a constant (because it is the integral of a force). For potential wells, we set that constant so the potential at infinity is zero.

Yah I know, you think you can actually calculate infinity....

Yes, the potential is negative everywhere for gravity. Where is it the largest absolute value? So where is the curvature largest?

Your confused. We are not discussing Kinetic energy but potential energy. Which is why it is called “potential”. The greatest potential is at the top of the well. The smallest potential is at the bottom of the well. It can’t move from the bottom, it has no potential to perform work.....

But that’s why you ignored the link I gave you. You got no choice but to ignore science.

At the center the greatest potential would be to move upwards, but the mass on all sides is preventing this. It can’t fall down because there is no down and therefore no potential energy.

There is only up and the possibility to gain potential energy, just as if you roll a ball up a hill the potential energy increases.

The Law of Conservation of Energy

“As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy.”

Stop ignoring science for pseudoscientific beliefs.

You won’t find one single source about potential energy that agrees with your claims. But that’s why you haven’t posted any, while ignoring the actual science I have cited....

 

[Subduction Zone]

We aren’t discussing force, but potential energy.



And yet science just told you that it looses potential energy as it goes downwards into the gravitational well, not gains it. If the potential energy was greater at the bottom then it would gain potential energy as it descended.



Yah I know, you think you can actually calculate infinity....




Your confused. We are not discussing Kinetic energy but potential energy. Which is why it is called “potential”. The greatest potential is at the top of the well. The smallest potential is at the bottom of the well. It can’t move from the bottom, it has no potential to perform work.....

But that’s why you ignored the link I gave you. You got no choice but to ignore science.

At the center the greatest potential would be to move upwards, but the mass on all sides is preventing this. It can’t fall down because there is no down and therefore no potential energy.

There is only up and the possibility to gain potential energy, just as if you roll a ball up a hill the potential energy increases.

The Law of Conservation of Energy

“As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy.”

Stop ignoring science for pseudoscientific beliefs.

You won’t find one single source about potential energy that agrees with your claims. But that’s why you haven’t posted any, while ignoring the actual science I have cited....

It appears that you are having quite a bit of trouble following the conversation. Perhaps if we broke it down to one point at a time you will be able to understand. Are you willing to go along?

 

[Polymath257]

We aren’t discussing force, but potential energy.

More specifically, we were discussing the relationship between force, potential, and curvature.

And yet science just told you that it looses potential energy as it goes downwards into the gravitational well, not gains it. If the potential energy was greater at the bottom then it would gain potential energy as it descended.

Which is why the potential is negative everywhere.

Yah I know, you think you can actually calculate infinity....

If you don't like the standard definition, that isn't my problem.

Your confused. We are not discussing Kinetic energy but potential energy. Which is why it is called “potential”. The greatest potential is at the top of the well. The smallest potential is at the bottom of the well. It can’t move from the bottom, it has no potential to perform work.....

And yes, the potential is negative everywhere. When discussing the size of that potential, we usually talk about the absolute value.

But that’s why you ignored the link I gave you. You got no choice but to ignore science.

At the center the greatest potential would be to move upwards, but the mass on all sides is preventing this. It can’t fall down because there is no down and therefore no potential energy.

There is only up and the possibility to gain potential energy, just as if you roll a ball up a hill the potential energy increases.

The Law of Conservation of Energy

“As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy.”

Stop ignoring science for pseudoscientific beliefs.

You won’t find one single source about potential energy that agrees with your claims. But that’s why you haven’t posted any, while ignoring the actual science I have cited....

*sigh* OK, the curvature is associated with the potential, not the force. That is why the curvature is largest when the potential is extremal, which is when the force is zero.

 

[Subduction Zone]

Let's start with why gravitational potential, or even potential energy is set at zero for infinity. It is because that is the one constant value one can use for everyone. For example you could set it at zero for the surface of the Earth, but then you would have positive and negative values for an object above the surface and one below. Worse yet how would you compare potential energy on the Earth compared to that on Mars using the Earth's surface as a zero point. It makes more sense to compare everything to the same value.

