Speed of light during acceleration

[Buckethead]

I'm a bit confused as to why the speed of light changes in an accelerating ship relative to an onboard observer. In other words, on a ship with a clock at the nose and a clock and observer at the tail, in an accelerating ship, the clock at the nose will tic faster. The reason (according to a Feynman lecture) is that tics (flashes of light for example from the nose clock) will be more closely spaced as the accelerating ship overtakes each flash pulse while they are traversing the ship from nose to tail.

If we view the acceleration as a series of constant velocities with instantaneous accelerations in between each different velocity, then at any given time (regardless of the velocity) the time it takes for light to travel from the nose to the tail will always be c. The problem with this of course is that instantaneous step in between velocities. But my difficulty with this is why does the light arrive sooner (in other words travel the distance faster than c) just because the ship is accelerating? If we start with the law that light always travels at c when measured in an inertial frame, why wouldn't light also travel at c in an accelerated frame. Is it just one of those "that's the way it is" situations?

 

[Dale]

If we view the acceleration as a series of constant velocities with instantaneous accelerations in between each different velocity, then at any given time (regardless of the velocity) the time it takes for light to travel from the nose to the tail will always be c.

This is not a valid way to construct a non inertial frame.

But my difficulty with this is why does the light arrive sooner (in other words travel the distance faster than c) just because the ship is accelerating?

In non inertial frames there is no requirement that light travel at c, and most non inertial frames do in fact violate that condition. However, the Dolby and Gull radar coordinates are an example of one way to form non inertial coordinates where light does travel at c.

why wouldn't light also travel at c in an accelerated frame.

Write down the transformation equation. That is why: because you chose that transformation. If you chose a different transformation (like radar coordinates) then it would.

 

[SiennaTheGr8]

I wonder if it isn't best to reserve the word "speed" for Lorentz coordinates (t, x, y, z).

 

[Vanadium 50]

I'm a bit confused as to why the speed of light changes in an accelerating ship

Let's back up. If you have something other than light traveling at constant velocity (i.e. in an inertial frame) and you are accelerating, what does it even mean for it to have a (single) velocity in your frame?

 

[Orodruin]

I wonder if it isn't best to reserve the word "speed" for Lorentz coordinates (t, x, y, z).

The best thing is to use coordinate independent quantities.

 

[jartsa]

I'm a bit confused as to why the speed of light changes in an accelerating ship relative to an onboard observer. In other words, on a ship with a clock at the nose and a clock and observer at the tail, in an accelerating ship, the clock at the nose will tic faster. The reason (according to a Feynman lecture) is that tics (flashes of light for example from the nose clock) will be more closely spaced as the accelerating ship overtakes each flash pulse while they are traversing the ship from nose to tail.

The clocks where ticking at the same rate. According to us. The observer inside the ship was fooled into thinking that the rates were different. He does not know about himself accelerating or moving towards light pulses. Observer's job is to observe, not to think or know anything.

At least this case it's better for us to think about an observer that does not think too much. The observer just looks at the clock and says "that clock seems to be ticking a bit too fast". We, inertial observers and thinkers, know the reason.

 

[Ibix]

The clocks where ticking at the same rate. According to us.

This is not correct. Viewed from an inertial frame, the clocks are moving at different speeds - the length of the rocket in this frame is shrinking.

The observer inside the ship was fooled into thinking that the rates were different.

No. And even of the effect were a purely frame dependent one, that does not make one frame's measurements "right" and another's "fooled into". Any accurate description is equally valid.

He does not know about himself accelerating or moving towards light pulses.

He knows about the acceleration. He has weight. It's one of the very few unambiguous "do it in a closed box" experiments you can do. He can't tell if the acceleration is due to being on a planet or in a rocket, but he knows he's accelerating.

 

[PeroK]

@Buckethead you've had a good few replies already but let me add a further perspective. If light is emitted by a source in an accelerating rocket, then when it is received the rocket is moving relative to the instantaneous IRF of the source at the time of emission.

