# A Time integral of worldsheet current

1. May 7, 2016

### ShayanJ

I'm reading this paper and have no problem with its calculations. But there is one little thing I'm not sure I understand. At some point, the author tries to find the amount of momentum carried across the spatial direction of the worldsheet in a time $\Delta t$ by calculating the following integral:

$\Delta P_1=\int dt \sqrt{-g} P^r_{\ \ x^1}$

where g is the induced metric on the worldsheet and r and t are the coordinates on the worldsheet which are also two of the background spacetime coordinates along with $x^1$.

My problem is, I'm not sure I understand the reason for the presence of $\sqrt{-g}$ in the integrand instead of integrating only $P^r_{\ \ x^1}$. Could anyone explain?

Thanks

2. May 12, 2016

### MathematicalPhysicist

It's the Jacobian, I guess they change variables in the integration. (I didn't read the paper).

3. May 12, 2016

### ShayanJ

Yeah, Its the Jacobian but there is no change of variable involved. The worldsheet coordinates are r and t which are also two of the coordinates of the background spacetime so no change of variable is needed.

4. May 12, 2016

### nrqed

It is necessary to make the expression reparametrization invariant. Look up for example the Lagrangian of general relativity, it is basically $\int \sqrt{-g} R$.

5. May 12, 2016

### ShayanJ

Thanks, that helps. But just to get it more clearly, I have one question.

In the case of the EH Lagrangian, We know that $\int R d^4 x$ is not a scalar because R is a scalar but $d^4 x$ is a weight $\pm 1$ scalar density, so we need to put $\sqrt{-g}$(which is a $\mp 1$ scalar density) there to make the whole thing a scalar.(I'm not sure about the weights of $d^4 x$ and $\sqrt{-g}$).

But here we're dealing with $P^r_{\ x^1}$ which is one of the components of a tensor which mixes with other components under a transformation and so is not a scalar. How should we come to the conclusion that putting $\sqrt{-g}$ besides it, makes the whole integral reparametrization invariant?

6. May 12, 2016

### nrqed

The "t" that is integrated over is not the physical time of the ambient space, it is just a parameter used to label points along the string. it is under reparametrization of that variable that the integral must be made invariant. With respect to that variable, the P is a scalar.

7. May 13, 2016

### Demystifier

What is still confusing is that $g$ is the determinant of the 2-dimensional metric (or is it?), while the integral is only 1-dimensional. Shouldn't one have $\int dt \sqrt{-g_{00}}$ instead? When the metric is diagonal, where does the additional factor $\sqrt{g_{rr}}$ comes from? Is there, perhaps, a hidden integration over $\int dr$ involved? (For instance, perhaps $P^r_1$ should really be $\int dr P^r_1$, or something like that.) In any case, the author seems to be rather sloppy in that equation.

Last edited: May 13, 2016