# Cant get units to work out for Momentum Density

Excuse the question but im stuck on this for the past hour.

Im studying Zweibach's "A first Course in String Theory".
From the Nambu-Goto Action he derives the wave equation which consists of the term

$$\frac{\partial \mathcal{L}}{\partial \dot{X} ^{\mu}} &= -\frac{T_{0}}{c}\frac{\dot{X} \cdot X')X'_{\mu}-(x')^{2}\dot{X}_{\mu}} {\sqrt{(\dot{X} \cdot X')^{2} - (\dot{X})^{2}(X')^{2}}} \equiv \mathcal{P}^{\tau}_{\mu}$$

Now this $$\mathcal{P}^{\tau}_{\mu}$$ is referred to as the momentum density and is used as such later on in the book:

$$p_\mu(\tau) = \int ^{\sigma _{1}}_{0} \mathcal{P}^{\tau}_{\mu}(\tau , \sigma) d\sigma$$

Im confused because I would expect $$\mathcal{P}^{\tau}_{\mu}$$ to have units of Mass divided by time, if \sigma has units of length, or units of momentum, if \sigma is dimensionless, so that when you integrate over \sigma you get units of momentum.

But as far as I can see $$\mathcal{P}^{\tau}_{\mu}$$ has units of mass times length. I get would think this becuase the lagrangian density has units of ML^2/T and it is differentiated with respect to velocity, thus it takes units of ML.

Can someone help with this?

Thanks.

B

## Answers and Replies

Im confused because I would expect $$\mathcal{P}^{\tau}_{\mu}$$ to have units of Mass divided by time, if \sigma has units of length, or units of momentum, if \sigma is dimensionless, so that when you integrate over \sigma you get units of momentum.

But as far as I can see $$\mathcal{P}^{\tau}_{\mu}$$ has units of mass times length. I get would think this becuase the lagrangian density has units of ML^2/T and it is differentiated with respect to velocity, thus it takes units of ML.

Can someone help with this?

Thanks.

B

The author is following the common mechanical notation that the momentum is equal to the velocity derivative of the Lagrangian. However, you error is that the Lagrangian has units of kg m^2/s^2 (not kg m^2/s), so taking the velocity derivative divides those units by m/s giving kg m/s, which are the correct units of momentum.

Thanks a lot for your reply!

Then what makes it a momentum density rather than the momentum itself?
I was under the impression that when a quantity is the "density" of something, that means that its units are lacking, but become the units of that something once we integrate over that missing variable.
Best example is mass density which is ks/m. and when we integrate it over a variable measured in m than we get back mass in kg.

can you explain this further to me?

Much appreciated!

I haven't studied String Theory, but I pulled that book out from my University's library and just skimmed a couple pages. It seems your answer lies on page 73, Equation (4.34) and (4.35). The item $\mathcal{L}$ is the Lagrangian density because you have yet to integrate over space to get the Lagrangian.
So I guess technically there should be an extra unit of length in what I said previously, though the idea/concept of the momentum being the velocity-derivative of the Lagrangian still holds true.

Hope that helps!

Yes! thanks for pointing out the page in the book. Ive forgotten that this whole section in the book exists that explains all of this specifically. Needed someone from the other side of earth to remind me.