Gravitational Potential Difference in Uniform Gravity Field

[vin300]

Moderation Note: Split from the original thread, which can be found https://www.physicsforums.com/showthread.php?p=680746#post680746"

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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.

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.

A uniform field does not have potential difference

 

[Doc Al]

A uniform field does not have potential difference

Really?

 

[vin300]

It doesn't. Since the force everywhere is the same

 

[Vanadium 50]

And force is the rate of change of potential.

 

[vin300]

Right

 

[vin300]

No Since the work done in bringing a unit mass from infinity to any point would be infinite, there is no potential difference

 

[vin300]

Intuition works

 

[Vanadium 50]

If the force is constant (and non-zero) and the potential is the rate of change of force, the potential cannot be constant.

No Since the work done in bringing a unit mass from infinity to any point would be infinite

That's simply not true.

Intuition works

I wouldn't trust it. It's leading you astray.

 

[vin300]

If the force is constant (and non-zero) and the potential is the rate of change of force, the potential cannot be constant.

The rate of change of potential is force
But not here


[qoute]
That's simply not true.[/quote]Why not true?The definition of potntial is the work done to bring a unit mass from infinity to a point, and is infinite here since the force is constant all over.In the classical case, force at infinity is zero

 

[vin300]

I wouldn't trust it. It's leading you astray.

Ah, you're too busy. I can see three in a line

 

[Doc Al]

Why not true?The definition of potntial is the work done to bring a unit mass from infinity to a point, and is infinite here since the force is constant all over.In the classical case, force at infinity is zero

The potential difference between any two points will be the work done to move a unit mass between those two points. (In some cases it makes sense to define the potential at infinity equal to zero, but not if the field is everywhere uniform.) Clearly the potential varies along the line of the force.

 

[vin300]

The potential difference between any two points will be the work done to move a unit mass between those two points.

True, but here the potential at every point in the field is infinite, so there's no difference

 

[Vanadium 50]

This is just plain wrong. You can't define the potential at one point to be infinity, then calculate that at another point it's also infinity, so the difference between them is zero. Apart from not being the way to solve the problem, this is mathematically incorrect.

A constant field has a potential growing linearly with distance.

 

[vin300]

It is mathematically incorrect, the answer is actually uncertain.
I made use of the fact that since everywhere in the field the force on a unit mass is the same, the potential everywhere is the same too, but this does not go well with the math
There is no imbalance in this field, and hence there should be no potential difference.
But this is not real, and hence the answer is not real

 

[Doc Al]

I made use of the fact that since everywhere in the field the force on a unit mass is the same, the potential everywhere is the same too, but this does not go well with the math

That "fact" is just plain wrong. Since the force = -dU/dx, the potential cannot be the same everywhere.

 

[Hootenanny]

I haven't read this thread in it's entirety, but I would like to comment on your last post:

I made use of the fact that since everywhere in the field the force on a unit mass is the same, the potential everywhere is the same too, but this does not go well with the math

As you say, this does not "go with the math" and is therefore incorrect! If there force on a unit mass is the same everywhere, that doesn't mean that potential is the same! In one dimension, the force is defined thus,

$$F = -\frac{dV}{dx}$$

We assume that the force is constant,

$$\frac{dV}{dx} = \text{const}$$

The potential is then (denoting the constant c₁),

$$V = c_1x + c_2$$

Hence, the potential is not the same everywhere!

EDIT: Doc Al beat me to it

 

[vin300]

That "fact" is just plain wrong. Since the force = -dU/dx, the potential cannot be the same everywhere.

Vanadium said that a lot earlier. And I replied "the potential at a point is the work done to bring a unit mass from infinity to that point and thus the potential everywhere in this field
is infinite"
Read my edited previous post.
There is no real answer to this because the situation itself is not real

 

[Hootenanny]

vin300 do you or do you not agree that the definition of force is the negative gradient of the potential? Please answer this question directly.

 

[DrGreg]

There is no real answer to this because the situation itself is not real

Exactly the problem. There is no such place as "at infinity", so trying to calculate the potential "at infinity" is where your logic goes wrong. You can't treat infinity like a real number and try to do maths with it. Think about what's really happening, without bringing infinity into it, and you should be able to make sense of it.

When we say "it takes an infinite amount of work to move to infinity" you can't take that literally. It's a shorthand for saying, "the further you go, the more work is required, without any upper limit".

Don't forget that you can always add a constant to a potential, you don't have to evaluate it "at infinity" to decide what it is elsewhere.

 

[vin300]

vin300 do you or do you not agree that the definition of force is the negative gradient of the potential? Please answer this question directly.

