# EmDrive Cavity modes

The EMDrive described in patent application GB 2493361 is a simple section of a sphere surrounded by a superconducting cooling system. I have attempted to solve the oscillation modes possible and find spherical Bessel functions for radial directions and Legendre polynomials for "theta". When I apply boundary conditions, $E_r$ must be zero at the side walls so $P_n^0$(cos($\theta_w$)) = 0 can only happen for specific angles. If the wall is not at the right angle, $E_r$ must be zero every where. Same is true for $E_\phi$, but that has zeros for $P_n^1$(cos($\theta_w$)) so it would be a different mode.

The claim in the patent is that putting RF into the cavity creates a force - but it does not say in what direction! The EmDrive web site talks about group velocity, but I don't see how that makes sense for a standing wave in a cavity. What am I missing? I can find $B$ from $\nabla\times E$), but is there an easy way to find energy density directly from the $E$ field solutions? Once I know energy density, finding forces should be straight forward.

What I expect is the net total force to be zero, but it would be nice to prove it.

• – rob Aug 12 '14 at 14:20
• Hi Mike, please use MathJax in your posts. See meta.math.stackexchange.com/questions/5020/… – Brandon Enright Aug 12 '14 at 14:39
• A patent is a legal document, it's not a scientific document. It is not peer reviewed for scientific accuracy and can, in general, contain total scientific nonsense, as long as it passes the usual tests for patents like novelty and non-obviousness. One can argue whether most "crank" patents satisfy either, since most of them are based on variations of the same pseudo-science myths. Having said that, of course one can generate a net force with a directional antenna... it's nothing else but a photon drive operating at a really low frequency, but the frequency of the photons doesn't matter. – CuriousOne Aug 12 '14 at 19:22
• I totally agree - but this does not even have an external antenna. The RF is completely contained. The other bizarre portion is that a few guys at NASA say they measure thrust from it (see related). When NASA does pseudo-science, we all have a problem. – Mike Rosing Aug 12 '14 at 20:36
• @Mike: I got that part and hence my use of the words "crank" and pseudo-science. Unfortunately, some NASA folks are doing pseudo-science. In every sufficiently large organization there are folks whose mind operates at the "fringes". We may not agree with that, but it is a fact. Now, if you look at employment law, there is no clause in there, which allows an employer, especially a public one like NASA, to terminate folks with a fringe mind without good reason, which means that everybody around them usually holds their nose and gets out of their way, but won't interfere. – CuriousOne Aug 12 '14 at 23:19

So there is not an easy way - I had to grind out all the algebra. Starting with $$\nabla^2 E +k^2E = 0$$ The Electric field forms are $$E_r = P_n^0(cos\theta)[\frac{A_n j_n(k r)}{k r} + \frac {B_n y_n(k r)}{k r}]$$ $$E_\phi = P_n^1(cos\theta)[C_n j_n(k r) + D_n y_n(k r)]$$ Using $\nabla \cdot E = 0$ and the symmetry over $\phi$ I find $$E_\theta = -\frac {P_n^1(cos\theta)} {n(n+1)} \{ A_n [\frac {j_n(k r)}{k r} + j'_n(k r)] + B_n[ \frac {y_n(k r)}{k r} + y'_n(k r)]\}$$ Clearly $n = 0$ is not allowed.
Using $\nabla \times E = i \omega B$ I find the following for the $B$ field $$B_r = \frac {-i} \omega k \frac {P_n^0(\cos\theta)} {n(n+1)}[C_n \frac {j_n(k r)} {k r} + D_n \frac {y_n(k r)} {k r}]$$ $$B_\theta = i \frac k \omega P_n^1(cos\theta) \{C_n [\frac {j_n(k r)}{k r} + j'_n(k r)] + D_n [\frac {y_n(k r)}{k r} + y'_n(k r)] \}$$ $$B_\phi = \frac {-i} \omega k \frac {P_n^1(cos\theta)} {n(n+1)}[A_n j_n(k r) + B_n y_n(k r)]$$ Note that the imaginary factor simply means that the B field is 90 degrees out of phase with the E field in time.
Now let's look at boundary conditions. Both $E_\phi$ and $E_\theta$ are zero at $r_1$ and $r_2$. We also have $E_\phi$ and $E_r$ are zero at $\theta_w$. The actual field is the sum over all $n$ but each mode is orthogonal to all other modes so each mode must match the boundary conditions independently. For $E_\phi$, given arbitrary $\theta_w$, $P_n^1(cos \theta_w) \neq 0$ so all coefficients $C_n$ and $D_n$ must be $0$. That knocks out $E_\phi$, $B_r$ and $B_\theta$. The same is true for $E_r$ - which says that unless the walls are at a specific angle which is a zero for $P_n^0$ or $P_n^1$, no fields will resonate.
In other words, if the EmDrive guys don't build the cavity to specific angles, it will simply reflect all power and won't have any RF in it at all! So let's suppose they DO build it to a specific angle so it will resonate, then the Poynting vector is $S = E \times B$. Only $E_\theta B_\phi$ is in the $\hat r$ direction, but this is $0$ at the $r_1$ and $r_2$ walls. So even if it resonates, it won't push. QED.
Life gets much more interesting if there is a center pipe so we have $\theta_1$ and $\theta_2$. Then we have both $P_n^m$ and $Q_n^m$ as angular solutions. It will be possible to find a set of coefficients which match all boundary conditions and the system will resonate. (Replace all $P_n^m$ above with ($P_n^m(cos \theta) + K_n Q_n^m(cos \theta))$