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In generic terms and expressions without going into specificity or nature of fields/forces in order to highlight the same, how exactly could we characterise the distinction between 'Potential' & 'Potential Energy'?
Difference Between Potential and Potential Energy
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- #2
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Potential is a property of a point in space, arising from a force field such as gravitational or electrical. Potential energy is a property of a body, or in some cases a system.
The potential energy of a body at a point in space, arising from force field F, equals the potential at the point from F, multiplied by the body's property relevant to F. For gravitational that will be mass and for electrical it will be charge.
For example, the Newtonian gravitational potential at distance R from the centre of a mass M is
$$V=-\frac{MG}{R}$$
and the the gravitational potential energy of a body of mass $m$ at that distance is
$$mV =-\frac{mMG}{R}$$
EDIT: Fixed denominator that had ##R^2## instead of the correct ##R##. Thanks to @Delta2 for pointing that out.
Also changed sign, so that potential and PE increase with distance from mass M.
The potential energy of a body at a point in space, arising from force field F, equals the potential at the point from F, multiplied by the body's property relevant to F. For gravitational that will be mass and for electrical it will be charge.
For example, the Newtonian gravitational potential at distance R from the centre of a mass M is
$$V=-\frac{MG}{R}$$
and the the gravitational potential energy of a body of mass $m$ at that distance is
$$mV =-\frac{mMG}{R}$$
EDIT: Fixed denominator that had ##R^2## instead of the correct ##R##. Thanks to @Delta2 for pointing that out.
Also changed sign, so that potential and PE increase with distance from mass M.
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- #3
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For a particle moving in an external field (electrostatic, gravitational) you have an equation of motion of the form
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}).$$
If ##\vec{F}## is conservative, i.e., if there is a scalar field ##V## such that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
you have
$$m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x}).$$
Multiplying with ##\dot{\vec{x}}## and integrating from ##t_1## to ##t_2## you get
$$\frac{m}{2} \dot{\vec{x}}^2(t_2) - \frac{m}{2} \dot{\vec{x}}^2(t_1)=-V[\vec{x}(t_2)]+V(\vec{x}(t_1),$$
or ordering expressions at ##t_1## and ##t_2##
$$\frac{m}{2} \dot{\vec{x}}^2(t_2) + V[\vec{x}(t_2)] = \frac{m}{2} \dot{\vec{x}}^2(t_1) + V[\vec{x}(t_1)],$$
i.e., the total energy
$$E=T+V=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}.$$
##T=m \dot{\vec{x}}^2/2## is called kinetic and ##V(\vec{x})## potential energy.
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}).$$
If ##\vec{F}## is conservative, i.e., if there is a scalar field ##V## such that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
you have
$$m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x}).$$
Multiplying with ##\dot{\vec{x}}## and integrating from ##t_1## to ##t_2## you get
$$\frac{m}{2} \dot{\vec{x}}^2(t_2) - \frac{m}{2} \dot{\vec{x}}^2(t_1)=-V[\vec{x}(t_2)]+V(\vec{x}(t_1),$$
or ordering expressions at ##t_1## and ##t_2##
$$\frac{m}{2} \dot{\vec{x}}^2(t_2) + V[\vec{x}(t_2)] = \frac{m}{2} \dot{\vec{x}}^2(t_1) + V[\vec{x}(t_1)],$$
i.e., the total energy
$$E=T+V=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}.$$
##T=m \dot{\vec{x}}^2/2## is called kinetic and ##V(\vec{x})## potential energy.
- #4
sophiecentaur
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Isn't it simpler than all this? Afaiaa, Potential is the Potential Energy per unit Mass (/Charge if you're talking Electrics)
- #5
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The potential of a force is by definition a scalar field, ##V##, such that ##\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})##. That's very simple, provided such a potential exists for a given ##\vec{F}##. You cannot make it simpler without confusing the subject unnecessarily.
- #6
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And a lot of textbook would call it potential energy.
- #7
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@andrewkirk the denominator in your formulas should be ##R## not ##R^2##.
- #8
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So what?
- #9
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Usually the potential is potential energy per unit charge or mass or something. For example in electrostatics it is ##\mathbf{E}=-\nabla V## and ##\mathbf{F}=\mathbf{E}q=-q\nabla V##.
However if we take the case of a spring then the potential is the same as the potential energy ##V=\frac{1}{2}kx^2## and ##F=-\nabla V=-kx##
However if we take the case of a spring then the potential is the same as the potential energy ##V=\frac{1}{2}kx^2## and ##F=-\nabla V=-kx##
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- #10
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Well, the OP asks about the difference between potential and potential energy. None of your posts adresses the issue.
