LED are current driven device, and batteries are voltage source. To convert a voltage source in a current source, in is enough to put a resistor in series... as high value as possible. This theory doesn't take the "efficiency" factor into consideration... so many of us are not using limiting resistors at all. This MAY be ok, if somehow you can ensure that the Luxeon doesn't exceed tha maximum rated temperature, by using an heatsink. Just to make an example, if a Luxeon is powered with 5 W, and considering an efficiency of 60%, the other 40% of the 5 W (2 W) is dispersed as heat, by increasing the temperature of the chip.
Now, the grade of temperature dispersion of an heatsink is measured in Celsius/W, or how many Celsius degree the surface temperature increases per watt of thermal energy dissipated. By simple arithmetics, if you do not want to exceed 60 Celsius at the junction between the Luxeon and the heatsink, and supposing the ambient temperature to be 30 Celsius, a 15 °C/W heatsink may suffice, even if a 10 °C/W is more advisable.
To test an heatsing, stick a wire wound resistor driven to dissipate 2 Watts (i.e. 10 ohm resistor with 2 volts across it) with some thermal compound, and after one minute measure the temperature at the junction by using a thermometer.
The absolute best way to drive the Luxeon is to use a switching power converter, like the chips built from Linear Technology, where is possible to include a diode as temperature sensor. Even a two-transistor switching power converter with current sensing (I call it "dirty" design) is OK; while proper current limiting, done with series resistor, can be rock-stable if correctly calculated. The disadvantage of passive drive is that the resulting circuit is too sensitive to the source voltage, so it may be necessary to resort to lithim disulphide battery or rechargeable alkaline. In any case, the heatsink is the most important component after the Luxeon itself.
Hope to have been clear enough.
Anthony