 

[Polymath257]

We aren’t discussing force, but potential energy.

And yet science just told you that it looses potential energy as it goes downwards into the gravitational well, not gains it. If the potential energy was greater at the bottom then it would gain potential energy as it descended.
.

I also notice that you shifted from the *potential* to the *potential energy*. They are different things: the potential energy is the potential times the mass. This is similar to the difference between the gravitational field and the gravitational force.

V=-GM/r

Notice that as r gets large, V goes towards zero.

Now, this is for gravity from a single point. If you want the potential for something like the Earth, this formula would apply for r>R, the radius of the Earth. For r<R, the specifics depend on the density profile of the Earth.

The deepest part of the potential well is at the center of the Earth. And that is where the curvature is largest. it is also where the force is zero.

Of course, we have never discussed exactly what is meant by the word 'curvature'. Do you want the Ricci curvature? Or do you have some other measure of curvature you might like to use?

 

[Justatruthseeker]

*sigh* OK, the curvature is associated with the potential, not the force. That is why the curvature is largest when the potential is extremal, which is when the force is zero.

The rest was pure hyperbole and not worth addressing....

Nooooooo. The potential is the largest when the curvature is the least.

What part of potential energy decreases as you go downhill do you fail to understand?

The curvature at the top of the hill is less than the curvature at the bottom of the hill. Likewise the gravitational force is less when the curvature is less.

As the curvature increases force increases and potential energy decreases.

How many times must you be shown with scientific definition after scientific definition that you are wrong.

The negative value means that work is done as potential energy is lost.

Potential energy - Wikipedia

“The negative sign follows the convention that work is gained from a loss of potential energy.”

Your belief that an object gains potential energy as it falls into a gravitational well is fiction.

They have told you this over and over and over....

“If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount.”

They also told you about your infinity and your r=0.....

“Given that there is no reasonable criterion for preferring one particular finite r over another, there seem to be only two reasonable choices for the distance at which U becomes zero: r = 0 and r = infinity.

So potential energy is zero at r = 0 and at infinity. Because at infinity the curvature can be considered virtually zero as can the gravitational force.....

Everyone can see you all trying to duck and dodge actual science just because your egos are too big to just say, yes, I was wrong.....

Get used to it, it’s been happening since the first conversation. All this is doing is showing everyone you are ignoring the science to keep your incorrect personal beliefs alive......

 

[Subduction Zone]

The rest was pure hyperbole and not worth addressing....

Nooooooo. The potential is the largest when the curvature is the least.

What part of potential energy decreases as you go downhill do you fail to understand?

The curvature at the top of the hill is less than the curvature at the bottom of the hill. Likewise the gravitational force is less when the curvature is less.

As the curvature increases force increases and potential energy decreases.

How many times must you be shown with scientific definition after scientific definition that you are wrong.

The negative value means that work is done as potential energy is lost.

Potential energy - Wikipedia

“The negative sign follows the convention that work is gained from a loss of potential energy.”

Your belief that an object gains potential energy as it falls into a gravitational well is fiction.

They have told you this over and over and over....

“If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount.”

They also told you about your infinity and your r=0.....

“Given that there is no reasonable criterion for preferring one particular finite r over another, there seem to be only two reasonable choices for the distance at which U becomes zero: r = 0 and r = infinity.

So potential energy is zero at r = 0 and at infinity. Because at infinity the curvature can be considered virtually zero as can the gravitational force.....

Everyone can see you all trying to duck and dodge actual science just because your egos are too big to just say, yes, I was wrong.....

Get used to it, it’s been happening since the first conversation. All this is doing is showing everyone you are ignoring the science to keep your incorrect personal beliefs alive......

Once again you are using the wrong terminology and not following the conversation. The term that was originally used was gravitational potential. You did not like that term, perhaps because you were not used to it. Rather than trusting me go to Wikipedia and search for "gravitational potential" you will find @Polymath257 was correct. When you demanded that he use potential energy he agreed but clearly stated that he was using the absolute value of the potential energy.

