Exploring the Effects of Speed on Time and Decay: Unraveling Einstein's Theory

[Mister T]

Sorry PAllen, I still fail to see the issue. You're mentioning the equivalence principal. So in a sealed free fall laboratory, this experiment doesn't work...neat physics.

Detecting a potential is not the same as detecting a potential gradient. If the time dilation effect you referenced is different when detected at sea level than at an altitude of 1.0 km, that would demonstrate tidal gravity. Just detecting the effect at sea level, or at the 1.0 km altitude, isn't enough.

In other words stand a meter stick up on end at sea level, and do the same with another at the top of a mountain. You can detect the time dilation effect between the ends of the sea-level meter stick, and the mountain-top meter stick. And perhaps note that it's not the same for each meter stick. Thus the time dilation effect is different at sea level than it is on the mountain top. That's a determination of tidal gravity.

 

[nitsuj]

Detecting a potential is not the same as detecting a potential gradient. If the time dilation effect you referenced is different when detected at sea level than at an altitude of 1.0 km, that would demonstrate tidal gravity. Just detecting the effect at sea level, or at the 1.0 km altitude, isn't enough.

In other words stand a meter stick up on end at sea level, and do the same with another at the top of a mountain. You can detect the time dilation effect between the ends of the sea-level meter stick, and the mountain-top meter stick. And perhaps note that it's not the same for each meter stick. Thus the time dilation effect is different at sea level than it is on the mountain top. That's a determination of tidal gravity.

lol i still don't get it...not at all an issue with your explanation...that reads easily.

 

[PAllen]

Detecting a potential is not the same as detecting a potential gradient. If the time dilation effect you referenced is different when detected at sea level than at an altitude of 1.0 km, that would demonstrate tidal gravity. Just detecting the effect at sea level, or at the 1.0 km altitude, isn't enough.

In other words stand a meter stick up on end at sea level, and do the same with another at the top of a mountain. You can detect the time dilation effect between the ends of the sea-level meter stick, and the mountain-top meter stick. And perhaps note that it's not the same for each meter stick. Thus the time dilation effect is different at sea level than it is on the mountain top. That's a determination of tidal gravity.

And to get really technical, time dilation per meter would be expected to change with position within a long uniformly accelerating rocket, per the Rindler metric, which is still flat spacetime. To truly distinguish this from Earth gravity you would need to measure the deviation in this change compared to Rindler. This is a second order effect.

 

[Dale]

Yes, you are right, but still its state changes every time. It doesn't make any progress over time but its state changes and that can be thought as a change in its structure I suppose.

Fundamental particles (muons, electrons, quarks, etc.) can certainly occupy different states. But regardless of what state they are in they have no internal structure. Internal structure and state are very different things.

In that sense the dice is a good analogy. It can be in different states, and those states may be stable, but it doesn't change its internal structure and has no memory.

 

[nitsuj]

And to get really technical, time dilation per meter would be expected to change with position within a long uniformly accelerating rocket, per the Rindler metric, which is still flat spacetime. To truly distinguish this from Earth gravity you would need to measure the deviation in this change compared to Rindler. This is a second order effect.

Wiki says the tensile force from tidal gravity for a 1 meter, 1kg rod on Earths surface is about 0.4uN...and apparently that's measurable. I wonder if ideally an experiment to measure tidal gravity could be done with these "things". I wonder if we can make a string that breaks at a specific tensile force value, where the only explanation for it breaking is tidal forces.

Same article says the tensile force on the same rod on the surface of a white dwarf (mass of sun, size of earth) is 0.24N lol tidal gravity is so subtle...they gave an extreme example of a neutron star (2 sun mass, 12km radius) where the tensile force on the rod is 10,000N 190km about the surface, that's some strong geometry!

 

[PAllen]

Wiki says the tensile force from tidal gravity for a 1 meter, 1kg rod on Earths surface is about 0.4uN...and apparently that's measurable. I wonder if ideally an experiment to measure tidal gravity could be done with these "things". I wonder if we can make a string that breaks at a specific tensile force value, where the only explanation for it breaking is tidal forces.

Same article says the tensile force on the same rod on the surface of a white dwarf (mass of sun, size of earth) is 0.24N lol tidal gravity is so subtle...they gave an extreme example of a neutron star (2 sun mass, 12km radius) where the tensile force on the rod is 10,000N 190km about the surface, that's some strong geometry!

