special – Page 2 – Physics Notes

Proper Acceleration

The proper acceleration of an observer is that observer’s own sensed acceleration. Its description in terms of felt force would require an excursion into dynamics, but it can be defined kinematically if the observer is aware of some other frame which is known for some reason to be inertial. After acceleration of an observer, the previous frame of that observer appears (to that observer) to be moving in the opposite direction. So if the previous inertial frame (which matched the observer at time [math]t[/math] on the traveller’s own clock) is observed at time [math]t+\Delta t[/math] to have relative velocity [math]-\Delta v[/math], then the limiting ratio [math]a(t)=\lim\limits_{\Delta t\to 0}\frac{\Delta v}{\Delta t}[/math] is what we take as the proper acceleration. (We will ensure later that this purely kinematic definition does indeed match the dynamical definition in terms of felt force but for now just take it as “the” definition of proper acceleration.)

Any observer, [math]\mathcal{O}[/math], lives through a sequence of inertial frames [math]\mathcal{F}_{\mathcal{O}}(t)[/math], labelled by the “proper time” [math]t[/math] showing on [math]\mathcal{O}[/math]’s clock, and each such frame associates any other frame with a relative velocity and a space-time displacement.

Let us use [math]v_{t}[/math], [math]t_{t}[/math], and [math]x_{t}[/math] to represent possible velocities and coordinates relative to the inertial frame [math]\mathcal{F}_{\mathcal{O}}(t)[/math], with [math]v_{t}(\mathcal{F})[/math], [math]t_{t}(\mathcal{F})[/math], and [math]x_{t}(\mathcal{F})[/math] being the values of those variables corresponding to some other frame [math]\mathcal{F}[/math] as seen from [math]\mathcal{F}_{\mathcal{O}}(t)[/math].

If [math]\mathcal{O}[/math] has proper acceleration [math]a(t)[/math], then the frame at time [math]t+\Delta t[/math] is thus seen from that at time [math]t[/math] as having velocity and coordinates given by [math]v_{t}=v_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\Delta t +O((\Delta t)^{2})\approx a(t)\Delta t[/math], [math]x_{t}=x_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\frac{(\Delta t)^{2}}{2}+O((\Delta t)^{3})\approx a(t)\frac{(\Delta t)^{2}}{2}[/math], and [math]t_{t}=t_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=\Delta t +O((\Delta t)^{2})\approx \Delta t[/math] [where the reason this last is not exactly [math]\Delta t[/math] is because, as soon as the velocity changes, the clock measuring that [math]\Delta t[/math] for the accelerated observer is moving (and so slowed down) relative to that of the observer at time [math]t[/math]].

[The inertial frame of the observer at time [math]t+\Delta t[/math] thus has time axis [math]x_{t+\Delta t}=0[/math] given in terms of the time [math]t[/math] coordinates approximately by [math]x_{t}-a(t)\frac{(\Delta t)^{2}}{2}=(a(t)\Delta t)(t_{t}-\Delta t)[/math], and the spacelike axis of simultaneity at [math]t_{t+\Delta t}=0[/math] is given by [math]t_{t}-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})[/math].

So the simultaneity space [math]t_{t+\Delta t}=0[/math] intersects [math]t_{t}=0[/math] at the event where [math]-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})[/math], i.e. [math]x_{t}=-\frac{1}{a(t)}+O((\Delta t)^2)[/math].

For the case of constant acceleration [math]a[/math] this makes all the simultaneity spaces pass through the same “fulcrum event” and this can be used to get a geometric proof that the worldline is hyperbolic and that events beyond the “Rindler horizon” (which is always at [math]x_{t}=-\frac{1}{a}[/math]) remain forever inaccessible to the accelerated observer.

This approach is discussed in more detail in the Wikipedia page on Rindler Coordinates (see also this version on the anonymously authored ‘mathpages’ website, and this discussion by Greg Egan). But in the discussion below I shall take what I think is a more intuitively direct path, and try to say something about the case of non-constant proper acceleration as well.]

As judged by the “birth” frame (corresponding to [math]t=0[/math]), a difference of [math]\Delta t[/math] in the value of [math]t_{t}(\mathcal{F})[/math] corresponds to a difference of [math]\gamma(t)\Delta t[/math] in [math]t_{0}(\mathcal{F})[/math](where [math]\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}[/math]).

