Some rf aspects that could be taken into account.
- Summary: no considerable impact
- but feel free to read the long version below
Check the answers that have covered the important physical and bandwidth aspects already.
1. Skin effect
rf current concentration at the surface are described by the skin effect.
As a rule of thumb the material thickness should be $>5* skin \, depth$.
Then the majority of the concentrated current (skin effect) is within the available material.
- current density follows a $e^{-z/\delta}$ relation, so with $z=5*\delta$ that's $<1\%$
- thicker material has no more considerable impact on losses
- for thinner material losses increase
Skin depth:
$\delta \, [m]= \sqrt{\frac{2\rho}{\omega\mu}}$
$\delta_{copper\_146MHz} = 5.4 \, \mu m$
In your example $>5* skin \, depth$ is well fulfilled, so the tube / wire has no difference.
2. Losses
By considering a two-wire transmission line analytical formulas provided by rf textbooks can be applied to assess the losses impact.
For sure that two-wire line isn't the same as a dipole antenna, but it can be used to provide some perspective without electromagnetic simulation.
Surface resistance: $R_s = \sqrt{\frac{\omega\mu\rho}{2}} = 3.1 \, m\Omega$
Series resistance in unit length of a two-wire line: $R' [\Omega/m]=\frac{R_s}{\pi * radius}$
- 19mm tube: $R'=104 \, m\Omega/m$
- 2mm wire: $R'= 992 \, m\Omega/m$
There is a considerable relative difference, but absolute it's still small especially concerning that not even a full wavelength is applied on a $\lambda/2 \, dipole$ (compared to a transmission line that's usually ${>> \lambda}$)
- For effectiveness that consideration also applies.
2.1 Putting it into perspective
With an arbitrary made up two-wire line example.
- assumptions: air with 0 conductane, 2mm wire, d>>a simplifaction
distance: $d=0,01m$
radius: $a=0,001m$ (2mm wire)
$j\omega C' = \frac{\pi \epsilon_0}{ln \,d/a}=0,011 \frac{F}{m}$
$j\omega L' = \frac{\mu_0}{\pi} \, ln \frac{d}{a}= 845 \frac{H}{m}$
$R'= 0.992 \, \Omega/m$ (from above 2mm wire)
2.1.1 Lossless
$Z_0=\sqrt{\frac{L'}{C'}} = 276.12 \, \Omega$
$\lambda=2.0533731 \, m$
2.1.2 Lossy line
$Z_0=\sqrt{\frac{R'+j\omega L'}{G'+j\omega C'}} = 276.12 -j 0.16 \, \Omega$
$\lambda=2.0533727 \, m$
Length of line to have $0.1 \, dB$ attenuation: $3.2 \lambda$
Even the 'higher' series resistance of the 2mm wire has a negligible effect on the impedance of a two-wire line compared to the lossless case (for short lenght like the case of a $\lambda / 2 dipole$).
So the difference in between tube / wire will be also negligible.
2.2 Losses impact on radiation pattern
The losses will affect the current distribution along the dipole length and therefore the radiation pattern.
Anyway based on the previous assessments I won't expect a recognizable impact.
3. Diameter impact on radiation pattern
To assess this without electromagnetic simulation programs:
In your case the air $\lambda$ is $2.05 \, m$ and therefore even the $19 \, mm$ tube diameter is $< 1\%$ of the wavelength.
Antenna elements that are applied to impact the radiation pattern usually are in the magnitude of the wavelength ($\lambda$, $\lambda /2$, $\lambda /4$ - but not $\lambda /10$ or $\lambda /100$).
Therefore I won't expect a recognizable impact on radiation pattern for your application.