Standard boost, particles with mass M > 0

• I
• kent davidge
In summary, Weinberg discusses the calculation of the rotation W(Λ,p) for particles with mass M > 0 and the Little Group of SO(3). He explains the need for a standard boost L(p) to map the four momentum from kμ = (0,0,0,M) to pμ and shows the expressions for the components of L(p). The choice of a specific set of expressions for (L(p))i j is not important, but the most convenient option is to use a rotation free Lorentz boost with velocity ##\vec{p}/p^0=\vec{p}/\sqrt{M^2+\vec{p}^2}##, as described in appendix B of his

kent davidge

Weinberg considers (p.68 QFT Vol. 1) particles with mass M > 0. The Little Group is SO(3). He wants to calculate the rotation
W(Λ,p) ≡ L-1(Λp) Λ L(p). He says that for this we need to choose a standard boost L(p) which carries the four momentum from
kμ = (0,0,0,M) to pμ. He then shows the expressions for the components of L(p). What I don't understand is that because the spatial components of kμ all vanish, i.e. ki = 0, why does it matter to choose a specific set of expressions for (L(p))i j?

You can choose any Lorentz transformation that maps ##k^{\mu}## into ##p^{\mu}##. The most convenient (standard) choice is to use the uniquely defined rotation free Lorentz boost, i.e., a boost with velocity ##\vec{p}/p^0=\vec{p}/\sqrt{M^2+\vec{p}^2}##.

For more details have a look at appendix B of my QFT manuscript:

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

kent davidge and dextercioby

Well, the English is a desaster. It was written when I just started to study QFT a long time ago...

vanhees71 said:
Well, the English is a desaster. It was written when I just started to study QFT a long time ago...

1. What is the definition of "Standard boost"?

Standard boost refers to a mathematical transformation used in physics to describe the motion of particles in different reference frames. It involves changing the coordinates of an object based on its velocity relative to an observer.

2. How is the mass of particles with M > 0 calculated?

The mass of a particle with M > 0 is typically calculated using the formula E=mc^2, where E is the energy of the particle and c is the speed of light. This equation takes into account the mass-energy equivalence described in Einstein's theory of relativity.

3. Are there any limitations or exceptions to the concept of "Standard boost"?

While the concept of Standard boost is a useful tool in describing the motion of particles, there are some limitations and exceptions. For example, it does not apply to particles traveling at speeds close to the speed of light, and it does not take into account the effects of gravity.

4. How does the mass of a particle affect its behavior under a Standard boost?

The mass of a particle plays a significant role in its behavior under a Standard boost. Heavier particles require more energy to accelerate and will experience a greater change in momentum compared to lighter particles under the same boost. This is known as the mass-energy equivalence principle.

5. Can the Standard boost be applied to particles with zero mass?

No, the Standard boost transformation only applies to particles with non-zero mass. Particles with zero mass, such as photons, follow different mathematical transformations, such as the Lorentz transformation, to describe their motion in different reference frames.