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

[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'.

 

[Mentz114]

This version agrees with the expressions for dx/dt and dy/dt that I obtained in#31 and you obtained in #20.

Your equations are different but that's no problem because mine don't produce a realistic curvature. I think the less said about my #20 and #21 the better .I did not have the benefit of the E&R paper then and I wrote in haste. I hope to have more convincing results soon. I'll PM you or post here.

 

[yuiop]

Your equations are different but that's no problem because mine don't produce a realistic curvature. ... I did not have the benefit of the E&R paper then and I wrote in haste. I hope to have more convincing results soon. I'll PM you or post here.

They are identical. Your version of dx/dt is:

$\frac{dx}{dt}=-\frac{\sinh\left( g\,t\right) }{{\cosh\left( g\,t\right) }^{2}}$

and my version is:

$\frac{dx}{dt} = \frac{ - \tanh(gt)}{cosh(gt)} = \frac{ - (\sinh(gt)/\cosh(gt))}{cosh(gt))} = \frac{ - \sinh(gt)}{{\cosh\left( g\,t\right) }^{2}}$

Eq 8 of the E&R paper gives K as $g \cos(\Theta) /c^2$ which is equal to g when the inclination $(\Theta)$of the path to the 'horizontal' is zero and using units of c=1, which is what we both got.

Additionally, using simple trigonmetry it can be obtained that

g \cos{(\Theta)} = g\cos{(\arcsin{(V_x/c) })}= g \sqrt{1-V_x^2}.

 

[Mentz114]

From my post #20, $\frac{dx}{dt}=-a\,x\,\tanh\left( a\,t\right)$ which is not $\frac{dx}{dt}=-\frac{\sinh\left( g\,t\right) }{{\cosh\left( g\,t\right) }^{2}}$.

Eq 8 of the E&R paper gives K as gcos(θ)/c2 which is equal to g when the inclination (θ)of the path to the 'horizontal' is zero and using units of c=1, which is what we both got

When $t=0$, yes. But the velocities I posted do not satisfy that. I get the same results as E&R when I boost a frame of $ds^2=-dt^2+dx^2+dy^2+dz^2$. I quoted that result above, though, without showing the calculation.

 

[yuiop]

From my post #20, $\frac{dx}{dt}=-a\,x\,\tanh\left( a\,t\right)$ which is not $\frac{dx}{dt}=-\frac{\sinh\left( g\,t\right) }{{\cosh\left( g\,t\right) }^{2}}$.

It is when $a=g$ and $gx = 1/cosh(gt)$.

In post #31 I stated

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

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

It follows that:

$\left(\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} \right) = \frac{1}{\cosh(gt)^2}$

We already know from the Minkowski metric with dtau =0 that:

$\left(\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} \right) = (gx)^2$

so $gx = 1/\cosh(gt)$ and our expressions are the same.

P.S. I added a bit to the end of my last post that you might of missed.

 

[Mentz114]

OK, I'll trust you haven't done anything circular. I'll have to leave this until tomorrow.

 

[Mentz114]

Your result $g \cos{(\varphi)} = g\cos{(\arcsin{(V_x/c) })}= g \sqrt{1-V_x^2}$ certainly agrees with E&R if that is the curvature at $t=0$.

Using frame fields, I can set my light beam off at an elevation angle $\varphi$ giving a worldline $\partial_t+\sin(\varphi)\partial_x+\cos(\varphi)\partial_y$ in the inertial frame. Now I use the vierbein to transform this into the accelerated frame and the result is $\frac{\beta\,\sin\left( \varphi\right) +1}{\sqrt{1-{\beta}^{2}}}\partial_t-\frac{\sin\left( \varphi\right) +\beta}{\sqrt{1-{\beta}^{2}}}\partial_x+\cos(\varphi)\partial_y$. Reparameterizing to $t$ gives

