B Potential Energy formula in Special Relativty

[Ibix]

Can you give a reference, where the Born coordinates are extended to the entire Minkowski space?

Wikipedia gives the Born coordinates in terms of cylindrical polars on Minkowski spacetime as$$\begin{eqnarray*}
t'&=&t\\
r'&=&r\\
\phi'&=&\phi-\omega t\\
z'&=&z
\end{eqnarray*}$$The associated coordinate basis is fine at any $r'$, and is clearly not a tetrad field at any $r'$. Do you mean something different by Born coordinates?

 

[PeterDonis]

that can be only valid in the case of infinite acceleration

No. It is valid for finite acceleration as well. The finite acceleration case is just as time symmetric as the infinite acceleration case.

But that can be only valid in the case of infinite acceleration. I am speaking here about finite acceleration (formula 2.5). When the (de-)acceleration starts after having traveled $4LY/\gamma$, then additional distance is needed for doing the de-acceleration.

You are wrong. Do the math and see. Note that the Gron paper's calculation using the Lagrangian gives the answer I am giving: see the paragraph right after equation 3.7.

 

[vanhees71]

Yes, sorry. I was in the obviously wrong opinion that we discuss accelerated observers with their reference frames, defined by tetrads with the time-like one the tangent-unit vector along his time-like worldline, but now obviously you want to discuss again something else with the Born coordinates (which have only the usual coordinate singularity at $r=r'=0$).

So what do you want to describe with Born coordinates for $\omega r' \geq 1$?

 

[PeterDonis]

what do you want to describe with Born coordinates

Spacetime. I was giving an example of a non-inertial coordinate chart that covers all of spacetime. I am not claiming that it gives a non-inertial tetrad field that covers all of spacetime.

 

[Ibix]

So what do you want to describe with Born coordinates for $\omega r' \geq 1$?

I thought it was just a counter example to the claim that non-inertial coordinates only cover part of Minkowski space. So nothing particularly practical.

 

[vanhees71]

Sure, that's of course right.

 

[pervect]

I'm curious what specific coordinates you are referring to.

I'm also curious. If Peter digs up the Dolby & Gull paper, and it has a line element, I'd like to see it.

If we require that the constant-time part of the metric be spatially flat, and that the metric coefficients be independent of time, we are led to consider line elements of the form

$$ds^2 = -f(x,y,z)*dt^2 + dx^2 + dy^2 + dz^2$$

Either of these restrictions could be relaxed, of course. But both are nice to have.

Then, if we insist that the acceleration only have components in the x-direction, i.e. we insist that $\Gamma^y{}_{tt} = \Gamma^z{}_{tt} = 0$, we are led to consider line elements of the form

$$ds^2 = -f(x)*dt^2 + dx^2 + dy^2 + dz^2$$

The solutions of this which have zero Riemann curvature, so that they represent the flat space-time of special realtivity, are limited. I am getting only the following possibility:

$$ds^2 = -(c1*x + c2)^2 dt^2 + dx^2 + dy^2 + dz^2$$

as the solution for

$$R_{txtx} = -\frac{1}{4 \, f(x)} \left( -2 \, f''(x)f(x) + f'(x)^2 \right) = 0$$

c1=0 corresponds to the Minkowskii metric, and has ##\Gamma^x{}_{tt} = 0, so it's not an accelerated frame of reference.

c2=0 corresponds to the Rindler metric.

Having both c1 and c2 non=zero doesn't change much. At some value of x, (c1*x+c2) will be zero, leading to the same coordinate singularity that we have with Rindler coordinates.

 

[pervect]

My question was not about releasing an object at height $h$, but about an object, that has at height $h$ already the velocity $0.8 c$. So it has at height $h$ a potential energy plus already a kinetical energy.

The question in this case is: Does the potential energy part of it need to include a factor $\gamma$?

One of the key differences of accelerated frames in SR as compared to Newtonian physics is the presence of pseudo-gravitational time dilation. This is part of your question, I think, but you haven't addressed the issue even conceptually.

If we use Rindler coordinates, we can write the peseudogravitational time dilation factor as being proportional to the height above the Rindler horizon, where one might be tempted to think that "time stops". Saying that time stops oversimplified, of course, time doesn't actually stop, we're just labelling our coordinates differently.

