1. Dec 26, 2013

### WannabeNewton

Hey guys. I have a question regarding exercise 6.5 in MTW ("Radar Distance Indicator"). The exercise asks the following: http://postimg.org/image/lkn2oujc5/; I did the problem several different ways and I always end up at the same final equation relating $L_0$ to $L$, an equation which doesn't yield the limiting case that the exercise wants. I'd really appreciate it if someone could tell me where I'm going wrong.

Let $p_0$ be the event on the observer's worldline at which the light signal is emitted by the observer; because we have the freedom to do so, we choose a global inertial frame with coordinates $(t,x)$ such that $p_0$ occurs at $\tau_0 = 0$ on the observer's clock i.e. $p_0 = (t_0,x_0) = (g^{-1}\sinh g\tau_0,g^{-1}\cosh g\tau_0) = (0,g^{-1})$.

Furthermore let $p_1$ the event at which the light signal is instantaneously reflected (to be precise we should really emit a new signal instantaneously so that we can use both ingoing and outgoing null geodesics), and let $p'_1$ the event on the observer's worldline that corresponds to the clock time $\frac{\tau}{2}$; note that by Einstein synchrony, $p'_1$ and $p_1$ are simultaneous in the momentarily comoving inertial frame of the observer at $p'_1$.

First we compute $L$; we work in the coordinates $(t,x)$ of the global inertial frame. Consider a momentarily comoving inertial frame at $p'_1 = (t'_1,x'_1)$ with 4-velocity $\xi^{\mu}$ and imagine a space-time displacement vector $\eta^{\mu}$ from $p'_1$ to $p_1 = (t_1,x_1)$. The ruler distance from $p'_1$ to $p_1$ relative to the momentarily comoving inertial frame is given by $L^2 = \eta^{\mu}\eta_{\mu}+ (\xi^{\mu}\eta_{\mu})^2$ which is the result of projecting $\eta^{\mu}$ onto the simultaneity slice of the observer at $p'_1$ and taking the length of the projection. Because we are in flat space-time, the simultaneity slice will be global.

We have $\eta^{\mu} = (t_1-t'_1,x_1-x'_1)$ and $\xi^{\mu} = (\cosh g\frac{ \tau}{2}, \sinh g\frac{\tau}{2}) = (gx'_1,gt'_1)$. Plugging this into the equation for $L^2$ and using the fact that $x^2 - t^2 = g^{-2}$ for events on the observer's worldline, we get $L = gx_1x'_1 - gt_1t'_1 - g^{-1}$.

Now note that $L_0 = \frac{\tau}{2} = g^{-1}\sinh^{-1}(gt'_1) = g^{-1}\ln (gt'_1 + (g^2t'^{2}_{1} + 1)^{1/2}) = g^{-1}\ln (gt'_1 + gx'_1)$.

Furthermore, $gt'_1 + gx'_1 = \frac{(gt'_1 + gx'_1)(x_1 - t_1)}{x_1 - t_1} = gL + 1 + g^2x_1 t'_1 - g^2t_1x'_1$ where I used the fact that $x_1 - g^{-1} = t_1$ which comes from the path of the light signal traveling between $p_0$ and $p_1$.

Thus we have $L_0 = g^{-1}\ln ( 1+gL + g^2x_1 t'_1 - g^2t_1x'_1)$.

As a sanity check, consider the coordinates $(\bar{t},\bar{x})$ of the momentarily comoving inertial frame. As noted above, in this frame we have $\bar{t}'_1 = \bar{t}_1$. Lorentz boosting the coordinates back to those of the global inertial frame, we have $\gamma t_1 - \gamma \beta x_1 = \gamma t'_1 - \gamma \beta x'_1$. Now $g^{-1}\gamma = g^{-1}\cosh g\frac{\tau}{2} = x'_1$ and $g^{-1}\gamma\beta = g^{-1}\sinh g\frac{\tau}{2} = t'_1$ so the previous equation implies that $t_1x'_1 - x_1t'_1 = x'_1 t'_1 - t'_1 x'_1 = 0$ hence $L_0 = x_0\ln ( 1+\frac{L}{x_0})$; this agrees with the equation in wiki: http://en.wikipedia.org/wiki/Rindler_coordinates#Notions_of_distance

Indeed one could have easily arrived at the above result by directly doing all of the relevant calculations in Rindler coordinates however since MTW never introduced Rindler coordinates, I wanted to do everything using the coordinates of a convenient global inertial frame, a method which I agree is more roundabout. Anyways, the problem is that if I take the limit $L\rightarrow g^{-1} = x_0$, I get $L_0 = x_0 \ln 2$ which is certainly not infinite. Could anyone tell me what's going wrong here? Thanks in advance!