The received light, therefore, must be Doppler shifted. Blue shifted for light emitted at the front and received at the rear, and red shifted for the other way round.

It's a small further step to attribute this shift to time dilation between the rear and the front.

I can't see much to be gained by trying to interpret the Doppler shift as some sort of non inertial variation of $c$.

 

[jartsa]

This is not correct. Viewed from an inertial frame, the clocks are moving at different speeds - the length of the rocket in this frame is shrinking.

I know. I ignored that effect as it is very small at non-relativistic speeds.

He knows about the acceleration. He has weight. It's one of the very few unambiguous "do it in a closed box" experiments you can do. He can't tell if the acceleration is due to being on a planet or in a rocket, but he knows he's accelerating.

So we know more than him. We know for sure that he's in a rocket.

 

[pervect]

I am interpreting this question as basically being about "what is an accelerated frame in special relativity". I would say that the usual description of an accelerated frame in special relativity is known as "Rindler coordinates".

Doing this without a lot of mathematics is challenging, the best "low math" answer I can come up with, after a bit of thought,is to say that an accelerated frame can be represented by the Rindler coordinate chart.

From the wiki on the topic, it looks like this.

https://en.wikipedia.org/wiki/Rindler_coordinates#/media/File:Rindler_chart.svg

(It doesn't seem like I can embed the .svg image in PF posts)

The chart shows lines of constant time (as measured in the accelerated frame), represented by straight lines, and lines of constant position (in the accelerated frame) , represented by hyperbolae. The straight lines of constant time are labelled on the chart with "t=-0, t=1", etc. The reason that the curves of constant position are hyperbolae can be derived in detail by considering the mathematics of the position of an observer with a constant proper acceleration as a function of time as seen in an inertial frame of reference. When one solves the equations, this curve is a hyperbola.

If one looks at the lines of simultaneity on the chart, labelled t=0, t=1, t=2, etc, one can see they are not parallel. This lack of being parallel is what is non-intuitive about the nature of an accelerated frame in special relativity, and is basically a result of the relativity of simultaneity. An accelerated observer changes their velocity as a function of time, and as a consequence of this change in velocity, their notion of simultaneity also changes with time. On a diagram, it winds up looking like the mentioned chart.

While this doesn't involve much math, it does require understanding space-time diagrams, and also some knowledge of the concept of the relativity of simultaneity. These are both areas that are not well understood by the lay audience, I've found, but I believe that they are vital to understanding the nature of an accelerated frame,

One may note that the diagram does not cover all of space-time. A vertical line through the origin represents a boundary that is called the "Rindler horizon" which is another non-intuitive feature of the accelerated frame of reference in special relativity.

 

[Buckethead]

Thank you all so much for the detailed responses.

This is not a valid way to construct a non inertial frame.

I was falling back here on the knowledge that acceleration of a ship (for example) does not cause time dilation, only the speeds at any given instant during acceleration affect it as has been discussed in other threads such as those regarding the twin paradox for example. I thought it might be a simple matter to just take it a step further and apply that same principle here.

In non inertial frames there is no requirement that light travel at c, and most non inertial frames do in fact violate that condition.

OK got it!

Let's back up. If you have something other than light traveling at constant velocity (i.e. in an inertial frame) and you are accelerating, what does it even mean for it to have a (single) velocity in your frame?

Since an instant in time during the acceleration has no duration, I suppose it would be impossible to say what the velocity of something (not light) would be during accleration, but there is one important difference here. With something other than light, different moments on the acceleration curve would imply different speeds for the object in question since that object is inertial and we are accelerating making the combined speed change. However, with light, at any point on the acceleration curve, the speed of light should be c. So if light is traveling at c at V1 and light travels at c at V2, then it seems light should also be traveling at c at some velocity between those two velocities.

@Buckethead you've had a good few replies already but let me add a further perspective. If light is emitted by a source in an accelerating rocket, then when it is received the rocket is moving relative to the instantaneous IRF of the source at the time of emission.