For an object accelerating at 1g, i.e. 10 m/s^2, the Rindler horizon is roughly one light year below the object.

Denoting the gravitational time dilation as $\gamma_g$, we can write:

$\gamma_g = \alpha x$

When x = 1/$\alpha$, the time dilation factor is 1, there is no time dilation. At x=0, we're at the Rindler horizon.

The expression I am getting for a conserved relativistic quantity E that represents energy in this case is:

$$E^2 = \gamma_g^2 \left( \left( m \,c \frac{dx}{d\tau} \right)^2 +\left( m\,c^2 \right)^2\right)$$

or

$$E = \gamma_g \,m\,c^2 \sqrt{ \left( \frac { \frac{dx}{d\tau}} {c} \right)^2 +1}$$

Here $\tau$ is proper time. This is done by methods similar to how we define a conserved energy in a Schwarzschild metric. Unfortunately, it's my own calculation, not something from a textbook.

I don't see anyway to motivate this by a Newtonian discussion. However, we can take the non-relativistic limit. If we chose a height where $\gamma_g \approx 1$, and let $v \approx \frac{dx}{d\tau}$, the expression reduces to

$$E \approx m\,c^2 + \frac{1}{2} m\,v^2$$

Note that the numerical value of our conserved energy depend on our reference height in the accelelerated frame.

 

[PeterDonis]

If we require that the constant-time part of the metric be spatially flat, and that the metric coefficients be independent of time

These are very restrictive assumptions, as your analysis of their implications shows.

 

[PeterDonis]

Note that the numerical value of our conserved energy depend on our reference height in the accelelerated frame.

Better yet, if you translate the coordinates so $x = 0$ corresponds to proper acceleration $\alpha$ (in "natural" units), then you have

$$
\gamma_g = 1 + \frac{a x}{c^2}
$$

where $a$ is the proper acceleration in conventional units (i.e., $\alpha = a / c^2$), and your conserved energy for $|x| \ll 1 / \alpha$ becomes

$$
E \approx m c^2 + \frac{1}{2} m v^2 + m a x
$$

which of course is easily recognizable as analogous to the Newtonian formula including gravitational potential energy as a function of height.

 

[pervect]

In order to do a simple check of the proposed energy formula, I decided to consider a simple case. Suppose we set up an accelerating frame, based on a reference rocket acclelerating at 1 light year/year^2, approximately 9.5 meters/second^2. What velocity do we need to launch a projectile at in order to reach a "height" of 1 light years above our reference rocket?

This is an example of geometric units. We measure all distances in light years, and all times in years. As a consequence, the numerical value of "c" is 1, as is the numerical value of "a".

We predict from the formula that $dx/d\tau$ must equal $\sqrt{3}$ for an object to have the same energy as an object moving with $dx/d\tau=0$ at a height of 1. At a height of 1, $\gamma_g = 1+ax/c^2 = 1+1 = 2$, meaning an object at that height has twice its rest energy.

In order for $dx/d\tau$ to be $\sqrt{3}$, we find $v = dx/dt = \sqrt{3}/2$ , giving $dx/d\tau$ = (dx/dt) \, (dt/d$\tau$) = $\sqrt{3}$.

Here's the plot I did on a Rindler chart. The chart charts the course of the two rockets, our reference rocket starting at d=1, and a second rocket 1 light year "higher" in the accelerated frame. These plots are done in the inertial frame co-moving with the rockets at t=0.

These two rockets are plotted on the graph in black. The projectile launched at a velocity of $\sqrt{3} / 2$ follows a straight line inertial trajectory of $d = \frac{\sqrt{3}}{2}\,t+1$, plotted in green.

The plot confirms the prediction. The green line representing the launched projectile just catches up to the black line of the higher rocket.

The equations are fairly simple if one wants to go into more depth. The reference rocket follows a hyperbola $d^2-t^2=1$. The rocket one light year above the reference rocket follows a hyperbola of $d^2-t^2 = 4$. There are various articles on "hyperbolic motion" that should explain this more - I'm not going to look for any further references unless there are questions. The green line does indeed meet the black line at d=4, t=$2 \,\sqrt{3}$

 

[PeterDonis]

Here is proper time.