$\frac{dx}{dt} = -\frac{\sin\left( \phi\right) +\beta}{\beta\,\sin\left( \phi\right) +1},\ \ \frac{dy}{dt} = \frac{\sqrt{1-{\beta}^{2}} \,\cos\left( \varphi\right) }{\left( \beta\,\sin\left( \varphi\right) +1\right) }$

which gives the x-y curvature $\mathcal{K}=-\frac{\cos(\varphi)\dot{\beta}}{\sqrt{1-{\beta}^{2}}\left( \beta\,sin\left( \varphi\right) +1\right) }$ which is with $\beta=gt$ , $\mathcal{K}=-\frac{g\,\cos\left( \varphi\right) }{\left( g\,\sin\left( \varphi\right) \,t+1\right) \,\sqrt{1-{g}^{2}\,{t}^{2}}}$

I don't know what to make of this because it looks like the curvature will increase with time, which seems wrong.

 

[yuiop]

...
which gives the x-y curvature $\mathcal{K}=-\frac{\cos(\varphi)\dot{\beta}}{\sqrt{1-{\beta}^{2}}\left( \beta\,sin\left( \varphi\right) +1\right) }$ which is with $\beta=gt$ , $\mathcal{K}=-\frac{g\,\cos\left( \varphi\right) }{\left( g\,\sin\left( \varphi\right) \,t+1\right) \,\sqrt{1-{g}^{2}\,{t}^{2}}}$

I don't know what to make of this because it looks like the curvature will increase with time, which seems wrong.

They define $\varphi$ as being the inclination of the path to the horizontal in F'. This implies that it is not a constant, otherwise they would have defined $\varphi$ as the initial inclination at time time t=t'=0 or they would have defined $\varphi$ as the inclination in F where it is constant. So with constantly changing $\varphi$ it might well be that K is constant over time if it changes by the right compensating amount although I have not worked that out. Then again, if you look at the path of the photon in the plot in post #24, the curvature is continually changing in the accelerating frame, starting out steep and then flattening out. Just for reference, $V_x$ is a constant as it is defined as the x component of the straight light path in the inertial reference frame.

P.S. If $\varphi$ is a constant then your dy/dt is a constant. Is that reasonable? Also you have not defined $\phi$ in your equations. Is it $\pi/2 -\varphi$?

 

[Mentz114]

They define $\varphi$ as being the inclination of the path to the horizontal in F'. This implies that it is not a constant, otherwise they would have defined $\varphi$ as the initial inclination at time time t=t'=0 or they would have defined $\varphi$ as the inclination in F where it is constant. So with constantly changing $\varphi$ it might well be that K is constant over time if it changes by the right compensating amount although I have not worked that out. Then again, if you look at the path of the photon in the plot in post #24, the curvature is continually changing in the accelerating frame, starting out steep and then flattening out. Just for reference, $V_x$ is a constant as it is defined as the x component of the straight light path in the inertial reference frame.

Yes, if $\varphi$ is not constant then the curvature calculations are way off. I need to work that through. It could take a few months.

P.S. If $\varphi$ is a constant then your dy/dt is a constant. Is that reasonable? Also you have not defined $\phi$ in your equations. Is it $\pi/2 -\varphi$?

The $\phi$ is a typo. Should be $\frac{dx}{dt} = -\frac{\sin\left( \varphi\right) +\beta}{\beta\,\sin\left( \varphi\right) +1},\ \ \frac{dy}{dt} = \frac{\sqrt{1-{\beta}^{2}} \,\cos\left( \varphi\right) }{\left( \beta\,\sin\left( \varphi\right) +1\right) }$. dy/dt wth $\cosh(\varphi)=1$ is not constant because $\beta$ is not constant, it is a function of time. This should be clear because $\dot{\beta}$ appears in the curvature.

I integrated my equations for $\varphi=0$, $\beta=gt$ to get

$x(t)=-(1/2)gt^2,\ \ y(t)=-\frac{asin\left( g\,t\right) }{2\,g}-\frac{t\,\sqrt{1-{g}^{2}\,{t}^{2}}}{2}$

and did a parametric plot with $g=1$. It looks OK, in spite of my doubts ( I reversed the sign of y(t) to get a left to right beam).