2. Dec 26, 2013

### Staff: Mentor

I think it's actually easier to choose the global inertial frame in which $p_1$ (the reflection event) occurs at coordinate time $t = 0$. That makes everything symmetric about the line $t = 0$ (i.e., the coordinate $x$ axis): the light signal is emitted at an event $(-t_1, x_1)$ and the reflected signal is received at the event $(t_1, x_1)$. The coordinates of $p_1$ (the reflection event) are $(0, \bar{x})$. The equation for the worldline of the accelerated observer is $x^2 - t^2 = x_0^2 = 1 / a^2$, where $a$ is the observer's proper acceleration, and the observer's proper time at any point on the worldline is $x_0 \sinh^{-1} \left( t / x_0 \right)$.

Given the above, we have $L = | \bar{x} - x_0 |$ (note that we take the absolute value for the same reason we have a $+ / -$ sign in the formula for $t_1$ below), and $L_0 = x_0 \sinh^{-1} \left( t_1 / x_0 \right)$. But since the light ray's path is null, we must have $\bar{x} = x_1 +/- t_1$ (note that there are two possible signs depending on which way the light goes). So we have $L = | x_1 - x_0 +/- t_1 |$, and we can rearrange this to obtain two solutions for $t_1$:

Ingoing (light emitted in the $-x$ direction): $t_1 = L - x_0 + x_1$.

Outgoing (light emitted in the $+x$ direction): $t_1 = L + x_0 - x_1$.

We then substitute into the equation for $L_0$ to obtain:

Ingoing: $L_0 = x_0 \sinh^{-1} \left( L/x_0 - 1 + x_1 / x_0 \right)$.

Outgoing: $L_0 = x_0 \sinh^{-1} \left( L / x_0 + 1 - x_1/x_0 \right)$.

The ingoing limiting case is now easy to see: as $L \rightarrow x_0$, $x_1 \rightarrow \infty$ (because the reflection point moves closer and closer to the origin of the global inertial frame, so the light signals must be emitted and received closer and closer to the asymptotes of the hyperbolic worldline of the observer), so $L_0 \rightarrow \infty$ as well. I'm not sure about the outgoing limiting case; that appears to yield a finite answer when $L = x_0$. Possibly MTW actually meant to refer only to the ingoing case.

Last edited: Dec 26, 2013
3. Dec 26, 2013

### Staff: Mentor

Just to add a brief comment, these simplify further if we use the equation for the worldline to get $x_1^2 - t_1^2 = x_0^2$. The ingoing solution then becomes $x_1^2 - (L - x_0 + x_1)^2 = x_0^2$, which works out to

$$\frac{x_1}{x_0} = \frac{1}{2} \left( \frac{x_0}{x_0 - L} + \frac{x_0 - L}{x_0} \right)$$

and gives

$$L_0 = x_0 \sinh^{-1} \left[ \frac{1}{2} \left( \frac{x_0}{x_0 - L} - \frac{x_0 - L}{x_0} \right) \right]$$

which obviously diverges as $L \rightarrow x_0$. But a similar procedure for the outgoing case gives

$$L_0 = x_0 \sinh^{-1} \left[ \frac{1}{2} \left( \frac{x_0 + L}{x_0} - \frac{x_0}{x_0 + L} \right) \right]$$

which is, again, finite as $L \rightarrow x_0$.

4. Dec 26, 2013

### WannabeNewton

Thanks Peter! I appreciate the help; for some reason I completely forgot about the ingoing case and focused only on the outgoing case

Also I agree with what you said: MTW must have forgotten to mention that they wanted the reader to only focus on the ingoing case. I can't see any conceivable physical reason for why $L_0$ would diverge in the outgoing case.

