Doppler & Gravitational Redshift: Non-Inertial Observer in SR?

[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.

 

[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.

 

[pervect]

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.

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).

 

[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?

 

[bernhard.rothenstein]

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?

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!)

 

[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 $g(C'(\tau),C'(\tau)) < 0$.

 

[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

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

 

[George Jones]

• 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 $g(C'(\tau),C'(\tau)) < 0$.

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).

Something like

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

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

 

[pervect]

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

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.

 

[bernhard.rothenstein]

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.

Please have a look at
"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,

 

[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

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

[tex]t(\tau) = -\frac{1}{A}\left(-Ae^{A\tau} - Ae^{-A\tau} + t_0\right)[/itex]

then can I simplify these into

$$z(\tau) = \frac{1}{A}\cosh(A\tau) + z_0$$

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

Is this what I should be expecting? It doesn't quite look right to me. I think it should be: (1/A)cosh(at - 1)+z_0 but I don't know.

 

[Ich]

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

 

[George Jones]

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

 

[bernhard.rothenstein]

doppler,radar echo, acceleration

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

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

 

[pervect]

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

Yes, that's what you should expect.

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

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

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

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.

then can I simplify these into

$$z(\tau) = \frac{1}{A}\cosh(A\tau) + z_0$$

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

Is this what I should be expecting? It doesn't quite look right to me. I think it should be: (1/A)cosh(at - 1)+z_0 but I don't know.

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

[add]You might compare to

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

just set c=1

 

[George Jones]

I have a minor quibble with

$$z(\tau) = \frac{1}{A}\cosh(A\tau) + z_0.$$

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

 

[pervect]

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.

 

[George Jones]

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.

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.

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

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

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

Regards,
George

 

[pervect]

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

George

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.

 

[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)

[tex]\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]

$$\mbox{velocity} = \frac{\mbox{velocity in z direction}}{\mbox{velocity in t direction}} = \tanh(A\tau)$$

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...)

 

[George Jones]

There are still some things wrong. Consider your expression for $z(\tau)$:

[itex]z(\tau) = \frac{1}{A}\left(Ae^{A\tau}+Ae^{-A\tau}\right) - z_0.[/tex]

This means that $z(0) = 2 - z_0$. Typically, $z_0$ is shorthand for $z(0)$, so you have chosen a non-standard usage for $z_0$. In post #16, $z_0$ is cosen such that $z_0 = z(0)$, which is different than yiour choice of $z_0$.

Also, for example, your expession for $z(\tau)$ gives

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

This means that

$$\frac{dz}{d\tau} = 2A \sinh(A \tau),$$

which is not what you have given for $dz / d \tau$.

Regards,
George

 

[pervect]

What you should be trying to show is that z(tau)^2 - t(tau)^2 is constant.

This requires that the origin be located correctly. Suppose that

z^2 - t^2 = a^2

and then z -> z+k, we get

z^2 + 2kz + k^2 - t^2 = a^2, so we see that the formula is not translation invariant. (Similar remarks apply to time translation).

We then proceed to set z0 and t0 in your expression equal to zero, making it the same as George's expression, setting a specific choice of origin. Doing so we get

z^2 - t^2 = (1/A)^2 * (cosh^2 - sinh^2) = 1/A^2

 

[Oxymoron]

I have a problem. I've just gone to all this trouble to show that a uniformly accelerating object traverses a hyperbolic worldline in Minkowski spacetime, so what if the object was traveling in a circle? An observer A in a circular orbit around an observer B has uniform acceleration does it not? Yet the distance between A and B is constant. So how would I draw this on a Minkowski diagram? A must be hyperbolic, B is a straight line, yet the distance between the two must remain the same => A must not be hyperbolic?!? Help me!

 

[pervect]

I have a problem. I've just gone to all this trouble to show that a uniformly accelerating object traverses a hyperbolic worldline in Minkowski spacetime, so what if the object was traveling in a circle? An observer A in a circular orbit around an observer B has uniform acceleration does it not?

Not really. The magnitude of the acceleration of an object moving in a circle is constant, but the direction of the acceleration is constantly changing.

I'm not sure how you derived the equations, but the versons I posted rely on the assumption that the acceleration is always in the same direction.

 

[Oxymoron]

That's the assumption I was missing! The equations are for one direction only. Thanks pervect. :) Incidentally, what would their worldlines look like on a Minkowski diagram?

 

[George Jones]

That's the assumption I was missing! The equations are for one direction only. Thanks pervect. :) Incidentally, what would their worldlines look like on a Minkowski diagram?

At least 3 dimensions are needed - one for time and two for space, say t, x, y.

In B's inertial frame, B's worldline is just the t axis. A's worldline is a spiral that goes up and around B's worldline. this is best drawn using a little bit of perspective.

Regards,
George

 

[robphy]

Here's an animation showing this spiral in spacetime:
http://physics.syr.edu/courses/modules/LIGHTCONE/anim/orb1.mpg

 

[Oxymoron]

If I look at this problem from the point of view of the accelerating observer, O', then is it true to say that an inertial observer, O, (with proper time $\tau$) measures A to have instantaneous velocity $v = tanh(A\tau)$. Then I can set up another inertial observer, O'', who at this point in time measures O' to be at rest. The worldline of O'' coincides with O so can I say that they each have a choice of frame? I am not sure I know what the benefit of setting up O'' in this problem? Does it mean we can set up a good basis with which to measure O' by?