- #11
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The potential is the potential of the vector field under consideration. There's an electrostatic potential for the electrostatic field, another potential for the gravitational field, and also a potential for all other kinds of "conservative forces", namely those forces which have a potential ;-).
- #12
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I don't understand, what's unclear. If the force has a potential, that's the potential energy part of the total, conserved energy. That's what I said above. Making many words without math leads to such nonsensical confusions!
- #13
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Yes it is just that the electrostatic field for example gives the force per unit charge, not just the force. So it is ##E=-\nabla V## but for the electrostatic force on a point charge q it is ##F=-q\nabla V##. Just saying because in your general approach in post #3 you equate ##\vec{F}## to ##-\nabla V## (and to ##m\ddot x##) which seems to imply that F is just the force, and not the force per unit charge (or mass).
- #14
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Sigh. This is really a superfluous discussion about semantics. There are many situations, where vector fields can be described as gradients of a scalar field. Since often these fields obey the Poisson equation, they are called potentials.
In mechanics there are potentials for many forces. In the energy balance they are called potential energy to distinguish them from the other part, the kinetic energy. These are just words. What's important is to understand the mathematics given in #3.
Of course, it's also important to study "potential theory", including Helmholtz's fundamental theorem of vector calculus too, but that's only rarely really needed in mechanics. There you usually find the potentials for the usually treated forces easily by direct calculation.
In mechanics there are potentials for many forces. In the energy balance they are called potential energy to distinguish them from the other part, the kinetic energy. These are just words. What's important is to understand the mathematics given in #3.
Of course, it's also important to study "potential theory", including Helmholtz's fundamental theorem of vector calculus too, but that's only rarely really needed in mechanics. There you usually find the potentials for the usually treated forces easily by direct calculation.
- #15
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No it is not just semantics. There is a difference of a factor of q (or m if we talk about gravitational field) between what you call potential energy and what is the potential energy.
- #16
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Read my postings again carefully!
- #17
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I think that you are the one who should start reading what people are asking about. At the high school and undegrad level potential and potential energy are not the same, and thus should be carefully distinguished. Throwing around Poisson and Helmholtz helps no one.
- #18
PeterDonis
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Why do you think there is such a distinction?
I don't see how this can be done. Without some specific examples of the distinction you think you see, how is anyone supposed to answer your question?
- #19
PeterDonis
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Then my question to you is the same as my question to the OP: why do you think they are different? A specific reference to a "high school" or "undergrad level" textbook that describes the distinction you are talking about would be helpful.
- #20
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That's interesting. I learned already in high school what the potential of a force is and that this potential is called the potential energy in the expression for the total energy, admittedly restricted to 1D motion, where you don't need gradients and line integrals but usual derivatives and 1D integrals do. It cannot be made simpler than that. Otherwise it leads to useless confusion and endless discussions about imprecisely defined words.
- #21
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Because that is what I have been tought in high school and at the university and that is what I teach all my students beacuse that is what they have to know during their high-school exams. I believe Resnick&Halliday talk about the two. And only high-school textbooks I can reference are in polish.
- #22
PeterDonis
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Can you give an example of a "potential" that you have been taught in high school that is not associated with a potential energy in the way @vanhees71 describes?
- #23
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Well, mixing electric potential and electric potential energy (which is potential times charge) is definitely not confusing...
- #24
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No one is saying that potential is not associated with potential energy, I'm saying that it is not the same (at least in high-school). E.g. electric potential is defined as ##\frac{E_p}{q}##.
- #25
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I have not even mentioned the electric potential, which is the potential of an electrostatic field and not of a force. This confusion has been brought in by somebody else. Once more: the potential of the force (!) IS the same as potential energy.
- #26
PeterDonis
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Ok, but saying the two are "not the same" with this definition seems like a quibble. Why is it important to make this distinction between the energy and the energy per unit charge?
- #27
PeterDonis
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I don't think you have carefully considered your position.
The "electric potential" that has been defined is the electrostatic potential energy per unit charge. Its gradient is the electrostatic force per unit charge. The only difference between these quantities and what you are calling "potential of the force" and the "force" itself is the "per unit charge" part. To me that doesn't seem like enough of a difference to warrant the position you are taking here.
- #28
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Oh come on! The OP was about mechanics, and there the potential is the potential of a force (or an interaction force between particles). I have clearly defined in #3 what I mean by potential in the context of the question. How can it be misunderstood that I mean different quantities, which where not talked about before in the thread?
- #29
PeterDonis
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Perhaps, if we include electrostatics in "mechanics".
