Why does the frequency of a light clock change when it is accelerated?

[yuiop]

This is an analysis of how a classic Einstein light clock behaves when accelerated orthogonal to its main axis. The results might be considered slightly controversial, so I am putting the details of the calculation here so that they can be checked.Consider a light clock that has its long axis parallel to the y-axis that is being accelerated in the x direction as per this diagram:

https://www.physicsforums.com/attachment.php?attachmentid=62152&stc=1&d=1380051964Let the emitter end of the light clock start at x=y=t=0 and accelerate with constant proper acceleration g.
At times t1, t2, t3 the light clock is at x1, x2, x3 respectively. To find the round trip time of the photon (or the time of one tick of light clock) we need to find $(\Delta t_2 + \Delta t_1) = (t_3-t_1)$

Given t1 we can find t2 starting from the obvious Pythagorus relationship:

$L^2 + (\Delta x_1)^2 = c^2( \Delta t_1)^2$

$L^2 + (x_2-x_1)^2 = c^2 (t_2-t_1)^2$

Using the equations of relativistic accelerating motion given http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html the spatial coordinates can be expressed in terms of the time coordinates via the relationship (using units such that c=1), $x =\sqrt{1/g^2 + t^2}$ giving :

$L^2 + \left(\sqrt{1/g^2+t_2^2}-\sqrt{1/g^2+t_1^1}\right)^2 = (t_2-t_1)^2$

This can rearranged to give t2 in quadratic form:

$t_2^2 - t_2*t_1(2+L^2g^2) + (t_1^2- L^4g^2/4 - L^2) = 0$

and solved in the usual way to give:

$t_2 = t_1(1+L^2g^2/2) \pm \sqrt{t_1^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}$

There are two possible solutions. When the sign of the second term is positive t2>t1 and when it is negative t2<t1 so the positive term is chosen. (Note that by swapping t2 for t1 and vice versa we can find t1 using the solution with the negative second term, if we are given t2.)

The above expression is generic and we can find t3 by substituting t3 for t2 and t2 for t1 in the above expression:

$t_3 = t_2(1+L^2g^2/2) + \sqrt{t_2^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}$

The above calculations show that given t1 we can find t2 and t3 and therefore the round trip time (t3-t1). We can find t1 given any initial conditions such as initial position x1 or initial velocity v1 at the time the photon is emitted at the start of a tick, using the relativistic equations of accelerating motion, given in the link above.

The round trip time (t3-t1) is the coordinate time in the initial inertial frame. The time interval of the light clock (its tick rate) according to a comoving (accelerating) observer is given by:

$(t_3' - t_1') = (1/g)\left(\sinh^{-1}(g*t_3) - \sinh^{-1}(g*t_1)\right)$

Assuming I have done the calculations correctly, the proper frequency of the light clock does not keep time with a co-accelerating ideal clock. The equations are time dependent and therefore also dependent on the instantaneous velocity or position of the clock. This is a slightly curious result, and demonstrates that while the ruler length of a measurement orthogonal to the direction of motion is constant and not subject to length contraction, the radar length is constantly changing, (except for the special case when g/c=1). It turns out that when g/c<1 the frequency of the light clock increases with increasing velocity and conversely the frequency slows down with increasing velocity of the light clock when g/c>1.

Before anyone gets too excited and thinks the accelerating light clock indicates absolute motion, bear in mind that it only records its velocity relative to the time it started accelerating. If the light clock stops accelerating and starts again it restarts from v=0. Effectively it does nothing more than multiplying the output of a traditional accelerometer by the elapsed time to estimate the instantaneous velocity.

Anyway, I won't dwell too long on the implications of the results found here, just in case I have made a glaring mistake. Hopefully someone here will do the honours and check the calculations.

P.S. From some old discussions on PF, we know that a light clock that has constant proper acceleration parallel to its main axis (undergoing Born rigid motion), does maintain constant proper frequency. It follows that a Michelson-Morley type arrangement of two light clocks at right angles to each other will show an interference fringe shift under constant proper acceleration. There is no reason to think that a MMX type arrangement would show a fringe shift when stationary in a gravitational field due to the acceleration of gravity.

This is an interesting exception to the equivalence principle which is not due to tidal effects. How does a light clock distinguish between acceleration in flat space, which causes an actual change in location and velocity, and gravitational acceleration when there is no actual change in location and velocity relative to the gravitational field?

 

[PAllen]

This is an interesting exception to the equivalence principle which is not due to tidal effects. How does a light clock distinguish between acceleration in flat space, which causes an actual change in location and velocity, and gravitational acceleration when there is no actual change in location and velocity relative to the gravitational field?

This implies, with certainty, that there is a mistake in your analysis. I don't have time to look for it now, but the only sort of exceptions to the equivalence principle involve charges or certain special devices that directly sense curvature at a point (which really means they can sense tidal gravity no matter how small the device). A light clock would not be one of these devices.

(An example of a special device of this type is a free falling drop of perfect fluid. It is not perfectly spherical in gravity, and the ratio of its longer dimension to it shorter is constant, independent of size [ as long as it is not too big]. It thus directly measures the second derivative of potential.)

 

[jartsa]

The distance between the two mirrors should be in the equation. (And as already mentioned, the velocity the of mirrors should not be there)

When the distance is large an observer that is co-moving with the mirrors will say something like: "The light takes the fastest route in this curved space-time. Light saves quite a lot of time by taking the route that visits quite high altitude."

 

[Mentz114]

I don't know where the rocket equations you linked to come from. I've done this instead.

