Mathematical model for differential signals and relation with reality
I will answer your question using - as an example - the classical differential amplifier (with an ohmic resistor RE in the common emitter path).

This circuit reacts upon both input voltages V1 and V2 at both input nodes. That means: We are interested in the output voltage Vo at one or at both signal voltages at the output node(s). 

Of course, for calculating the corresponding gains we could start from the beginning (without using well-known formulas) on the basis of Kirchhoffs laws (KCL and KVL). However, it is much more simple and elegant to split the calculation into two parts which are very simple to treat. 

For this purpose the two arbitrary input signal voltages V1 and V2 are split into the following signals:

Arithm. mean value: Vcm=(V1+V2)/2

Differential (symmetric): Vdd=(V1-V2)/2=Vd/2.

As we can see: V1=Vcm+Vdd and V2=Vcm-Vdd.

(Because Vcm appears in both parts, it is a common mode value. As a general rule: Two arbitrary voltages - positiv or negative - can always be split into a common mode part Vcm and symmetric diff. part Vdd. The amplified signals of both parts can easily be found and superimposed at the output).

Now - it is a simple task to find the gain for both parts and to calcuate the output voltages using superposition.

(1) Common mode gain: Gcm=Vocm/Vcm=-gmRc/(1+2gmRE)

(2RE because for common mode signals we have two equal emitter currents through RE)  

(2) Symm. diff. gain: Gdd=Vodd/Vdd=(+-)gmRc 

(We have used the formula for an emitter stage without RE-feedback, because both emitter signal currents have opposite signs and cancel each other).  

Using these gain expressions, we can find the output signals (with |Gd|=gmRc/2):

(1) Vo1=-Vcm*|Gcm|-Vdd*|Gdd|=-Vcm*|Gcm|-Vd*|Gd|

(2) Vo2=-Vcm*|Gcm|+Vdd*|Gdd|=-Vcm*|Gcm|+Vd*|Gd|

Comment 1: Please note, that we have a common mode input voltage Vcm also in case one of the input is grounded !) 

Comment 2: Since we want that the diff. amplifiers reacts upon differential signals Vd only, we want to have a very small common mode output. Therefore, we select a very large resistor RE or - as a better solution - we are using the large dynamic output resistance of a third transistor in the common leg of the circuit.