EDIT: just for clarification, is $x_0$ in your calculation the spatial coordinate of the event $p_0$ on the observer's worldline simultaneous with the reflection event, with simultaneity being relative to the momentarily comoving inertial frame of the accelerating observer at $p_0$? Furthermore, since you wrote $L = |\bar{x} - x_0|$, that is you had $x_0$ simultaneous with $\bar{x}$, are you working in a global inertial frame that is at rest with respect to the momentarily comoving inertial frame of the observer at $p_0$ but displaced by the amount $x_0$ from the momentarily comoving inertial frame?

Last edited: Dec 26, 2013
5. Dec 26, 2013

### Staff: Mentor

Yes. And it's also $1 / a$, where $a$ is the proper acceleration, so you can substitute $1 / a$ for $x_0$ in my formulas to eliminate the dependence on the particular frame I chose. See further comments below.

Yes, I should have clarified that. Putting in the offset from the MCIF is convenient for this calculation because it makes the asymptotes of the observer's hyperbolic worldline cross at the origin. (Of course this is the same thing that's done when converting to Rindler coordinates.) But, as I noted above, you can equate $x_0$ with $1 / a$ and thereby remove the artifact of this particular frame choice from the formulas.

6. Dec 26, 2013

### WannabeNewton

Great thanks again Peter! I had another reason for making this thread, and that was to ask about the usefulness of radar coordinates in a general context. I was reading a paper that defined radar coordinates for arbitrarily accelerating observers in general curved space-times. Assuming the radar echo experiment is done in a small enough neighborhood of the emission event so that there exists a unique null geodesic to propagate from the emission event to reflection event, the radar distance is defined exactly as above.

Now in flat space-time, which is where the above exercise takes place, the uniqueness of light signals between emission and reflection events isn't an issue. Regardless, we saw above with the ingoing light signals, the radar distance gives a divergent value even for uniformly accelerating observers given that the momentarily comoving ruler distance between the reflection event and its simultaneous pair on the observer's worldline approached the same distance as that of the observer from the Rindler horizon. Thus if we used the same radar distance prescription that works impeccably for inertial observers by averaging over the time it takes the light signal to complete the radar echo experiment then we get a nonsensical measure of distance as determined by the accelerated observer.

As such, when exactly would one use radar coordinates for non-inertial observers? Clearly if the reflection event is close enough to the worldline of the observer, the radar distance agrees with the local ruler distance in the MCIF but if the goal of radar distance is to plot events that are non-locally separated then how does one actually go about using it whilst avoiding divergent radar distance results as in the above exercise?

7. Dec 26, 2013

### Staff: Mentor

I don't think you can. Any accelerated observer has a Rindler horizon, and there are no null paths to or from events behind the Rindler horizon.

8. Jan 3, 2014

### WannabeNewton

Hi Peter. I had another question crop up in relation to the above. Consider the following exercise: http://postimg.org/image/3oo33w2gt/

For starters, even if we restrict ourselves to the back mirror of the light-clock, the statement "by looking at the time it takes the photon to make one to-and-fro bounce in the instantaneous rest frame" is nonsensical as far as I can tell since the back mirror has a different instantaneous rest frame at each instant of time during the round-trip bounce of the photon between the two mirrors.

So what does the exercise mean by this? I think the poor wording is really trying to say the following: calculate the time it takes the photon to make one to-and-fro bounce in the inertial frame instantaneously at rest with respect to the back mirror at the very start of the photon's round-trip bounce and compare it to the time it takes the photon to make one to-and-fro bounce according to a clock carried by an observer comoving with the back mirror. Or is it saying something else?

Furthermore since $\gamma = \gamma(t)$, which $\gamma$ is the exercise referring to? Is it $\gamma(0)$ in whatever inertial frame we choose to work in throughout the round-trip bounce of the photon, so that if we do indeed work in the initial instantaneous rest fame $\gamma = 1$?

The approximation that the proper acceleration $\alpha$ is constant across the light-clock is equivalent to the condition $\frac{\alpha l}{c^2}<<1$; in this approximation the back mirror moves according to $x_1(t) = \frac{1}{2}\alpha t^2$ relative to the initial instantaneous rest frame and the front mirror moves according to $x_2(t) = \frac{1}{2}\alpha t^2 + l$ relative to said frame.