 

[pervect]

Coordinates set up by O'' are very useful inertial coordinates. While O is initally at rest in these coordinates, they area not cooridnates in which O is stationary for all time - O is still accelerating.

O often sets up coordinates in such a manner that O is always at the origin, by using a set of different inertial coordinates of which O'' is just one member.

These coordinates have the property that O is always at the origin (unlike the coordinates of O''), but have the disadvantage that they are not inertial.

 

[Oxymoron]

Suppose I have a worldline of a light source $A(t)$ parameterized by proper time such that $g(A',A')=-1$. As the light source travels forward through time it continuously emits photons in all directions. At each of these emitting events there is a corresponding light cone from $A(t)$.

Now suppose there is an inertial observer on the worldline $B(\tau)$ parameterized by his own proper time $\tau$, and such that $g(B',B') = -1$. The worldline of B intercepts the light cones from A transversely (geometrically obvious). But what this means is that I should be able to "join" the two worldlines, A and B, with intersecting null rays.

Q: Does this mean that I have a Jacobi field, $J$? Does it mean that I have

$$J(0)=A'(t_0)$$
$$J(1)=\lambda B'(\tau_1)$$

Where $\lambda$ is just some constant?

Q: Then, can I interpret $\lambda$ as the delay factor in the frequency of a light signal observed by B from A? That is, if A sends out a light signal at a constant frequency toward B (where the frequency is respect to the unit time interval of A's proper time) then the light signals reach B, $\lambda$ times the unit time interval of B's proper time.

So, does the $\lambda$ which arises naturally from the Jacobi field have an intimate relationship with what we observe as frequency shifting?

 

[gptejms]

Oxymoron,in your calculation of dz/dtau &. dt/dtau are you doing this:-starting from the Rindler metric find out the christoffel symbols and use those in the geodesic equation.If yes,pl. write out the details of your calculation i.e. christoffel symbols etc.

Have you shown that the relativistic doppler effect formula follows from the GR calcution during the period of acceleration?

 

[bernhard.rothenstein]

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.

The paper
Frequency measurements by uniformly accelerating observers
R.Neutze and W.Moreau
Physics Letters A 179 (1993) 389,390
offers an answer to your questions.

 

[vanesch]

Oxymoron,in your calculation of dz/dtau &. dt/dtau are you doing this:-starting from the Rindler metric find out the christoffel symbols and use those in the geodesic equation.If yes,pl. write out the details of your calculation i.e. christoffel symbols etc.

If it can be of any help, I worked out with Mathematica the equations of motion for a geodesic in a general diagonal metric (of which the Rindler metric is a special case), with the connection coefficients (called gammanormal) and all that.
You can find it as the attached document "geodesic1.nb" in
this post:
https://www.physicsforums.com/showpost.php?p=994335&postcount=138

(where I had to do it for another reason, which was a counter argument to Hossenfelder's anti-gravity)

Small detail: I write the metric as: ds^2 = g11 dt^2 + g22 dx^2 +...
note that there is no minus sign before g11, so it has to be included in the definition of g11 itself.

 

[pervect]

For the line element

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

(one form of the rindler metric)

it's fairly easy to compute that the only non-zero Christoffel symbols

$$\Gamma_{ztt} = -\Gamma_{tzt} = -\Gamma_{ttz} = g(1+gz)$$

from the defintion of the Christoffel symbols, and the fact that only g_{tt} has any non-zero derivatives. (This uses the usual form, where the last two indexes commute).

see for example
http://en.wikipedia.org/wiki/Christoffel_symbols$$G_{ijk} = \frac{1}{2} \left( - \partial g_{jk}/ \partial x^i + \partial g_{ik} / \partial x^j + \partial g_{jk} \partial x^i \right)$$

(This is the usual form where symmetric in the last two indices.) Usually it is desired to raise the first index, though.

Raising the index with g^{ij} (easy to calculate and do) then gives:

$$\Gamma^z{}_{tt} = (1+gz)g \hspace{.5 in} \Gamma^t{}_{zt} = \Gamma^t{}_{tz} = \frac{g}{1+gz}$$

 

[vanesch]

For the line element

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

(one form of the rindler metric)

it's fairly easy to compute that the only non-zero Christoffel symbols

$$\Gamma_{ztt} = -\Gamma_{tzt} = -\Gamma_{ttz} = g(1+gz)$$

from the defintion of the Christoffel symbols, and the fact that only g_{tt} has any non-zero derivatives. (This uses the usual form, where the last two indexes commute).

see for example
http://en.wikipedia.org/wiki/Christoffel_symbols


$$G_{ijk} = \frac{1}{2} \left( - \partial g_{jk}/ \partial x^i + \partial g_{ik} / \partial x^j + \partial g_{jk} \partial x^i \right)$$

(This is the usual form where symmetric in the last two indices.) Usually it is desired to raise the first index, though.

Raising the index with g^{ij} (easy to calculate and do) then gives:

$$\Gamma^z{}_{tt} = (1+gz)g \hspace{.5 in} \Gamma^t{}_{zt} = \Gamma^t{}_{tz} = \frac{g}{1+gz}$$

Indeed, that's what I have too.

For fun, I attach a quite general mathematica notebook that does this kind of calculations...