Understanding Relativistic Momentum in Special Relativity

DRC12
Messages
41
Reaction score
0
Last week in physics we were learning about special relativity and we got the equation p=λmv. When writing the equation the teacher also put the equations for regular momentum p=mv and regular force F=ma. I noticed that momentum is the integral of force where v is the integral of a and mass is constant. The problem is if momentum is the integral of force then how is the relativistic momentum derived is mass isn't constant anymore
 
Physics news on Phys.org
In the formula for relativistic momentum, p=λmv, m is the invariant mass and is constant.


[Edit: That should be γ not λ.]
 
Last edited:
but if m is constant an object could theoretically go faster then the speed of light all you need is to supply a constant force on it and it would accelerate forever the changing m making it harder to accelerate the object once it starts going too fast. As m increases the force does less in less until it can't accelerate no matter the force on it as in the mass eventually reaches infinity
 
DRC12 said:
but if m is constant an object could theoretically go faster then the speed of light all you need is to supply a constant force on it and it would accelerate forever the changing m making it harder to accelerate the object once it starts going too fast. As m increases the force does less in less until it can't accelerate no matter the force on it as in the mass eventually reaches infinity
Note that I said that m is constant. But λm (sometimes called the 'relativistic mass') is not constant.

[Edit: That should be γ not λ.]
 
Last edited:
DRC12 said:
but if m is constant an object could theoretically go faster then the speed of light all you need is to supply a constant force on it and it would accelerate forever the changing m making it harder to accelerate the object once it starts going too fast. As m increases the force does less in less until it can't accelerate no matter the force on it as in the mass eventually reaches infinity

There are two different conventions.

An older convention is to consider mass as changing. Then F=ma and p=mv, and a constant force produces less and less acceleration as you approach the speed of light, because m is increasing.

The more common convention these days is to consider mass as constant. Then F=m\gamma a, F=dp/dt, and p=m\gamma v. Now a constant force produces a decreasing acceleration because of the factor of gamma. (If your teacher is using lambda instead of gamma, that would be an unusual notation.)

You can't mix the two systems.
 
DRC12 said:
but if m is constant an object could theoretically go faster then the speed of light all you need is to supply a constant force on it and it would accelerate forever the changing m making it harder to accelerate the object once it starts going too fast. As m increases the force does less in less until it can't accelerate no matter the force on it as in the mass eventually reaches infinity

You can show that a constant force would take an infinite amount of time to accelerate a massive object to the speed of light. In fact,

\int_0^T F dt = \int_0^c \frac{m dv}{(1-\frac{v^2}{c^2})^{3/2}} = mc \lim_{v\rightarrow c} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}},

which diverges. The LHS gives FT so T must be infinite.
 
bcrowell said:
(If your teacher is using lambda instead of gamma, that would be an unusual notation.)
He did use gamma i just mixed them up
 
DRC12 said:
He did use gamma i just mixed them up
And I didn't even notice--in my mind I was thinking gamma (γ) not lambda (λ). :redface:

Sorry about that!
 
Doc Al said:
And I didn't even notice--in my mind I was thinking gamma (γ) not lambda (λ). :redface:

Sorry about that!

No problem I'd been looking at light waves earlier today and had been using lambda as frequency and used it without thinking
 
  • #10
Thanks to everyone I understand it now I didn't think about how gamma wasn't a constant or how gamma was expressing the change in the rest mass and I wasn't thinking of it as rest mass
 
Back
Top