Time integral of worldsheet current

[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

 

[MathematicalPhysicist]

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

 

[ShayanJ]

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

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.

 

[nrqed]

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.

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

 

[ShayanJ]

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

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?

 

[nrqed]

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?

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.

 

[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.