 

[yuiop]

I integrated my equations for $\varphi=0$, $\beta=gt$ to get

$x(t)=-(1/2)gt^2,\ \ y(t)=-\frac{asin\left( g\,t\right) }{2\,g}-\frac{t\,\sqrt{1-{g}^{2}\,{t}^{2}}}{2}$

and did a parametric plot with $g=1$. It looks OK, in spite of my doubts ( I reversed the sign of y(t) to get a left to right beam).

I notice that the Rindler horizon appears at x = -1/(2g) with your equations when it should be at -1/g in a Rindler metric. I tried plotting it with various values of g. I checked your integrations and they seem OK. I will have to do a x,y plot with my equations and see how it compares.

 

[Mentz114]

I notice that the Rindler horizon appears at x = -1/(2g) with your equations when it should be at -1/g in a Rindler metric. I tried plotting it with various values of g. I checked your integrations and they seem OK. I will have to do a x,y plot with my equations and see how it compares.

There are differences in our approaches, such as your $g$ being a parameter of the metric, and mine being in the boost parameter $gt$. I expect your plot will be very similar because the model falls into the class that satisfies the E&H general prescription.

 

[yuiop]

I integrated my equations for $\varphi=0$, $\beta=gt$ to get

$x(t)=-(1/2)gt^2,\ \ y(t)=-\frac{asin\left( g\,t\right) }{2\,g}-\frac{t\,\sqrt{1-{g}^{2}\,{t}^{2}}}{2}$

and did a parametric plot with $g=1$. It looks OK, in spite of my doubts ( I reversed the sign of y(t) to get a left to right beam).

My equations for x and y in terms of t as given earlier were:

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

The attachment below shows the parametric plot with $V_x=0$ and $g=1$. I have reversed the x and y-axis as you have, so that the acceleration x-axis is up the page. I have also subtracted 1/g so that my coordinate origin coincides with yours. This is equivalent to adding a constant of integration to your value of x to make yours coincide with mine. As you can see from the attached plot my curve appears to be perfectly circular and the Rindler horizon appears at x= -1/g. This circular shape of the curve remains for any value of g.

It can easily be shown that the constant $V_x$ in my equations is equivalent to $sin(\varphi)$ which is therefore also a constant. Substituting this value into the above equations and correcting the offset of the origins to x gives:

$x = \frac{\sqrt{1-\tanh^2(gt)}}{g(1-\sin(\varphi)\tanh(gt))}-\frac{1}{g}, \quad y = \frac{\cos(\varphi) \tanh(gt)}{g(1-\sin(\varphi)\tanh(gt))}$

These equations should now be directly comparable with yours.

If the equations are parametrised in terms of the Minkowski coordinate time T, so that:

x = \sqrt{(\sin(\varphi)+1/g)^2 - T^2} - 1/g, \quad y = T \cos(\varphi)

they are much simpler and the parametric plot remains identical.

 

[Mentz114]

Interesting. My curvature has retained a relativistic factor $1/\sqrt{1-g^2t^2}$. The Taylor series in $t$ around $t=0$ is $-g-\frac{{g}^{3}\,{t}^{2}}{2}-\frac{3\,{g}^{5}\,{t}^{4}}{8}-\frac{5\,{g}^{7}\,{t}^{6}}{16}$. For small $t$ we agree.

This has been an instructive ( and for me error-strewn) exercise which has had a satisfactory result. I'm glad you raised the topic.

 

[yuiop]

For what its worth, I get a curvature value of:

$K = g (-2 \cosh(g*t)+2 /\cosh(g*t) (1/\cosh(g*t)+1)-1)$

This is using Eq13 of http://mathworld.wolfram.com/Curvature.html and setting $\varphi =0$

 

[Mentz114]

The Taylor expansion of the curvature above is $g-4\,{g}^{3}\,{t}^{2}+\frac{5\,{g}^{5}\,{t}^{4}}{3}+...$.