If the exercise really does mean what I said above then the time $\Delta t_1$ for the photon to go from the back mirror to the front mirror relative to the initial instantaneous rest frame is given by $c\Delta t_1 = l + \frac{1}{2}\alpha \Delta t_1^2$ and the time $\Delta t_2$ for the photon to go from the front mirror to the back mirror is given by $l(1 - \frac{\alpha^2 \Delta t_1^2}{c^2})^{1/2} - c\Delta t_2 = \alpha \Delta t_1 \Delta t_2 + \frac{1}{2}\alpha \Delta t_2^2$.

By doing a perturbation series expansion of both of these equations in powers of $\frac{\alpha l}{c^2}$ and working to $O(\frac{\alpha^2 l^2}{c^4})$ we get $\Delta t = \Delta t_1 + \Delta t_2 = \frac{2l}{c}(1 - \frac{1}{2}\frac{\alpha l}{c^2})$.

According to the exercise we're supposed to get $\frac{\Delta t}{\Delta t_0} = \gamma (1 + \frac{1}{2}\frac{\alpha l}{c^2})^{-1} = (1 - \frac{1}{2}\frac{\alpha l}{c^2})$ where $\gamma = 1$ because we're working in the initial instantaneous rest frame (so that $\Delta t = \Delta t_0$ if $\alpha = 0$).

However this would require $\Delta t_0 = \frac{2l}{c}$ which doesn't make sense to me because the time it takes the photon to make one to-and-fro bounce according to a clock carried by an observer comoving with the back mirror is not $\frac{2l}{c}$. Sorry for all the questions and thanks in advance :)

9. Jan 3, 2014

### PAllen

Note that there are two senses of radar coordinates I've seen used. Ones is using radar simultaneity to define foliation, while using proper radial distance within a spatial slice as a radial coordinate. The other is to simply use proper time between signal events over 2 (multiplied by c, as desired) as the radial coordinate. This means that radial coordinate speed of light is forced to be c. This is nice for some purposes but not others (and this is what I think is normally meant by radar coordinates). However, radar simultaneity combined with proper distance for spatial coordinate has its own advantages. In particular, in this case, there is no distance divergence in the ingoing direction (there is still a horizon beyond which the coordinates cannot be extended - but the same is true for Fermi-Normal coordinates).

[Edit: Another point of comparison is to consider inertial motion, then uniform acceleration, then inertial motion. Fermi-Normal coordinates fail to cover a region of spacetime in this case. Radar coordinates (either flavor) cover all of spacetime in such a case. ]

Last edited: Jan 3, 2014
10. Jan 3, 2014

### Staff: Mentor

I think this is more or less right; let me specify more explicitly how I would model the scenario to get the answer that is given. I'll use units in which $c = 1$ as usual since I'm too lazy to type $c$ everywhere.

In the inertial frame we'll use, the back mirror follows the worldline $x^2 - t^2 = 1 / a^2$, and the front mirror follows the worldline $x^2 - t^2 = (1/a + l)^2$. If we consider a null ray that bounces off the back mirror at $t = 0$, then after the bounce (before the bounce is just the time reverse of this, so calculating one half of the trajectory is enough) the null ray follows the worldline $x = 1/a + t$, and it will reach the front mirror where the two worldlines cross, i.e., at a time $t_f$ given by

$$\left( \frac{1}{a} + t_f \right)^2 - t_f^2 = \left( \frac{1}{a} + l \right)^2$$

Expanding out the squares, cancelling terms, and multiplying by $2 / a$ gives

$$t_f = l + \frac{a}{2} l^2 = l \left( 1 + \frac{1}{2} a l \right)$$

If we observe that $l$ is also the time it would take the light signal to go from the back mirror to the front mirror if there were zero acceleration, which we can call $t_0$, we have

$$t_f = t_0 \left( 1 + \frac{1}{2} a l \right)$$

which is starting to look like what the problem wants. For the final step involving $\gamma^{-1}$, see below.

I think this is just a quick and dirty way of converting the coordinate time in the instantaneous rest frame into the proper time of the observer comoving with the back mirror. Given the coordinates $t$ and $x$, the proper time is given by

$$\tau = \frac{1}{a} \tanh^{-1} \left( \frac{t}{x} \right)$$

and we can divide this by $t$ to give a ratio $\gamma$ between coordinate time and proper time. So to convert $t_f$ above into $\tau_f$, we need to include an extra factor of $\gamma$ (or $\gamma^{-1}$; see below) on the RHS.

