Does the self inductance and velocity factor of the elements of a dipole antenna change its resonant frequency?

Question

Below are well known equations from Wikipedia for the resistance and reactance in the impedance of a dipole antenna.

where a is the radius of the conductors, k is the wave number 2πf/c, η0 denotes the impedance of free space = 377Ω, and γe is Euler's constant = 0.57721566.

Wikipedia doesn't specify where the feed point is, but since the feed point impedance is a function of the position of the feed point along the antenna, and there are no variable terms in the equations for feed point position, I assume the equations are for a feed point in the center.

It seems there are no terms anywhere in the equations for anything related to the self inductance of the elements, or related to the fact that radio waves travel slower in metal than in air.

Every one knows that a center fed dipole exactly 1/2 λ in length has about + 45 Ω of reactance at the feed point impedance so to make it resonant it has to be shortened about 5 %.

I proved this by writing a program using Microsoft Visual Studio .NET which allows me to input different values for frequency, length and radius and which then calculates R and X using these equations. The program showed that a dipole which is exactly 1/2 λ in length does in fact have + 45 Ω of reactance in the feed point impedance. I also added the reactance caused by self inductance and subtracted that from the rest of the reactance to see if they are the same, which they are not except for one value of radius.

It's undeniable that the presence of self inductance in the elements of a dipole will affect the reactance in the feed point impedance, and similarly the velocity factor of the metal will affect the resonant length and so also affect the feed point reactance, so it seems strange that the above equation for reactance doesn't appear to include terms for these factors.

In order to be resonant, a dipole must have to be shortened also to allow for self inductance of the elements and for the fact that the radio waves travel slower in the elements than in air, right?

So considering the above equations, does the self inductance and velocity factor of the elements of a dipole change its resonant frequency?

Addition. -----------------------------------------------------------

I suppose what i'm trying to point out is this.

Either one of the two following statements must be true.

1. The above equations are correct and take into account the self inductance and velocity factor of the dipole elements, and in this case the extra + 45 Ω of reactance present in the impedance for a dipole of exactly 1/2 λ in length can be attributed at least in part to those characteristics.

2. The above equations assume an ideal dipole with no self inductance and a velocity factor of 1, and so the reactance caused by these characteristics for a real dipole is present in addition to the + 45 Ω and the cause for this rogue 45 Ω to me remains a mystery.

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Answer 1

The wikipedia article states that this approximate impedance is calculated assuming a sinusoidal current distribution on a centre-fed dipole.

There are no fields in the conductors as this analysis assumes perfect conductors.

This formula is an approximation, and the term

\$Ci(\frac{2ka^2}{L}) \$

which goes as \$log(\frac{2ka^2}{L}) \$ as \$a \rightarrow 0\$

and will dominate the reactance as \$a \rightarrow 0\$ for all element lengths not \$ \lambda/2\$, and the fact that it doesn't have any affect at precisely \$ \lambda/2\$ I think should be seen as a limitation of the approximation.

Edit:

1. This is an approximation. Read the references as to the approximations used in both the assumed current distribution and in constructing the integral equation.
2. In most antenna work with good conductors there are minimal fields in the conductors. The reactance that seems to concern you is still there with perfect conductors, which have no fields in the conductors.
3. The generally accepted intuitive answer as to why a half-wave dipole resonates slightly shorter than half a wavelength is the capacitive effect of the wire ends. For discussion on this, google 'dipole end effect' (no quotes)

Apart from that I'm going to leave you to the guys over at amateur radio stack exchange where you have posted the identical question.

amateur radio stack exchange question