Cant get units to work out for Momentum Density

In summary, the conversation discusses the confusion regarding the units of momentum density in the context of the Nambu-Goto Action and the Lagrangian density in Zweibach's "A First Course in String Theory." The author clarifies that the Lagrangian density has units of kg m^2/s^2, while the momentum density has units of kg m/s. The conversation also mentions the concept of integrating over a missing variable to obtain the correct units for a quantity.
  • #1
benbenny
42
0
Excuse the question but I am 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

[tex] \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} [/tex]

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

[tex] p_\mu(\tau) = \int ^{\sigma _{1}}_{0} \mathcal{P}^{\tau}_{\mu}(\tau , \sigma) d\sigma [/tex]

Im confused because I would expect [tex] \mathcal{P}^{\tau}_{\mu} [/tex] 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 [tex] \mathcal{P}^{\tau}_{\mu} [/tex] 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
 
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  • #2
benbenny said:
Im confused because I would expect [tex] \mathcal{P}^{\tau}_{\mu} [/tex] 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 [tex] \mathcal{P}^{\tau}_{\mu} [/tex] 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.
 
  • #3
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!
 
  • #4
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 [itex]\mathcal{L}[/itex] 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!
 
  • #5
Yes! thanks for pointing out the page in the book. I've 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.
 

What is momentum density?

Momentum density is a physical quantity that describes the amount of momentum per unit volume of a substance. It is typically denoted as p or ρp, where p is the momentum and ρ is the density of the substance.

Why is momentum density important?

Momentum density plays a critical role in understanding the motion and behavior of matter. It is used in various fields such as fluid dynamics, solid mechanics, and quantum mechanics to describe the movement and interactions of particles and systems.

How is momentum density calculated?

Momentum density is calculated by taking the product of the velocity and mass of a particle or system. In mathematical terms, it can be expressed as p = mv, where p is the momentum, m is the mass, and v is the velocity.

What are the units for momentum density?

The units for momentum density depend on the system of measurement being used. In the International System of Units (SI), momentum density is measured in kilograms per cubic meter (kg/m³). In the British system, it is measured in slugs per cubic foot (slug/ft³).

Why might units not work out for momentum density?

Units may not work out for momentum density if the values used in the calculation are not consistent. For example, if the mass is given in kilograms but the velocity is given in miles per hour, the resulting momentum density may not have a consistent unit. It is important to convert all values to the same unit before performing calculations.

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