The final issue is that the problem's formula is stated in terms of frequency $\nu$ instead of time. The usual convention for $\gamma$ is that it is larger than 1 if there is relative motion; but $\tau$ should be smaller than $t$, which would indicate that the above formula involving time should have $\gamma^{-1}$ on the RHS. (Note that there are two competing effects involved: the $\gamma^{-1}$ factor reduces $\tau$ relative to $t_0$, as expected for time dilation; but the other factor *increases* $\tau$ relative to $t_0$, because the front mirror is moving away from the light signal so it takes more time to travel even in the inertial frame than it would if the mirrors were at rest.)

But frequency is time inverted, so having $\gamma^{-1}$ on the RHS in the formula for time would put $\gamma$ on the RHS in the corresponding formula for frequency. That doesn't match what's given in the problem, so either they are defining $\gamma$ in the opposite way, or they are defining $\nu$ and $\nu_0$ in the opposite way. I haven't tried to figure out which it is, but I think it has to be one or the other.

11. Jan 3, 2014

### Staff: Mentor

I should also mention that this observation brings up a significant wrinkle that the problem statement does not address. Suppose we had picked the front mirror as the "fiducial" mirror and done the analysis in terms of a null ray that bounces off the front mirror at time $t = 0$, and computed the travel time $\tau_b$ for the null ray to reach the back mirror, in terms of the front mirror's proper time. What would we have found? The two worldlines of the mirrors would be $x^2 - t^2 = 1/a^2$ for the front mirror, and $x^2 - t^2 = \left( 1/a - l \right)^2$ for the back mirror, and a similar analysis to my previous post would give

$$\tau_b = t_0 \gamma^{-1} \left( 1 - \frac{1}{2} a l \right)$$

Note, of course, the key minus sign.

In other words, for this case, both of the effects mentioned in my quote above work in the *same* direction, to make $\tau$ *smaller* than $t$; the $\gamma^{-1}$ factor works as before, but since the back mirror is moving *towards* the light signal, it takes *less* coordinate time to travel than it would if the mirrors were at rest.

So which is the "right" answer? I think the right answer is, if the correction term in $a l$ is not negligible, then you can't ignore the asymmetry between the mirrors, which means you have to specify which mirror is the "standard" one to use when determining the clock's reading. To put it another way, if you have a light clock and you don't pay attention to this, you are implicitly assuming that the correction term in $a l$ is negligible, so you'd better not subject to clock to an acceleration $a$ large enough to violate that assumption.

12. Jan 4, 2014

### WannabeNewton

Thank you very much Peter. That was very instructive because your method of approaching the problem is from the very general context in which the back mirror and front mirror proper accelerate at different rates in order to maintain a constant proper length of the light clock; but that being said the problem said to assume that the proper acceleration across the light-clock was constant. So both the back mirror and front mirror should have the same proper acceleration $a$ (the book uses $\alpha$ but we can stick to $a$). As noted this approximation amounts to requiring that $\frac{a l}{c^2} << 1$ because this amounts to saying that the variation of $a$ across the characteristic length of the light-clock (which is $l$) is much less than the invariant $c^2$ which makes this entire quantity dimensionless.

First let me note from the outset that the purpose of this problem was to evaluate to what extent an accelerating clock (in this case the light-clock) would be equivalent to an inertial clock (resp. inertial light-clock). For an inertial light-clock, if $\Delta t_0$ represents the time measured between the round-trip bounce of the photon between the mirrors with respect to an observer comoving with the light-clock and the light-clock is moving with some velocity $v$ with respect to an inertial frame then we know that $\Delta t = \gamma \Delta t_0$ where $\Delta t$ is the time between the round-trip as measured in the inertial frame.

The purpose of this problem was to show that when we have an accelerating light-clock, if $\Delta t_0$ is the time as measured by an observer comoving with the back mirror of the light-clock (which in this approximation means the entire light-clock itself since all parts of the light-clock have the same proper acceleration and initial velocity if $\frac{a l}{c^2}<<1$) and $\Delta t$ is the time as measured in an inertial frame through which the light-clock accelerates then $\Delta t = \gamma \Delta t_0 (1 - \frac{1}{2}\frac{al}{c^2})$ i.e. we don't get back the exact time-dilation formula because the accelerating light-clock is a non-ideal clock unless we calibrate it by making $l$ arbitrarily small.

