The Gravity of Laser Light: How Does It Impact Acceleration?

[PeterDonis]

The other one accelerates so that he experiences at every point of time the same accelleration as the other person experiences gravitational acceleration

"At every point of time" is frame dependent.

Also, neither of the accelerations are "gravitational acceleration"; neither one is at rest in a gravitational field. The observer inside the spherical shell Earth is accelerating in a flat region of spacetime, just like (to a good approximation) the observer far away. Neither of them is standing at rest on the surface of the spherical shell, which is what they would have to do to be at rest in a gravitational field.

In the end one is at rest in the center of Earth and the other one moves in free space with a constant relative velocity relative to earth.

What is this supposed to show?

 

[DrDu]

I just did a nonrelativistic calculation: The subject experiencing "true" acceleration, but no gravitation potential, will end up with a speed of $v_0=\sqrt{gr_0}$ relative to earth, where g is the acceleration at the surface of Earth (radius $r_0$). The gravitational potential of the second observer, which ends up inside the "shell" Earth is $-gr_0=-v_0^2$. This looks suspiciously as if $g_{00}$ can be expressed in terms of $1-v_0^2/c^2$ upon a more exact relativistic calculation.

 

[PeterDonis]

The gravitational potential of the second observer

Could just as well be that of an observer standing on the surface of the shell Earth, since the gravitational potential is the same there as everywhere inside the shell (the gravitational potential everywhere inside the shell is constant).

is $-gr_0=-v_0^2$

The $g r_0$ part is correct, for the humdrum reason that the gravitational potential at the surface of any spherically symmetric massive body with mass $M$ and radius $R$ is $U = - GM / R$, and the acceleration $g$ given by the gradient of that potential is $- GM / R^2$, so $g = U / R$.

The $v_0^2$ part, however, is wrong. If you check the math, you will see that $g r_ 0 = v_0^2 / 2$ for your scenario. Which tells us nothing whatever about the gravitational field of the Earth or anything else; all it tells us is that kinetic energy per unit mass is $v^2 / 2$ for motion in a straight line in flat spacetime. In other words, it's just the work-energy theorem in disguise.

This looks suspiciously as if $g_{00}$ can be expressed in terms of $1-v_0^2/c^2$ upon a more exact relativistic calculation.

Numerology is not physics. If you actually do the relativistic calculation, you will see that your suspicion is wrong.

 

[DrDu]

You are quite negative. In Peter T. Landsbergs, "Thermodynamics and Statistical Mechanics", he mentions that, at least at the time when the book was written, three possible transformation laws for temperature between intertial systems were under discussion: 1) The Planck-Einstein theory $T(v) =T(0)/\gamma$, 2) $T(v)=T(0)\gamma$ and 3) $T(v)=T(0)$.
All I am trying to say is that from the Tolman temperature one should be able to show which of these relations is compatible on grounds of the principles of general relativity.

 

[PeterDonis]

You are quite negative.

I am actually being charitable; your posts are verging on personal speculation, which is off limits at PF.

three possible transformation laws for temperature between inertial systems

Which has nothing to do with Tolman's Law, since "inertial systems" implies flat spacetime and SR; in GR in curved spacetime there are no global inertial systems. Any comparison between temperatures at different locations in a gravitational field can't be done using inertial systems, because any comparison between any properties at different locations in a gravitational field in curved spacetime can't be done using inertial systems.

All I am trying to say is that from the Tolman temperature one should be able to show which of these relations is compatible on grounds of the principles of general relativity.

Sorry, but this is simply incorrect, no matter how plausible it appears to you. See above.

 

[Demystifier]

the observed Unruh temperature will also vary from bottom to top. But this has nothing to do with Tolman's law, because there is no gravitational potential; spacetime is flat. Tolman's law is about the effects of curved spacetime.

That's wrong. The variation of Unruh temperature is exactly a manifestation of Tolman law. It appears because $g_{00}$ is not a constant in an accelerated frame. Tolman law does not require a curvature.

 

[DrDu]

I just looked up Einstein's "Über die spezielle und allgemeine Relativitätstheorie". Methinks he uses exactly my line of argument to justify the redshift for a rotating disk from the special relativistic transformation of time and then argues that this argument carries over to general gravitational field. I don't see why this line of argument should not be applicable to temperature.

 

[vanhees71]

The difference is that the frequency of a wave is the time-component of the wave-four vector, $(\omega/c,\vec{k})$, and temperature is a scalar (field). Also note that relativistic thermodynamics before around the mid 1960ies was a mess with several different conventions! For an early review, see

N. G. van Kampen, Relativistic thermodynamics of moving systems, Phys. Rev. 173, 295 (1968),
https://doi.org/10.1103/PhysRev.173.295

 

[DrDu]

In Landau/Lifshetz Vol. 5, they derive as condition for thermodynamical equilibrium the condition $T\sqrt{g_{00}}=const.$. In the following they remark that "this can also be written as $T=const. \cdot \frac{dx_0}{ds}$ which allows to consider the body not only in a system where he is at rest, but also in a system where he is moving (rotates as a whole). To that end, the derivative has to be taken along the world line passing through a given point of the body." So he goes exactly the same way I was proposing to extend the Tolman relation to moving bodies.

 

[DrDu]

I managed to get a copy of the van Kampen paper and also found this quite interesting discussion:
https://hal.archives-ouvertes.fr/hal-02299780/file/Black_Body_Radiation_in_Moving_Frames.pdf

 

[PeterDonis]

The variation of Unruh temperature is exactly a manifestation of Tolman law. It appears because $g_{00}$ is not a constant in an accelerated frame. Tolman law does not require a curvature.