The acceleration is 1/x in the x direction. The light clock extends into the y-direction at constant x.

Using $ds^2=x^2\ dt^2 - dx^2 -dy^2 - dz^2$ with $ds^2=0$ we get $x_0dt = dy$ (for some fixed $x_0$). For a clock following the stationary (or comoving) worldline $u^\mu=\partial_t/x$ we get $d\tau=x_0dt$. Combining this with $x_0dt = dy$ gives $d\tau = dy$ which on integration gives $\tau=y_2-y_1$. Which, I think, shows that the light clock will tick as the same rate as any other clock at $x_0$.

 

[Ibix]

$t_2 = t_1(1+L^2g^2/2) \pm \sqrt{t_1^2((1+L^2g^2/2)^2 -1) + L^2(1+L^2g^2/4)}$

For ease of computation, you can simplify this expression to:
$$t_2=\left(L^2g^2/2+1\right)t_1+L\sqrt{\frac{1}{4}L^2g^2+1}\sqrt{g^2t_1^2+1}$$
where I have selected the positive solution, as you did. Also as you say, the expression for $t_3$ as a function of $t_2$ is the same form.

You then give the following formula for the proper time interval:

$(t_3' - t_1') = (1/g)\left(\sinh^{-1}(g*t_3) - \sinh^{-1}(g*t_1)\right)$

Have you tried plotting this? The ancient version of OpenOffice I've got on this thing won't open xlsx, and I'm not going to upgrade an half past midnight... I've only done it in said ancient version of OpenOffice's spreadsheet, but the graph is flat to as many decimal places as that will provide - see the attachment. Since that is in accordance with the expectations of relativity I'm not going to bother going through the algebra to prove that it's constant. I suspect that taking the sinh of both sides will get you there if you want; otherwise differentiate with respect to $t_1$ and see if that is zero.

 

[yuiop]

The distance between the two mirrors should be in the equation. (And as already mentioned, the velocity the of mirrors should not be there)

It is! It is the length L shown next to the red line (representing the light click) in the diagram. L is the initial and final equations. You are right to expect that v should drop out of the final equations, but it does not ...

When the distance is large an observer that is co-moving with the mirrors will say something like: "The light takes the fastest route in this curved space-time. Light saves quite a lot of time by taking the route that visits quite high altitude."

I was not quite sure what you meant by this at first, but I think I do now. (See below).

I don't know where the rocket equations you linked to come from.

That page has been referred to often in these forums and no one has had any issues with before. The equations agree with equations given for Rindler coordinates from other sources except for notation differences and the fact that the rocket equations have a spatial offset of 1/d to make the accelerating particle start from the origin. For example "en.wikipedia.org/wiki/Rindler_coordinates" gives the spatial coordinate in the accelerating frame as $x = \sqrt{X^2-T^2}$. Since the light clock is stationary in the accelerating frame we can replace x with 1/g to obtain $X = \sqrt{1/g^2 + T^2}$ where X is the distance traveled in the inertial frame. This is essentially the equation given in the rocket page when you subtract the offset. Also note that the rocket page uses lower case for the Minkowski coordinates and upper case for the Rindler coordinates and Wikipedia use exactly the opposite convention.

For the time coordinate transformation, Wikipedia gives $T =x \sinh(gt)$ which solves to $t = \sinh^{-1}(T/x)*(1/g) = \sinh^{-1}(gT)*(1/g)$ which is equivalent to the equation on the rocket page except for the reversed notation.

Equation 2.4 of this paper on accelerated motion is also equivalent to the equation I used, except for the reversed notation again and is the same as the rocket page equation except the 1/g offset.

I've done this instead.

The acceleration is 1/x in the x direction. The light clock extends into the y-direction at constant x.

Using $ds^2=x^2\ dt^2 - dx^2 -dy^2 - dz^2$ with $ds^2=0$ we get $x_0dt = dy$ (for some fixed $x_0$). For a clock following the stationary (or comoving) worldline $u^\mu=\partial_t/x$ we get $d\tau=x_0dt$. Combining this with $x_0dt = dy$ gives $d\tau = dy$ which on integration gives $\tau=y_2-y_1$. Which, I think, shows that the light clock will tick as the same rate as any other clock at $x_0$.

I agree that your calculation is correct. I think the disagreement in our conclusions comes about because in your method, the photon is traveling precisely along the y-axis of the accelerating reference frame, while in my method the photon follows a curved path in the accelerating reference frame. In my method the photon only travels a straight path in the Minkowski frame. This straight path transforms to a curved path in the Rindler frame and does not remain on the y axis. I think this is what jartsa was alluding to when he said the photon takes a route that spends more time at 'high altitude'.

<EDIT> I have deleted he last part of this post while I analyse the spreadsheet uploaded by Ibix. His method appears to work (i.e constant frequency) for the accelerated light clock and he is using very similar methods to mine. Looks like I will have to find out where mine has gone wrong. :(

 

[PeterDonis]

the inertially accelerated (unguided channel) transverse light clock in flat space will have a radar length that varies over time.

I haven't worked through this calculation in detail (though I hope to have time to soon), but this doesn't make sense, because the worldlines of both ends of the light clock are integral curves of Killing vector fields, so you can pick any pair of "comoving" events along those worldlines you like and call the time of that pair of events "t = 0", and do a local analysis of the light clock centered on that pair of events, and the analysis will come out the same. It will also come out the same as doing it in a local inertial frame in a gravity field, so if the clock doesn't change in a gravity field, it shouldn't change in the accelerated frame here either.

