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The main question I would like to ask is whether quantities such as density are dependent on the frame of reference. I have searched several forums and the answer is somewhat controversial. Some answers use the concept of relativistic mass to justify that it is invariant. Some of the answers say that relativistic mass is not a correct concept (given in Classical Mechanics by John R. Taylor pg 633) and that mass is invariant and hence the density must be an observer dependent quantity. This is a little odd because of the following thought experiment:

Imagine that a container filled with liquid is made to travel at relativistic speeds. In the frame of the container, the density is d at temperature T and pressure P. To a person in the ground frame the volume of the liquid will decrease because of length contraction. At the critical density there must be a phase transition from liquid to solid. In the moving frame, the container has a liquid but in the rest frame, the container is filled with a solid.

So based on the above I have the following questions:

  1. Is there anything wrong in the above argument and why is it incorrect? (eg. Phase diagram changes depending on the velocity of the object)
  2. If the density is observer-dependent, could it mean that thermodynamics is observer-dependent?
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  • $\begingroup$ It is really subtle in my opinion. Even without entering the relativistic mass / no rel. mas debate, it is a good point that length contraction is at work. So there is something mixed up in your question and/or the answer you got before posting. I personally think that phase transition diagrams must be shifted according to the moving observer. It is not that you as observer magically observe a gas nebula contracting (yes geometrically) into a giant liquid drop. This is independent of what mass is considered, even taking rest mass the observed density value rises, but it does not carry any info $\endgroup$ – Alchimista Sep 16 at 9:57
  • $\begingroup$ On internal processes, that will always happen within a rest frame. This is a point similar to the fact that nothing change within a body that acquires relativistic mass. Though again, the two are different aspects. $\endgroup$ – Alchimista Sep 16 at 10:03
  • $\begingroup$ one would expect that phase transitions depends only on functions of temperature, pressure, density, chemical potential, etc. that form Lorentz invariants (i.e. scalars) $\endgroup$ – lurscher Sep 16 at 19:58
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    $\begingroup$ I think this comes down to the fact that phase diagrams themselves transform under Lorentz transformations. We measure the transformed state variables $(P', T', \rho ' , V')$ and try to plot them on the untransformed phase diagram, leading to the apparent contradiction that phase is not a relativistic invariant. But if we were to plot $(P', T', \rho' , V')$ on the appropriately transformed phase diagram, we would find that the system is in exactly the same state we would find it in if we performed the measurements in the rest frame and plotted them on the rest phase diagram. $\endgroup$ – Charles Hudgins Sep 16 at 23:55
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    $\begingroup$ Forget about phase diagrams for a moment. What actually determines whether a bulk of stuff is solid or liquid? How does that change with relativistic transformation? What would happen to the substance if you made a hole in the box from the point of view of the relativistic observer? $\endgroup$ – Luaan Sep 17 at 6:39
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Every time someone comes up with an apparent "paradox" in Special Relativity, it is always related to some application of the "Lorentz contraction" or "time dilation" concepts in some questionable contexts. Over the years of many similar discussions, I've worked out the recipe for dissolution of such "paradoxes": just reduce your "paradox" down to the Lorentz transformations. They are fundamental to the Special Relativity -- both time dilation and the Lorentz contraction are derived from the Lorentz transformations. So, whatever you get from Lorentz transformation is what Special Relativity actually predicts. Any contradiction with the result you get from using Lorentz transformation is a mistake somewhere in your reasoning.

In order to reduce your "paradox" to these fundamental equations of Special Relativity - you first have to formulate your problem in terms of space-time events and world lines. When dealing with thermodynamic systems you'll have to get down to individual particles making up your bodies. Extract their worldlines and Lorentz-transform them to get your trajectories in moving reference frames.

To illustrate this, I've randomly generated a 1000 particles bouncing in a 3x1 box with Maxwell-distributed velocities:

enter image description here

I've saved the worldlines of all the particles and applied the Lorentz transform with $\gamma=2$ to each of the particles. Plotting the resulting coordinates, I've got this:

enter image description here

Above I've kept the scale size of both scales and made sure that the center of the scale is at the center of the resulting particle cloud. You can see that the density has, indeed, increased. Notice that the velocities of the particles along the x-axis are changed in respect to the bounding volume -- it is not a Maxwell distribution anymore.