[Note: It is also true that, as judged by the frame at time [math]t[/math], a difference of [math]\Delta t[/math] in the value of [math]t_{0}(\mathcal{F})[/math] corresponds to a difference of [math]\gamma(t)\Delta t[/math] in [math]t_{t}(\mathcal{F})[/math] (since the question of which frame appears to be time dilated depends on which frame is making the comparison). And if this still seems confusing then you may need to re-visit the earlier discussion of time dilation.]

So, if we use [math]t_{b}(t)[/math] as shorthand for [math]t_{0}(\mathcal{F}_{\mathcal{O}}(t))[/math], and similarly for [math]x_{b}(t)[/math] and [math]v_{b}(t)[/math], then the relations above can be written in terms of the birthframe coordinates as follows:

[math] t_{b}(t+\Delta t)=t_{b}(t)+\gamma(t) \Delta t+O((\Delta t)^{2})[/math],

[math] x_{b}(t+\Delta t)=x_{b}(t)+v_{b}(t)(\gamma(t) \Delta t)+O((\Delta t)^{2})[/math], and

[math] \begin{align}v_{b}(t+\Delta t)&=v_{b}(t)[+]a(t)\Delta t =\frac{v_{b}(t)+a(t)\Delta t}{1+(v_{b}(t))(a(t)\Delta t)}\\&=(v_{b}(t)+a(t)\Delta t)(1 – (v_{b}(t))(a(t)\Delta t))+O((\Delta t)^{2})\\&= v_{b}(t)+a(t)\Delta t – (v_{b}(t))^{2}(a(t)\Delta t)+O((\Delta t)^{2})\end{align}[/math]

(where [+] is just shorthand for the relativistic velocity sum).

So we have the differential equations

[math]t_{b}'(t)=\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}[/math],

[math]x_{b}'(t)=\gamma(t)v_{b}(t)=\frac{v_{b}(t)}{\sqrt{1-{v_{b}(t)}^{2}}}[/math], and

[math]v_{b}'(t)=a(t)(1-{v_{b}(t)}^{2})[/math].

The last of these is separable and so is easily solved (by cross-multiplying and integrating) to get [math]\int \frac{v_{b}'(t)dt}{(1-{v_{b}(t)}^{2})}=\int a(t)dt[/math].

This then gives [math]\ln{\sqrt{\frac{1+v}{1-v}}}=\int a(t)dt[/math], which gives [math]\frac{1+v}{1-v}=e^{2\int a(t)dt}[/math] and so [math]v_{b}(t)=\frac{e^{2\int a(t)dt}-1}{e^{2\int a(t)dt}+1}=\tanh(\int a(t)dt)[/math].

Note: Here the integrated proper acceleration [math]\int a(t)dt[/math] is the subjective velocity change, which from now on we’ll call [math]v_{s}(0,t)[/math], and [math]v_{b}(t)=\tanh(v_{s}(0,t))[/math] is the apparent velocity of the traveller as inferred by the birthframe observer at the birthframe time when the traveller’s clock reads time [math]t[/math].

For the coordinates, with respect to the birthframe, of [math]\mathcal{O}[/math]’s worldline event at proper time [math]t[/math] we now get

[math]\begin{align}t_{b}(t)&=\int\frac{dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\frac{dt}{\sqrt{1-{\tanh(v_{s}(0,t)}^{2}}}\\&=\int\cosh(v_{s}(0,t))dt\end{align}[/math]

and

[math]\begin{align}x_{b}(t)&=\int\frac{v_{b}(t)dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\cosh(v_{s}(0,t))\tanh(v_{s}(0,t))dt\\&=\int\sinh(v_{s}(0,t))dt\end{align}[/math].

Note: By FTC&ChainRule, [math]x_{b}'(t)=\sinh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)[/math] and [math]t_{b}'(t)=\cosh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)[/math], so the apparent velocity as seen by the birthframe observer is given by [math]v_{b}=\frac{dx_{b}}{dt_{b}}=\frac{x_{b}'(t)}{t_{b}'(t)}=\tanh(v_{s}(0,t))[/math].