Put another way, a key characteristic of Killing motions is that nothing changes along them, but your calculation purports to show that something is changing along a Killing motion, so there must be a mistake somewhere.

I do agree that, in the accelerated frame, the light's path should appear curved; that's essentially the same argument that is used to show that gravitational fields must bend light. So Mentz114's analysis, which assumes that dt and dy are the only nonzero differentials along the light's path in the accelerated frame, can't be right as it stands either.

On a quick skim of your OP, I did notice one other thing that's worth asking about: how are you accounting for length contraction along the x axis? As the velocity of the light clock in the x direction increases, it will appear more and more length contracted in the inertial frame in which you are doing the calculation; but that means a length contraction factor should appear somewhere in your Pythagorean theorem formula, and I don't see one.

 

[yuiop]

On a quick skim of your OP, I did notice one other thing that's worth asking about: how are you accounting for length contraction along the x axis? As the velocity of the light clock in the x direction increases, it will appear more and more length contracted in the inertial frame in which you are doing the calculation; but that means a length contraction factor should appear somewhere in your Pythagorean theorem formula, and I don't see one.

Hi Peter. See the edit at the end of my last post. ibix appears to have cracked this. I notice he only uses times t1 and t3 in his spreadsheet, whereas i was using the velocities at time t1 and t3. I am still looking for the differences. You are right that I did not take length contraction into account and I don't think that is what is causing the problem. All the measurements for the Pythagorean calculation are made entirely in the inertial reference frame so length contraction should not be a factor.. i think .. still scratching my head...

P.S. Agree with everything else you said.

 

[yuiop]

OK, I have found the problem, thanks to the input from Ibix. All the calculations in the OP are correct as is the alternative more compact version given by Ibix. The bug was in my spreadsheet.

Now the frequency of the accelerating light clock is not time (or velocity) dependent, but the radar length is generally shorter than the ruler length. This seems to confirm the idea that the photon in the clock is traveling along a curved path in the accelerated frame, so it is not tracking exactly along the y axis. I assume this is why the discrepancy between radar length and ruler length does not show up in Mentz's method.

Thanks guys for all the input. Problem solved (I think).

I have attached the corrected spreadsheet in Open Office format.

 

[yuiop]

For ease of computation, you can simplify this expression to:
$$t_2=\left(L^2g^2/2+1\right)t_1+L\sqrt{\frac{1}{4}L^2g^2+1}\sqrt{g^2t_1^2+1}$$
where I have selected the positive solution, as you did. Also as you say, the expression for $t_3$ as a function of $t_2$ is the same form.

Thanks for checking the calculations Ibix. The equations were basically sound and you pointed me to the bug in my spreadsheet. As a point of interest that might be of use to someone, the negative solution of the quadratic gives the time of the preceding event. If the emission event is time t1, the reflection event is time t2 and the return event is t3 then we can obtain t1 and t3 if we know the time of the reflection event from:

$$t_1=\left(L^2g^2/2+1\right)t_2-L\sqrt{\frac{1}{4}L^2g^2+1}\sqrt{g^2t_2^2+1}$$

and

$$t_3=\left(L^2g^2/2+1\right)t_2+L\sqrt{\frac{1}{4}L^2g^2+1}\sqrt{g^2t_2^2+1}$$

 

[jartsa]

It is! It is the length L shown next to the red line (representing the light click) in the diagram. L is the initial and final equations.

Oh yes, the distance is there. Good.

As the equations in the OP are actually OK, I guess, I have used them find out that:

When the acceleration increases, these kind of light clocks increase their ticking rate, according to an ideal co-accelerating clock.

Larger light clocks differ from ideal clocks more than smaller ones. (large means distance between mirrors is large)

 

[yuiop]

When the acceleration increases, these kind of light clocks increase their ticking rate, according to an ideal co-accelerating clock.

Larger light clocks differ from ideal clocks more than smaller ones. (large means distance between mirrors is large)

Hi jartsa. That does appear to be the case. Just for info, the form of the equations presented by Ibix and as used in his spreadsheet is less prone to computational errors by the software for extremes such as small L.

 

[harrylin]

[..] When the acceleration increases, these kind of light clocks increase their ticking rate, according to an ideal co-accelerating clock. [..]

I'm really puzzled. As already mentioned, doesn't that break the equivalence principle? But wait, if this is correct then it may be interpreted as a prediction of difference in behaviours of different types of clocks in a gravitational field.

 

[PeterDonis]

if this is correct then it may be interpreted as a prediction of difference in behaviours of different types of clocks in a gravitational field.

Yes; changing the acceleration corresponds to changing "height" in the field; changing the ticking rate corresponds to changing the gravitational time dilation factor.

 

[yuiop]

I'm really puzzled. As already mentioned, doesn't that break the equivalence principle? But wait, if this is correct then it may be interpreted as a prediction of difference in behaviours of different types of clocks in a gravitational field.

Yes. A long light clock in a strong gravitational field will also tick at a slightly different rate than an ideal clock, because the light will travel in a curved path in the gravity field as well. It was only when the light clock appeared to be varying it frequency with constant acceleration (due to an error in my spreadsheet) that the equivalence principle was being violated. That is fixed now.