In the moving frame, the container has a liquid but in the rest frame, the container is filled with a solid.

As for your thought experiment, you can apply this reduction to Lorentz transformations to realize that there is no way you can get dynamic motions of individual particles of a liquid and then Lorentz-transform them to get static (relative to each other) positions of the same particles in a solid body.

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    $\begingroup$ @ChrisCunningham Take a look at physics.stackexchange.com/questions/83488/… . To me, the bottom line is that the whole debate is terminological - depends on your definition of "temperature" and a practical problem you are solving. My apporach of reducing it to Lorentz transformations works and lets you obtain unambiguous results in either case. $\endgroup$ – Kostya Sep 16 at 16:14
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    $\begingroup$ I don't think that successfully avoids the question; if the whole debate is terminological, then that means there is no such thing as an observer-independent phase diagram, because it depends on your definition of temperature... what am I missing here? I (and I think, the OP) am talking about the kind of diagram on the right side of this image: soft-matter.seas.harvard.edu/index.php/… $\endgroup$ – Chris Cunningham Sep 16 at 18:13
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    $\begingroup$ Phase is about more than static vs. dynamic. The particles in your second simulation seem to be much more closely packed and less energetic, i.e. closer to the solid state. Yet shouldn't state (including how close you are to being a solid or a liquid) be a relativistic invariant? $\endgroup$ – Charles Hudgins Sep 16 at 23:41
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    $\begingroup$ @CharlesHudgins Solids aren't defined by being closely packed or less energetic. They're instead defined by being rigidly locked together, so that the system can support a shear stress. The point is that the particles are still moving freely around. $\endgroup$ – knzhou Sep 17 at 6:12
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    $\begingroup$ "In order to reduce your "paradox" to these fundamental equations of Special Relativity - you first have to formulate your problem in terms of space-time events and world lines." Thank you for writing that. By explaining this clearly you've made this website better :-) $\endgroup$ – DanielSank Sep 18 at 15:19
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Is there anything wrong in the above argument and why is it incorrect?

You have implicitly assumed that the pressure remains constant, and therefore that since density is increasing, at some point as the speed of the medium increases you will hit a "critical point" and undergo a phase transition. This does not follow, since the pressure of the matter will increase along with the density.

Specifically, the mass/energy density, pressure, and momentum density of a bulk solid all form part of the stress-energy tensor of the material, where $$ T_{00} = \text{mass/energy density} \\ T_{0i} = \text{momentum density vector} \\ T_{ij} = \text{stress tensor} $$ (working in units where $c = 1$.) In particular, let's suppose that we have an isotropic fluid at rest, so that $T_{00} = \rho$, $T_{ij} = P \delta_{ij}$, and all other components of the stress-energy tensor vanish. When we transform into another frame moving at speed $\beta$ in the $x$-direction relative to the rest, we find that $$ T'_{00} = \gamma^2 (\rho + \beta^2 P) \\ T'_{11} = \gamma^2 (P + \beta^2 \rho) $$ The key realization here is while the observed density of the medium ($T_{00}$) increases as the speed of the medium increases, the pressure ($T_{11}$) will also increase. The argument that "since the density increases, you'll eventually get a phase transition" is therefore not necessarily correct; the density at which a phase transition occurs depend on the pressure, which is also changing.

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    $\begingroup$ I understand what you're saying and you're correct that I did not talk about the pressure changing as well. However, is it still possible for the substance to be in 2 different phases in 2 different reference frames? $\endgroup$ – Sparsh Mishra Sep 17 at 14:27
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A phase transition should occur when the attracting forces between particles becomes high relative to their relative motion (broadly speaking).

Thought experiment: We have two objects with a certain opposite electrical charge. Now we move with respect to the objects, so that the distance between them becomes smaller (and charge won't). Does this change the attractive forces between them? The laws of nature don't change for a moving observer (the fundamental assumption in relativity), so I would say yes. With a factor $\gamma^2$, according to Coulomb's law, the attractive force would increase. However, would it be in disagreement with what an observer that doesn't move (w.r.t. the objects) would see?

