An engineer wants to find the net electromagnetic force on a charged capacitor. Instead of integrating the Lorentz force density ρE over the volume of the charges, she draws a spherical surface around the capacitor and integrates T·dA. Why can she choose any closed surface enclosing the capacitor, not just the one geometrically closest to it?
AThe stress tensor is constant in free space, so the surface shape does not matter
BBy the divergence theorem, the surface integral of the stress tensor over any closed surface gives the same total force as long as the same sources are enclosed — just as Gauss's law gives the same flux for any surface enclosing the same charge
CThe electric and magnetic fields cancel outside the capacitor, so the integrand is zero away from the surface
DThe stress tensor is only defined at the surface of the object, so any enclosing surface maps to the same values
This is the key advantage of the stress-tensor method. In static fields, ∇·T̄ = f (force density), so the total force on enclosed sources equals ∮ T̄·dA for any closed surface by the divergence theorem — analogous to how ∮ E·dA = Q_enc/ε₀ for any Gaussian surface. This freedom to choose a convenient surface (where the fields may be simpler or more symmetric) is exactly what makes the method powerful in practice. The Lorentz force method requires integrating over all the sources; the stress tensor method only requires integrating over the field at an arbitrary enclosing surface.
Question 2 Multiple Choice
A plane electromagnetic wave with intensity I strikes a surface. If the surface is perfectly absorbing, the radiation pressure is I/c. What is the radiation pressure if the surface is perfectly reflecting?
AStill I/c — the total energy deposited per unit time is the same regardless of reflection
B2I/c — perfect reflection reverses the wave's momentum, so the impulse delivered to the surface is twice as large
CI/(2c) — only half the wave's momentum is available for transfer to a reflecting surface
D0 — a perfectly reflecting surface returns all the wave's energy, so it exerts no net force
Momentum transfer to the surface equals the change in the wave's momentum. An absorbed wave delivers momentum flux I/c (its full forward momentum is lost). A perfectly reflected wave reverses direction: its momentum changes by 2 × (I/c), so the force per unit area is 2I/c. This factor-of-two difference is directly computable from the Maxwell stress tensor: the stress on an absorbing surface includes only the incoming wave's momentum flux, while the stress on a reflecting surface includes both the incoming and outgoing wave contributions. This doubling is why reflective solar sails are more efficient than absorbing ones.
Question 3 True / False
In the Maxwell stress tensor, the diagonal components (T_xx, T_yy, T_zz) represent electromagnetic shear stresses, while the off-diagonal components represent electromagnetic pressure.
TTrue
FFalse
Answer: False
This is reversed. The diagonal components represent electromagnetic *pressure* — force per unit area perpendicular to a surface — while the off-diagonal components represent *shear stress* — force per unit area parallel to a surface. The physical intuition is that T_ij is the flux of the i-th component of momentum in the j-th direction; when i = j (diagonal), the momentum flux is perpendicular to the surface (pressure); when i ≠ j (off-diagonal), the momentum flux is parallel to the surface (shear).
Question 4 True / False
To use the Maxwell stress tensor to calculate the force on a physical object, you should integrate over a surface that lies exactly on the object's boundary — you can seldom use a more distant enclosing surface.
TTrue
FFalse
Answer: False
The freedom to choose any convenient closed surface enclosing the object is one of the method's main practical advantages. In static cases, ∮ T̄·dA is the same for every closed surface enclosing the same sources, by the divergence theorem. You might choose a distant spherical surface where the fields have simpler form (e.g., pure radiation fields), a surface where boundary conditions are known analytically, or a surface aligned with coordinate symmetries. Choosing the boundary surface as the object's own surface can sometimes be the hardest choice because the fields may be complex there.
Question 5 Short Answer
How is using the Maxwell stress tensor to calculate force on an object analogous to using a Gaussian surface in electrostatics? What does this analogy reveal about the physical meaning of the stress tensor?
Think about your answer, then reveal below.
Model answer: In electrostatics, Gauss's law says ∮ E·dA = Q_enc/ε₀: the total electric flux through any closed surface equals the total enclosed charge. You can compute the charge enclosed without knowing the detailed charge distribution — just the field at the surface. The Maxwell stress tensor works the same way for force: ∮ T̄·dA = F_enc (in static fields), so the total electromagnetic force on everything inside the surface equals the stress-tensor flux through the surface. You can compute the force without knowing the detailed current/charge distribution inside — just the field at the surface. This reveals that T_ij is a *momentum flux tensor*: it tells you how fast electromagnetic momentum is flowing through unit area in a given direction. Force is the rate of momentum transfer, so integrating T over a surface gives the total rate at which the field delivers momentum to enclosed sources.
The analogy is deep: both Gauss's law and the stress-tensor method convert a volume integral (over sources) into a surface integral (over fields). Both exploit the divergence theorem. Both let you choose the most convenient surface rather than integrating over the actual source distribution. The stress tensor is essentially a '3×3 Gauss's law' for electromagnetic momentum rather than charge.