Actually, for the observer at $x = 1 / \alpha$ in Rindler coordinates, the observer's proper time is the same as Rindler coordinate time (since this observer has no time dilation in Rindler coordinates). So in the approximation you are using for your energy formula, Rindler coordinate time can be substituted for proper time.

 

[pervect]

Actually, for the observer at x=1/α in Rindler coordinates, the observer's proper time is the same as Rindler coordinate time (since this observer has no time dilation in Rindler coordinates). So in the approximation you are using for your energy formula, Rindler coordinate time can be substituted for proper time.

I don't think this is true.

For instance, a clock near the Rindler horizon will be ticking much differently than the coordinate time there. In general one needs to calculate $d\tau$ from the line element.

The formula for energy was originally derived in Rindler line eleement, but when we modified the energy formula to

$$E = \left(1 + \frac{a\,x}{c^2} \right) \sqrt{1 + \left( \frac{1}{c} \frac{dx}{d\tau} \right)^2}$$

(which I do prefer) the line element in the accelerated coordinate system (t,x,y,z) becomes:

$$ds^2 = -c^2\,d\tau^2 = -(c^2 + a^2\,x^2) dt^2 + dx^2 + dy^2 + dz^2$$
or with the +--- convention, which may be easier for this case,

$$d\tau^2 = (1+ a^2\,x^2 / c^2)dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

 

[pervect]

I had a few more thoughts.

First of all I wanted to give an overview of the concepts used to derive the energy I worked on earlier, formula, now that I'm reasonably hapy with said formula. Previously, I had simply suggested that it was similar to methods used to calculate the energy of test masses orbiting in the Schwarzschild geometry of GR without giving more details.

The basic idea behind the calculation is Noether's theorem. Noether's theorem says, roughly speaking, that a conserved energy is associated with a time-translation symmetry. Fun historical fact - it was concerns over the concept of energy in General Relativity raised by Hilbert, Felix Klein, and others that led to the development of the theorem. It's quite an interesting story, also interesting is the sexism that Noether had to deal with in her everyday work, such as not being able to lecture or be paid for her work by the university.

The time-translation symmetry in this case is the time-translation symmetry of the metric. So writing down the metric (or line element) allows us to calculate the energy simply from the time-translation symmetry associated with it. Well, perhaps the details get a bit complex, but the idea is simple.

A notable feature of the energy relation is that the energy - including the rest energy - goes to zero at the Rindler horizion, where (1+ax/c^2) goes to zero. And below that, it goes negative. What does this mean, physically? My argument is that it means that the concept of an accelerated frame is breaking down at the Rindler horizon. Certainly there are problems about thinking of an object "at rest" at said horizon, it's a null surface. Light could have a constant Rindler coordinate, but it cannot meaningfully be said to be "at rest".

Furthermore, I will go so far as to suggest that since the expression for energy basically only depends on the existence of the time translation symmetry, any sort of coordinates that describe an accelerated frame that explicitly have this symmetry are going to experience similar problems. It has been suggested that there may be a way around this. I suppose I can't absolutely rule out the possibility, but I won't be convinced that there is a way around the issue, which I have called a size constraints on the size of an acclerated frame, until I see a well-behaved line element associated with such a frame. So far, we haven't seen any such thing.

 

[PeterDonis]

a clock near the Rindler horizon will be ticking much differently than the coordinate time there

I meant the observer at $x = 1 / \alpha$ in your original formulation, where the Rindler horizon is at $x = 0$. In the revised formulation I gave, which you have now adopted, this observer is at $x = 0$, and the Rindler horizon is at $x = - c^2 / a$.

 

[Sagittarius A-Star]

$$d\tau^2 = (1+ a^2\,x^2 / c^2)dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

 

[kent davidge]

I thought it was just a counter example to the claim that non-inertial coordinates only cover part of Minkowski space. So nothing particularly practical.

You say this, but the mentioned coordinates (Born) were shown to be singular at $r = 0$. So let me ask: when you physicists say that a coordinate system covers all of spacetime, is it fine to neglect a single pole like the one in this case?

 

[vanhees71]

Well, the coordinate singularity of Born coordinates at $r=0$ are really not an issue. You can simply reintroduce "Cartesian coordinates" via $x=r \cos \varphi$, $y=r \sin \varphi$, and you have a singularity free chart everywhere. From a mathematical point of view it's a chart over all spacetime.

