Gravitational Potential Difference in Uniform Gravity Field

[D H]

A uniform field does not have potential difference

A uniform gravitational field \mathbf a = -ma\hat {\mathbf x} where a is some constant has a gravitational potential field U=gx, for example. Add any arbitrary constant to the potential and you get exactly the same uniform gravitational field.

 

[Hootenanny]

This thread was derailed at post #31 and hasn't come back on track since.

Good point - I've split the two threads.

 

[vin300]

A very simple question:
If
F=-GMm/r^2
is constant at two points inthe field,why is
E(pot.)=-GMm/r
not constant at those points?
I'm pretty sure the formulae are correct, aren't they?

 

[Born2bwire]

Neither would be constant, but they could be the same value (disregarding the direction of the force).

 

[vin300]

Neither would be constant,.

Good.Now look at Janus' post#19, in which he says two points in a field at different altitudes both at 1g have different potentials, and explain it to me

Moderation Note: Post number #19 is located in https://www.physicsforums.com/showthread.php?t=82309".

 

[Born2bwire]

Good.Now look at Janus' post#19, in which he says two points in a field at different altitudes both at 1g have different potentials, and explain it to me

Now that I am already in the trap I'll answer once and let the mods move these posts as well. The force is constant, but the potential is not. Obviously, for a constant force over all space, you would need a potential that scales linearly since force is the gradient of the potential.

 

[Dadface]

Currently we cannot define an absolute potential at a point,we can only calculate potential differences.We choose a reference location of infinity and ascribe to this a potential of zero but these are arbitary choices only and are made for convenience.In the majority of practical problems close to the earth, Earth's surface is taken as a reference location and this is given the value of zero potential.Again these are arbitary choices.If we remain fairly close to the Earth's surface the field is approximately uniform.

The force per unit mass is given by dV/dx but in a uniform field we can use non calculus notation and write F=V/x(here x stands for the distance between two places where the potential difference=V).If unit mass is moved between the two places the energy transferred is given by E=Fx=V/x*x therefore E=V.Since the weight is inversely proportional to the distance between the points this distance cancels when the equation Fd is used and the work done is finite.The same result drops out more simply from the definition of potential.

 

[D H]

Good.Now look at Janus' post#19, in which he says two points in a field at different altitudes both at 1g have different potentials, and explain it to me

Example: you have a uniform gravity field of one g. (one that does not decrease in strength with height. You have two clocks of equal mass sitting at different heights in that field. The two clocks will have different gravitational potentials even though they both experience exactly 1g, and according to GR, the one that is above the other will run faster.

He is talking about a uniform gravity field of one g and how clocks at different heights will run at different rates according to general relativity. As the Earth does not have a uniform gravity field, Janus obviously is not talking about the Earth in this paragraph. He is using a uniform gravity field to illustrate a point: That gravitational time dilation is a function of gravitational potential, and not gravitational force.

Thus a clock sitting at the center of the Earth is not at the same potential as one in space, even though they experience the same gravitational force locally.

Now Janus obviously is talking about the Earth. At the center of the Earth the gravitational force toward the Earth is zero. The only (significant) gravitational force at the center of the Earth is that toward the Sun. Yet a clock located 1 AU from the Sun but well away from the Earth will run at a different rate than a clock at the center of the Earth because of the difference in gravitational potential between the two points.

 

[vin300]

Good God! I thought since nobody liked my posts these were deleted

 

[Hootenanny]

Good God! I thought since nobody liked my posts these were deleted

Apologies, that was my fault. I should have sent you a PM letting you know that they had been moved. For future reference, you can find the location of all the threads you started (such as this one) and all the posts you made by looking at your profile.

 

[vin300]

He is talking about a uniform gravity field of one g .

How much can this uniform field practically extend?

 

[D H]

For crying out loud! Forever.

 

[Dadface]

For crying out loud! Forever.

Only in a thought experiment.

 

[vin300]

For crying out loud! Forever.

OK Believe me, now you are in a trap
The constant field extends forever, so not a function of distance.
F=-kGMm
Now the body is at rest, its potential energy is the negative of the energy required to escape this field
That is infinite
Thus I'm right, at every point it has infinite potential energy, and the potential difference is indeterminate.
If potential difference is indeterminate,
-dP/dx=F, F is also indeterminate which contradicts the original assumption

 

[Hootenanny]

OK Believe me, now you are in a trap
The constant field extends forever, so not a function of distance.
F=-kGMm
Now the body is at rest, its potential energy is the negative of the energy required to escape this field
That is infinite
Thus I'm right, at every point it has infinite potential energy, and the potential difference is indeterminate.
If potential difference is indeterminate,
-dP/dx=F, F is also indeterminate which contradicts the original assumption

I do wish that you would read some of the posts properly. Here you are simply repeating yourself. I have already address this point https://www.physicsforums.com/showpost.php?p=2327044&postcount=27"

Please re-read the post I have linked to and tell me exactly where you disagree with me.

 

[vin300]

I have read all those posts properly.But can you say something in the above post is wrong?
Isn't the potential infinite?

 

[Hootenanny]

But can you say something in the above post is wrong?

Yes and I have explained it in the post that I linked to as well as previous ones.

 

[D H]

Now the body is at rest, its potential energy is the negative of the energy required to escape this field

Wrong. That is but one definition of gravitational potential energy. While it is very useful in the case of a gravitational field toward some massive body, this definition has no use in the case of a uniform field. The obvious solution is to use a different definition.

All that matters with potential energy is the difference in potential energy between two points. The specific values assigned to those points is completely arbitrary. We use the energy needed to escape a field for the simple reason that the point at infinity is a convenient reference. There is absolutely nothing wrong with setting the Earth's gravitational field to be zero at the Earth's geoid. The potential at infinity would then be a finite positive number rather than zero. The potential difference between the surface of the Earth and the point at infinity is exactly the same regardless of which point is arbitrarily chosen as the zero reference.

 

[Dadface]

Vin,I repeat a previous post:
In a uniform field the weight of unit mass can be given numerically by:

Weight =V/x..Here x is the distance between two points where the potential differs by V.

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

 

[vin300]

definition has no use in the case of a uniform field.

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

All that matters with potential energy is the difference in potential energy between two points.

Yes

. There is absolutely nothing wrong with setting the Earth's gravitational field to be zero at the Earth's geoid.

No problem

The potential at infinity would then be a finite positive number rather than zero.

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.
Now hootenanny will say that isn't possible,
but that is perfectly possible, for you derive the formula of potential at a point using this very definition
Potential energy is the negative of the work done to bring the mass from infinity to a point
Here, the lower limit is infinity and the upper is the distance from the surface(r)

If you integrate, the term with r=infinity in the denominator disappears, and -GMm/r is the potential energy

 

[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.