# Is it possible to have an anisotropic temperature to a Brownian motion?

Resolving the Langevin equation. Tenperature is a scalar, is there a way to make it into vector?

• Temperature can have a gradient, which is a vector. But temperature itself is a scalar. Commented Apr 2 at 13:12

Absolutely Not!

But don't let that stop you from musing:

For an ideal gas, the temperature is based on the average energy of an ideal gas, via:

$$\frac 1 2 m\bar v^2 = \frac {3}{ 2} kT$$

so

$$T = \frac {1} {3} \frac {m \bar v^2}{k}$$

Since a squared velocity is a scalar, there's really no way to back out a vector...

...but, we can make things more complicated, by considering a non-zero average velocity $$\vec{\bar v}\rightarrow \bar v_i$$. Abandoning all common sense, that allows us to construct the (it's hard to say it): The (symmetric) temperature tensor:

$$T_{ij} = \frac 1 3 \bar v_i \bar v_j$$

with the trace being the normal temperature:

$$T = T_{ii}$$

Now if we pick the right coordinate ($$\hat v = \hat z$$), you can express that in spherical tensors with two non-zero components (scalar and longitudinal polarization):

$$T_{ij} \rightarrow T^{(l=0,m=0)} + T^{(l=2,l=0)}$$

where the magnitude of each is proportion to temperature (by $$\frac 1 {\sqrt 3}$$ and $$\frac 2 {\sqrt 3}$$, respectively).

So, what we've done is turn normal temperature into an isotropic tensor, $$T\delta_{ij}$$, and a traceless velocity alignment "natural form$$^1$$" rank-2 tensor (proportional to $$T$$) looking like $$v_iv_j - \frac 1 3 v^2\delta_{ij}$$, and we've thrown out the velocity direction information.

So it wasn't a fruitful exercise. Also: I won't be accepting downvotes on this answer...consider it a belated April $$1^{\rm st}$$.

[1] A natural form rank-$$N$$ tensor (at least in 3-dimensions) is symmetric in all indices and traceless (on contraction of any-pair)...which takes the $$3^N$$ DoF down to $$2N+1$$....which is also the number of spherical harmonics with $$l=N$$. (For me: that was a Eureka moment regarding the geometric significance of tensors, which is a concern for many physics students).