So what? The question isn't about what you mean by "potential", it's about what the OP means by "potential". If you're going to just adopt a different meaning, you at least need to explain why the OP's meaning isn't a good choice. And doing that means doing what I did--suggesting that the difference between "energy" and "energy per unit charge" is not very important. And if that's the case, then neither is the difference between "potential of a force" and "potential of a force per unit charge". You have not responded to this point at all.
- #30
sophiecentaur
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I have already made this point but it got no reaction. I was assuming that the OP was basically about High school level physics. I am still convinced that the distinction is between Potential, an Intrinsic quantity (Energy per unit charge / mass) and Potential Energy, an extrinsic quantity (Energy for a particular mass or charge).
This is the equivalent sort of distinction to comparing Density with Mass - and many other pairs of quantities.
Some of the above posts imply that the Maths is the whole story. This is true except that dimensional analysis can be hard when complicated maths is involved.
- #31
Doc Al
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Absolutely. A student who cannot distinguish potential from potential energy will have trouble with simple problems. Such as: Calculate the KE a given charge gains when it falls through a given potential difference.
The two concepts are intimately related but are not the same. Especially at the lower levels.
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- #33
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You tend to forget that there are different levels of physics. For a high school teacher that is obvious - there are two potentials considered in high-school: gravitational and electric. No gradients and other fancy stuff. And yes, you can teach electrostatics without using gradients and vector calculus explicitly, just look at some of the physics olimpiad problems.
- #34
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Maybe it's another convention you use. A potential of a vector field is a scalar field, whose gradient is this field. In the context of the question posed in #1 it can only be the potential of a force. In #3 I've chosen the most simple case of the motion of a particle in an external force field, i.e., you have
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
In high school, of course, you don't have vector calculus at your disposal and that's why it's usually simplified to the motion along a straight line, i.e., you have
$$F(x)=-V'(x),$$
and also here one calls ##V## the potential of the force.
There is no way to understand electrostatics without the use of vector calculus. In high school we usually used the integral form and only very symmetric situations to calculate electrostatic fields. Of course also the potential of the electric field was introduced using the application of the motion of a charge in this field. Then of course ##V(\vec{x})=q \Phi(\vec{x})##, where ##q## is the charge of the particle under consideration.
As I said, you make things more complicated in avoiding a minimum of math!
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
In high school, of course, you don't have vector calculus at your disposal and that's why it's usually simplified to the motion along a straight line, i.e., you have
$$F(x)=-V'(x),$$
and also here one calls ##V## the potential of the force.
There is no way to understand electrostatics without the use of vector calculus. In high school we usually used the integral form and only very symmetric situations to calculate electrostatic fields. Of course also the potential of the electric field was introduced using the application of the motion of a charge in this field. Then of course ##V(\vec{x})=q \Phi(\vec{x})##, where ##q## is the charge of the particle under consideration.
As I said, you make things more complicated in avoiding a minimum of math!
- #35
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My 2 cents and personal opinion.
However, it does not mean that it has to be the only valid point of view. Au contraire, it is the reason why STEM fields have the reputation they have and why students are forced to relearn concepts over and over again instead of treating them scientifically in the first place in my opinion. I do not like the underlying assumptions that students at school are not capable of learning things right. Such an assumption does not match my experiences.
Teach them right instead of teaching them twice!
We as a website are committed to the way subjects are learned at universities. There is a reasonable exception in the homework forums. I can not see that the homework standard should be applied everywhere else as some members try to achieve. (Evidence: this thread and https://www.physicsforums.com/threads/epsilon-delta-proof-and-algebraic-proof-of-limits.1016299/.) We should not lower our standards needlessly.
'Keep them stupid' is the reason students hate STEM. And those who don't are damned to relearn things over and over again, excused by "complicated math".
This is certainly the main opinion and the given fact at our schools.
However, it does not mean that it has to be the only valid point of view. Au contraire, it is the reason why STEM fields have the reputation they have and why students are forced to relearn concepts over and over again instead of treating them scientifically in the first place in my opinion. I do not like the underlying assumptions that students at school are not capable of learning things right. Such an assumption does not match my experiences.
Teach them right instead of teaching them twice!
We as a website are committed to the way subjects are learned at universities. There is a reasonable exception in the homework forums. I can not see that the homework standard should be applied everywhere else as some members try to achieve. (Evidence: this thread and https://www.physicsforums.com/threads/epsilon-delta-proof-and-algebraic-proof-of-limits.1016299/.) We should not lower our standards needlessly.
'Keep them stupid' is the reason students hate STEM. And those who don't are damned to relearn things over and over again, excused by "complicated math".