# B Does my speedy spaceship (.999+C) have a temperature?

#### hmmm27

So, the doppler effect will make the ship

- get (to the occupants) and appear (to a stationary observer) as being hotter at the front and colder at the rear...

while time dilation makes the ship

- get hotter while appearing colder, both all 'round.

#### Orodruin

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get (to the occupants) and appear (to a stationary observer) as being hotter at the front and colder at the rear...
No. Occupants will not notice a thing. For the occupants, the ship is stationary.

#### jbriggs444

Homework Helper
So, the doppler effect will make the ship

- get (to the occupants) and appear (to a stationary observer) as being hotter at the front and colder at the rear...

while time dilation makes the ship

- get hotter while appearing colder, both all 'round.
If I understand the thinking, the notion is of a photon gas which is isotropic in a particular "rest" frame. Within this gas, the moving ship sees an anisotropy -- blue-shifted high energy forward and red shifted low energy aft.

Special relativity is not responsible for the anisotropy. The setup of the universe is.

The thinking about time dilation is trickier. If you appear colder to the other guy, the other guy also appears colder to you. However there is some sense to be made of the conundrum.

Consider the situation with two parallel, infinitely long trains running on parallel tracks in opposite directions at high speed. Can the trains reach thermal equilibrium with one another? It seems to me that each will gain heat from the other due to relativistic beaming and a resulting net blue shift. There is no free lunch, of course. The gain in heat comes at the expense of each train's kinetic energy -- it is a fancy way of engaging in frictional mutual slow-down.

#### russ_watters

Mentor
Again, this depends. If you mean the rest temperature, sure. If you mean the temperature that you would associate with the spectrum radiated from the body that will be affected by the Doppler effect and correspondingly red or blue shifted.
That's fine, I've just never heard of doppler shift being referred to as a temperature alteration. E.G., we don't say redshifted stars/galaxies are colder than blueshifted ones, do we?

But yeah, I guess the OP's wording implies that word usage.

#### Heikki Tuuri

Suppose that your spaceship is a black body in temperature T in the co-moving frame.

Let a static observer use an infrared telescope and measure the black body spectrum at a temperature T' with the correct surface brightness. Then we could say that the static observer "sees" the ship having the temperature T'.

I did some quick calculations, which might contain an error. If the spaceship is moving straight toward the static observer, then he does see the correct black body radiation for a temperature T' > T. If the spaceship is receding, T' < T. The spectrum is shifted because of the Doppler effect and time dilation.

If the spaceship is moving by, then the static observer sees it length-contracted. My calculations suggested that the surface brightness might appear too high for a black body at a temperature T / gamma.

Staff Emeritus
I hesitate to jump in, because the detour suggests that this is really the start of an anti-relativity screed.

However....

Classically, there are several different definitions of temperature. They are equivalent in that they all give the same value for the same system. In relativity, these transform differently so do not agree. They will never agree.

It also doesn't matter. If I have a rocket moving at .99c with respect to my heat bath, it is not in thermal equilibrium with that heat bath. There's no way to say what its temperature "really is" since it doesn't fulfill the conditions to even have a temperature.

#### pervect

Staff Emeritus
I gather there are a couple of relativistic treatments of temperature. One of them, as I recall, had inverse temperature as a 4-vector. So $\Delta Q$ became the change in a 4-vector (not just a scalar energy), and the change in inverse temperature also became a 4-vector. And $\Delta S$, which was still a scalar, became the dot-product of these two 4-vectors, the change in energy-momentum, and the inverse temeprature.

However, I was more comfortable with the treatments (which I've also seen) where temperature was always specified in the rest frame of whatever had the temperature.

The reference I recall was https://arxiv.org/abs/physics/0505004. Apologies if I messed up anything in my recollections, it's been a while since I read it. I don't know what the impact factor of this paper was, it looked like a decent place to start to me, but I'm not that familair with thermodynamics.

"Does my speedy spaceship (.999+C) have a temperature?"

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