That is valid detection of tidal gravity. I suspect the smallest scale detection would be achieved with SQUID based accelerometers. Time dilation is just not a very practical way to detect tidal gravity.

 

[Mister T]

And to get really technical, time dilation per meter would be expected to change with position within a long uniformly accelerating rocket, per the Rindler metric, which is still flat spacetime. To truly distinguish this from Earth gravity you would need to measure the deviation in this change compared to Rindler. This is a second order effect.

So in GR you can have a potential gradient that is not due to tidal gravity? This is not so in the Newtonian approximation, correct?

 

[jbriggs444]

So in GR you can have a potential gradient that is not due to tidal gravity? This is not so in the Newtonian approximation, correct?

Is it just me? The term "potential gradient" sounds like a NOOP based on the fundamental theorem of calculus.

 

[Ibix]

So in GR you can have a potential gradient that is not due to tidal gravity? This is not so in the Newtonian approximation, correct?

I think @PAllen's point is that if you built a 1km long rocket and accelerated it at 1g, then got two clocks attached to opposite ends of a 1m rule then you would always find the same relative tick rate between the clocks as you climbed deck by deck. However, if you built a 1km tall tower on Earth and climbed it with the same pair of clocks you would notice different relations between the tick rates as you climbed.

With today's clock technology we could not repeat the experiment with a 1m rocket/tower and two clocks 1mm apart, although in principle the same type of results would be expected. This is why he's saying that the 1m measurement is of a potential gradient (present in Rindler coordinates in the rocket and in the tower) and not of tidal gravity (present in the tower and not the rocket).

 

[PAllen]

So in GR you can have a potential gradient that is not due to tidal gravity? This is not so in the Newtonian approximation, correct?

Yes, a born rigid rocket will have lower proper acceleration at the front compared to the back, under constant thrust = constant proper acceleration of the back. In equlibrium, a real rocket would be expected to exceedingly well approximated by born rigidity.

 

[PAllen]

I think @PAllen's point is that if you built a 1km long rocket and accelerated it at 1g, then got two clocks attached to opposite ends of a 1m rule then you would always find the same relative tick rate between the clocks as you climbed deck by deck. However, if you built a 1km tall tower on Earth and climbed it with the same pair of clocks you would notice different relations between the tick rates as you climbed.

With today's clock technology we could not repeat the experiment with a 1m rocket/tower and two clocks 1mm apart, although in principle the same type of results would be expected. This is why he's saying that the 1m measurement is of a potential gradient (present in Rindler coordinates in the rocket and in the tower) and not of tidal gravity (present in the tower and not the rocket).

No, that's not true. See my reply above.

[edit: I should say, most of what you say is clearly correct and what I meant. However, born rigidity leads to change in 'g' with position in a uniformly accelerating rocket, but with a different, slower, functional dependence than 'g' as a function of altitude near earth.]

 

[Ibix]

No, that's not true. See my reply above.

[edit: I should say, most of what you say is clearly correct and what I meant. However, born rigidity leads to change in 'g' with position in a uniformly accelerating rocket, but with a different, slower, functional dependence than 'g' as a function of altitude near earth.]

Ah - because (in an inertial frame) the rocket is doing something like length contracting more and more, so the acceleration is non-uniform even in the steady state. Right?

 

[PAllen]

Ah - because (in an inertial frame) the rocket is doing something like length contracting more and more, so the acceleration is non-uniform even in the steady state. Right?

Right.

 

[Dale]

So in GR you can have a potential gradient that is not due to tidal gravity? This is not so in the Newtonian approximation, correct?

In Newtonian gravity the usual mgh formula is the potential gradient in a uniform field (no tidal effects).

Edit: oops, mgh is the potential energy, the potential is gh, and the potential gradient is g. See below

 

[Mister T]

In Newtonian gravity the usual mgh formula is the potential gradient in a uniform field (no tidal effects).

Right. I knew I had gotten something wrong in my thinking when I read @PAllen 's response in which he talked about the potential gradient in a rocket. So the issue is whether or not the potential gradient is uniform. One would need to measure how the potential gradient changes as one ascends that mountain. If it changes uniformly there's no tidal gravity, but if the decrease is nonlinear then that would be a detection of tidal gravity.

