Why the voltage across emitter resistor comes out to be Zero while doing the ac analysis of a differential amplifier
You haven't posted a schematic but I think I know what you're asking. 

In short, what you're seeing is what's called a "virtual ground". That simplifies the math a lot for a differential amplifier small signal analysis. You want to keep the differential input close to zero so that the transistor stays in the amplification region.

This a differential pair:



And when doing the small signal analysis, you get:



See how the DC sources have become zero for the small signal anaylis. Also, you mentioned an emitter resistor. For this example there was a current source, but the important part is to keep an eye to what happens to the node denoted as 'P'.

Apply KVL around the \$v_{in1}\$ and \$v_{in2}\$ loops to get:
$$ v_{in1}-v_{\pi 1}=v_P=v_{in2}-v_{\pi 2}$$

Anc KCL at the P node:
$$ \dfrac{v_{\pi 1}}{r_{\pi 1}}+g_{m1}v_{\pi 1}+\dfrac{v_{\pi 2}}{r_{\pi 2}}+g_{m2}v_{\pi 2}=0$$

And if the transistors are closely matched (pi resistors and transconductance approximately equal), \$v_{\pi 1}\approx -v_{\pi 2}\$. Then you could use those equations to find that,

$$v_P=0$$

You can find full derivations in the Fundamentals of Microelectronics book by Razavi.

You could also get a graphical feeling for this:



The Y-axis shows \$V_X\$ and \$V_Y\$, these are simply the output signal of both transistors. The differential output is then \$V_X-V_Y\$. They are capitalized because the represent the bias or DC state of the transistor, you don't want to disrupt the bias point. That's why the input signals (\$v_{in1}\$ and \$v_{in2}\$ should be small enough so that it won't steer the bias point somewhere not intended. 

Back to the graphic, note that the X-axis represents the differential input. So, right the center of the plot, where \$V_{in1}-V_{in2}=0\$ you can see how I drew a red line. You want to stay close to that point so that you get a linear response or better said, amplify your differential input in a linear fashion. Your gain would be the slope of that red line. The farther away from the center of the plot, the less linear range you get.

You would wisely keep the difference \$V_{in1}-V_{in2} =0\$ or \$V_{in1}=V_{in2}\$ so that you take full advantage of the linear range. Again notice the capital letters here this a DC components of the input signal. 

For differential amplifiers, you superimpose a small signal to each of the DC components in this manner \$V_{in1}+v_{in1}\$ and \$V_{in2}-v_{in2}\$. As you can see, these signals are complementary (you increase one and decrease the other by the same amount). 

The signals being complementary become obvious when you try to find the difference again: \$V_{in1} + v_{in1} - (V_{in2}-v_{in2})=0\$. If \$V_{in1}=V_{in2}\$ then \$v_{in1}=-v_{in2}\$.

After this, you could take advantage of utilizing the half circuit and simplify the analysis:



Just like the input signals, the voltage at node P has a DC and an AC component, \$V_P+v_P\$ and what becomes zero is the AC component, \$v_P\$. That's a virtual ground or ac ground. For this example, I had a current source but the same applies to the node if there had been a resistor.