This is in agreement with our previously derived formula for [math]v_{b}(t)[/math], but to get it in terms of the [math]t_{b}[/math] coordinate we need to compose with the function giving the proper time [math]t[/math] of [math]\mathcal{O}[/math] when it is at birthframe time [math]t_{b}[/math] – which is the composition inverse of the function [math]t_{b}(t)=\int\cosh(\textstyle\int a(t)dt)dt=\int\cosh(v_{s}(0,t))dt[/math] .

In general, the above integral is not easy to compute explicitly, but in the special case that [math]a(t)=a[/math] is constant, then we have [math]v_{s}(0,t)=at+C_{v}[/math], so [math]v_{b}(t)=\tanh(at+C_{v})[/math] (with [math]v_{b}(0)=0[/math] giving [math]C_{v}=0[/math]), and in that case we can complete the integrals for the coordinates to get

[math]t_{b}(t)=\int\cosh(at+C_{v})dt=\frac{1}{a}\sinh(at+C_{v})+C_{t}[/math] with [math]t_{b}(0)=0[/math] giving [math]C_{t}=0[/math], so [math]t_{b}(t)=\frac{1}{a}\sinh(at)[/math], and

[math]x_{b}(t)=\int\sinh(at+C_{v})dt=\frac{1}{a}\cosh(at+C_{v})+C_{x}[/math] with [math]x_{b}(0)=0[/math] giving [math]C_{x}=-\frac{1}{a}[/math], so [math]x_{b}(t)=\frac{1}{a}\cosh(at)-\frac{1}{a}[/math].

And in this case we can invert to get [math]t(t_{b})=\frac{1}{a}[\sinh^{-1}(at_{b})][/math] which gives

[math]\begin{align}v_{b}(t_{b})&=\tanh(a[\frac{1}{a}[\sinh^{-1}(at_{b})])\\&=\tanh(\sinh^{-1}(at_{b}))=\frac{at_{b}}{\sqrt{1+(at_b)^2}}\end{align}[/math]

Which, as expected, remains less than [math]c[/math] (which is 1 in our units) and so the acceleration observed by the birthframe observer is decidedly NOT constant but rather decreasing to zero as the speed gets closer and closer to [math]c[/math].

[insert brief discussion of Rindler coords and horizon for the case of const accel for all time]

For the case of piecewise constant proper acceleration, say [math]a(t)=a_{i}[/math] between proper times [math]t_{i}[/math] and [math]t_{i+1}[/math], we can use the foregoing analysis to follow how the frame changes in each interval. But in this case the period of constant acceleration is starting at [math]t=t_{i}[/math] rather than [math]t=0[/math], so the conditions for the constants of integration (which depend on [math]i[/math]) are now as follows:

[math]\tanh(a_{i}t_{i}+C_{v,i})=v_{b}(t_{i})[/math] gives [math]C_{v,i}=\tanh^{-1}(v_{b}(t_{i}))-a_{i}t_{i}=v_{s}(0,t_{i})-a_{i}t_{i}[/math], but we’ll often just use [math](a_{i}t_{i}+C_{v,i})=\tanh^{-1}(v_{b}(t_{i}))=v_{s}(0,t_{i})[/math] directly;

[math]\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})+C_{t,i}=t_{b}(t_{i})[/math] gives [math]\begin{align}C_{t,i}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(v_{s}(0,t_{i}))\\ \text{[also}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(\tanh^{-1}(v_{b}(t_{i})))\\&=t_{b}(t_{i})-\frac{1}{a_{i}}v_{b}(t_{i})\sqrt{1+v_{b}(t_{i})^2}\;\; \text{];}\\ \text{and}\end{align}[/math]

[math]\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})+C_{x,i}=x_{b}(t_{i})[/math] gives [math]\begin{align}C_{x,i}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(v_{s}(0,t_{i}))\\ \text{[also}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(\tanh^{-1}(v_{b}(t_{i})))\\&=x_{b}(t_{i})-\frac{1}{a_{i}}\frac{v_{b}(t_{i})}{\sqrt{1+v_{b}(t_{i})^2}}\;\; \text{].}\end{align}[/math]

