Exercise 26 in Schutz's First course in GR

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SUMMARY

The discussion focuses on calculating the energy required to accelerate a particle with rest mass from speed v to speed v + δv, where δv is significantly smaller than v. The participant initially uses the 4-momentum approach but realizes a mistake in their calculations. The correct expression for energy is identified as E = γ(v + δv)mc², where γ is the Lorentz factor. The conclusion drawn is that as the speed approaches the speed of light (c), the energy required diverges to infinity.

PREREQUISITES
  • Understanding of special relativity concepts, particularly 4-momentum.
  • Familiarity with the Lorentz factor (γ) and its implications in relativistic physics.
  • Knowledge of Taylor series expansions and their applications in physics.
  • Basic principles of energy-mass equivalence as described by Einstein's theory.
NEXT STEPS
  • Study the derivation and implications of the Lorentz factor (γ) in special relativity.
  • Learn how to apply Taylor series expansions in physics problems, particularly in relativistic contexts.
  • Explore the concept of relativistic momentum and its relationship to energy calculations.
  • Investigate the implications of approaching the speed of light on energy requirements in particle physics.
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Students of physics, particularly those studying special relativity, as well as educators and anyone interested in the energy dynamics of particles at relativistic speeds.

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Homework Statement


The question as follows:

Calculate the energy that is required to accelerate a particle of rest mass ##m\ne 0## from speed ##v## to speed ##v+\delta v## (##\delta v \ll v##).
Show that it would take an infinite amount of energy to accelerate the particle to the speed of light.

Homework Equations

The Attempt at a Solution



Here's what I have done so far, the 4-momentum before is ##(m,mv)## and the 4-momentum after is: ##(E,m(v+\delta v))##, the square of the 4 momentum is conserved, i.e:
$$E^2 - m^2(v+\delta v)^2 = m^2-m^2v^2$$

After rearranging I get the following equation for the energy:

$$E=mv\sqrt{1/v^2+2\delta v /v +(\delta v/v)^2}$$

I think I have a mistake somewhere, since I don't know how to expand this in a Taylor series, I have the expansion ##\sqrt{1+x} \approx 1+1/2 x##, but here I have ##1/v^2## inside the sqrt.

Perhaps I am wrong with the 4-momentum or something else, any tips?

Thanks in advance.

[/B]
 
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I think I got it wrong it should be ##E=\gamma(v+\delta v)mc^2##, and $$\gamma(v+\delta v) \approx 1+1/2 (v+\delta v)^2$$

For ##v +\delta v = c$, we get the E diverges.
 

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