Gravitational Potential Difference in Uniform Gravity Field

[vin300]

When you use ..Work= Force*Distance the x cancels so the work is finite not infinite.

That is what everyone has been saying on this thread forever, but I showed that this approach leads to a contradiction(if you believe the potential is infinite)

 

[Dadface]

That is what everyone has been saying on this thread forever, but I showed that this approach leads to a contradiction(if you believe the potential is infinite)

The work done per unit mass is given,quite simply, by E=V so if in your thought experiment V is infinite then the work done is infinite.Sorry but I cannot see a contradiction and nor can I envisage a real situation where V is infinite.

 

[Hootenanny]

There's no such thing as "a uniform field"

Practically no, but theoretically, yes. In any case, that isn't the principle point here.

The potential at infinity is a finite positive number.OK.The potential energy anywhere in this field would be the work done to bring it from infinity to this point.

No it wouldn't. In this case, defining the Earth's geoid to be the point of zero potential, the potential of the field would be the work done to bring the unit mass from the Earth's geoid to this point.

 

[vin300]

What is the definition of V?

 

[Doc Al]

What is the definition of V?

What matters is the difference in potential, which is the work done to move a unit mass against the field between two points.

 

[D H]

For crying out loud!

That bears repeating. Only this time I'll say what I really meant last time:

Stop being intentionally thick.

Given a force field \mathbf F(\mathbf x), a function U is a potential function for the given force field if \mathbf F(\mathbf x) = \mathbf \nabla U(\mathbf x). That gradient means that potential energy is only defined to within an arbitrary constant. At which point the potential is chosen to be zero is arbitrary. You are hung up on an arbitrary point, the point at infinity. That point is very convenient in the case of a force field that follows an inverse square law. One must choose the point at which the potential is defined to be zero wisely as some choices yield nonsense results. For example, choosing the location of a point mass to be the zero point of the potential function for that point mass's gravitational potential yields nonsense results. Choosing the point at infinity in the case of a uniform force field similarly leads to nonsense results.

The solution is simple: Choose a different point. Any point that does not yield nonsense results will do. The origin, for example, is a reasonable choice. Given a uniform gravitational field \mathbf a=g\hat{\mathbf u}, a reasonable choice for a potential function is \mathbf U(\mathbf x) = -a \mathbf x\cdot \mathbf u.

 

[vin300]

No it wouldn't. In this case, defining the Earth's geoid to be the point of zero potential, the potential of the field would be the work done to bring the unit mass from the Earth's geoid to this point.

Yes but all practical and theoretical definitions of a concept must remain the same, or the concept changes.
Even in this case, the potential energy is the negative of the energy supplied to relieve the mass of this field!

 

[Hootenanny]

I will echo what DH said, stop being intentionally dense!

Even in this case, the potential energy is the negative of the energy supplied to relieve the mass of this field!

I agree. Note that in both cases - (a) defining the point of zero potential to be the limit as r approaches infinity - and (b) defining the point of zero potential to be the limit as r approaches zero; we take the potential energy at a point to be the work done to move the unit mass from the point of zero potential, which is 'infinity' in (a) and r=0 in (b), to that point.

Now, what don't you understand about that?

 

[vin300]

What matters is the difference in potential, which is the work done to move a unit mass against the field between two points.

If the potential at a point according to the definition is infinite, and the potential at another point is infinite, then according to the definition the potential difference between these points is infinity subtracted from infinity

 

[Hootenanny]

If the potential at a point according to the definition is infinite, and the potential at another point is infinite, then according to the definition the potential difference between these points is infinity subtracted from infinity

It seems to me that you aren't bothering to read anything that anyone else is writing, either that, or you are intentionally ignoring the parts that don't suite your argument.

I have already posted the definition of the potential difference between two points. Please take a look at this definition (https://www.physicsforums.com/showpost.php?p=2327044&postcount=27"). Do you or do you not agree that that is the definition of the potential difference between two points? Please answer this question directly.

As an aside, infinities are not real numbers (they do not belong to the real number line) and therefore do not have the same additive properties of real numbers. I.e. infinity - infinity is not simply zero.

 

[D H]

then according to the definition the potential difference between these points is infinity subtracted from infinity

Which is mathematical nonsense. You are not using the correct definition of potential, and you are being very stubborn about it. Stubbornness can be a useful trait at times, but not when it impedes your learning.

 

[Doc Al]

If the potential at a point according to the definition is infinite, and the potential at another point is infinite, then according to the definition the potential difference between these points is infinity subtracted from infinity

That should tell you that using "infinity" as a reference point is silly. All that physically matters is the change in potential between two points, which is well defined and trivially calculated. For a uniform field, choosing infinity as a reference is asinine.

You are hung up on a definition of gravitational potential that uses infinity as a reference, which is of limited use. Time to move on.