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Can a fluid approach the speed of light according to the equation of continuity?

Physics Asked on January 26, 2021

We all are well aware with the equation of continuity (I guess) which is given by:

$$A_1V_1= A_2V_2$$

Where $A_1$ and $A_2$ are any two cross sections of a pipe and $V_1$ and $V_2$ are the speeds of the fluid passing through those cross sections.


Suppose $A_1$ is $1 m^2$ and $V_1$ is $2 m/s$ and $A_2$ is $10^{-7} m^2$. This will mean that $V_2$ will be $2×10^{7} m/s$ which is much closer to the speed of light.

But my teacher just said it is not possible. He didn’t give a reason.


Is he saying this because density of the liquid will change (given mass increases and length contracts with higher speed)?

Why can’t we use this equation to push fluids to a higher speed?

If there is a limit on the maximum speed we can get to with this equation, what is it?

For simplicity of calculations, you may take water.

5 Answers

That simplified form of continuity equation assumes that the fluid is incompressible. That is only a valid assumption at low Mach numbers. I think a typical “rule of thumb” is that a Mach number less than 0.3 is required for the assumption to hold. So for the continuity equation to hold in that form requires a speed which is much less than the speed of sound which in turn is much less than the speed of light. You cannot use the continuity equation to achieve supersonic flow, let alone relativistic flow.

Note, that is not to say that supersonic flow is impossible, but rather that it is not possible simply by application of that form of the continuity equation. You need a form that accounts for compressibility. Superluminal flow is fundamentally impossible as no massive particle can reach c with finite energy.

Correct answer by Dale on January 26, 2021

You can check your hypothesis by closing the mouth of a pipe (from which water is flowing out) to different degrees. You will find that the flow velocity increases initially as the outlet area is decreased, reaches a maximum and then falls off to zero as you proceed to completely close the outlet. This is because, when the outlet area is very small, viscous forces and surface tension effects become dominant and retard the flow velocity.

Answered by Deep on January 26, 2021

That is one version of the continuity equation for sure. But that is not the most general one. As a matter of fact, that condition can be deduced from the incompressibility of a fluid as in a limit of the Navier-Stokes equation:

$$rho(partial_t + v cdot nabla)v + nabla p = 0,$$

where $rho$ is the density, $p$ is the pressure and $v$ is the velocity field in the fluid, assumed to satisfy $nablacdot v = 0$. But even if you take everything under consideration in this description, the formalism will allow for fluids to have a speed higher than the speed of light.

The reason for this is that even the Navier-Stokes equation is a limit of its relativistic counterpart, the conservation of the stress-energy tensor:

$$nabla^mu T_{munu} = 0.$$

And from this equation, it is easy to show that your fluid will not propagate faster than light under the assumption that your $T_{munu}$ is physical.

Answered by Tales Rick Perche on January 26, 2021

One must look at all relevant equations when discussing what's possible or what's doable. In this case, you must look also at the momentum equation, which is Newton's Law, and when you do, you see that the pressure required for high Mach number in fluids becomes impossible to achieve in practice.

In talking about incompressible flow, from a fundamental point of view, it's not possible to achieve even Mach 1 in an "incompressible" fluid. The criterion by which a fluid can be considered incompressible is that the Mach number squared be much less than unity. That means the Mach number itself should be less than about 0.3, and above that, the fluid is compressible; i.e. changes in it's density become important. Thus, one can argue that it's impossible for an incompressible fluid to have a velocity close to the speed of sound in that fluid, simply because it must experience compressibility for Mach number above about 0.3.

Also, when you go through the numbers, you will find that it takes very large pressures to accelerate say, water to high Mach number. It's thus pure fantasy to even imagine speeds approaching the speed of light.

Answered by ttonon on January 26, 2021

One thing not pointed out in the above answers is that if a fluid was truly "incompressible", it would violate special relativity. This is because the speed of sound in a fluid can be derived from its equation of state $rho(P,T)$ which gives the density as a function of the pressure and temperature. Specifically, we have $$ frac{1}{v^2} = left( frac{partial rho}{partial P} right)_s, $$ where $v$ is the speed of sound in the material and the derivative is taken while holding entropy constant. If $rho$ were really & truly "incompressible", then it would be constant with respect to $P$, and the above derivative would vanish. Thus, $v^2 to infty$, which is faster than light.[citation needed]

The resolution to this is that "incompressible" doesn't literally mean incompressible. It just means "the range of pressures we're looking at is so small that the compression of the fluid is negligible."

Answered by Michael Seifert on January 26, 2021

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