The received light, therefore, must be Doppler shifted. Blue shifted for light emitted at the front and received at the rear, and red shifted for the other way round.

It's a small further step to attribute this shift to time dilation between the rear and the front.

This is a good point. And I agree, if Doppler shift is experienced here, then indeed the clocks must be moving at different rates. The sticky point for me is just why should there be Doppler shift if the source and receiver are not moving relative to each other. The answer of course is because of the acceleration, and I have no doubt that this is all true, it's just that in a way this "smells" of a "medium for light" that the ship is accelerating through making the speed of light change. In other words, its not relative, it's absolute. The ship is accelerating past a moving beam of light and the apparent speed of the light as a result is changing. This seems to defy the constantcy of the speed of light.

I am interpreting this question as basically being about "what is an accelerated frame in special relativity". I would say that the usual description of an accelerated frame in special relativity is known as "Rindler coordinates".

Doing this without a lot of mathematics is challenging, the best "low math" answer I can come up with, after a bit of thought,is to say that an accelerated frame can be represented by the Rindler coordinate chart.

From the wiki on the topic, it looks like this.

https://en.wikipedia.org/wiki/Rindler_coordinates#/media/File:Rindler_chart.svg

Thanks for the link. I'll look at this more closely and see if I can't glean some insight from this.

 

[PeroK]

This is a good point. And I agree, if Doppler shift is experienced here, then indeed the clocks must be moving at different rates. The sticky point for me is just why should there be Doppler shift if the source and receiver are not moving relative to each other. The answer of course is because of the acceleration, and I have no doubt that this is all true, it's just that in a way this "smells" of a "medium for light" that the ship is accelerating through making the speed of light change. In other words, its not relative, it's absolute. The ship is accelerating past a moving beam of light and the apparent speed of the light as a result is changing. This seems to defy the constantcy of the speed of light.

Not at all. If we simply use the instantaneous IRF of the source at the time of emission. In that IRF, the receiver is moving at some speed when it receives the light. It makes no difference how the receiver came to attain that velocity in the IRF we are using. Remember that the Doppler shift is not about a changing speed of light, but a changing frequency/wavelength.

If you want to cover all the bases, you could have two sources and two receivers:

One source is accelerating and the second source is inertial, and they are relatively instantaneously at rest at the time they both emit a pulse of light.

The same with the two receivers: one receiver is accelerating and the second is inertial, and they are also relatively at rest when the light is received.

In all cases you must get the same Doppler shift, as it depends only on the relative velocity of the source and the receiver. It has nothing to do with acceleration, per se, except that the acceleration creates a relative velocity between them.

 

[jartsa]

I was falling back here on the knowledge that acceleration of a ship (for example) does not cause time dilation, only the speeds at any given instant during acceleration affect it as has been discussed in other threads such as those regarding the twin paradox for example. I thought it might be a simple matter to just take it a step further and apply that same principle here.

Let me list the possible principles you can apply:

1: Be in an inertial frame and use the principle that acceleration does not cause time dilation.

2: Be in a non-inertial frame and use these two simple principles:

A: Clocks with lot of potential energy tick at high rate.
B: Light pulse next to a clock that ticks at high rate changes position at high rate.
​

If you are in an accelerating rocket then you can think that clocks at the nose of the rocket have potential energy.
If the acceleration varies then the potential energy varies.

If a light pulse's leading and trailing edges are in different potentials, what happens to the light pulse? Or if two consecutive light pulses are in different potentials, what happens to the distance between the pulses?

 

[Dale]

I was falling back here on the knowledge that acceleration of a ship (for example) does not cause time dilation, only the speeds at any given instant during acceleration affect it as has been discussed in other threads such as those regarding the twin paradox for example. I thought it might be a simple matter to just take it a step further and apply that same principle here.

A coordinate chart has very few requirements. It must be smooth and it must be a 1-to-1 mapping between points in the chart and events in the manifold.