If you have read all my posts,:
Yes, I agree the force is the negative gradient of potential, this is a fact
Here there is a fixed force with an uncertain difference of potential, which is what makes the thing unreal

 

[Hootenanny]

If you have read all my posts,:
Yes, I agree the force is the negative gradient of potential, this is a fact

Good, I'm glad we can agree on something.

Here there is a fixed force without a difference of potential, which is not real

This is where your error lies. If there is a constant (non-zero) force, then there must be a potential field that is a function of position. Please, disregard all your 'intuition' and simply follow the mathematics.

Now, concerning the mathematics, do you disagree with anything that I wrote in https://www.physicsforums.com/showpost.php?p=2326940&postcount=46" post?

 

[Doc Al]

If you have read all my posts,:
Yes, I agree the force is the negative gradient of potential, this is a fact

OK.

Here there is a fixed force without a difference of potential, which is not real

This contradicts the previous sentence. (You are hung up on defining potential as work done from infinity. That certainly doesn't apply here.)

 

[vin300]

Ok Now here I prove that the math you use for this problem is wrong
The definition of potential at a point that I already wrote twice is true.From that, you derive the formula of potential you used hitherto this way:
Integrate -(GM/r^2)dr with a lower limit of infinity and upper limit of r
But here, the force is independent of distance
Integrate -kGMdr with a lower limit of infinity and upper limit of r(where k=constant)
P=infinite, for any value of r
If you use this infinite potential to determine the PD you get an unreal answer
That because the situation is unreal
The beauty of it

 

[Hootenanny]

Ok Now here I prove that the math you use for this problem is wrong
The definition of potential at a point that I already wrote twice is true.From that, you derive the formula of potential you used hitherto this way:
Integrate -(GM/r^2)dr with a lower limit of infinity and upper limit of r
But here, the force is independent of distance
Integrate -kGMdr with a lower limit of infinity and upper limit of r(where k=constant)
P=infinite, for any value of r
If you use this infinite potential to determine the PD you get an unreal answer
That because the situation is unreal
The beauty of it

Why is [negative] infinity an unreal answer here? Surely one would expect it to be infinite?

Your second integrand is equivalent to a constant force acting on an object. Now, suppose that object is moved through an infinite distance whilst being acted upon by a constant force. Surely it make sense that this would take an infinite amount of energy?

Or do you disagree?

 

[vin300]

Now, suppose that object is moved through an infinite distance whilst being acted upon by a constant force. Surely it make sense that this would take an infinite amount of energy?

Exactly, but what is unreal is potential difference between two points in this field
oo-oo
I know you participated in that thread

 

[Doc Al]

Exactly, but what is unreal is potential difference between two points in this field

Nonsense. A potential difference between two points in a uniform field is perfectly well defined. (And routinely used in every intro physics class.) The only thing "unreal" is having a uniform field extending to infinity.

 

[Hootenanny]

The point, as it always has been, is that the concept of infinity is complex and cannot be blindly applied. Loosely, every point in an infinite space is an infinite distance away from the 'edge' or infinity.

Exactly, but what is unreal is potential difference between two points in this field
oo-oo

No it is NOT! The potential difference between two points, a and b, in a [1D] field is,

$$V = \int_a^b F \; dx$$

Since F is constant,

$$V = \left[c_1x\right]_a^b$$

$$V = c_1\left(b-a\right)$$

Which is finite for finite a and b! What you are computing is the potential difference between infinity and any point on the real number line!

 

[D H]

You are stuck at infinity, vin300. The obvious solution is to use some other reference as the reference plane.

For example, the Earth's gravitational field appears to be a uniform gravitational field for points sufficiently close to the surface of the Earth. When this approximation is used, we do not use the potential difference between a point and that at infinity. We use the potential difference between a point and the surface of the Earth: U=mgh.

 

[Dadface]

Sorry if the following is irrelevant but I am in a rush and I need to read the thread again more carefully.

Firstly I am unable to envisage a situation where you can get a perfectly uniform gravitational field.You can get a uniform electric field between two oppositely charged metal plates.The force on a charge between the plates is independant of the position of that charge and the work done in taking the charge from one plate to the other is given by
=F*d=(V/d)q*d=Vq (V=pd,d=distance between plates,q= charge).In other words the work done is independant of the distance moved.Of course the electric field and gravitational field are different but the analogous features they share can be carried over to the problem being discussed here.

 

[D H]

Sorry if the following is irrelevant but I am in a rush and I need to read the thread again more carefully.

This thread was derailed at post #31 and hasn't come back on track since. Edit That is post #1 now. See post #32 below.

Firstly I am unable to envisage a situation where you can get a perfectly uniform gravitational field.

An infinite plane will do the trick. How you can find an infinite plane is a difference issue.

 

[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".