So, for [math]t_{i}<t<t_{i+1}[/math] we have

[math]\begin{align}v_{b}(t)&=\tanh[\tanh^{-1}(v_{b}(t_{i}))+a_{i}(t-t_{i})]\\&=\frac{v_{b}(t_{i})+\tanh(a_{i}(t-t_{i}))}{1+(v_{b}(t_{i}))(\tanh(a_{i}(t-t_{i})))}=v_{b}(t_{i})[+]\tanh(a_{i}(t-t_{i}))\end{align}[/math]

with

[math]\begin{align}t_{b}(t)&=\int\cosh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\sinh[a_{i}t+C_{v,i}]+C_{t,i}\\&=\frac{1}{a_{i}}\sinh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{t_{b}(t_{i})-\frac{1}{a_{i}}\sinh[v_{s}(0,t_{i})]\}\\&=t_{b}(t_{i})+\frac{1}{a_{i}}\{\sinh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\text{,}\end{align}[/math]

(giving

[math]\begin{align}t_{b}(t_{i+1})-t_{b}(t_{i})&=\frac{1}{a_{i}}\{\sinh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\sinh[v_{s}(0,t_{i+1})]-\sinh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\cosh(v_{s}(0,t))dt\end{align}[/math] since [math]\frac{d}{dt}\sinh[v_{s}(0,t)]=\cosh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)[/math] and [math]\frac{d}{dt}v_{s}(0,t)=a_{i}[/math] for [math]t_{i}<t<t_{i+1}[/math] ),

and

[math]\begin{align}x_{b}(t)&=\int\sinh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\cosh[a_{i}t+C_{v,i}]+C_{x,i}\\&=\frac{1}{a_{i}}\cosh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{x_{b}(t_{i})-\frac{1}{a_{i}}\cosh[v_{s}(0,t_{i})]\}\\&=x_{b}(t_{i})+\frac{1}{a_{i}}\{\cosh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\text{.}\end{align}[/math]

(giving

[math]\begin{align}x_{b}(t_{i+1})-x_{b}(t_{i})&=\frac{1}{a_{i}}\{\cosh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\cosh[v_{s}(0,t_{i+1})]-\cosh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\sinh(v_{s}(0,t))dt\end{align}[/math] since [math]\frac{d}{dt}\cosh[v_{s}(0,t)]=\sinh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)[/math] and [math]\frac{d}{dt}v_{s}(0,t)=a_{i}[/math] for [math]t_{i}<t<t_{i+1}[/math]).

Spacetime diagrams from POV of both twins in the finite acceleration versions of the twin “paradox”.

If properly worded this would have been a good question. From the comments attached to the question we see that the questioner is really asking for two diagrams, one showing the point of view of each of the twins rather than a single diagram showing the coordinate systems of both. And by the ambiguous condition of “constant acceleration” he means constant acceleration as perceived by the stationary observer rather than constant proper acceleration as felt by the traveler.

Of course the case of constant proper acceleration would be more realistic in the sense that it just requires the traveler to experience a constant g-force, whereas constant observed acceleration requires an increasing applied force (which would actually become unbounded as the speed got closer and closer to c). But for a limited time it is possible to keep adjusting the applied force so as to create a constant acceleration relative to the Earth’s frame and in that case the relevant part of the world line (in any inertial frame) is a simple parabolic segment (rather than the hyperbolic segment that would correspond to constant proper acceleration).

With the assumption of constant accelerations in the stay-at-home inertial frame, the spacetime diagram in terms of stay-at-home coordinates is just this:

Here we have a parabolic segment taking the traveler from the start event to where he reaches a cruising speed of , followed by a straight line segment or the bulk of the trip, then a parabolic segment for deceleration, a vertical segment for time spent at the destination, another parabolic segment for acceleration back towards home, straight line for the cruise, and the final parabolic deceleration phase.

