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In Griffiths electrodynamics, The maxwell stress tensor is used to determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q. To do this, we solve the integral

$$\vec{F}=\int_S\vec{T}\cdot d\vec{a}$$ where S is the closed surface enclosing the entire northern hemisphere and $\vec{T}$ is the Maxwell stress tensor. The result we get is that $$\vec{F}=\int_S\vec{T}\cdot d\vec{a}=\frac{1}{4\pi \epsilon_o}\frac{3Q^2}{16R^2}\hat{z} \, .$$ This is clearly a non-zero net force in the $\hat{z}$ direction. Later in the text though, Griffiths goes on to say that $\int_S\vec{T}\cdot d\vec{a}$ represents the "momentum per unit time flowing in through the surface". But if we have already calculated that $$\vec{F}=\int_S\vec{T}\cdot d\vec{a}=\frac{1}{4\pi \epsilon_o}\frac{3Q^2}{16R^2}\hat{z}$$ in the case of the hemisphere, then that means that a non-zero amount of momentum is flowing in through the surface of the hemisphere at any giving time. But how can this be if we can easily make the assumption that the uniformly charged sphere is static. That is, if we assume that there is some force that is holding the charges together, counteracting their mutual repulsion then the charged sphere remains intact and does not change in time and so its total momentum remains constant. If this is the case, then how can momentum be flowing into the surface enclosing the northern hemisphere? It seems to me like the interpretation of $\vec{F}=\int_S\vec{T}\cdot d\vec{a}$ as the momentum per unit time flowing in through the surface is rendered untenable by this very example? Am I missing something in my interpretation or is griffiths interpretation actually incorrect here?

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That is, if we assume that there is some force that is holding the charges together, counteracting their mutual repulsion then the charged sphere remains intact and does not change in time and so its total momentum remains constant.

Absolutely correct. The problem lies in the assumption. The Maxwell stress tensor directly derives from the Lorentz force (or in your simplified example, from the Coulomb force) that a charge distribution exerts on itself. And this force would drive the charge distribution apart if no other forces are present, that hold it together. These other forces are, in your example, outside the description of the referenced Maxwell stress tensor. A complete description would also include another stress tensor $\mathbf{T}_2$, that describes the binding forces. In this complete description, the net force over the whole hemisphere would be zero because the configuration is static, and hence, no momentum flux through the surface would occur.

If you do not know the whole stress tensor, as is the case in your example (unknown contribution of binding forces), it is not valid to consider the given (partial) stress tensor as a representation of net momentum flux.

This can be most lucidly expressed simply by citing Newton's second law (augmented by the superposition principle, "lex quarta"): $$\frac{d\mathbf{p}}{dt}=\sum_i \mathbf{F}_i$$ If the sum of all forces on the system equals zero, then the momentum of that system is conserved. Nevertheless, nobody keeps you from computing or measuring only $\mathbf{F}_1$, but then you should not label this a momentum change.

For example, you fix a rope to the wall, and then you pull at the rope. The rope doesn't move, but if you erroneously ignore the force that the wall exerts on the rope, and just identify the force you apply to the rope, you get to the false conclusion that the rope were changing momentum.

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  • $\begingroup$ Thanks for the great response! Everything makes sense now. Just one related question though: In my example, when we calculate $\vec{F}=\int \vec{T}\cdot d\vec{a}$ for only the bottom surface of the hemisphere (a circle surface that bisects the sphere) we get a non zero answer. My intuitions tell me that the integral should vanish on this surface due to symmetry though. My explanation for this non-vanishing integral is that this integral alone does not convey the net force on the surface. Only when the integral is performed over a closed surface does it convey the net EM force on thecharges? $\endgroup$ – SalahTheGoat Apr 21 at 6:36

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