 

[jartsa]

Right. I knew I had gotten something wrong in my thinking when I read @PAllen 's response in which he talked about the potential gradient in a rocket. So the issue is whether or not the potential gradient is uniform. One would need to measure how the potential gradient changes as one ascends that mountain. If it changes uniformly there's no tidal gravity, but if the decrease is nonlinear then that would be a detection of tidal gravity.

So we measure the g, right?
So, potential gradient = g
The g does not change uniformly in an accelerating rocket. How fast the g changes is
proportional to g squared. In an accelerating rocket and in an uniform gravity field the g is not uniform, and the change of g is not uniform either.

(g = gravitational acceleration)

 

[David Lewis]

So, potential gradient = g

It might be gh. (It takes energy to move in the direction of motion in an accelerating rocket.)

 

[jbriggs444]

It might be gh. (It takes energy to move in the direction of motion in an accelerating rocket.)

Gravitational potential is approximately gh near the surface of the Earth.

Gravitational force is the gradient of the potential and is, in the Newtonian approximation, $\frac{GM}{r^2}$ which is equal to g at the Earth's surface. Tidal gravity is about how the gravitational force changes with position -- i.e. whether the gravitational force field has a non-vanishing curl.

The term "potential gradient" in this context is a synonym for the gravitational force field -- the gradient of the gravitational potential. So the "potential gradient" is indeed equal to g, not gh.

 

[Dale]

Gravitational potential is approximately gh near the surface of the Earth.

Gravitational force is the gradient of the potential and is, in the Newtonian approximation, $\frac{GM}{r^2}$ which is equal to g at the Earth's surface. Tidal gravity is about how the gravitational force changes with position -- i.e. whether the gravitational force field has a non-vanishing curl.

The term "potential gradient" in this context is a synonym for the gravitational force field -- the gradient of the gravitational potential. So the "potential gradient" is indeed equal to g, not gh.

Oh, you are right. I have added an edit to my incorrect post above. Sorry for any confusion I caused!

 

[vanhees71]

Gravitational potential is approximately gh near the surface of the Earth.

Gravitational force is the gradient of the potential and is, in the Newtonian approximation, $\frac{GM}{r^2}$ which is equal to g at the Earth's surface. Tidal gravity is about how the gravitational force changes with position -- i.e. whether the gravitational force field has a non-vanishing curl.

The term "potential gradient" in this context is a synonym for the gravitational force field -- the gradient of the gravitational potential. So the "potential gradient" is indeed equal to g, not gh.

In Newtonian theory the gravitational field is a potential field, i.e., its curl is vanishing. The field equation of motion reads
$$\vec{\nabla} \cdot \vec{G}=-4 \pi \gamma \rho, \quad \vec{G}=-\vec{\nabla} g,$$
where $\vec{F}$ is the gravitational field, $\gamma$ Newton's gravitational constant, $\rho$ the mass density of the gravitating body and $g$ the potential of the gravitational field. The first equation says that the mass density of matter is the source of the gravitational field, and the 2nd equation says that it's a curl-free field. Finally you get
$$\Delta g=4 \pi \gamma \rho.$$
The solution is given by the Green's function of the Laplace operator, leading to
$$g(\vec{x})=-\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
For a point source in the origin, $\rho(\vec{x})=M \delta^{(3)}(\vec{x})$ you get, of course,
$$g(\vec{x})=-\frac{\gamma}{r}, \quad \vec{G}=-\vec{\nabla} g=-\gamma \frac{\vec{x}}{r^3}, \quad r=|\vec{x}|.$$
The gravitational force on a test particle of mass $m$ is
$$\vec{F}_{\text{G}}(\vec{x})=m\vec{G}(\vec{x})=-m \vec{\nabla} g(\vec{x}).$$

 

[Mister T]

Gravitational force is the gradient of the potential and is, in the Newtonian approximation, $\frac{GM}{r^2}$ which is equal to g at the Earth's surface.

That's the gravitational field, the force is $mg$.

 

[jbriggs444]

That's the gravitational field, the force is $mg$.

Yes. I tend to speak a little fast and loose because there is no good pithy term for the force-per-unit-mass.