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For a given scalar potential $V$, it is known that the corresponding force field $\mathbf{F}$ can be computed from

$$ \mathbf{F} = -\nabla V $$

Suppose a rigid body is placed inside this potential. The torque on the body $\mathbf{T}$ exerted by the scalar field will be

$$ \begin{align} \mathbf{T} &= \int_M \left( \mathbf{r} \times \mathbf{F}(\mathbf{r}) \right) dm \\ &= \int_M \left( \mathbf{r} \times -\nabla V(\mathbf{r}) \right) dm \\ &= \int_M \left( \nabla V(\mathbf{r}) \times \mathbf{r} \right) dm \end{align} $$

with $\mathbf{r}$ the position vector to the mass element $dm$, and the integration carried out over the entire body $M$.

So, being not too familiar with rigid body dynamics, I was wondering -- does something like a (vector/scalar) potential $P$ exist, such that the local torque induced by the potential $V$ can be expressed as

$$ \matrix{ \mathbf{T} = I \cdot \nabla P & & &\text{(or some similar form)} } $$

with $I$ the moment of inertia tensor of the rigid body?

If such a thing exists:

  • what is its name?
  • where should I start reading?
  • What is the proper expression for the torque $\mathbf{T}$?
  • how does $P$ relate to $V$?

If such a thing doesn't exist:

  • why not? :)
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You should read here and here and anything else you find on screw theory and rigid body mechanics. –  ja72 Jul 9 '13 at 13:44
    
If you want the full treatment of rigid body dynamics using screw theory look at this great lecture here or in Eucledian Space here –  ja72 Jul 9 '13 at 13:50
    
@ja72 Great links BTW thanks –  WetSavannaAnimal aka Rod Vance Jul 10 '13 at 0:55
    
@RodyOldenhuis Rody, if you're interested (you mentioned you might like to go and revise vector analysis) in the kinds of ideas explored in screw theory (I wasn't aware of it as such), then <a href="amazon.com/Geometric-Algebra-Physicists-Chris-Doran/dp/… Algebra for Physicists" by Doran and Lasenby</a> is a good reference. In this setting, I notice that the "screw" defined on ja72's link is the wholly analogous to a quaternion or member of the algebra of physical space. –  WetSavannaAnimal aka Rod Vance Jul 10 '13 at 1:02
    
@WetSavannaAnimalakaRodVance notice that the linked lecture from Copenhagen has some math-typo's but the ideas are correct. –  ja72 Jul 10 '13 at 12:02
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2 Answers 2

up vote 2 down vote accepted

Great question. A little background first.

Note that any force $\boldsymbol{F}$ moment $\boldsymbol{M}$ system on a point A can be equipollently translated into the screw axis S leaving only the components of $\boldsymbol{M}$ that are parallel to $\boldsymbol{F}$. The location is found by

$$ \boldsymbol{r} = \frac{\boldsymbol{F} \times \boldsymbol{M}}{\boldsymbol{F}\cdot\boldsymbol{F}} $$

Also the moment components parallel to $\boldsymbol{F}$ are described by a scalar pitch value $h$ found by

$$ h = \frac{ \boldsymbol{M} \cdot \boldsymbol{F}}{\boldsymbol{F} \cdot \boldsymbol{F}} $$

In reverse, a moment is defined by a force vector $\boldsymbol{F}$ passing through an axis located at $\boldsymbol{r}$ with pitch $h$

$$ \boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F} + h \boldsymbol{F} $$

Have you noticed how difficult it is to apply a pure moment on a rigid body, without applying a force? This is because you cannot have one without the other. A moment is really a result of the line of action of forces. So the scalar potential of a moment is really the same as the one for forces with

$$ \boldsymbol{M} = - \boldsymbol{r} \times \nabla V - h \nabla V = -\left( \left[1\right] h + \boldsymbol{r}\times \right) \nabla V $$

The problem is that in rigid body mechanics forces are not treated as scalar fields, but spatially constant, and temporally varying. Furthermore, I cannot think of a case where spatially varying moments arise that are NOT due to a force at a distance. I suppose you can come up with a tensor pitch $h$ instead of a scalar which is spatially varying for a definition like $\boldsymbol{M} = -\left( H + \boldsymbol{r}\times \right) \nabla V$, but then you will be making things up that do not have any physical meaning that I know of.

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Your answers and comments taught me the most, so I'll accept your answer. Thanks! –  Rody Oldenhuis Aug 22 '13 at 7:45
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Nice question. Wish I were bright enough to think of things as out of left field as this one. So I'm sorry to say the answer is no, for two reasons:

  1. The scalar field, call it $P(\mathbf{r})$ you seek has to be, by definition of a field, a function of the point in space $\mathbf{r}$ alone, whereas the torque on the body depends both on $\mathbf{r}$, the body's shape, mass distribution AND the body's orientation.

