# Accelerating Observer

1. May 12, 2006

### Oxymoron

In Special Relativity,

If an inertial photon emitter, A, emits photons every second and an observer, B (moving away from A at a constant velocity v), intercepts these photons and instantaneously reflects them back to A then we would expect to see the Doppler redshift effect.

My question I would like to pose is this: suppose that B is non-inertial, suppose that B is accelerating away from A (where A remains inertial). If B is accelerating then we could also say that B is inertial in a gravitational field and therefore we should see the Gravitational redshift effect.

Does the accelerating observer still see regular doppler effect? Is the expected gravitational redshift just the Doppler effect in disguise? Is there gravitational redshift at all?

PS. I understand that SR does not work for curved spaces and just because I said that B expects to see grav. redshift does not mean I think there is curvature. What gravitational effects B sees would be, I guess, pseudo-gravitational.

Last edited: May 12, 2006
2. May 12, 2006

### Ich

I´ll give it a shot:
Yes, you will see "gravitational" redshift. Even more, behind B emerges an event horizon (I think it´s called Rindler Horizon). When A reaches it, the same things happen as in a black hole (without the spaghetti effect, of course): B sees A fade away at the horizon and receives no more signals.

3. May 12, 2006

### pervect

Staff Emeritus
The short answer is this. If you work the problem in inertial coordinates, there is no gravitational red-shift.

When you work the problem in non-inertial coordinates, there is gravitational red-shift.

So whether the red-shift is "doppler" or "gravitational" depends only on the coordinate system used. Everyone will agree on the amount of redshift, of course.

Thus observer A, in an inertial frame of reference, using inertial coordinates, sees only doppler shift. (At least if he works the problem correctly!).

Observer B, in a non-inertial frame of reference, using non-inertial coordinates, does see gravitational red-shift. To be more specific, B's metric has a g_00 that varies with position, the redshift associated with this variation of g_00 is mathematically the same as that associated with the g_00 in any other gravitating system (say the Schwarzschild geometry, for instance).

4. May 14, 2006

### Oxymoron

If an observer, B, is accelerating then they are not an inertial observer, right? But, is it possible to align with each point along B's worldline an inertial observer (i.e. diagramatically, a straight line drawn tangent to each point along B's curved worldline) whose speed is identical to B's instantaneous velocity at that point?

5. May 14, 2006

### bernhard.rothenstein

If we follow the procedure you propose in the case of the Doppler Effect with accelerating observers, there are some problems of non locality associated with the fact that two successive wave crests of the wave are not received by the same observer. The approach you propose works only in the case of very small periods. I could furnish references.
(sine ira et studio!)

6. May 14, 2006

### Oxymoron

I would like to now set up (more for my own purposes than anything else) exactly what I am talking about.

• The "trajectory" of an inertial observer $O_1$ is a parameterized curve $C(t)$ on a 4-dimensional manifold, $M^4$ with an everywhere timelike tangent vector $C'(t)$ such that $g(C'(t),C'(t)) < 0$.
• Measurements will be made by this inertial observer in flat Minkowski spacetime. Its coordinates, $\{x^{\mu}\,:\,\mu = 0,1,2,3\} \equiv \{t,x,y,z\}$ will be a coordinate chart for the whole manifold.
• For any other timelike worldline we choose the affine parameter along that curve to be the proper time, $\tau$, defined so that its tangent vector, $C'(\tau)$ is such that $g(C'(\tau),C'(\tau)) < 0$.

Last edited: May 14, 2006
7. May 14, 2006

### Oxymoron

Now that we have set up this tangent vector $C'(\tau)$ how can we define the instantaneous acceleration at some time, $\tau = \tau_i$? One (incorrect, I think) way to define instantaneous acceleration would be to measure how much the tangent vector varies over some time period. But this takes into account velocity over a range of times, not one instant. So what would be a good way to define instantaneous acceleration? Would it involve limits? Something like

$$\lim_{\Delta\tau\rightarrow 0}C'(\Delta\tau) - C'(\tau)$$

8. May 14, 2006

### George Jones

Staff Emeritus
While this is true, one can be more specific, i.e., if the affine parameter along a curve is proper time, $\tau$, then, $C'(\tau)$ is such that $g(C'(\tau),C'(\tau)) = -1$ (for your signature).