 

[Mentz114]

I hope yuiop won't mind me posting this question here because it's related to the accelerating light clock. I'm trying to find the null geodesic that starts at (x₀,y₀) in the y-direction and intersects y=L at (x₁,y₀+L) . The metric is $ds^2=-a^2x^2dt^2+dx^2+dy^2+dz^2$. Starting with the Lagrangian

$\mathcal{L} = -a^2x^2 \dot{t}^2 + \dot{x}^2+\dot{y}^2$ where the dot means d/dλ, λ being an affine parameter. The Euler-Lagrange equations for x are

$\frac{\partial \mathcal{L}}{\partial x}=-2a^2 x\dot{t}^2$ and $\frac{d}{d\lambda}\frac{\partial \mathcal{L}}{\partial \dot{x}}=2\ddot{x}$ so $\ddot{x}=-a^2\dot{t}^2x$.
Because $\mathcal{L}$ does not depend explicitly on t, y nor z, $\ddot{t}=\ddot{y}=\ddot{z}=0$.

Applying the constraint that this is a null geodesic, $\dot{x}^2+\dot{y}^2=1$ we can write $\dot{t}^2=1/(ax)^2$ so that $\ddot{x}=-1/x$. Integrating this gives $\dot{x}=\lambda(1/x_0-1/x)$.

This looks like $v=at$ which is what I'd expect on a time-like trajectory but not a null one.

At this point I'm stuck. How did I go wrong ?

 

[yuiop]

I hope yuiop won't mind me posting this question here because it's related to the accelerating light clock. ...

You're welcome I would be interested in the answer too and I think it can only enhance this thread.

 

[pervect]

I only skimmed the work, but my first thought is that you assumed that light travels in a straight line. This is only true in inertial coordinates.

What I think you need to compute would be called in optics the "optical path length", to do this I would suggest first to compute the curve that the light actually follows (which will be a null geodesic).

You can get this from the metric, but the path will be a null path, so it won't have a direct length you can compute via the Lorentz interval.

Wiki says http://en.wikipedia.org/wiki/Optical_path_length

n optics, optical path length (OPL) or optical distance is the product of the geometric length of the path light follows through the system, and the index of refraction of the medium through which it propagates.

You can get the "index of refraction" from the metric for an accelerated obserer. Unlike any normal optical material, the effective index of refraction is going to be greater than 1.

I haven't carried through this calculation, or even thought of it much, so I might be missing something or even making a conceptual error. This is my current thinking though.

Intuitively, light will follow an approximately parabolic path. You'll have to angle the beam upwards. As the floor of the spacecraft accelerates, the light will first gain altitude, reach some maximum altitude, then loose altitude.

Wiki's definition seems to be coordinate dependent, which bothers me a bit. I haven't had any luck finding further defitions of optical path length. But using the wiki defintion, you'd take the 3-space length of the parabolic path (ignoring the time part of the metric) then adjust it for the time dilation.

[add]
The basic idea is to find the round-trip time for the light beam as measured by a clock present at the start and end of the round trip. There may be other approaches, this wiki-inspired one seems the easiest.

 

[Mentz114]

I only skimmed the work, but my first thought is that you assumed that light travels in a straight line. This is only true in inertial coordinates.

Taking the light beam in the accelerating elevator scenario I assumed is that the light will fall in a curve. I have not made any explicit assumptions by defining $\dot{t}, \dot{x}$ and $\dot{y}$. These can be any functions of the affine parameter.

What I think you need to compute would be called in optics the "optical path length", to do this I would suggest first to compute the curve that the light actually follows (which will be a null geodesic).

I stated that my aim was to find a null geodesic.

You can get this from the metric, but the path will be a null path, so it won't have a direct length you can compute via the Lorentz interval.

I am using the metric to get the Lagrangian. Is there another way ?

Wiki says http://en.wikipedia.org/wiki/Optical_path_length

You can get the "index of refraction" from the metric for an accelerated obserer. Unlike any normal optical material, the effective index of refraction is going to be greater than 1.

I haven't carried through this calculation, or even thought of it much, so I might be missing something or even making a conceptual error. This is my current thinking though.

Intuitively, light will follow an approximately parabolic path. You'll have to angle the beam upwards. As the floor of the spacecraft accelerates, the light will first gain altitude, reach some maximum altitude, then loose altitude.

Wiki's definition seems to be coordinate dependent, which bothers me a bit. I haven't had any luck finding further defitions of optical path length. But using the wiki defintion, you'd take the 3-space length of the parabolic path (ignoring the time part of the metric) then adjust it for the time dilation.

Thanks, I'll check it out.

[edit]
I think I have a solution. In the local coords on the lift the light appears to travel straight, so the geodesic has no $\partial_x$ component. But when we use the tetrad that converts between the two frames on this geodesic, a $\partial_x$ component appears. This shows that the light will appear to follow a curved path in the other (non-inertial) coords.

I'm still working out the details.

 

[Mentz114]

Carrying on from my previous post. The frame basis of the geodesic $u^\mu=f_0$ is

$f_0=\frac{\cosh\left( a\,t\right) }{a\,x}\partial_t-\sinh\left( a\,t\right) \partial_x,\ f_1= -\frac{\sinh\left( a\,t\right) }{a\,x}\partial_t+\cosh\left( a\,t\right) \partial_x,\ f_2=\partial_y,\ f_3=\partial_z$

from which we get the tetrad $\left(f_a\right)^\mu$ which transforms from the local frame basis to the coordinate basis, i.e. $\left(f_a\right)^\mu\left(f_b\right)^\nu\eta^{ab}=g^{\mu\nu}$.

In the local frame the light clock is set up by shining the light across the elevator along the y-axis, and the local null geodesic is just $u^a=\partial_\hat{t}+\partial_\hat{y}$. Transforming $u^a$ gives

$u^\mu = \left(f_a\right)^\mu u^a=\frac{\cosh\left( a\,t\right) }{a\,x}\partial_t-\sinh\left( a\,t\right) \partial_x+\partial_y$. The norm is still zero, $g_{\mu\nu}u^\mu u^\nu = 0$.

