So:
1. E = k⋅T, with k = Boltzmann constant, T = absolute temperature
The thermal voltage can then be defined as the voltage potential that the conduction band needs to experience in order for electrons in it to overcome the thermal energy k⋅T. (Valence band electrons are a totally different matter.) This only applies to semiconductor junctions where there is no long-range electric force felt, as there is in copper wires, for example. So this applies to the conduction band near PN junctions, but it does NOT apply to the conduction band of metal wire, where the force has very long range instead.
This voltage is simply equal to the ambient thermal energy per particle, divided by the electronic charge, q, of the electron. Hence:
2. Vt = k⋅T ⁄ q, with q = electron charge (in Coulombs, usually.)
Here are the two constants:
3. k = 1.3806503⨯10⁻²³ m²⋅kg ⁄ s²⋅K
4. q = 1.60217646⨯10⁻¹⁹ coulombs
Ambient temperature is often taken to be about 27℃, or 300.15 K. Call it 300 K. If you plug that in to equation (2) you get ≈ 0.025852 volts. Most folks just round that to 26mV and call it square. But you need to keep in mind that it changes with temperature, because then the conduction band electrons are moving in a more energetic medium and have more to overcome in motion.
Bottom line is that this basic bit of physics cannot be ignored when computing BJT gain. It's always present. So if you ground out the emitter, there is still a tiny k⋅T ⁄ q voltage there and because of it, you can just as well imagine that there is a "tiny re" present there -- if you also know the current flowing. So this is why "little re" depends upon the emitter current. You just divide Vt by the emitter current to get the effective "little re" and then use that as a mental device when computing gain with various emitter resistors. Since Vt is in the tens of millivolts and since most quiescent BJT currents are on the order of a few milliamps, the "little re" will be a few tens of Ωs.