If $C'$ is continuous, then

$$\lim_{\Delta\tau\rightarrow 0}C'(\Delta\tau) - C'(\tau) = C' \left( \lim_{\Delta\tau\rightarrow 0} \Delta \tau \right) - C'(\tau) = C'(0) - C'(\tau).$$

Actually the definition is fairly simple - if the curve is parametrized by proper time, the instantaneous 4-accleration at proper time $\tau$ is simply $C''(\tau)$.

Regards,
George

9. May 14, 2006

### pervect

Staff Emeritus
I think what you want is:

$$\lim_{\Delta\tau\rightarrow 0}(\frac{C'(\tau + \Delta\tau) - C'(\tau))}{\Delta\tau}$$

This is almost right - but you have to parallel transport the first vector from time $\tau + \Delta\tau$ to time $\tau$ before performing the subtraction.

After inclinding parallel transport into the definition, you wind up with the 4-acceleration being the covariant derivative of the 4-velocity. (Your tangent vector is just the 4-velocity of a particle following the curve, in case you didn't notice).

The metric associated with whatever coordinate system you chose defines a unique notion of parallel transport - at least according to standard GR. There are a couple of ways to arrive at this conclusion, one of the more intiutive is the geometric construction of Schild's ladder. Giving a metric, you can define the "length" of the side of a quadrilateral. The sides of an infinitesimal quadrilateral are then defined to be "parallel" only when opposite sides of the parallelogram have equal lengths.

Last edited: May 14, 2006
10. May 15, 2006

### bernhard.rothenstein

"Frequency shifts for accelerated sources and observers:an illustration of non-locality in frequency measurements. Eur.J.Phys. 19 569-574 (1998)
At request I could send a copy. Your question offers a possibility to extend the paper according to your scenario,

11. May 15, 2006

### Oxymoron

Ive just finished calculating that a uniformly accelerating observer has hyperbolic motion. Does this sound correct?

I managed to get it down to a system of two first order O.D.E.'s:

$$\frac{\mbox{d}z(\tau)}{\mbox{d}\tau} = At(\tau)+b$$

$$\frac{\mbox{d}t(\tau)}{\mbox{d}\tau} = Az(\tau)+c$$

where b and c are integration constants. Then I solved them using Maple and got

$$z(\tau) = \frac{1}{A}\left(Ae^{A\tau} - Ae^{-A\tau} - z_0\right)$$

$$t(\tau) = -\frac{1}{A}\left(-Ae^{A\tau} - Ae^{-A\tau} + t_0\right)[/itex] then can I simplify these into [tex]z(\tau) = \frac{1}{A}\cosh(A\tau) + z_0$$

$$t(\tau) = \frac{1}{A}\sinh(A\tau) + t_0$$

Is this what I should be expecting? It doesnt quite look right to me. I think it should be: (1/A)cosh(at - 1)+z_0 but I dont know.

Last edited: May 15, 2006
12. May 15, 2006

### Ich

Take a look at http://www.mathpages.com/home/kmath422/kmath422.htm" [Broken]

Last edited by a moderator: May 2, 2017
13. May 15, 2006

### George Jones

Staff Emeritus
I don't have time to take a close look, but you have is not quite right.

For example,

$$\begin{equation*} \begin{split} z(\tau) &= \frac{1}{A}\left(Ae^{A\tau} - Ae^{-A\tau} - z_0\right)\\ &= e^{A\tau} - e^{-A\tau} - \frac{z_0}{A}\\ &= 2 \sinh(A\tau) - \frac{z_0}{A}. \end{equation*} \end{split}$$

Regards,
George

14. May 15, 2006

### bernhard.rothenstein

I think that the second equation in the paper ou popose holds only in the case of very small periods.
sine ira et studio

Last edited by a moderator: May 2, 2017
15. May 15, 2006

### pervect

Staff Emeritus
Yes, that's what you should expect.