The three components of $u^\mu$ are 'affine' velocities, ie $dx^\mu/d\lambda$ and the final step is to convert them to coordinate velocites by dividing by $dt/d\lambda=\frac{\cosh\left( a\,t\right) }{a\,x}$

This gives $\frac{dx}{dt}=-a\,x\,\tanh\left( a\,t\right)$ and $\frac{dy}{dt}=\frac{a\,x}{\cosh\left( a\,t\right) }$

This looks plausible. At $t=0,\ x=x_0$ the velocity is $ax_0$ in the y-direction and zero in the x-direction. ($ax$ is the coordinate speed of light). Also $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=a^2x^2$, again this is what we require.

The worldline $u^\mu$ satisfies the geodesic equations

$\frac{d^2t}{d\lambda^2}+{\Gamma^t}_{\mu\nu}u^\mu u^\nu=0$ and $\frac{d^2y}{d\lambda^2}=0$ but so far (I could be making a mistake) not the $x$ equation.

I have to say, I'm not content with some aspects of this. But I can't find any other approach to what seems a pretty simple problem.

(I've noticed that I've written $\sinh(at)$ etc when it should be $\sinh(a\tau)$ or $k\sinh(a\lambda)$. I don't think it matters, because the final step of dividing the spatial velocities by $\dot{t}$ cancels the affine factors. )

 

[Mentz114]

I just want to correct some obvious mistakes in my notation, that do not affect the results. The frame basis vectors should be written with a raised index, so $u^\mu=f^0$ (doh!)

$f^0=\frac{\cosh\left( a\,\tau\right) }{a\,x}\partial_t-\sinh\left( a\,\tau\right) \partial_x,\ f^1= -\frac{\sinh\left( a\,\tau\right) }{a\,x}\partial_t+\cosh\left( a\,\tau\right) \partial_x,\ f^2=\partial_y,\ f^3=\partial_z$

and the tetrad is $\left(f^\mu\right)_a$ so that $\left(f^\mu\right)_a \left(f^\nu\right)_b \eta^{ab}=g^{\mu\nu}$. The tranformation of $u^a$ is $\left(f^\mu\right)_a u^a=u^\mu$ and this is what was calculated.

 

[yuiop]

This looks plausible. At $t=0,\ x=x_0$ the velocity is $ax_0$ in the y-direction and zero in the x-direction.

This tells me that what you are analysing is not the light path of a photon in a clock where the source and the mirror are both on the y axis. In your scenario, the photon starts out parallel to the y-axis and falls away towards the Rindler horizon. In the charts below the left chart is the point of view of a Minkowski observer, that observers a light beam (red arrow) emitted at X=x=2, T=t=0. The beam travels vertically upwards on this chart as X is constant. The paths of the accelerating Rindler observers are depicted by the dashed curves and the Rindler horizon by the solid diagonal lines. The right hand chart is the point of view of the Rindler observers and the red path of the light beam can be seen falling away to the vertical Rindler horizon at X=x=0. Approaching x=0 the velocity (dx/dt) of the light beam tends to zero in these coordinates just as the falling coordinate velocity of light in the Schwarzschild metric tends to zero as the event horizon is approached.

This appears to be the situation you are analysing, which is not exactly the light clock situation. In my next post I will post some equations for the light path when it intercepts the mirror located on the same x coordinate as the source, as in the OP. My method is not as elegant as yours and uses my brute force approach, but it may provide some insights.

 

[yuiop]

In the OP, the coordinate time of reflection of the photon ($t_2$) is given in terms of the time of the emmision of the photon ($t_1$) and if we set $t_1=0$ for convenience, the equation significantly reduces to:

$t_2 = L \sqrt{1+L^2g^2/4}$

In that post I was using the rocket equations that have a spatial offset that ensure the Rindler observer is co-located with the Minkowski observer at time T=t=0. From here on I will be using the Rindler transformations and conventions given in Wikipedia. They use lower case for the Rindler coordinates and upper case for the the Minkowski coordinates so using this convention,

$\Delta T = L \sqrt{1+L^2g^2/4}$

This time interval is independent of any spatial offsets so does not need modifying.

$\Delta X/\Delta T$ is the X component of the light velocity in the Minkowski coordinates and will be abreviated to $V_x$. Using Wikipedia, $x = \sqrt{X^2-T^2}$ so $X=\sqrt{x^2 +T^2}$ and:

$V_x = \frac{\Delta X}{\Delta T} = \frac{\sqrt{x_0^2+\Delta T^2}}{\Delta T} = \sqrt{x_0^2/\Delta T^2+1} = \sqrt{1/(g\Delta T)^2 +1}$

Given that the light clock is co-located with a Rindler observer at $x_0=1/g$ and since the photon starts and finishes at $x_0=1/g$, the X coordinate of the photon at any time T is:

$X = 1/g+V_xT$

Substituting this expression for X into the Wikipedia expression for x given above, yields:

$x = \sqrt{(1/g+V_xT)^2 - T^2}$

Differentiating with respect to T gives:

$\frac{dx}{dT} = \frac{V_x(1+gTV_x)-gT}{g\sqrt{(1/g+V_xT)^2 - T^2}} = \frac{V_x(1+gTV_x)-gT}{gx}$

At the apogee of the light path, dx/dT=0 and this occurs at $T = V_x/(gV_x^2-g)$.