  2. Let's suppose you try to overcome this first problem by defining a scalar field $P(\mathbf{r})$ that is meaningful only for a particular body (a "test standard") that is in a "standard" orientation that is always the same. Then, the torque "field" for this standard body and orientation as a function of its centre of mass (or some other "standard" point within the body defining its position) $\mathbf{r}$ will be $\mathbf{T}\left(\mathbf{r}\right) = \int_M \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) \wedge \mathbf{r}^\prime \rho\left(\mathbf{r}^\prime\right) dV^\prime$ where the primed position vectors are the dummies for the integration and the position in the field $\mathbf{r}$ "offsets" them. $\rho$ is the body's mass distribution. Now, if this field is to be derivable from a scalar potential ($\mathbf{T}\left(\mathbf{r}\right) = \nabla P(\mathbf{r})$) then a necessary condition for this is that $\nabla \wedge \mathbf{T} = \mathbf{0}$ (the curl always annihilates the gradient where $\nabla \wedge \nabla$ exists). So, if you work out the curl we get $\nabla \wedge \mathbf{T}\left(\mathbf{r}\right) = \int_M \left[\mathbf{r}^\prime . \nabla\, \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) - \mathbf{r}^\prime \nabla^2 V\left(\mathbf{r} + \mathbf{r}^\prime\right)\right]\rho\left(\mathbf{r}^\prime\right) dV^\prime$, which, unless I am very much mistaken, is not identically nought for a general scalar force potential $V$ and so, unfortunately, the necessary condition is not fulfilled and the torque field is not derivable from a potential.

Again, great idea and hope this helps.

Edit after Rody's question: "how about a vector potential".

A vector potential IS possible, but, as stated in 1. above, it is only meaningful for the particular body in question in a particular constant orientation.

To show this, we form the divergence of the torque "field":

$\nabla . \mathbf{T}\left(\mathbf{r}\right) = \int_M \left[\mathbf{r}^\prime . \nabla \wedge \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) -\nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) . \nabla \wedge \mathbf{r}^\prime\right]\rho\left(\mathbf{r}^\prime\right) dV^\prime = 0$. So there is always a field $\mathbf{A}$ such that $\mathbf{T}\left(\mathbf{r}\right) = \nabla \wedge \mathbf{A}$. The easiest way to visualize this is in three dimensional Fourier space. In Fourier space the operations $\nabla . ()$ and $\nabla \wedge ()$ are replaced by $i\,\mathbf{k} . ()$ and $i\,\mathbf{k} \wedge ()$ where $\mathbf{k}$ is the "wavevector" (triple of the three Fourier transform variables $k_x$, $k_y$, $k_z$). So, a divergenceless (aka solenoidal) vector field in Fourier space is always orthogonal to $\mathbf{k}$ (i.e. orthogonal to the position vector in Fourier space); otherwise put, it is tangent to spheres centred at the origin. For such a vector field, in Fourier space we have $\mathbf{k} \wedge (\mathbf{k} \wedge \mathbf{\tilde{T}}) = k^2 \mathbf{\tilde{T}})$ and the curl is invertible for this special case. Therefore, to find the vector field $\mathbf{A}$ we transform $\mathbf{T}$ to Fourier space to get $\tilde{T}$, then the Fourier transform of $\mathbf{A}$ must be $\mathbf{\tilde{A}} = \frac{i}{k^2} \mathbf{k} \wedge \mathbf{\tilde{T}}$, then transform back to get the vector potential.

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Thanks, much appreciated. follow-up to your points: 1) true, however, you could concoct something akin to $\mathbf{r} \times I\cdot\mathbf{r}$ to capture the body's orientation and mass-distribution...or some other "magical" combination of matrix/vector operations :) 2) Hmm...haven't worked out the details myself, but I assume what you mean by $\nabla\nabla$ does not equal the Laplace operator $\nabla^2$? (otherwise, it would be identically nought :)) Anyway, how about a vector potential then? –  Rody Oldenhuis Jul 9 '13 at 11:58
    
No, written out in full $(\mathbf{r}^\prime . \nabla) \nabla V$ is $\left(x^\prime\partial_{x} + y^\prime\partial_{y} + z^\prime \partial_{z}\right) \left(\partial_x V \mathbf{\hat{x}}+\partial_y V \mathbf{\hat{y}} + \partial_z V \mathbf{\hat{z}}\right)$, so the "cross terms" are left in the whole integrand. To check whether or not a vector potential can be defined, try working out the divergence of the integral (the divergence annihilates the curl) and see whether that is nought. If it is, you might be in business! –  WetSavannaAnimal aka Rod Vance Jul 9 '13 at 12:11
    
Dammit, I should really get those vector identities back in my head; it's been too long! Thanks again! I'll leave the question open for the weekend, though, to be sure ^_^ –  Rody Oldenhuis Jul 9 '13 at 12:16
    
@RodyOldenhuis: euhm...I mean, leave it open for a few days :) –  Rody Oldenhuis Jul 9 '13 at 12:55
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