This looks right, because it imples that the 4-acceleration is minkowski-perpendicular to the 4-velocity, as it should be, i.e. in geometric units

$$\frac{d t}{d \tau}\frac{d^2 t}{d \tau^2} = \frac{d z}{d \tau} \frac{d^2 z}{d \tau^2}$$

If you evaluate the above using your equations, you find that the equation is satisfied.
The 4-acceleration must be perpendicular to the 4-velocity because the 4-velocity always has a constant magnitude of 1 - if the 4-accelreation were not perpendicular, this could not be true.

Your intermediate results appear to have a typo, but your final results look OK to me.

If we write down the 4-velocities, we get

$$\frac{dz}{d\tau} = \mathrm{sinh} \, A \tau \hspace{.5 in} \frac{dt}{d\tau} = \mathrm{cosh} \, A \tau$$

which is a standard result.

sanity checks:

the magnitude of the 4-velocity is 1, and timelike
--this is OK, since cosh^2 - sinh^2 = 1

the 4-velocity is perpendicular to the 4-acceleration
--this is OK

the magnitude of the 4-acceleration is A

-- this checks, the 4-acceleration is
$$\frac{d^2 t}{d\tau^2} = A \, \mathrm{sinh} \, A \tau \hspace{.5 in} \frac{d^2 z}{d \tau^2} = A \, \mathrm{cosh} \, A \tau$$

so the magnitude squared is A^2 (cosh^2 - sinh^2) = A^2

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

just set c=1

Last edited by a moderator: May 2, 2017
16. May 15, 2006

### George Jones

Staff Emeritus
I have a minor quibble with

Usually, $z_0$ is defined by $z_0 = z(0)$, but this doesn't work for the above equation. Changing $z_0$ to $B$ and setting $\tau = 0$ in the above gives

$$z(0) = \frac{1}{A}\cosh(0) + B$$

so $B = z_0 - A^{-1}$. This results in

$$z(\tau) = \frac{1}{A} \left[ \cosh(A\tau) - 1 \right]+ z_0.$$

More seriously, your first-order equations are symmetrical in z and t, so it's impossible to know which variables sinh and cosh should be assigned. Some important information must have been lost in your derivation.

Regards,
George

Last edited: May 15, 2006
17. May 15, 2006

### pervect

Staff Emeritus
Interchanging z and t yields a space-like geodesic rather than a time-like geodesic, so it's not a valid solution to the problem.

MTW derives a similar solution from the equations below, on pg 166 (without going through the details of the derivation or claiming that their solution is completely general) - MTW omits the additive constants t0 and z0, the only difference from the offered solution.

$$u^i u_i = -1 \hspace{.5 in} u^i a_i = 0 \hspace{.5 in} a^i a_i = g^2$$

where u is the 4-velocity and a is the 4-acceleration. The metric g_ab is Minkowskian (the signature can be inferred from the fact that the norm of a 4-velocity is negative).

we should add for completness to the above equations the fact that
$$a^i = \frac{d u^i}{d \tau}$$

I believe that the above system of equations is equivalent to the system of equations given by Oxymoron.

The solution presented by Oxymoron is not the most general solution, because u(0) = {1,0,0,0}. Therfore somewhere along the line in simplifying the results an assumption was made that the 4-velocity at time 0 was 0. I believe that the general results would add in an additional constant given by performing the following substutution in oxymoron's solution.

$$\tau \rightarrow \tau + k$$

Note that it's fairly easy to solve the equations as presented by Oxymoron by hand, we just take the derivative of both sides of the first equation, substitute the value for dt/dtau on the right hand side by the second equation, and we have a simple linear second order equation in one variable for z.

Last edited: May 15, 2006
18. May 16, 2006

### George Jones

Staff Emeritus
Yes, but the system of first-order equations that Oxymoron wrote down do not reflect this. This is probably why Maple gave the "wrong" solution. Something is missing.

The first equation, assuming that motion takes place along the z-axis, is

$$1 = \left(u^{0}\right)^{2} - \left(u^{3}\right)^{2},$$

which is of the form

$$1 = \cosh^2 \alpha - \sinh^2 \alpha,$$

with $u^0 = \cosh\alpha$ and $u^3 = \sinh\alpha$.