Now bear with me here Peter because I'm still very confused by the problem statement (here it is again for reference: http://postimg.org/image/3oo33w2gt/ )

1. I reread the problem statement and it first said that the light-clock is accelerating through an inertial frame and only then mentioned "the" instantaneous rest frame. So now I'm even more confused as to what "the" instantaneous rest frame the problem is referring to and whether or not $\nu$ (and the associated round-trip time $\Delta t$) actually refers to quantities in this background inertial frame as opposed to quantities related to "the" instantaneous rest frame. How would you interpret this?

2. I don't know if you can access the following paper (http://pra.aps.org/abstract/PRA/v53/i6/p3761_1) but in it an almost identical calculation is done wherein a light-clock is accelerated longitudinally with all parts of the light-clock accelerated by the same magnitude $a$ in the same approximation $\frac{a l}{c^2} << 1$ and the round-trip time $\Delta T$ for the photon is calculated in an inertial frame in which the light-clock has an initial velocity $v$ at the time $t = 0$ when the photon leaves the back mirror.

The result the paper gets is $\Delta T = \frac{2 l}{c}[1 - \frac{a l}{2c^2} + \frac{v^2}{2c^2} + \frac{3av l}{2 c^3}]$. The paper then says that $\Delta T' \equiv \frac{2l}{c}$ "is the period registered in the frame of the accelerating vertical clock" (they use vertical to mean longitudinal), which should be the same as the period measured by an observer comoving with the accelerating light-clock. Is the the claim that $\Delta T' \equiv \frac{2l}{c}$ actually valid, in the approximation that the back and front mirrors have the same proper acceleration, to first order in $\frac{al}{c^2}$?

EDIT: As a side note, if we let $\gamma = \sqrt{1 - \frac{v^2}{c^2}}$ in $\frac{\nu}{\nu_0} = \gamma^{-1}(1 + \frac{al}{2c^2})$, then we get $\frac{\Delta T}{\Delta T_0} = (\frac{\nu}{\nu_0})^{-1} = (1 - \frac{al}{2c^2} + \frac{v^2}{2c^2} + O(\frac{a^2 l^2}{c^4}))$ which is almost like what the paper referenced above has :tongue:

Thanks again!

Last edited: Jan 4, 2014
13. Jan 4, 2014

### Staff: Mentor

Yes, I know the problem said that, but to be honest I don't see why they bothered. It doesn't make the math any easier; you just end up writing one worldline (say the front mirror's, assuming the back mirror's is as I wrote it before) as $\left( x - l \right)^2 - t^2 = 1/a^2$, and when you work things out you end up with pretty much the same equation anyway, with the same amount of computation. So if the math is just as easy doing it right from the start, why not just do so?

Yes, I agree with all of this.

I think this is just sloppy wording; but then again, I think the problem as a whole is formulated in a sloppy manner. To get the answer they're asking for, you have to pick a single inertial frame in which to do the analysis (or, I suppose, you could do it in Rindler coordinates, but let's not go there in this thread... ), and you have to pick which event on which mirror's worldline is the one at which the mirror is momentarily at rest in the chosen inertial frame. That amounts to picking an instantaneous rest frame of the mirror in which to work the problem, so I think "inertial frame" and "instantaneous rest frame" basically mean the same thing. (Put another way, if by "inertial frame" they mean some other frame, so what? You don't use that frame for the analysis, so who cares what frame it is?)

Unfortunately I can't, it's behind a paywall.

Hmm. I think this may be the two effects I referred to in my previous post offsetting each other: the front mirror is moving away from the light reflected from the back mirror, so in an inertial frame a round-trip light pulse covers a distance greater than $l$ each way; but the proper time of an observer comoving with the mirror runs slow compared to the time in an inertial frame, by time dilation. So this may amount to saying that the two effects just cancel each other.

The wrinkle with this, as I mentioned before, is that it only works if you pick the back mirror as the one the light pulse gets reflected off of; if you pick the front mirror instead, the two effects can't cancel (because the back mirror is moving towards the light reflected from the front mirror, so in an inertial frame the round-trip light pulse covers a distance *less* than $l$ each way).

I think this is a reasonable sanity check on the results, but again, I think they're only valid as given if we take the back mirror as the one the light pulse gets reflected off of.