Relativistic Particles: Reaching Absolute Temperature?

[Somali_Physicist]

Hello fellow physicists.Recently I imagined a particle far away from everything (infinitely away) such that there is no potential energy affecting the particle.

E_p = 0
E_k = mc²δ : δ = (1-v²/c²)^-1/2
L = E_P- E_k
= -mc²δ
∂L/∂v = vE₀/c² *δ^3 where E₀ = mc²
d( ∂L/∂x)/dt = 0
∴
vE₀/c² *δ^3 = 0
mvδ³ = 0
I'm not 100% sure if next lines are correct
p = mvδ
∴ pδ² = 0
∴
p = 0 ⇒ m = 0 or v = 0
δ = 0
(1-v²/c²)^-1/2 = 0
v = c ⇒ m =0

Inconclusion a particle with zero potential is at velocity = 0 , what does this mean? a particle unmoving means absolute temperature has been reached!

v_rms = (3RT/m)^1/2
T = mv_rms*1/3R
T ⇒ 0 as E_p ⇒ 0

Other solution states
Light or massless particles travel in a purely kinetic environment. I will update this with reference to both its wave like instances as well as its photon like instances.

Please criticize this as I wish to learn.
-SomaliPhysicist.

 

[mfb]

d( ∂L/∂x)/dt = 0
∴
vE0/c2 *δ^3 = 0
mvδ³ = 0

You lost a time derivative here.
All the following equations should have a time derivative. The expressions are all constant in time, which means the velocity is constant for a single particle with no external potential. This is just inertia.

 

[Orodruin]

d( ∂L/∂x)/dt = 0
∴
vE0/c2 *δ^3 = 0

Apart from you trying to apply classical reasoning to an a priori relativistic particle, this is just wrong.

Also, the Lagrangian of a relativistic particle is not given by $E_k - E_p$. See https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechanics

 

[Somali_Physicist]

You lost a time derivative here.
All the following equations should have a time derivative. The expressions are all constant in time, which means the velocity is constant for a single particle with no external potential. This is just inertia.

Time derivative of 0 = 0 right?
∂L/∂x = ∂(-mc²δ)/∂x = 0

 

[Somali_Physicist]

Apart from you trying to apply classical reasoning to an a priori relativistic particle, this is just wrong.

Also, the Lagrangian of a relativistic particle is not given by $E_k - E_p$. See https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechanics

Thank you!

 

[Orodruin]

Time derivative of 0 = 0 right?
∂L/∂x = ∂(-mc²δ)/∂x = 0

Yes, but the time derivative of any constant is also 0. You therefore cannot conclude that what you are differentiating to get zero necessarily is zero. You are missing the integration constant.

 

[Somali_Physicist]

Yes, but the time derivative of any constant is also 0. You therefore cannot conclude that what you are differentiating to get zero necessarily is zero. You are missing the integration constant.

Oh , so there is a constant?
∂L/∂x = C

Also could you give a quick summary on the difference between relativistic langragian and classical.

Is it simply because:
S= ∫Lδdτ

 

[Orodruin]

Also could you give a quick summary on the difference between relativistic langragian and classical langragian.Is it just taking into account coordinates?

This is all described on the Wikipedia page I linked to in post #3. I suggest that you read it carefully and come back if you have further questions. We cannot write you a textbook.

 

[PeterDonis]

This is all described on the Wikipedia page I linked to in post #3. I suggest that you read it carefully and come back if you have further questions. We cannot write you a textbook.

And with that, this thread is closed as the OP is speculation based on mistaken premises.