The problem with the construction you describe is that in the direction away from the acceleration you get overlapping lines of simultaneity. This leads to a single event in spacetime mapping to multiple coordinates. So it violates the 1-to-1 condition.

 

[jartsa]

This is not correct. Viewed from an inertial frame, the clocks are moving at different speeds - the length of the rocket in this frame is shrinking.

I know. I ignored that effect as it is very small at non-relativistic speeds.

At low speeds that effect that the shrinking motion causes the rear observer to receive extra light pulses is larger than the effect that the shrinking motion causes the rear clock to tick a reduced number of times.

And that might cause the guy in the accelerating ship to get fooled into thinking that there is a much larger difference between the rates of the clocks at the rear and at the front than there really is.

I mean, if he wasn't already fooled by the much bigger Doppler shift effect caused by the much bigger total velocity change, then he might get fooled by the much smaller Doppler shift effect caused by the tiny velocity change caused by the shrinking.

@Ibix You see, I still think that it's quite okay to say that the guy in the accelerating rocket gets fooled.

 

[pervect]

A coordinate chart has very few requirements. It must be smooth and it must be a 1-to-1 mapping between points in the chart and events in the manifold. to a single event in spacetime mapping to multiple coordinates. So it violates the 1-to-1 condition. The problem with the construction that Dale is criticizing here is that in the direction away from the acceleration you get overlapping lines of simultaneity.

I want to expand on this a bit.

This behavior of the Rindler coordinate chart is visible on the diagram I mentioned.

Much of the diagram, enough to show this basic behavior, can be done by the method of first finding the trajectory of an accelerating observer. This is well known to be hyperbolic motion.

After drawing the trajectory of the rocket , which is the curved bold line on the space-time diagram, one needs to draw the "lines of simultaneity" on the diagram. These are the straight, but slanted lines, on the diagram. Each line represents the concept of "simultaneous events" as judged by the instananeously co-moving inertial frame at different times (events) on the worldline of the rocket.

The key issue here is that each of these different inertial frames has a different notion of simultaneity - a different slanted line.

This has already been done in the diagram I copied from the wiki article.

In this diagram, time, as measured in a non-accelerated inertial reference frame, which is labelled $X^0$ runs up the page - the bottom of the page is the past, the top of the page is the future. Space runs from left to right. So, $X^0$ for the time, and $X^1$, running from left to right, for the space, are the space and time coordinates in some reference inertial frame.

There is a similar diagram, with different and perhaps better labelling , from the wiki, that I mentioned , at https://en.wikipedia.org/wiki/Rindler_coordinates#/media/File:Rindler_chart.svg

In the diagram above, the curved bold line is the set of events in space-time that correspond to an accelerated rocket. The straight, slanted lines are the lines of simultaneity that represent the concept of events that are "at the same time" in the various instantaneous co-moving inertial frames. The instantaneous co-moving inertial frames are different from the reference frame. Each point on the trajectory of the rocket has a set of events that, in that instantaneous frame, occur "at the same time" . These events are the slanted, diagonal lines on the space-time diagram.

You can see that the slanted lines intersect at a single point. This point is not on the accelerated worldline (the curved bold line), but to it's left. This means that it is "behind" the accelerated rocket, as judged in the reference inertial frame that serves as the background for the diag This

The basic prerequisite for understanding the diagram is to understand what a space-time diagram is, and how to interpret them. Usually, being able to interpret someone else's diagram requires that one be able to draw at least some basic diagrams oneself. For some reason this seems to be difficult to get people to do - but it's really not that hard.

The fundamental reason the results are not intuitive is that the diagram illustrates the relativity of simultaneity. This is something that many, many people have difficulty in understanding. If the OP (Buckethead) is familiar with some other concept of the relativity of simultaneity, then perhaps he can apply it to see where his construction went wrong. If the OP is not familiar with the relativity of simultaneity, I doubt that there will be an understanding reached until this is rectified :(. My suspicion is that the notion of the "relativity of simultaneity" is one of the main blockages that prevent people from grasping the idea of space-time diagrams, but I could be mistaken.

