# Relativistic RMS Speed

by Vodkacannon
Tags: relativistic, speed
 P: 38 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. $v = \sqrt{\frac{3KT}{m}}$ How could we manipulate this equation?
 Sci Advisor P: 5,942 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.
 P: 38 Yes that may have been a more appropriate question for me to ask in hopes of connecting relativity and kinetic gas theory
Mentor
P: 11,255

## Relativistic RMS Speed

 Quote by Vodkacannon How could we manipulate this equation?
We don't.

Instead, we start from scratch and re-derive the Maxwell speed distribution using relativistic kinetic energy instead of non-relativistic 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.
P: 38
 Quote by jtbell We don't. Instead, we start from scratch and re-derive the Maxwell speed distribution using relativistic kinetic energy instead of non-relativistic 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.
Interesting. thanks. I had known about Maxwells Distribution laws before as it's in my AP textbook. I can see how the formula sort of resembles the plain old relativistic equations.
PF Gold
P: 4,863
 Quote by jtbell We don't. Instead, we start from scratch and re-derive the Maxwell speed distribution using relativistic kinetic energy instead of non-relativistic 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.
It's interesting that this makes no mention of the Maxwell-Juttner distribution, which should be the applicable distribution. Do you know if this analysis is consistent with that? At a glance, it doesn't look like it would be.
Emeritus
PF Gold
P: 5,500
 Quote by mathman However, increasing energy (heating particles up) close to relativistic speeds leads to increase in m rather than v.
This is the relativistic mass convention, which is very old-fashioned. Physicists today use the convention that mass is invariant, so, e.g., rather that writing p=mv with an m that is a function of v, they write $p=m\gamma v$, where m is a constant.
Thanks
P: 3,864
 The Maxwell Speed Distribution for Relativistic Speeds (PDF)
I could be wrong, but I think the formula given in this paper is incorrect. Basically he's taking the density of states to be uniform in velocity space with a sharp cutoff imposed at v=c.

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.
P: 5,942
 Quote by bcrowell This is the relativistic mass convention, which is very old-fashioned. Physicists today use the convention that mass is invariant, so, e.g., rather that writing p=mv with an m that is a function of v, they write $p=m\gamma v$, where m is a constant.
Nitpicker - what would you call $m\gamma$?
PF Gold
P: 4,863
 Quote by mathman Nitpicker - what would you call $m\gamma$?
m$\gamma$

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$\gamma$a anyone?
KE = (1/2)m$\gamma$v^2 anyone?

Another observation is that $\gamma$ is really part of the normalization of 4-velocity to be a unit 4-vector. The correct 4-force equation for constant mass particle is simply m * 4-acceleration.
P: 5,942
 Quote by PAllen m$\gamma$ 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$\gamma$a anyone? KE = (1/2)m$\gamma$v^2 anyone? Another observation is that $\gamma$ is really part of the normalization of 4-velocity to be a unit 4-vector. The correct 4-force equation for constant mass particle is simply m * 4-acceleration.
It may be old fashioned, but my understanding was that the total energy is mc2, which can be expanded as a series in v to get m0c2 +m0v2/2 + ... = rest mass energy + kinetic energy.
PF Gold
P: 4,863
 Quote by mathman It may be old fashioned, but my understanding was that the total energy is mc2, which can be expanded as a series in v to get m0c2 +m0v2/2 + ... = rest mass energy + kinetic energy.
But KE, exactly, = mc^2(γ-1) [m being rest mass] which is completely different from substituting mγ into the Newtonian formula, getting (1/2)mγv^2. The point is that there is only one relativistic analog of Newtonian kinematic formulas for which you can pretend mγ plays the role of mass. So thinking there is a general concept of relativistic mass which is analogous to the m in Newtonian formulas leads only to a large number of common errors

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