What are e and V in exponent of the equation I = I0(eeV/kT - 1)? is it really one variable "eV", as in electron volts, or is e just 2.718... and V for volts?
It's confusing. The first e is just the base of the natural logarithms (2.718...), but the e in the exponent is the charge on the electron. I prefer to write it as:
e^\frac{qV}{kT} to eliminate this confusion. Note that if you measure kT in electron volts (eV), then the electron charges cancel, and you can just use the voltage V directly. Note that at room temperature, kT is .026 eV, so you can write:
I = I_0 e^\frac{V}{.026}
Hi,
I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem.
Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$
Where ##b=1## with an orbit only in the equatorial plane.
We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$
Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units,
According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##,
## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units.
So is this conversion correct?
Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?