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I Potential energy in the presence of two fields

  1. Nov 30, 2016 #1
    I'm trying to understand potential energy, especially with regard to gravity. There's a double negative that's difficult to grasp: The direction of the force is negative because the vector points radially outward and the force of gravity is directed inward, and the work done to bring an object from infinity is negative because gravity is assisting in moving the object. Seeing how this works with two overlapping fields will help me understand how this works.

    Let's say I've got a very concentrated form of vacuum energy that creates an anti-gravity field (any repulsive field will do). The field generates an acceleration of ##a_{0}## directed radially outward that doesn't change with distance. If I place the source of the vacuum energy in the center of the Earth, having a mass of ##M##, what is the potential energy of an object with a mass of ##m## at a distance ##r## from Earth's center with the two fields?
    My initial take on the formula is:

    ##PE = -m\left(\int_{\infty}^{r}-\frac {G M}{r^2}dr+\int_{r}^{0}a_0 dr\right)##
    ##PE = -m\left(\frac {G M}{r}-a_0 r\right)##

    The rub (for me, anyway) is understanding how to integrate the second field. I'm guessing you'd integrate in the same direction as gravity (from infinity to 0) in order for the potential to work along the same vector.
     
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  3. Nov 30, 2016 #2

    A.T.

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    The gradient of the potential?
     
  4. Nov 30, 2016 #3
    Eventually I'm working towards a single field potential, but in the context of the statement, I'm just talking about the force vector. The choice of a unit vector directed outward appears to be a convention when dealing with gravitational potential energy.
     
    Last edited: Nov 30, 2016
  5. Nov 30, 2016 #4

    Dale

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    A.T. is making the point that the force is the negative gradient of the potential. So what function has a gradient that gives you the desired force?

    Once you have the two potentials then you can just add them.
     
  6. Nov 30, 2016 #5
    OK. This was actually more helpful than I first gave it credit. You're saying:$$F=-\nabla\Phi$$
    So I know the answer is:$$F=m\left(-\frac{G M}{r^2}+a_0\right)=-\nabla\Phi$$So, working backwards to find ##\Phi##, we need to change the reference for the ##a_0## force to:$$PE = -m\left(\int_{\infty}^{r}-\frac {G M}{r^2}dr+\int_{0}^{r}a_0 dr\right)$$Which yields:$$PE = -m\left(\frac {G M}{r}+a_0 r\right)$$
    And if I understand the concept of the resulting (combined) field, it's the acceleration felt by a unit of mass, so we get:$$\Phi = -\frac {G M}{r}-a_0 r$$Is this correct?
     
    Last edited: Nov 30, 2016
  7. Nov 30, 2016 #6
    As an aside, you seem to be saying (which I've seen in other sources as well):$$F=-\nabla\Phi$$Could you explain how this works? Force is in units of ##kg## ##m## ##s^{-2}## yet potential field is in units of Force per unit mass, or ##m## ##s^{-2}##.
     
    Last edited: Nov 30, 2016
  8. Nov 30, 2016 #7

    vanhees71

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    It cannot be correct since a force is a vector, and you didn't write down one. Concerning units: In the SI the force is measured in Newton, which is kg m/s^2. If the force has a potential the force is given by
    $$\vec{F}=-\vec{\nabla} V.$$
    This means that the potential is measured in units of ##\text{N m}=\text{kg} \text{m}^2/\text{s}^2=J## (Joule).
     
  9. Nov 30, 2016 #8

    Dale

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    Yes, exactly. You can check to make sure that it is correct simply by taking the gradient. So in spherical coordinates for the gravitational part ##-\nabla\Phi=-\nabla(-GM/r) =(-GM/r^2,0,0)## which is a ##1/r^2## magnitude force pointing towards the origin and for the other part ##-\nabla\Phi=-\nabla(-a_0 r)= (a_0,0,0)## which is a constant magnitude force pointing away from the origin.
     
