How Do You Calculate the Magnitude of a 4-Vector?

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SUMMARY

The magnitude of a 4-vector, specifically the momentum-energy 4-vector of a photon, is calculated using the formula \( (P_\mu P^\mu)^{1/2} \). In the case of a photon traveling in the positive x-direction, represented as \( P_\mu = (E, E, 0, 0) \), the computation yields \( (E^2 - E^2)^{1/2} = 0 \). This result is accurate because the photon is massless, confirming that its 4-momentum magnitude is indeed zero.

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  • Understanding of 4-vectors in physics
  • Familiarity with the concept of four-momentum
  • Basic knowledge of special relativity
  • Mathematical proficiency in tensor notation
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  • Study the properties of massless particles in special relativity
  • Learn about the implications of 4-momentum in particle physics
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Jonsson
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Hello there,

Given any 4-vector ##x = (x_0,x_1,x_2,x_3)##, hvor do I compute its magnitude? I couldn't find a simple explanation online.

For example I want to compute the magnitute of the momenergy 4-vector of a photon traveling in the positive x-direction where ##E## is the energy of the photon and ##P_\mu = (E,E,0,0)##. What is incorrect in the following computation of its magnitude?
$$
(P_\mu\,P^\mu)^{1/2} = (E^2-E^2)^{1/2} = 0
$$

Surely the magnitude of the vector should not be ##0##? Can you help? Thank you for your time.

Kind regards,
Marius
 
Physics news on Phys.org
It is 0. You did it correctly. The magnitude of the four momentum is the mass and light is massless.
 

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