Let's say the objects have equal mass and the acceleration between the objects in the 'standing still'-frame is $g/r^2$ (relative acceleration), and they start with a velocity such that they will make a circular motion with radius $R$. For the moving observer this should Lorentz-contract to an ellipse. When the objects align according to your line of motion, the attraction would be bigger, but the velocity of the objects change with a factor $\gamma$ too (addition of perpendicular velocities. So the centripetal force should change with a factor $\gamma^2$, which we have seen it does, so there it tempts to keep a circular motion. When the objects align perpendicular to your line of motion, the force doesn't change, but the velocity decreases, so there they would fall a bit to each other. If someone is very skeptic whether their movement will be elliptic, you can do it explicitly by finding an expression for their attraction as a function of their angle and solve the differential equation with the right starting conditions. But this broadly explains why there would be no issue with the attracting forces being bigger.

Now a liquid in a container is the same problem with more bodies. So generalizing this result, they should move in a way that would give same liquid when Lorentz-transformed. So if my reasoning is correct, for a moving observer it is possible for a liquid to have a higher density than would be possible in a resting frame and still be a liquid, i.e.: the phase transition shifts.

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The frame of reference of a system is exactly that: the frame in which we refer to the system. Whenever we say something like "The wavelength of the lowest electron orbital of a hydrogen atom is such-and-such", or The speed of sound is 343 m/s", we are speaking with regard to the frame of reference of the hydrogen atom, or the air the sound is moving through, respectively. Likewise, when we say that a substance has a certain phase diagram, that's in the frame of reference of the substance. if you throw boiling water into a black hole, then when it's right above the event horizon it will be blue-shifted into looking like it's so cold it should be freezing, but it will still be boiling.

Special relativity says that the coordinates we use to describe a system change with the reference frame, but the actual events don't. The measured velocities of molecules will depend on reference frame: velocity is distance divided by time, and both of those are coordinates. But whether it's freezing or boiling or whatever will not depend on the reference frame, because those are actual events.

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  • $\begingroup$ To concur with this answer, many quantities/terminology in relativity are simply understood to apply in the proper frame. For example, if we say a fluid has zero expansion (the expansion tensor is zero), this means the expansion is zero in the proper frame $\endgroup$ – Colin MacLaurin Sep 17 at 18:34
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A rough sketch of an answer: first, how do you tell whether something is a liquid or a solid? Say you measure viscosity: put the substance between concentric cylinders, apply a mild torque on the inner cylinder. If it moves a bit, and sticks, you have a solid, if it reaches a constant angular velocity, you have got a liquid.

Clearly, even the fastest observer looking at the experiment being performed, will agree with the observer at rest concerning its outcome. Say it is a liquid. Both will agree that the molecules have enough room to slip around and let the inner cylinder rotate.

But on the other hand, the molecule's trajectories will appear quite different: let both observers have molecular vision and follow the trajectories of their favourite molecules. The observer at rest will see an ordinary dynamics, all molecules randomly distributed, with no direction singled out. The moving observer, on the other hand, will see molecules with a rapid global motion and additionally packed much closer together in the direction of that motion than in the transverse direction. Further, the action of the central forces between the molecules needs to be transformed appropriately.

What I am getting at, is that the moving observer's take on the system cannot at all be adequately described in simple terms of pressure and temperature: the system is not isotropic, it possibly has non trivial strains etc... The questionable assumption you make in your question is exactly this, as far as I can see.

So the moving observer's best bet, in order to compute the behaviour of the moving system, is simply to transform his observations to the system's rest frame, thereby duplicating the rest observer's results.

Note that this approach is useful for a lot of problems in relativity. Look at the problem as one of trajectories, or events in space-time, and do the Lorentz transform.

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The literature contains four (4) different formulas for the transformation of temperature and the history of the subject is itself quite interesting; I suggest you read the quite recent and openly available paper by Jiri J. Mares, Pavel Hubik, and Vaclav Spicka: "On relativistic transformation of temperature", DOI 10.1002/prop.201700018 and references therein. Einstein himself seems to have held all four at least once in his lifetime, so do not feel bad if you are confused.

Denote the variables in the moving coordinate system of velocity $w$ with primes, and those in the one moving with center of mass without primes. Then the controversy is about what should the coefficient of transformation $\alpha $ be in the formula $T'=\alpha T$. The various results show that $\alpha = (\sqrt{1-w^2/c^2})^a$ where $|a| = \mathcal O(1)$ and supposedly depends on the heat transfer situation in the moving system, but this is also controversial.