Since the motion is parametrized by proper time, let $\alpha = f(\tau)$, where $f(\tau)$ is a function of proper time that is determined by the third equation in your (pervect's) system,

$$\begin{equation*} \begin{split} -A^2 &= \left( \frac{du^0}{d\tau} \right)^{2} - \left( \frac{du^3}{d\tau} \right)^{2}\\ &= \left[ \sinh^{2} \left( f(\tau) \right) - \cosh^{2} \left( f(\tau) \right) \right] \left( \frac{df}{d\tau} \right)^{2}\\ &= - \left( \frac{df}{d\tau} \right)^{2}. \end{split} \end{equation*}$$

This is easily solved for $f(\tau)$, and then.

$$u^0 = \frac{dt}{d\tau}, \hspace{.5 in} u^3 = \frac{dz}{d\tau}$$

also are easily solved.

I do not believe that Oxymoron's system of equations is equivalent to your system.

Regards,
George

Last edited: May 16, 2006
19. May 16, 2006

### pervect

Staff Emeritus
Well, if we take the original eq's

$$\frac{\mbox{d}z(\tau)}{\mbox{d}\tau} = At(\tau)+b$$
$$\frac{\mbox{d}t(\tau)}{\mbox{d}\tau} = Az(\tau)+c$$

re-writing them in my notation

$$u^0 = A x^1 + c \hspace{.25 in} u^1 = A x^0 + b$$

By differentiating both sides we can say that
$$a^0 = A u^1 \hspace{.25 in} a^1 = A u^0$$

which means that

$$a^0 u^0 = a^1 u^1 = A u^0 u^1$$

Thus we can derive the Minkowski-perpindicular equation from his equations

We can also derive
$$a^1 a^1 - a^0 a^0 = A^2 (u^0 u^0 - u^1 u^1)$$

which sets the magnitude of the 4-acceleration, given that we know the magnitude of the 4-velocity is 1.

so the only thing missing is the relationship

$$u^1 u^1 - u^0 u^0 = -1$$

Offhand, I dont' seem to see how this can be derived from his equations. So it's probably the "missing piece" you are complaining about.

However, this condition comes from assuming that $\tau$ is proper time rather than an affine parameter, and that the path is timelike, both things that were stated.

Last edited: May 16, 2006
20. May 16, 2006

### Oxymoron

First of all there is a small typo in what I wrote. The solutions to the system of ODE's that Maple calculated is

$$z(\tau) = \frac{1}{A}\left(Ae^{A\tau}+Ae^{-A\tau}\right) - z_0$$

$$t(\tau) = -\frac{1}{A}\left(-Ae^{A\tau} + Ae^{-A\tau}\right) + t_0$$

and I just assumed that $C_1/A = z_0$ and $C_2/A = t_0$, and both equations are parameterized according to proper time $\tau$

I believe (after looking in a book) that George's equation for $z(\tau)$ in post #16 is what I am after. But if I just take $z(\tau)$ as I have just written, then (as Pervect pointed out)

$$\textrm{velocity in z direction} = \frac{\mbox{d}z(\tau)}{\mbox{d}\tau} = \sinh(A\tau)[/itex] [tex]\textrm{velocity in t direction} = \frac{\mbox{d}t(\tau)}{\mbox{d}\tau} = \cosh(A\tau)[/itex] [tex]\mbox{velocity} = \frac{\mbox{velocity in z direction}}{\mbox{velocity in t direction}} = \tanh(A\tau)$$

which looks fine to me. Perhaps it is just a question of parameterization?

My aim at this stage is to express what I have in the form:

[tex]X^2 - Y^2 = \mbox{Constant}[/itex]

So that I may proclaim that the shape of uniformly accelerating motion is hyperbolic. Apparently I can work from my two solutions, $z(\tau)$ and $t(\tau)$ and "eliminate" the parameter?? But I do not know how to do this at this moment (I assume I just rearrange things...)

Last edited: May 16, 2006