Using the Rindler metric with $dS=0$ for light the following relationship is obtained:

$\frac{1}{dt} = \frac{gx}{\sqrt{dx^2+dy^2}}$

$\frac{dT}{dt} = \frac{gx}{\sqrt{(dx/dT)^2+ (dy/dT)^2}}$

Using the chain rule:

$\frac{dx}{dt} = \frac{dx}{dT}*\frac{dT}{dt} = \frac{gx }{\sqrt{1+dy^2/dx^2}}$

Now we need dy/dx for the above equation to be useful. Since y=Y the calculations for dy/dT are much simpler and it can quickly be obtained that:

$\frac{dy}{dT} =\frac{dY}{dT}=\frac{\Delta Y}{\Delta T} =\frac{L}{\Delta T}= \frac{1}{\sqrt{1+L^2g^2/4}}$

Since we now know dx/dT and dy/dT, dy/dx can be obtained from (dy/dT)*(dT/dx) and this value can be inserted in the equation above for dx/dt.

 

[yuiop]

While working out the equations to plot the graphs below, I noticed an error in the equation for $V_x$ in the previous post:

$V_x = \frac{\Delta X}{\Delta T} = \frac{\sqrt{x_0^2+\Delta T^2}}{\Delta T} = \sqrt{x_0^2/\Delta T^2+1} = \sqrt{1/(g\Delta T)^2 +1}$

This should be (with substitution of the expression for $\Delta T$):

$V_x = \frac{\Delta X}{\Delta T} = \frac{\sqrt{x_0^2+\Delta T^2}-x_0}{\Delta T} = \frac{Lg/2}{\sqrt{1+L^2g^2/4}}$

The rest of the previous post is unaffected as $V_x$ is only referred to by its name rather than its expanded form.

To plot the light path (red curve) of the light clock in Minkowski coordinates (the chart on the left in the diagram below) the equation for X in terms of T is used:

$X = 1/g + V_x T$

For the graph, the length of the light clock is L=1.5 and the Rindler observer is at $x_0 = 1/g =1$

In the diagram (e) is the point of emission and (m) is the point of intersection of the light path with the mirror.

The chart on the right shows the light path as seen by the Rindler observer. To plot this path in (x,t) coordinates, a pair of parametric equations with T as the common parameter is used:

$x = \sqrt{X^2-T^2}$
$t = 1/g + \tanh^{-1}(T/X)$

In the above equations, $X = (1/g +V_x T)$ is used so that T is the only variable on the right hand side of the equations. ($g, L$and $V_x$ are constants for the path.)

In the Minkowski chart the light path is not at the usual 45 degree angle that we normally expect for a light path. This is because there is a component of the light velocity in the y direction that is not visible in the x,t coordinates of the chart.

Also note that the light path is curved in Rindler coordinates, but to the Rindler observers, the visual appearance is that the floor of the Einstien elevator and the horizontal light clock are bent upwards and the light appears to travel in a straight path. The fact that the radar length of the light clock is shorter than the ruler length of the apparently curved light clock, tends to support the visual appearance.

I think it would be interesting to create a metric and set of coordinates for an accelerating system where the orthogonal axes are curved to match the light path. In other words the accelerated observers define a straight line as the path followed by a light beam. I know there are people here who have the mathematical ability to do that sort of thing.. .

 

[yuiop]

I have found a way to calculate dx/dt and dy/dt in a more a more useful form that I derive here. As usual I start with a correction to the last post.

The chart on the right shows the light path as seen by the Rindler observer. To plot this path in (x,t) coordinates, a pair of parametric equations with T as the common parameter is used:

$x = \sqrt{X^2-T^2}$
$t = 1/g + \tanh^{-1}(T/X)$

In the above equations, $X = (1/g +V_x T)$ is used so that T is the only variable on the right hand side of the equations. ($g, L$and $V_x$ are constants for the path.)

The equation for t should read $t = (1/g) * \tanh^{-1}(T/X)$ which can be rearranged to:

$T = X * \tanh(gt)$

The correct equation was plotted but a typo got into the text.

Substituting the expression for T into the expression for X gives:

$X = 1/g + V_xX\tanh(gt)$

Solving for X gives:

$X = \frac{1}{g(1-V_x\tanh(gt))}$

Substituting this expression for X into the earlier expression for T gives:

$T = \frac{\tanh(gt)}{g(1-V_x\tanh(gt))}$

The transformation for x is $x=\sqrt{X^2-T^2}$ so:

$x = \frac{1-\tanh^2(gt)}{\sqrt{g^2(1-V_x\tanh(gt))^2}}$

Differentiating x wrt to t gives:

$\frac{dx}{dt} = \frac{(V_x - \tanh(gt))*\sqrt{1-\tanh^2(gt)}}{(V_x \tanh(gt)-1)^2}$

Plotting dx/dt against t with L=1.5 and g = 1 as in the previous post, gives:

https://www.physicsforums.com/attachment.php?attachmentid=62556&stc=1&d=1381114828

it can be seen that dx/dt goes to zero at about t=0.7 which is in good agreement with the time of the photon arriving at apogee in the chart in the last post.

Now for dy/dt,

In the Minkowski reference frame $Y = \Delta Y/ \Delta T * T$ and we know Y=y. Substituting the expression for T given above and $2V_x/(Lg)$ for $\Delta Y/ \Delta T$ gives:$y = Y = \frac{2V_x \tanh(gt)}{Lg^2(1-V_x\tanh(gt))}$

Differentiating y wrt t gives:

$\frac{dy}{dt} = \frac{2 V_x}{Lg (V_x \sinh(g t)-\cosh(g t))^2}$

Plotting dy/dt versus t with the same values of L and g as before:

https://www.physicsforums.com/attachment.php?attachmentid=62557&stc=1&d=1381114828

It can be seen that dy/dt reaches a maximum of about 1.25c near (or at?) the time of apogee.

