frequency domain phase shift
Suppose one signal is $a_1(t) = \cos(\omega t)$ and the other is 
$a_2(t) = \cos(\omega t - \theta)$. Note the
phase shift $\theta$ between the two signals, and note also that we can write
$$a_2(t) = \cos(\omega t - \theta) 
= \cos\left(\omega\left(t-\frac{\theta}{\omega}\right)\right) 
= \cos(\omega (t - t_0)) = a_1(t-t_0)$$
where $t_0 = \frac{\theta}{\omega}$ showing that $a_2(t)$ is a delayed version of $a_1(t)$.  Suppose that $a_1(t)$ and $a_2(t)$
are transmitted from the two separate antennas simultaneously, that is,
no additional time delay between them, 
and reach the receiver over paths of delays $\tau_1$ and $\tau_2$
respectively and gains $A_1$ and $A_2$ respectively.
Then the received signals are 
$$\begin{align}
A_1a_1(t-\tau_1) &amp;= A_1\cos(\omega (t -\tau_1)) = A_1\cos(\omega \hat{t})\\
A_2a_2(t-\tau_2) &amp;= A_2\cos(\omega (t -t_0 - \tau_2))\\
&amp;= A_2\cos(\omega (\hat{t} +\tau_1 -t_0-\tau_2))\\
&amp;= A_2\cos(\omega (\hat{t} - (t_0 + \tau_2-\tau_1))\\
&amp;= A_2\cos(\omega \hat{t} -\hat{\theta})
\end{align}$$
where $\hat{t} = t-\tau_1$ is the time as measured by the receiver clock.
Thus we see that the second signal is delayed 
with respect to the former by $t_0+\tau_2-\tau_1$, that is,
the delay introduced at the transmitter plus the difference in the
delays between the two paths, and this can be expressed as a phase
shift of $\hat{\theta}= \frac{t_0+\tau_2-\tau_1}{\omega}$.


If $\tau_1 = \tau_2$ so that the
two paths are of identical lengths, the second signal
looks like a multipath signal with respect to the first with a delay
of $t_0$ (equivalently the same phase shift $\theta$ as existed at
the transmitter).
If $t_0 = 0$ so that effectively only one signal is
transmitted, but is received over paths of different lengths,
the second signal is a multipath signal with respect to the first
with a delay of $\tau_2-\tau_1$ and it too can be expressed
as a phase shifted version of the first signal.