If you want to describe observers by worldlines given by three of the coordinates held fixed and the fourth to be used as the time coordinate, you are restricted to the cylinder $\omega r<1$. Then $t$ is the time coordinate of course. At $\omega r=1$ you have a light-like curve, and for $\omega r>1$ the coordinate lines are all spacelike.

 

[Ibix]

You say this, but the mentioned coordinates (Born) were shown to be singular at $r = 0$.

Fair point. Examples without this issue are available - simply define $x'=x-x_0\sin(t/t_0)$ and otherwise keep Einstein coordinates. Or you could use radar coordinates based on a non-inertial but not eternally accelerating worldline, if you prefer a less contrived example.

 

[vanhees71]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

I think Rindler coordinates can be introduced most easily by defining it via a congruence of time-like world lines, considering points in constant proper acceleration along the $x$ axis of a Minkowski coordinate system $(t,x,y,z)$ with the world-line element
$$\mathrm{d}s^2=\eta_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d}^{\nu}=\mathrm{d} t^2-\mathrm{d} \vec{x}^2,$$
using natural units with $c=1$. In the following we only need to consider the $(t,x)$ plane, because in the change to Rindler coordinate nothing happens with $y$ and $z$. So we can set the corresponding new coordinates simply $y=\eta$ and $z=\zeta$.

Then the congruence of timelike world lines in the planes parallel to the $tx$ plane is defined by the EoM. of observers starting with 0 velocity at $x=\xi$. The solution in terms of a convenient affine parameter $\lambda$ is
$$t=\xi \sinh(\lambda \alpha), \quad x=\xi \cosh(\lambda \alpha).$$
That it's indeed an affine parameter is easily seen from
$$\dot{t}^2-\dot{x}^2=(\xi \alpha)^2,$$
and indeed $\lambda$ is the proper time of the observer on the worldline starting at $\xi=1/\alpha$.
The line element in the Rindler coordinate $(\lambda,\xi,\eta,\zeta)$ is
$$\mathrm{d} s^2 =\xi^2 \alpha^2 \mathrm{d} \lambda^2-\mathrm{d} \vec{x}^{\prime 2}$$
with $\vec{x}'=(\xi,\eta,\zeta)$.
The alternative form mentioned in #76 is also correct. It just defines $\xi$ differently with the initial point of the congruence of uniformly accelerated observers at $x=0$:
$$t=(1/\alpha +\tilde{\xi}) \sinh(\alpha \lambda), \quad x=(1/\alpha+\tilde{\xi}) \cosh(\alpha \lambda)-1/\alpha$$
leading to the mentioned line element
$$\mathrm{d} s^2=(1+\alpha \tilde{\xi})^2 \mathrm{d} \lambda^2-\mathrm{d} \tilde{x}^2$$
where $\tilde{x}^2=(\tilde{\xi},\eta,\zeta)$.

 

[pervect]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

Good catch. I'm pretty sure your psoposal is correct - I'll doulbecheck in the morning, if I remember.

 

[pervect]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

I looked it up - MTW uses a similar line element, but in geometric units. And you're right. There's a simpler argument for why you're right though that doesn't need to be looked up.

Basically, the metric coefficient is $-\gamma_g^2 \, dt^2$. This follows directly from the concept of gravitational time dilation . $\gamma_g$ used to be $\alpha x$ in the (geometrized) Rindler coordinates, and the point at which the factor $\gamma_g$ was unity was at $x=1/\alpha$. With the new coordinates, $\gamma_g = 1 + ax/c^2$ and $\gamma_g$ is unity at x=0.

 

[Sagittarius A-Star]

$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

With the new coordinates, $\gamma_g = 1 + ax/c^2$ and $\gamma_g$ is unity at x=0.

Yes. Gron derives in his book the following equation (4.50):
$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$
This expresses the combined effect of the gravitational and the kinematic time dilation.

• While the traveling twin travels inertially, then $g=0$ and the left term under the square root is "1". That gives the normal time-dilation formula for inertial frames.
• While traveling accelerated, then $g\neq 0$ and the right term under the square root can be neglected for the influence on age difference, if the turnaround-time is defined to be arbitrarily small compared to the inertial travel times.