P.S. Actually, using the Rindler metric with dS=0 for a light path, it is easy to see from $(dy/dt)^2 = (gx)^2 - (dx/dt)^2$ that dy/dt reaches a maximum exactly at the apogee when dx/dt is zero.

 

[yuiop]

Once again, no one found the deliberate error I put the in the last last post, to see if anyone is checking the calculations ... (that is my excuse and I am sticking to it) :tongue:

$X = \frac{1}{g(1-V_x\tanh(gt))}$

Substituting this expression for X into the earlier expression for T gives:

$T = \frac{\tanh(gt)}{g(1-V_x\tanh(gt))}$

The transformation for x is $x=\sqrt{X^2-T^2}$ so:

$x = \frac{1-\tanh^2(gt)}{\sqrt{g^2(1-V_x\tanh(gt))^2}}$

Differentiating x wrt to t gives:

$\frac{dx}{dt} = \frac{(V_x - \tanh(gt))*\sqrt{1-\tanh^2(gt)}}{(V_x \tanh(gt)-1)^2}$

In the above, the expression for x should read:

$x = \frac{\sqrt{1-\tanh^2(gt)}}{g(1-V_x\tanh(gt))}$

Fortunately the derivative following the error was done on the correct expression, so the error does not affect any other calculations or plots in the last post.

As a further error check, (dy/dt)^2 + (dx/dt)^2 should equal (gx)^2 as pointed out by Mentz114.

It turns out that if I calculate (dy/dt) from $\sqrt{(gx)^2 - (dx/dt)^2}$ I get:

$\frac{dy}{dt} = \frac{\sqrt{1-V_x^2}}{ (V_x \sinh(g t)-\cosh(g t))^2}$

which is algebraically equivalent to the result in the previous post:

Differentiating y wrt t gives:

$\frac{dy}{dt} = \frac{2 V_x}{Lg (V_x \sinh(g t)-\cosh(g t))^2}$

Hopefully, that is all the bugs ironed out!

 

[Mentz114]

I'm struggling a bit to follow all this. As the subject has changed to light bending I can recommend this paper ( if you haven't read it )

Juergen Ehlers and Wolfgang Rindler
Local and Global Light Bending in Einstein’s and other Gravitational Theories
General Relativity and Gravitation 29 (1997), p. 519-529.

available here http://pubman.mpdl.mpg.de/pubman/item/escidoc:152994:1/component/escidoc:152993/330688.pdf

In the Einstein elevator they show that the 2D curvature in the plane in which the light propagates is related to the proper acceleration. They use Fermi normal coordinates so the result is very general but approximate in orders of t.

Perhaps you can do the curvature calculation here http://mathworld.wolfram.com/Curvature.html which only requires dy/dx for the curve.

 

[Mentz114]

This tells me that what you are analysing is not the light path of a photon in a clock where the source and the mirror are both on the y axis. In your scenario, the photon starts out parallel to the y-axis and falls away towards the Rindler horizon.

This baffles me. In my earlier posts the acceleration is in the x-direction and the light starts parallel to the y-axis.

I've compared the curvature of my bending light path with Ehlers and Rindler's result and it agrees.

I have to say that I still don't know what you are calculating. What is $V_x$ ?

 

[yuiop]

This baffles me. In my earlier posts the acceleration is in the x-direction and the light starts parallel to the y-axis.

I've compared the curvature of my bending light path with Ehlers and Rindler's result and it agrees.

I have to say that I still don't know what you are calculating. What is $V_x$ ?

$V_x$ is the component of the light velocity in the inertial frame in the x direction (the direction of acceleration). For non zero $V_x$ the light does not start parallel to the y axis. This is required because to hit a target on the y-axis (the mirror) the light has to be be initially aimed upward, like firing a cannonball at a distant target at the same altitude in a gravitational field. The diagram in #22 illustrates what happens if the light has $V_x=0$ and therefore starts parallel to the y axis. The diagram in #24 shows the case where $V_x$ is positive so that the photon rises above the floor of the accelerating elevator, reaches apogee and falls back towards the mirror that is also on the y-axis (or the floor of the elevator if you like).

Due to correcting some errors in subsequent posts, it is probably easier to read or skim the posts in reverse order first, so that you are aware of the corrections when you read the earlier posts.

 

[Mentz114]

$V_x$ is the component of the light velocity in the inertial frame in the x direction (the direction of acceleration). For non zero $V_x$ the light does not start parallel to the y axis. This is required because to hit a target on the y-axis (the mirror) the light has to be be initially aimed upward, like firing a cannonball at a distant target at the same altitude in a gravitational field. The diagram in #22 illustrates what happens if the light has $V_x=0$ and therefore starts parallel to the y axis. The diagram in #24 shows the case where $V_x$ is positive so that the photon rises above the floor of the accelerating elevator, reaches apogee and falls back towards the mirror that is also on the y-axis (or the floor of the elevator if you like).

Due to correcting some errors in subsequent posts, it is probably easier to read or skim the posts in reverse order first, so that you are aware of the corrections when you read the earlier posts.

OK. E&R introduce the angle $\varphi$ between the light ray and the x-axis which introduces a factor of $\cos(\varphi)$ in the curvature and the acceleration. There's no loss of generality by setting $\varphi$ to zero. I'll try to work out the curvature with $V_x=0$.