 

[Justatruthseeker]

I also notice that you shifted from the *potential* to the *potential energy*. They are different things: the potential energy is the potential times the mass. This is similar to the difference between the gravitational field and the gravitational force.

Don’t think you can double talk your way out. The potential when discussing curvature, the gravitational field is gravitational potential energy.

V=-GM/r

Notice that as r gets large, V goes towards zero.

Because as r gets large, curvature is less, therefore there is less force. But as the citation in my post above also told you as r gets small U goes towards zero.... and as r gets large U also goes to zero.

This is where you fail to understand the physics. At r=0, U is zero. Then it increases until at some arbitrary point it begins to decrease until at infinity it is also zero.

Now, this is for gravity from a single point. If you want the potential for something like the Earth, this formula would apply for r>R, the radius of the Earth. For r<R, the specifics depend on the density profile of the Earth.

What, now I got to listen to your false beliefs in increasing density as well, despite the scientific data that has already falsified that belief????

The deepest part of the potential well is at the center of the Earth. And that is where the curvature is largest. it is also where the force is zero.

No, there is zero curvature, which is why there is no force. Stop treating it like a point singularity....

Of course, we have never discussed exactly what is meant by the word 'curvature'. Do you want the Ricci curvature? Or do you have some other measure of curvature you might like to use?

Depends if you want to discuss the degree that matter will tend to converge or diverge in time.....

Or if you would just like to ignore any mass above your head.....

 

[Justatruthseeker]

Once again you are using the wrong terminology and not following the conversation. The term that was originally used was gravitational potential. You did not like that term, perhaps because you were not used to it. Rather than trusting me go to Wikipedia and search for "gravitational potential" you will find @Polymath257 was correct. When you demanded that he use potential energy he agreed but clearly stated that he was using the absolute value of the potential energy.

Only you are confused.....

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

V = U/m

So as the unit mass decreases with radius...... the gravitational potential at that location decreases which is the gravitational potential energy at that location per the unit of mass.....

Try to keep up please.....

The gravitational potential IS the gravitational potential energy per the unit of mass. Which just shows you never understood to begin with.

 

[Subduction Zone]

Only you are confused.....

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

V = U/m

So as the unit mass decreases with radius...... the gravitational potential at that location decreases which is the gravitational potential energy at that location per the unit of mass.....

Try to keep up please.....

The gravitational potential IS the gravitational potential energy per the unit of mass. Which just shows you never understood to begin with.

Wrong source. But I did make a small error. What you have missed is that we have always been discussing the absolute value, something that you could not understand.

Gravitational potential - Wikipedia

Tell me, where is the curvature greatest in that image, not the slope. The slope is not the same as the curvature.

 

[Polymath257]

They also told you about your infinity and your r=0.....

“Given that there is no reasonable criterion for preferring one particular finite r over another, there seem to be only two reasonable choices for the distance at which U becomes zero: r = 0 and r = infinity.

And those are mutually exclusive choices.

So potential energy is zero at r = 0 and at infinity. Because at infinity the curvature can be considered virtually zero as can the gravitational force.....

OK, add potential energy to the list of things you don't understand. As described in *your* article, there is an arbitrary constant that arises when doing the Newtonian calculation (which is what is being done in *your* article). We can choose that constant in any way that is convenient. But we only can make that choice *once*.

But, as *your* article goes on to state:

"The singularity at r = 0 {\displaystyle r=0}

in the formula for gravitational potential energy means that the only other apparently reasonable alternative choice of convention, with U = 0 {\displaystyle U=0}

for r = 0 {\displaystyle r=0}

, would result in potential energy being positive, but infinitely large for all nonzero values of r, and would make calculations involving sums or differences of potential energies beyond what is possible with the real number system. Since physicists abhor infinities in their calculations, and r is always non-zero in practice, the choice of U = 0 {\displaystyle U=0}

at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first."