In this diagram the coordinates are [math]t_H[/math] for the time on the stay-at-home clock and [math]x_H[/math] for the position in the stay-at-home coordinate system, and we will use the name [math]x^{I}_{HT}[/math] for the function which gives the traveler’s position in stay-at-home coordinates in terms of the time [math]t_H[/math] that the stay-at-home observer perceives as concurrent with the traveler’s arrival at that position. (The superscript I on the function name is to indicate that this is what he infers rather than what he actually sees). So the graph of [math]x_H=x^{I}_{HT}(t_H)[/math] shows what the stay-at-home thinks is the position of the traveler when his (stay-at-home) clock shows time [math]t_H[/math]. This is one interpretation of the homie’s “point of view” but it is not what he actually sees.

What the homie actually sees is delayed by the light travel time from the traveler (just as what we see of a distant star many light years away is not what it is actually happening there now but what happened that many light years ago).

So to get the graph of what the homie actually sees we must look at the point on the previous graph that is the source of a light signal reaching home at time [math]t_H[/math].

We can get a graph of what the homie actually sees by tracing down each light-line from the [math]t_H[/math] axis to where it meets the [math]x_H=x^{I}_{HT}(t_H)[/math] graph and plotting the [math]x_H[/math] value of that event as [math]x^{O}_{HT}(t_{H})[/math] (with the superscript [math]O[/math] identifying the position actually observed at time [math]t_{H}[/math] rather than that which was inferred to be simultaneous).

Now let’s look at things from the point of view of the traveler.

The vertical axis now corresponds to the traveler’s clock time [math]t_T[/math] and the horizontal lines either to distances that we want to associate with that time. If we want to plot what is actually seen by the traveler then for each [math]t_T[/math] we plot the position coordinate corresponding to the distance from which the signal is coming (as determined, eg, by parralax). and if we want to plot where the traveller infers that the homie actually is at the time [math]t_T[/math] we attribute the distance of the source seen at time [math]t_T[/math] to the earlier time [math]t_T-\frac{|x_T|}{c}[/math]

What the traveler actually sees at any event on his worldline is exactly the same as what is seen by an inertial traveler whose world line passes through that event with zero relative velocity (ie for which the worldline is tangent to that of the traveler at that event). Such a tangential traveler sees the values of [math]x^O_{TTE}[/math] and [math]t^O_{TTE}[/math] corresponding to a time [math]x^O_{TTE}/c[/math] earlier in his own frame – so that [math]x^O_{TTE}=[/math] and [math]t^O_{TTE}[/math]

events that are seen by him at the time his clock shows time [math]t_T[/math] with position along that line corresponding to the distance he measures (eg by parallax) to that event; or, in the case of the inferred view those that are inferred to be happening simultaneously with that [math]t_T[/math] click of the clock or to those distance [math]x_T[/math] which he measures (eg by parallax) to whatever event we are talking about.

When the traveler’s clock reads time [math]t_T[/math] he is at the event for which [math]t_H[/math] is such that [math]\int_{0}^{t_H}\frac{1}{1-v(t)^2}dt=t_T[/math] where [math]v(t)=at,v_f,v_f-a(t-t_f),0,-a(t-t_r),-v_f,-v_f+a(t_h-t)…[/math]

To find [math]x^{I}_{TH}(t_T)[/math] we have to use the [math]t_T=[/math]constant simultaneity space for the traveler and find its intersection with the [math]x_H=0[/math] worldline of the homie

The light signal that the traveler is receiving from homie at this event can be seen from the above diagram to come from [math]t_H=[/math] and we have [math]x_H=[/math]

And

Source: (1) Can you draw a spacetime diagram from the POV of both twins in the CATP (constant acceleration twin paradox? Assume constant acceleration during launch and landing. See comment attached to the question for details. – Quora

Is special relativistic time dilation a real effect or just an illusion? Given two inertial frames each observer finds that the clock of the other runs slower than that observer’s own clock. So who is right?

This is a pretty good answer except that I wouldn’t say either of them is right if they think that their perception of relative slowness represents something that is objectively true for all observers.

Time dilation is a real effect on the perceptions of observers (with regard to the rates at which one another’s clocks are ticking). Neither of them is “right” if they think there is any real sense in which the other’s clock is objectively slower. But neither of them is wrong about how it appears to them, so it’s not really an illusion any more than the fact that if they are looking at one another then their ideas of the “forward” direction are opposite to one another. What turns out to be more of an illusion is the sense we all have that there is some absolute standard of time which determines which of two spatially separated events occurs before the other.