E&R say that the bending is visible in the accelerating frame and the curvature of the path is qualitatively the same for light and matter. It certainly is true for matter, as you can verify by throwing something horizontally.

The curvature formula for light has $c^3$ in the denominator so it is very small.

I can see two values of dy/dt above. I assume the one with $\sqrt{1-V_x^2}$ is correct.

 

[yuiop]

I can see two values of dy/dt above. I assume the one with $\sqrt{1-V_x^2}$ is correct.

Both values are correct although it is not obvious.

$\sqrt{1-V_x^2} = \frac{2V_x}{Lg}$

For $V_x = 0$, $\sqrt{1-V_x^2} = 1$ and $\frac{2V_x}{Lg} = 0/0$ since the length of the light clock is L=0 when $V_x =0$.

Setting $V_x =0$ in my expressions I get:

$\frac{dx}{dt} = \frac{ - \tanh(gt)}{cosh(gt)}$ and $\frac{dy}{dt} = \frac{1}{ \cosh^2(g t)}$

Now using the above values, it works out that $dx^2/dt^2 + dy^2/dt^2 = g^2x^2 = 1/\cosh^2(gt)$.

Using the value obtained above for $g^2x^2$ and substituting into dx/dt and dy/dt

$\frac{dx}{dt} = -gx \tanh(gt)$ and $\frac{dy}{dt} = \frac{gx}{\cosh(g t)}$

This is the same as the result you got in post#20 for the limited case of a light clock of zero length.

 

[Mentz114]

I've done two calculations of the curvature with different velocities, which you need to check. For a time parameterized curve the x-y curvature is
\begin{align}
\mathcal{K} &= \frac{\dot{\phi}}{(\dot{x}^2+\dot{y}^2)^{3/2}}
\end{align}
where dots indicate differentiation wrt $t$ and $\phi=arctan(\dot{y}/\dot{x})$.

Using $dy/dt=\frac{1}{{\cosh\left( g\,t\right) }^{2}}$ and $dx/dt=-\frac{\sinh\left( g\,t\right) }{{\cosh\left( g\,t\right) }^{2}}$, gives $\mathcal{K}=g\,{\cosh\left( g\,t\right) }^{2}$ so $\mathcal{K}(0)=g$

Using $dy/dt=\frac{1}{{\cosh\left( g\,t\right) }}$ and $dx/dt=--g\,\tanh\left( g\,t\right) \,x$, gives

$\mathcal{K}=\frac{{g}^{2}\,{\cosh\left( g\,t\right) }^{4}\,x}{{\left( {g}^{2}\,{\sinh\left( g\,t\right) }^{2}\,{x}^{2}+1\right) }^{\frac{5}{2}}},\ \ \mathcal{K}(0)=g^2x$

My own calculation for the Minkowski spacetime gives $\mathcal{K}=\gamma g$, $\gamma=1/\sqrt{1-g^2t^2}$ but I used the simple rectilinear chart of the Minkowski spacetime and boosted a static frame with $\beta= gt$.

 

[yuiop]

Now its my turn to ask why you have two different expressions for dy/dt (and dx/dt)? Are your two different velocities two values of $V_x$?

 

[Mentz114]

Now its my turn to ask why you have two different expressions for dy/dt (and dx/dt)? Are your two different velocities two values of $V_x$?

I must apologise because I haven't worked which I should use. The trajectory of the light beam starting out in the y-direction at time $t=0$ in the accelerated frame coordinates is required. I think you did set your $t_0$ to zero so the time part should be OK.

 

[yuiop]

I've done two calculations of the curvature with different velocities, which you need to check. For a time parameterized curve the x-y curvature is
\begin{align}
\mathcal{K} &= \frac{\dot{\phi}}{(\dot{x}^2+\dot{y}^2)^{3/2}}
\end{align}
where dots indicate differentiation wrt $t$ and $\phi=arctan(\dot{y}/\dot{x})$.

Using $dy/dt=\frac{1}{{\cosh\left( g\,t\right) }^{2}}$ and $dx/dt=-\frac{\sinh\left( g\,t\right) }{{\cosh\left( g\,t\right) }^{2}}$, gives $\mathcal{K}=g\,{\cosh\left( g\,t\right) }^{2}$ so $\mathcal{K}(0)=g$

This version agrees with the expressions for dx/dt and dy/dt that I obtained in#31 and you obtained in #20, (when expressed purely as functions of t).

$\dot{y}/\dot{x} = -1/\sinh(gt)$

$\dot{\phi}= g/\cosh(gt))$

${(\dot{x}^2+\dot{y}^2)^{3/2}} = (gx)^3 = 1/\cosh(gt)^3$

so $K = g \cosh(gt)^2$ and K(0) = g.

So yes, everything seems OK here.

Using $dy/dt=\frac{1}{{\cosh\left( g\,t\right) }}$ and $dx/dt=--g\,\tanh\left( g\,t\right) \,x$, gives

$\mathcal{K}=\frac{{g}^{2}\,{\cosh\left( g\,t\right) }^{4}\,x}{{\left( {g}^{2}\,{\sinh\left( g\,t\right) }^{2}\,{x}^{2}+1\right) }^{\frac{5}{2}}},\ \ \mathcal{K}(0)=g^2x$

I think this version fails because when you take the derivative with respect to t, it does not take account of the fact that x is a function of t. Also the expression for dy/dt is missing a factor of gx and the expression for dx/dt should be negative because the photon is 'falling'.