  10. Nov 30, 2016 #9

    vanhees71

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    No! The gradient of a scalar field is a vector. If ##r## is ##|\vec{x}|##, as usual, then the gravitational field is
    $$\vec{g}=-\vec{\nabla} \phi=-\frac{GM \vec{x}}{r^3}$$
    and
    $$|\vec{g}|=GM/r^2.$$
    This is indeed the correct gravitational field outside of any spherically symmetric mass distribution (with the center at the origin of the reference frame).

    It is of utmost importance to distinguish clearly the various types of quantities in classical mechanics. For point mechanics you usually deal with scalars and vectors. A scalar is just a quantity with some magnitude (like mass, time, etc.), a vector has both a magnitude and a direction (like the position vector with respect to some origin chosen arbitrarily in space, velocities, accelerations, forces, etc.).
     
  11. Nov 30, 2016 #10

    Dale

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    So in SI force is in units of ##N## and energy is in units of ##J=N m##. The gradient involves derivatives with respect to position, so if the potential function has units of ##Nm## then the gradient of the potential has units of ##N##.

    Usually, of course, you use a potential which is given per unit mass or per unit charge so you have a potential in SI units of ##Nm/kg## or ##Nm/C## and then the gradient gives a force field in units of ##N/kg## or ##N/C## respectively.
     
  12. Nov 30, 2016 #11

    vanhees71

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    Units must (by law!) be set in upright (roman) letters. So it's ##\text{N}## not ##N##, etc.
     
  13. Nov 30, 2016 #12

    Dale

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    I agree. I think that we were cross posting and you are responding to an older edit. I believe that my current post is correct.
     
  14. Nov 30, 2016 #13

    vanhees71

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    Now it's fine!
     
  15. Nov 30, 2016 #14
    It is my understanding that a Newton, ##N##, is the amount of force needed to change the velocity of a 1 ##kg## mass by 1##m## ##s^{-2}##. Why can't you replace ##N## with ##kg## ##m## ##s^{-2}##? To me it makes a lot more sense to think of a potential field in terms of the acceleration it imparts on a unit mass.
     
  16. Nov 30, 2016 #15

    vanhees71

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    Again: units must (by legal law!) be set in upright (Roman) letters!!!

    The SI has 7 base units, the kg, m, s, A, K, mol, and cd. You can express any units in terms of these units, but it's often more convenient to introduce other derived units. So the unit of force is named N (Newton), and it is ##1\text{N}=1 \text{kg} \, \text{m}/\text{s}^2##. So your definition is correct. Also you have the unit of energy called Joule, and in terms of the base units it's ##1 \text{J}=1 \text{kg} \, \text{m}^2/\text{s}^2##.
     
  17. Nov 30, 2016 #16

    Dale

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    You certainly can make that replacement if you like. If I am talking about force then I like to use units of N, but that is strictly a personal preference and you are free to use base units if your preference is different. It is of no physical consequence whatsoever.

    Be aware that not all potentials are for a unit mass. The most common counterexample is the electrostatic potential which would be for a unit charge. So don't ingrain the "per unit mass" too deeply into your thought process.
     
  18. Dec 1, 2016 #17

    vanhees71

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    From the point of view of Newton's Laws it's most convenient to use a potential for the force itself, because then the equation of motion for a point particle subject to a force reads
    $$m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x}).$$
    Now multiply this with ##\dot{\vec{x}}## (scalar product) and use the product rule on the left-hand side and the chain rule on the right-hand side, and you get
    $$\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t} \dot{\vec{x}}^2=-\frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).$$
    Now combine the to sides of the equation to
    $$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x}) \right)=0,$$
    and this means that the expression in the bracket is conserved during the entire motion of the particle, and the corresponding conserved quantity is the total energy (kinetic and potential energy):
    $$E=\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=\text{const}.$$
     
    Last edited: Dec 2, 2016
  19. Dec 1, 2016 #18
    This is very helpful to think of the gradient of potential energy this way.
    I think your second formula should read:$$\mathrm{m}\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} \dot{\vec{x}}^2=-\frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).$$
     
    Last edited: Dec 1, 2016
  20. Dec 2, 2016 #19

    vanhees71

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    Yes sure, thanks for the hint. That was a typo. I've corrected it in the original posting.
     
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