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### Special Relativity and Temperature

```Date: 12/21/2005 at 19:39:29
Subject: special relativity and temperature

Hey Dr. Math...

If temperature is the total amount of kinetic energy possessed by the
particles in a given substance, and special relativity holds that
nothing can move faster than the speed of light, does that mean that
there is an upper limit for how hot a substance can become (e.g. the
temperature where the particles begin to move at the speed of light)?
Is there some kind of weird relativistic effect as the individual
particles become time dilated and so forth to varying degrees?  At
what temperature, if any, would this become significant, and what
would an observer of an object approaching the "temperature limit" see?

I'm not sure whether I've made some huge conceptual error in even
asking this question or not, but if it is valid, it seems that an
object would become time dilated and experience an increase in
equivalent mass as its temperature increased.  Probably, the
temperature would have to be extremely high for these effects to
become noticeable.  Especially hard to conceptualize is what would
happen if some parts of the object became time dilated and length
contracted, but not others, a scenario that is certainly possible with
heat.

```

```
Date: 12/22/2005 at 08:20:54
From: Doctor Rick
Subject: Re: special relativity and temperature

What you are missing is that kinetic energy is not (1/2)mv^2 in the
relativistic case.  If it were, then it could be no greater than (1/2)
mc^2.  But that's not the case.

The energy of a particle in special relativity is

E = mc^2/sqrt(1-v^2/c^2)

where m is the mass of the particle, v is its velocity, and c is the
speed of light.  Note that when the particle is at rest, the energy
is not 0, but

E = mc^2

which you have probably seen before: it is Einstein's most famous
equation.  We call this the "rest energy" of the particle.  The total
energy can then be broken into two parts:

Energy = rest energy + kinetic energy (energy of motion)

E =      mc^2        + mc^2(1/sqrt(1-v^2/c^2) - 1)

When v is much less than c, the kinetic energy is very close to

KE = approximately (1/2)mv^2

which you can verify if you are familiar with the expansion

(1+x)^n = 1 + nx + n(n-1)x^2/2! + ...

This shows that the relativistic kinetic energy is really kinetic
energy.  However, as v increases toward c, KE increases WITHOUT
LIMIT.  You can check that out too: as v/c gets close to 1, the
square root gets closer and closer to zero, and 1/sqrt(1-v^2/c^2)
gets larger and larger.  There is no limit to how great the energy
can be.

Another way of looking at this is that the "relativistic mass" of
the particle increases without limit as the velocity nears c.
Therefore it takes more and more input of energy to increase the
velocity another notch; it would take an infinite amount of energy
to raise the velocity all the way to c.  That's why a particle can't
be accelerated to the speed of light or beyond: it takes too much
work.

You're right that temperature is a function of the kinetic energy of
the particles in a substance.  What I have shown is that the
temperature is not limited by relativity.  Of course, as you pump
more and more energy into a substance, eventually it will cease to
be that substance: first molecules will be dissociated, then atoms
(forming a plasma), then even the nuclei will be dissociated.  But
that's another story.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 12/22/2005 at 22:39:09
Subject: Thank you (special relativity and temperature)

Thank you very much for this informative answer; it was exactly what I
was looking for.  You run a good site here!
```
Associated Topics:
College Physics
High School Physics/Chemistry

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