 

[Buckethead]

Thank you all so much for your continued and in depth replies. It's really helped me to sort this out.

In all cases you must get the same Doppler shift, as it depends only on the relative velocity of the source and the receiver. It has nothing to do with acceleration, per se, except that the acceleration creates a relative velocity between them.

I couldn't see before how two accelerating ships with an unvarying distance could have a relative velocity between them but I've worked that out. I imagined a tape recorder with a separate write head and read head. If the tape is accelerated across the heads, The write is is writing a constant tone. The read head will read a constant but higher frequency tone. Analogies are great!

I want to expand on this a bit.

This behavior of the Rindler coordinate chart is visible on the diagram I mentioned.

Thank you for this chart. It's very helpful and I can clearly see how the planes of simultaneity change slope as the velocity changes. I keep forgetting the importance of relativity of simultaneity and must always keep it in mind.

There is just one small lingering question and that is the question of how does a light beam get its energy when it is emitted as explained in the following analogy:

Two ships are traveling at a constant velocity. Ship A is traveling at V and ship B is coming up from behind at 2V. At the very moment they are side by side, they both emit a blue pulse of light to their rear. Sometime later ship C retraces the path of ship A and B and is also traveling at V. It has 2 receivers on board and the two light pulses are received both at the same time. The pulse from ship A is still blue but the pulse from ship B is red. Some assumptions can be made. The two light beams must have been traveling in parallel and side by side the whole time since they hit ship C at the same time. Which means they must have been traveling the same speed through space. So the only conclusion one can make of these two pulses is that while traveling through space they must have had different energies (or frequencies?). The question is why? The emitters on both ships are blue. But clearly when the light left ship B it left as red light. Is there a physical explanation for this?

(EDIT) I think the answer might be that if the two light beams are traveling side by side at the same velocity, then the relative velocity between ship B and the released light beam is greater than the relative velocity between ship A and the released light. This seemed problematic at first but when seen from ship C, then the ships are indeed releasing their light (relative to each ship A and B) at different velocities which would make their frequencies different.

 

[Orodruin]

So the only conclusion one can make of these two pulses is that while traveling through space they must have had different energies (or frequencies?

Energy is not a property of a light beam. It is a property of a system of an observer and a light beam.

 

[Buckethead]

Energy is not a property of a light beam. It is a property of a system of an observer and a light beam.

OK, thanks. And also, I just realized that relative to ship C, ship B's light emitter is time dilated so of course according to ship C ship B actually has a red emitter on board, not a blue one.

 

[Orodruin]

OK, thanks. And also, I just realized that relative to ship C, ship B's light emitter is time dilated so of course according to ship C ship B actually has a red emitter on board, not a blue one.

It is a combination of that and the same classical Doppler shift you would observe in the sound from an ambulance siren. The net effect when you have something approaching you woud be a blue shift even though the source woud be time dilated in your frame.

 

[PeroK]

I couldn't see before how two accelerating ships with an unvarying distance could have a relative velocity between them but I've worked that out. I imagined a tape recorder with a separate write head and read head. If the tape is accelerated across the heads, The write is is writing a constant tone. The read head will read a constant but higher frequency tone. Analogies are great!
.

It's really just a matter of kinematics. If a ship is accelerating in an IRF:

At time $t_0$ the ship (all of it) has a speed of $v_0$. At some later time the ship (all of it) has a velocity of $v_1 = v_0 + a\Delta t$.

In particular, if the ship has height $h$ and a pulse of light is emitted from the front of the ship and received at the rear, then (assuming $v_0$ is small):

$\Delta t \approx \frac{h}{c}$

$v_1 \approx v_0 + \frac{ah}{c}$

In the IRF in which we are analysing this:

The source had speed $v_0$ when the light was emitted; and, the receiver had speed $v_1$ when the light was received. That implies a Doppler shift.

