The formula you want is called Planck's Law. Copying Wikipedia:

The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it
gives off as radiation of different frequencies. It is measured in
terms of the power emitted per unit area of the body, per unit solid
angle that the radiation is measured over, per unit frequency.

$$ B_\nu(\nu, T) = \frac{ 2 h \nu^3}{c^2} \frac{1}{e^\frac{h\nu}{k_\mathrm{B}T} - 1} $$

Now to work out the total power emitted per unit area per solid angle by our lightbulb in the X-ray part of the EM spectrum we can integrate this to infinity:

$$P_{\mathrm{X-ray}} = \int_{\nu_{min}}^{\infty} \mathrm{B}_{\nu}d\nu,
$$

where $\nu_{min}$ is where we (somewhat arbitrarily) choose the lowest frequency photon that we would call an X-ray photon. Let's say that a photon with a 10 nm wavelength is our limit. Let's also say that 100W bulb has a surface temperature of 3,700 K, the melting temperature of tungsten. This is a very generous upper bound - it seems like a typical number might be 2,500 K.

We can simplify this to:

$$
P_{\mathrm{X-ray}} = 2\frac{k^4T^4}{h^3c^2} \sum_{n=1}^{\infty} \int_{x_{min}}^{\infty}x^3e^{-nx}dx,
$$

where $x = \frac{h\nu}{kT}$. wythagoras points out we can express this in terms of the incomplete gamma function, to get

$$
2\frac{k^4T^4}{h^3c^2}\sum_{n=1}^{\infty}\frac{1}{n^4} \Gamma(4, n\cdot x)
$$

Plugging in some numbers reveals that the n = 1 term dominates the other terms, so we can drop higher n terms, resulting in

$$
P \approx 10^{-154} \ \mathrm{Wm^{-2}}.
$$

This is **tiny**. Over the course of the lifetime of the universe you can expect on average no X-Ray photons to be emitted by the filament.

More exact treatments might get you more exact numbers (we've ignored the surface area of the filament and the solid angle factor for instance), but the order of magnitude is very telling - there are no X-ray photons emitted by a standard light bulb.

$$ [P_i,L_j]=\varepsilon_{jkl} ([P_i, X_k]P_l+X_k[P_i,P_l])=-\mathrm{i}\hbar \varepsilon_{jkl}\delta_{ki}P_l = -\mathrm{i} \varepsilon_{jil} P_l = \mathrm{i}\varepsilon_{ijl} P_l$$

Which shows that as expected $P_k$ is a vector.

The lesson here is that you should not use the same letter twice as an external index and a dummy index. Here $i, j$ are external indices, so when writing the cross product with implied sum you should use something like $L_j=\sum_{kl}\varepsilon_{jkl} X_kP_l=\varepsilon_{jkl} X_kP_l$.

## Best Answer

Newton's second law states the force on a system is proportional to its acceleration. For linear restoring force $-kx$, we have

$$F = ma$$ $$ -kx = m\ddot{x}$$

where $\ddot{x}$ is the second time-derivative of position, i.e. acceleration. Then you have

$$ \ddot{x} = a = -\frac{k}{m}x = -\omega^2 x$$

where we define $\omega = \sqrt{\frac{k}{m}}$.