Register to reply 
Relativistic RMS Speed 
Share this thread: 
#1
Jan313, 03:05 PM

P: 39

How can we restrict the RMS speed of molecules to comply with relativity?
They obviously can't go at or faster than the speed of light. If you are dealing with particles inside of very hot stars for example you may get very high erroneous speeds. [itex]v = \sqrt{\frac{3KT}{m}}[/itex] How could we manipulate this equation? 


#2
Jan313, 03:12 PM

Sci Advisor
P: 6,070

I am not familiar with the equation you gave. However, increasing energy (heating particles up) close to relativistic speeds leads to increase in m rather than v.



#3
Jan313, 03:23 PM

P: 39

Yes that may have been a more appropriate question for me to ask in hopes of connecting relativity and kinetic gas theory



#4
Jan313, 03:29 PM

Mentor
P: 11,778

Relativistic RMS Speed
Instead, we start from scratch and rederive the Maxwell speed distribution using relativistic kinetic energy instead of nonrelativistic kinetic energy. This appears to have what you are looking for: The Maxwell Speed Distribution for Relativistic Speeds (PDF) See equation 21 on page 4. 


#5
Jan313, 03:41 PM

P: 39




#6
Jan313, 03:48 PM

Sci Advisor
PF Gold
P: 5,059




#7
Jan313, 03:53 PM

Emeritus
Sci Advisor
PF Gold
P: 5,597




#8
Jan313, 04:58 PM

Sci Advisor
Thanks
P: 4,160

The correct approach is to use a uniform density in momentum space. For nonrelativity it doesn't matter since p = mv is linear, but it does matter in our case. In momentum space there is no cutoff, one integrates over all p out to infinity. 


#9
Jan413, 04:00 PM

Sci Advisor
P: 6,070




#10
Jan413, 04:10 PM

Sci Advisor
PF Gold
P: 5,059

The reason I discourage calling it a mass is the temptation for those learning SR to substitute it into Newtonian formulas. There is only one common formula (momentum) for which this works. F = m[itex]\gamma[/itex]a anyone? KE = (1/2)m[itex]\gamma[/itex]v^2 anyone? Another observation is that [itex]\gamma[/itex] is really part of the normalization of 4velocity to be a unit 4vector. The correct 4force equation for constant mass particle is simply m * 4acceleration. 


#11
Jan513, 02:20 PM

Sci Advisor
P: 6,070




#12
Jan513, 03:09 PM

Sci Advisor
PF Gold
P: 5,059

[edit: Thinking more, I see the main reason for the shift is the wide adoption of 4vectors for SR. In this scheme you have mU. for 4momentum; m is rest mass, U is 4 velocity. No mγ in sight (it is internal to the 4velocity). Then, 4force is naturally mA, using 4acceleration; no nonsensical transverse and longitudinal relativistic mass as seen in some ancient relativity books.] 


#13
Jan713, 04:08 PM

Sci Advisor
P: 6,070




#14
Jan913, 04:36 AM

P: 3,187




Register to reply 
Related Discussions  
Speed of a Relativistic Particle?  Introductory Physics Homework  1  
Ender's Relativistic Speed  Special & General Relativity  11  
Speed of Pi Meson  Relativistic question  Introductory Physics Homework  5  
What happens after you travel at a relativistic speed  Special & General Relativity  15  
Relativistic speed  Advanced Physics Homework  4 