I finally got a proper plot of the light beam starting with a trajectory elevated by 30^o from the horizontal. The scales of x and y axes are not the same so the angle looks too big.

 

[yuiop]

I don't know what to make of this because it looks like the curvature will increase with time, which seems wrong.

Actually, it might be constant over time.
While trying to find where and why we differ, I found an error in the quoted formula for the curvature:

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 $\mathcal{K}(0)=g$

With reference to equation 3 of http://mathworld.wolfram.com/Curvature.html the formula should be:
\begin{align}
\mathcal{K} &= \frac{\dot{\phi}}{\sqrt{(\dot{x}^2+\dot{y}^2)}}
\end{align}
Using $\dot{y}=\frac{1}{\cosh\left( g\,t\right) ^{2}}$, $\dot{x}=-\frac{\sinh\left( g\,t\right) }{\cosh\left( g\,t\right)^{2}}$ and $\phi=arctan(\dot{y}/\dot{x})$,

so $\dot{\phi}= -g/\cosh(g\,t)$ and

$\mathcal{K} = -g$

This means the curvature is independent of time and has constance radius, so the curve is that of a circle.

The above results are for the special case when $\varphi =0$. It turns out after doing the full calculation that the more general result for the curvature is:

$\mathcal{K} = -g \cos(\varphi)$

where $\varphi$ is a constant equal to the initial angle to the horizontal of the trajectory. This last result agrees with the result (Eq 8) obtained in the Ehlers & Rindler paper.

 

[Mentz114]

That was careless of me ( unless the web page has been changed). It makes no difference to my result because $\dot{x}^2+\dot{y}^2=1$. I don't think the curvature can be constant in time because that would imply a circular path for the falling light beam.

 

[yuiop]

... It makes no difference to my result because $\dot{x}^2+\dot{y}^2=1$.

I think this points to the problem. $\dot{x}^2+\dot{y}^2=1$ is the instantaneous tangential velocity of the light particle and the velocity of light is not c according to an observer accelerating in the x direction, when the x coordinate is not constant. By analogy, the coordinate velocity of a radially moving light particle in the Schwarzschild metric is not equal to c, but c(1-2M/r).

I don't think the curvature can be constant in time because that would imply a circular path for the falling light beam.

The path is circular, but the light does not eventually return to the observer, because the path is cut off by the Rindler horizon, so the path is actually a semi circle.

 

[Mentz114]

I think this points to the problem. $\dot{x}^2+\dot{y}^2=1$ is the instantaneous tangential velocity of the light particle and the velocity of light is not c according to an observer accelerating in the x direction, when the x coordinate is not constant. By analogy, the coordinate velocity of a radially moving light particle in the Schwarzschild metric is not equal to c, but c(1-2M/r).

My analysis is done in flat spacetime so the analogy is inapposite, but you're right about the difference between the accelerated and unaccelerated frame. Remember that we 'Newtonized' the velocities in the x,y plane (following E&R ). The 4-velocity tells a different story. In fact, the curvature of $dx/d\lambda$ and $dy/d\lambda$ agrees with your result ! The curvature is -g, always.

The path is circular, but the light does not eventually return to the observer, because the path is cut off by the Rindler horizon, so the path is actually a semi circle.

I'm not sure if this agrees with E&R bcause they say that the acceleration depends on $t$ but is $g$ at $t=0$

Remember that we can't compare our results directly. ** It seems we can, see below. Your acceleration comes from the $g$ parameter in the metric, mine comes from a boost with $\beta=gt$. I'm working in local frames.

I'm confident that my calculations are correct and with the framework specified, make sense physically.

[edit warning] - I redid the curvature of the velocities parametrized with an affine parameter and it agrees with your result.

 

[yuiop]

It is good to see we are converging on some sort of agreement. As you point out I think we were analysing slightly different cases. Mine was from the point of view of a stationary Rindler observer where g and $\varphi$ are constants. I know from the methods I used in the analysis that no approximations were used and the result is exact. Your analysis is for the point of view of successive inertial reference frames that are momentarily co-moving with the accelerating elevator (I think).