Hence, the choice of U=0 at r=0 is NOT allowed for a singular potential.

Now, it is a common exercise in the mechanics courses in physics to determine the gravitational potential if, say, we assume a constant density sphere. Again, we typically set U=0 at infinity, which makes the potential energy negative for all values.

The result (again, for constant density, M is the total mass of the sphere, R is the radius of the sphere):

V= -GM/r if r>R

V=(GM/2R^3)(r^2 -3R^2) if r<R

The value of the potential at r=0 is then

V=-3GM/(2R)

which, note, is not zero.

Do you want me to give more details? Maybe details for a different density profile? If you want, I could take the graph of the force that you gave and do a numerical integration to get the potential. It would *still* not be zero at the center.


/E: Corrected an exponent.

 

[Polymath257]

Only you are confused.....

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

V = U/m

So as the unit mass decreases with radius...... the gravitational potential at that location decreases which is the gravitational potential energy at that location per the unit of mass.....

*sigh* NOT the mass of the gravitating object. The mass of the object being exposed to the force. See below.

Try to keep up please.....

Please go learn some basics.

The gravitational potential IS the gravitational potential energy per the unit of mass. Which just shows you never understood to begin with.

Yes. As a simple case, if there are two masses, M (the mass we want to find the potential for) and m (the test mass), then the potential energy looks like

U=-GMm/r

for a singular well (the easiest case). Notice that this depends on *both* masses, M and m.

Now, divide that by the *test* mass to get

V=U/m = -GM/r

The whole point is that now the potential only depends on M and not on m. This is a common trick to get something that only depends on the gravitating mass.

In the case of a spherical mass M of constant density, see my previous post.

Here's a cute little question to test understanding: what are the dimensions for potential?

 

[Polymath257]

V = U/m

So as the unit mass decreases with radius...... the gravitational potential at that location decreases which is the gravitational potential energy at that location per the unit of mass.....

I just want to point out that in addition to dividing by wrong mass (you should divide by the test mass), your math is wrong here.

If you divide by something smaller, you get a *larger* result, not a smaller one.

 

[ecco]

If you divide by something smaller, you get a *larger* result, not a smaller one.

Golleee! Even I knew that. If I recall, I learned that in 1st grade.

 

[Justatruthseeker]

Wrong source. But I did make a small error. What you have missed is that we have always been discussing the absolute value, something that you could not understand.

Gravitational potential - Wikipedia

Tell me, where is the curvature greatest in that image, not the slope. The slope is not the same as the curvature.

Yes, I know, in that picture they are showing you a fiction of an imaginary sheet pulled downward by the force of gravity underneath the sheet.

They fail to take into account this is nothing as would be the reality since the curvature must be from all directions simultaneously..... Not a false representation of a one directional curve..... You do understand that gravity works in all directions at the same time, not just one, do you not? So knowing this you give a false misrepresentation of the reality????

I noticed you didn't include one with the earth in place in that false representation....

Oh that's right, because it encloses the entire earth as it indents the fabric of spacetime, not curves to the center - making your case even weaker....

Or maybe you would prefer NASA's fabricated sorry excuse for reality????

Weak representations follow a weak case......

So why are you not arguing for the picture you attached? That the curvature is greatest at the south pole at the surface???? That's what your picture represents after all.....

But that's the problem with those not being able to perceive nothing but simple representations versus what would be the reality.... Those people think they understand, but in reality understand nothing....

Versus the reality....

Where each line of force would bend into the object from all angles, then curve inwards and outwards as the mass below lessened and the mass above increased......

You simply can't picture the reality so all you can imagine is a ball placed onto a rubber sheet bending it downwards. It not only doesn't reflect reality, but gives a false misrepresentation of what is actually happening.....

Small error compounded by another error by showing me a picture that requires a ball placed on a sheet with the curvature going underneath the ball........ Oy vey!!!!