It is well-known fact that integral Dehn surgeries on $3$-sphere $S^3$ are viewed as the result on the boundary of attaching $2$-handles $B^2 \times B^2$ to the $4$-ball $B^4$.


Is there an analogue of rational surgeries relating handle attachment of rational framing? If not, which problem occurs?



Rational surgery and attaching $2$-handles
To attach a 4-dimensional 2-handle to the 4-ball, one requires an attaching region in $S^3=\partial B^4$ and a map from the attaching region of the handle (which has a natural parametrization as $S^1\times D^2\subset \partial(D^2\times D^2)$) to the attaching region in $S^3$. The attaching region in $S^3$ is determined by specifying a knot $K\subset S^3$ (and then convention dictates that the attaching region $\nu(K)\cong S^1\times D^2$ is parametrized by identifying the Seifert longitude $\lambda$ for $K$ with $S^1\times\{pt\}$). Thus the handle may be attached via any orientation reversing homeomorphism from the $S^1\times D^2$ in the boundary of the handle to the $S^1\times D^2$ neighborhood of $K$. There are *only* an integers worth of such maps up to isotopy (see eg Rolfsen *Knots and Links* 2D4 and 2E5); in particular $S^1\times \{pt\}$ has to be mapped to $\lambda+n\mu$ and $\{pt\}\times \partial D^2$ has to be mapped to $\mu$, where $\mu$ denotes a meridian of $K$. 


The resulting boundary after the handle attachment should be thought of as (the bits of the boundary of the handle that didn't get stuck to anything)$\cup$(the bits of the boundary of $S^3$ that didn't get something stuck to them) . That's $(D^2\times S^1) \cup (S^3\smallsetminus\mathring{\nu(K))}$, so the boundary is some Dehn surgery on $K$. And we can see which; we had to send $\partial D^2\times \{pt\}$ to $\lambda+n\mu$, so the only surgeries we can obtain are integral. 