Mathematically, if you assume spacetime is flat and that you have a perfect fluid whose worldlines are the Rindler congruence, yes, you will obtain Tolman's Law.

However, those assumptions are not really consistent, because "perfect fluid" implies nonzero stress-energy, and flat Minkowski spacetime is not a solution of the EFE for nonzero stress-energy. Similar remarks apply to Unruh radiation; giving the radiation a nonzero energy density, which is required for it to have a nonzero temperature, is not really consistent with spacetime being flat. (Calling the energy density "vacuum energy density" doesn't help; that just means you have a nonzero cosmological constant, and flat Minkowski spacetime is not a solution of the EFE for a nonzero cosmological constant either.)

Also, if you have a perfect fluid with accelerated worldlines, something has to be producing that acceleration. A static, spherically symmetric configuration of matter can do that without expending energy because of the particular way it curves spacetime. But in flat spacetime, you can't just magically say the fluid accelerates; something has to be making it accelerate, and that something has to be expending energy (like a rocket engine). And you have to include the energy involved in that in your analysis as well. That means there is even more stress-energy present in your model, which isn't compatible with flat spacetime, and it also means there is the possibility of other heat transfer that your model does not include.

I understand the use of the flat spacetime Rindler case as a heuristic; I just don't think it can do all the work that is being ascribed to it in this thread.

 

[DrDu]

But in flat spacetime, you can't just magically say the fluid accelerates; something has to be making it accelerate, and that something has to be expending energy (like a rocket engine). And you have to include the energy involved in that in your analysis as well. That means there is even more stress-energy present in your model, which isn't compatible with flat spacetime, and it also means there is the possibility of other heat transfer that your model does not include.

When you really fear that you need a rocket so heavy, that its gravity would disturb the system, which I consider quite unlikely, you could imagine other sources of acceleration like reflection of a laser beam.

 

[PeterDonis]

When you really fear that you need a rocket so heavy, that its gravity would disturb the system, which I consider quite unlikely, you could imagine other sources of acceleration like reflection of a laser beam.

Sure, a rocket of ordinary size, even counting all of its fuel and rocket exhaust, won't be enough stress-energy to curve spacetime--but it also won't provide enough acceleration to serve as any kind of illustration of Tolman's law; the temperature gradient due to the variation in $g_{00}$ with position will be many, many orders of magnitude too small to detect. If you're going to say we could in principle detect such small temperature differences, then we could just as well in principle detect such small amounts of spacetime curvature. The two go together; you can't just arbitrarily take one into account and ignore the other.

Similar remarks would apply to any other source of acceleration, like a laser beam.

 

[DrDu]

Sure, a rocket of ordinary size, even counting all of its fuel and rocket exhaust, won't be enough stress-energy to curve spacetime--but it also won't provide enough acceleration to serve as any kind of illustration of Tolman's law; the temperature gradient due to the variation in $g_{00}$ with position will be many, many orders of magnitude too small to detect. If you're going to say we could in principle detect such small temperature differences, then we could just as well in principle detect such small amounts of spacetime curvature. The two go together; you can't just arbitrarily take one into account and ignore the other.

Similar remarks would apply to any other source of acceleration, like a laser beam.

For a laser beam, it should be possible to estimate the effect of the energy density of the laser. The accelerated system can only see the gravity of the light already reflected. This intensity is maximal and constant in time (judged from the accelerated system) but proportional to acceleration directly at the reflector. The gravitational potential due to light reflected at earlier times should be integrable due to redshift. Maybe one could estimate whether it increases linearly or sub-linearly with acceleration?

 

[PeterDonis]

For a laser beam, it should be possible to estimate the effect of the energy density of the laser.

You don't have to estimate; the stress-energy tensor of an EM field is well known.

https://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

The accelerated system can only see the gravity of the light already reflected.

Not at all. The source of the laser, which must contain enough energy to supply it, will already be in the past light cone of the accelerated system, along with its gravity.

 

[DrDu]

The intensity of the laser will have to increase steadily to maintain constant acceleration. The accelerated observer will always see the reactor only as a point source of gravity (retarded) which is more and more far away. The laser he will only notice when it hits him and probably the gravity of the light reflected.

 

[PeterDonis]

The intensity of the laser will have to increase steadily to maintain constant acceleration.

So what?

The accelerated observer will always see the reactor only as a point source of gravity (retarded)

If you are using Newtonian gravity, this is not correct; there is no retardation in Newtonian gravity.

If you are using relativity, gravity is not a force, it's spacetime curvature, sourced by stress-energy. All of the stress-energy that is going to be pumped into the laser is already in the past light cone of the accelerated observer before the laser even turns on. So retardation is irrelevant.

The laser he will only notice when it hits him

He will only notice the acceleration due to the laser when it hits him, yes. But that does not mean he only notices the gravity due to the stress-energy that gets pumped into the laser when it hits him. See above.

and probably the gravity of the light reflected

As already stated, all of the stress-energy that causes the "gravity" the accelerated observer senses is already in his past light cone before the laser even turns on.

The reflection of the laser light does put some of that stress-energy closer to the accelerated observer than it would otherwise be, which I think (I haven't done the detailed math to check) will increase somewhat the gravity the accelerated observer measures, from what it would have been if the laser had not fired. But that doesn't mean the accelerated observer measures no gravity at all until the laser hits him. See above.