# What does the |-sign mean in this text?

1. Nov 25, 2014

### Nikitin

http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Exam_tfy4240_Dec_2013.pdf
solutions: http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Solution_tfy4240_Dec_2013.pdf

In problem 3 a) I'm supposed to reform the integrand from $\vec{J}(r',t_r)$ to something like $\vec{J}(r'|\omega)$. In practice, one just extracts the $t_r$ dependent part of the function and writes what remains as $\vec{J}(r'|\omega)$. But what does my professor mean with that $"|"$ bar? Why not just write the remaining part of the function as $\vec{J}(r')$?

And excuse me if this is physics notation and not math notation.. In that case pls move this thread to the physics section!

Last edited: Nov 25, 2014
2. Nov 25, 2014

### Doug Huffman

The most general meaning that I know is "evaluated at" or, with super and subscripts, evaluated between. I know it best as indicating conditional probability.

3. Nov 25, 2014

### Nikitin

I don't think it's evaluated at as the $\omega$ dependent part was taken out I think. oh well i guess I'll have to bother my prof and send him an email

4. Nov 25, 2014

### Staff: Mentor

I would interpret that to mean that $\vec{J}$ depends parametrically on $\omega$.

5. Nov 27, 2014

### Nikitin

That cannot be correct in my specific example. I think..

6. Nov 27, 2014

### Staff: Mentor

Why?

7. Nov 27, 2014

### Nikitin

Because in my example the only dependence $\vec{J}(\vec{r},t)$ has on $\omega$ is thru $e^{i \omega t}$, and $\vec{J}(\vec{r},t) = \vec{J}(\vec{r}|\omega) ~e^{i \omega t}$ according to my professor. This is physics stuff though, not math.

8. Nov 27, 2014

### Staff: Mentor

That is not correct. You have that "the only dependence $\vec{J}(\vec{r},t)$ has on $t$ is thru $e^{i \omega t}$." This form for $\vec{J}(\vec{r},t)$ is to give it an explicit dependence on $t$, but it certainly doesn't mean that the prefactor is independent of $\omega$.

9. Nov 27, 2014

### Nikitin

How can that be the case? $\vec{J}(\vec{r},t)$ is given by $I(t) = I_0 e^{i \omega t}$ in the problem.

10. Nov 28, 2014

### Nikitin

He responded to my mail; it means the same as conditional probability, "r given omega " in this case.

11. Dec 1, 2014

### Staff: Mentor

That's very unconventional. But it basically amounts to what I was saying: it means that $\vec J$ depends parametrically on $\omega$.