# Proof of relativistic energy/mass to grow to infinity for v approaching c?

by Vincentius
Tags: kaufmann, relativistic energy, relativistic mass
 P: 30 By SR, relativistic energy and mass are proportional to the Lorentz factor γ, therefore, grow to infinity for v$\rightarrow$c. This relationship for the relativistic mass has been confirmed by the Kaufmann experiment and its successors, via measuring the deflection of high velocity electrons by an electric and/or magnetic field. My question is: is this really unambiguous proof of energy/mass increase? Or, could as well the effective electric (and/or magnetic) deflection force on the electron go to zero for v$\rightarrow$c, such that the inertial mass (and energy) only appears to increase?
 Mentor P: 16,298 Hi Vincentius, welcome to PF! Generally, the concept of relativistic mass has been discarded by modern physicists. However, from particle accelerator experiments it is abundantly clear that the total energy increases to infinity for v->c. Or, if not to infinity, then at least for as high energy as we have been able to probe thus far. If it were just a decrease in the Lorentz force then there wouldn't be enough energy to make the reaction products that we observe.
 P: 30 Thanks DaleSpam. Yes, I know the particle accelerators run on high energy to make the reaction products. However, if an electron is being accelerated over a potential of 1 V, then the electron is considered to have acquired 1eV of energy in kinetic energy. But does this still hold at relativistic velocities? How can we be sure the effective force of the accelerating field isn't falling off with the velocity of the electron (like the deflection force in my original question)? On the basis of which experimental result can this possibility be ruled out?
Mentor
P: 11,052

## Proof of relativistic energy/mass to grow to infinity for v approaching c?

 Quote by Vincentius However, if an electron is being accelerated over a potential of 1 V, then the electron is considered to have acquired 1eV of energy in kinetic energy. But does this still hold at relativistic velocities?
Particle accelerators accelerate particles in many small steps. If the potential difference ΔV in the later steps didn't produce qΔV in additional kinetic energy, it would be pretty obvious at the end, after the last step.
P: 30
 Quote by jtbell Particle accelerators accelerate particles in many small steps. If the potential difference ΔV in the later steps didn't produce qΔV in additional kinetic energy, it would be pretty obvious at the end, after the last step.
Do you mean the energy of the particle after the last step is somehow measured, other than by adding up all ΔV's?
Mentor
P: 16,298
 Quote by Vincentius Thanks DaleSpam. Yes, I know the particle accelerators run on high energy to make the reaction products. However, if an electron is being accelerated over a potential of 1 V, then the electron is considered to have acquired 1eV of energy in kinetic energy. But does this still hold at relativistic velocities?
Yes.

 Quote by Vincentius How can we be sure the effective force of the accelerating field isn't falling off with the velocity of the electron (like the deflection force in my original question)? On the basis of which experimental result can this possibility be ruled out?
As I said above, the observed reaction products are more massive and energetic than you could get otherwise.
Mentor
P: 11,052
 Quote by Vincentius Do you mean the energy of the particle after the last step is somehow measured, other than by adding up all ΔV's?
Yes. For example, you can smack the particle(s) into a target to bring them to a stop, and then measure how much the target warms up (a calorimeter). This assumes the target is big and dense enough that it can absorb any secondary particles produced by interactions of the incoming particles inside the target.

Or you send the particles through a magnetic field and measure the curvature of their path, which gives you the momentum. From the momentum, you get the energy from

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

and then the kinetic energy from $K = E - m_0 c^2$.
P: 30
 Quote by jtbell Yes. For example, you can smack the particle(s) into a target to bring them to a stop, and then measure how much the target warms up (a calorimeter). This assumes the target is big and dense enough that it can absorb any secondary particles produced by interactions of the incoming particles inside the target. Or you send the particles through a magnetic field and measure the curvature of their path, which gives you the momentum. From the momentum, you get the energy from $$E^2 = (pc)^2 + (m_0 c^2)^2$$ and then the kinetic energy from $K = E - m_0 c^2$.
Thanks jtbell! Using a calorimeter, I suppose, also requires measurement of the number of incoming particles, right? Or isn't it a true temperature measurement, but some indirect method?
On the second method: I understand the principle, but this deflection method raises the exact same question of my original post.
P: 6
 Quote by Vincentius My question is: is this really unambiguous proof of energy/mass increase? Or, could as well the effective electric (and/or magnetic) deflection force on the electron go to zero for v$\rightarrow$c, such that the inertial mass (and energy) only appears to increase?
Try to elaborate a theory that instead of using Relativistic Newton's Second Law of Motion (1) and Lorentz force (2) uses Classical Newton's Second Law of Motion and that modified weak Lorentz force that you suspect might act on fast moving particles.

(1) http://en.wikiversity.org/wiki/Special_relativity
(2) http://en.wikipedia.org/wiki/Mass-to-charge_ratio

The predictions of you theory should fit the experimental results!
Mentor
P: 16,298
 Quote by Vincentius On the second method: I understand the principle, but this deflection method raises the exact same question of my original post.
If the Newtonian KE formula were correct then highly relativistic electron collisions would max out at about 1.5 MeV, which could never produce muons with a mass of over 100 MeV. You simply cannot explain that based on weakening EM forces.
 P: 6 Do two relativistic electrons form a non relativistic 100 MeV muon? http://en.wikipedia.org/wiki/Muon If two relativistic electrons form a relativistic muon then a mass spectrometer will behave as if the electrons and muon have large energies when in reality maybe just the Lorentz force is weaker at high speeds. Even if the electrons and muons have classical kinetic energies at relativistic speeds, if Lorentz force weakens then they will appear deviated as if they had relativistic energies and Lorentz force were classical. However, this theory of weak Lorentz forces seen by high speed particles likely leads to contradictions or huge complications. Likely physicists took it into account but something did not work.

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