All theories assume that entropy is a relativistic invariant because of its supposed "combinatorial" origin. Also, the pressure is always taken to be invariant. For example Planck says 1 that because $\frac {\partial H'}{\partial V'}=p'$ and both $\gamma H'$ and $\gamma V'$ are invariant, here $\gamma = \frac{1}{\sqrt{1-w^2/c^2}}$, so must the pressure $p$ be invariant as well. Since the total energy $E$ transforms as $\gamma E$ and $\frac {\partial S'}{\partial E'}=\frac{1}{T'} $ we also have that $T'/ \gamma$ is invariant, and $T'=\sqrt{1-w^2/c^2}T$, i.e., here $a=-1$; in other words, the moving system appears to be colder.

Write the Gibbs equation in the center of mass system as $$dU=TdS-pdV+\mu dN$$, where $\mu$ denotes the chemical potential per particle. Now divide both sides with $\gamma$ to get $$dU'=T'dS-pdV'+\mu'dN$$ where $\mu'=\mu/\gamma$. Next find the transformed version of the Clausius-Clapeyron equation 2 that is derived from the Gibbs equation and describes the condition for phase equilibrium: $$\frac{dp}{dT}=\frac{\Delta S}{\Delta V}$$ Here $\Delta S$ is the entropy of formation per molecule and $\Delta V$ is the volume difference between the two coexisting phases per molecule. Now divide both sides with $\gamma$ and we have the corresponding relationship in the moving coordinate system $$\frac{dp'}{dT'}=\frac{\Delta S'}{\Delta V'}$$ since $\Delta S'=\Delta S$. So the two CC equations show that whenever there is phase transition in the cm system we have transition in the moving system, as well.

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The density would be invariable

I'd like to answer this question in two parts. First I'll comment on the mass, then the volume and then the consequent effects.

Mass does not change due to frame change

Relativistic mass is a concept which is not at all fundamental and was just introduced to serve as a shortcut to perform calculations. Mass of an object is nothing but energy and energy can change without explicit heat exchange or work. Thus the mass of the object will remain constant provided that it is not exchanging heat and there is no involvement of work on the system. So if we say that something has a relativistic mass, i.e it's mass varies according to the frame we see it in, we are violating the energy conservation principle. Conclusions, mass is the same irrespective of which frame it is observed from.

Volume of the actual system won't change

The concept of lorentz contraction is what appears to you, as an observer in some other frame. It doesn't actually happen in the object. The distributions of particles in a systems do not change actually just because of some arbitrary observers are moving around with certain velocities. So since the mass and volume don't change, the density will remain the exact same. The fact that needs to be highlighted here is that internal process cannot be affected by external observations. If you need any more clarifications, feel free ask.

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  • $\begingroup$ I understand that this will all hold in the reference frame of the moving object. But in a stationary reference frame, why would the observed volume not change because of length contraction? $\endgroup$ – Sparsh Mishra Sep 16 at 7:44
  • $\begingroup$ Why would you have lorentz contraction in a stationary frame? And in a stationary frame why would you observe the particles of an object change their distributions is the questions you can ask. Be it any frame, the distribution of particles in a system cannot change because of external observations $\endgroup$ – Sanika Khadkikar Sep 16 at 7:46
  • $\begingroup$ I'm sorry, but I don't quite understand. Wouldn't we expect the length of the object measured in the stationary frame to be different from that in the moving frame? and if so, wouldn't the density be different? $\endgroup$ – Sparsh Mishra Sep 16 at 7:49
  • $\begingroup$ The length 'difference' which you'll observe will be because of relative motion between both the frames. You will ofcourse 'observe' a different volume. But the key point is the volume of the actual object hasn't changed. Because of relative motion between you two, space seems contracted to you. $\endgroup$ – Sanika Khadkikar Sep 16 at 7:53
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    $\begingroup$ I'm sorry, I hate to give downvotes, but in relativity, every observer is equally "right". When it seems to an observer that the volume changes, then it has changed. You cannot talk about an "actual" volume, just the volume according to the rest frame. $\endgroup$ – Antaios Sep 16 at 9:10

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