Energy-momentum tensor for a relativistic system of particles

Click For Summary
The discussion focuses on deriving the energy-momentum tensor for a system of non-interacting relativistic particles. The correct Lagrangian for such a system is the sum of the individual Lagrangians, expressed as L = ∑i Li, where Li = -m0i/γi. It is clarified that the total energy-momentum tensor is also the sum of the individual tensors, Tμν = ∑i Tiμν. The initial confusion regarding the covariant formalism in the Lagrangian is addressed by emphasizing the need to consider each particle's contributions separately. Understanding these relationships is crucial for correctly applying the principles of relativistic systems.
Frostman
Messages
114
Reaction score
17
Homework Statement
Construct the energy-momentum tensor for a relativistic system of non-interacting particles and show explicitly that it is conserved.
Relevant Equations
##T^{\alpha\beta}=\frac{\partial L}{\partial \varphi/_\alpha}\varphi/^\beta-g^{\alpha\beta}L##
##T^{\alpha\beta}/_\alpha=0##
I think it is quite simple as an exercise, following the two relevant equations, but at the beginning I find myself stuck in going to identify the lagrangian for a relativistic system of non-interacting particles.
For a free relativistic particle I know that lagrangian is:
$$L=-\frac{m_0}{\gamma}$$
But for a system of non-interacting particles I can use this one?
$$L=\sum_i-\frac{m_{0i}}{\gamma}$$
But when I step to energy-momentum tensor I don't have any covariant formalism in this lagrangian. Somebody can help me?
 
Physics news on Phys.org
A:No, the Lagrangian for a system of non-interacting particles is not just the sum of the Lagrangians for each individual particle. The Lagrangian for a system of non-interacting particles is simply the sum of the individual particle's Lagrangians.$$L = \sum_i L_i$$where$$L_i = -\frac{m_{0i}}{\gamma_i}$$The total energy-momentum tensor for such a system is then given by$$T^{\mu \nu} = \sum_i T_i^{\mu \nu},$$where $T_i^{\mu \nu}$ is the energy-momentum tensor of particle $i$.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

Similar threads

Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
7K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
702
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K