 

[JCMateri]

Let $S\sim T,X,Y,Z$ be stationary coordinates with a source of light at $(X,Y,Z)=0$ and let $S'\sim t,x,y,z$ be the coordinates in an accelerated frame with relative acceleration $=g$ along the $x$ direction for both systems.

$$\begin{eqnarray*}
T&=&xsinh(gt)\\
X&=&xcosh(gt)
\end{eqnarray*}$$
The inverse transformation is$$\begin{eqnarray*}
t&=&\frac{1}{g}tanh^{-1}\left(\frac{T}{X}\right)\\
x&=&\sqrt{X^2-T^2}
\end{eqnarray*}$$
Then $$\left(\begin{array}{c}
dT\\
dX
\end{array}\right)=\left(\begin{array}{cc}
gxcosh(gt)&sinh(gt)\\
gxsinh(gt)&cosh(gt)
\end{array}\right)\left(\begin{array}{c}
dt\\
dx
\end{array}\right)$$
$$dS^2=dT^2-dX^2$$
$$=(gxcosh(gt)dt+sinh(gt)dx)^2-(gxsinh(gt)dt+cosh(gt)dx)^2$$
$$=(gx)^2dt^2-dx^2$$

For light $dS^2=0$ so $(gx)^2dt^2-dx^2=0$. Then $dx^2/dt^2=(gx)^2$ This suggests the speed of light grows without bound. Can this be right?

 

[Orodruin]

You are confusing the coordinate speed with physical speed.

 

[Ibix]

This suggests the speed of light grows without bound. Can this be right?

Draw a Minkowski diagram in frame $S$ with Rindler coordinate lines and sketch a light ray crossing the $X$ axis. Then draw the same ray a bit later in a frame $S'$, which is boosted with respect to $S$ such that the ray is just crossing the $X'$ axis. (These two images should be identical, except with the light ray translated to the right.) In both diagrams, mark a small distance $\delta$ in the positive $X$ (or $X'$) direction from where the light ray crosses the axis. What do you notice about the Rindler coordinate time at which the light ray reaches that point?

 

[JCMateri]

I'm still working on the above. One thing I notice is that if $x=1/g$ we have $dx^2/dt^2=1$. Furthermore, we can see from the above that $X^2-T^2=x^2$. I'll return to this soon.

 

[Orodruin]

I'm still working on the above. One thing I notice is that if $x=1/g$ we have $dx^2/dt^2=1$. Furthermore, we can see from the above that $X^2-T^2=x^2$. I'll return to this soon.

Again, you need to realize that what you are talking about when you consider $dx/dt$ is just the coordinate speed of light. I believe you are putting way too much importance on what is a simple change of coordinates in Minkowski space.

 

[Ibix]

I'm still working on the above. One thing I notice is that if $x=1/g$ we have $dx^2/dt^2=1$. Furthermore, we can see from the above that $X^2-T^2=x^2$. I'll return to this soon.

Assuming that's in response to my suggestion, the only thing you need to note is that lines of constant Rindler time are straight lines through the origin of the Minkowski diagram. Then you should be able to see immediately whether your conclusion from your maths is correct or not.

 

[Dale]

This suggests the speed of light grows without bound. Can this be right?

When I get a result that seems weird or wrong, the first thing that I do is, of course, check the math. But then the next thing that I do is try to see if I can get the same result a different way.

Here, you know the equation for a pulse of light in the inertial coordinates and you also know the coordinate transformation, so you can directly calculate the worldline of the light pulse and get the speed from there. So for an upwards traveling light pulse in the inertial coordinates are $X=X_0 + T$

$$\begin{eqnarray*}
t&=&\frac{1}{g}tanh^{-1}\left(\frac{T}{X_0+T}\right)\\
x&=&\sqrt{(X_0+T)^2-T^2} = \sqrt{X_0^2+2 X_0 T}
\end{eqnarray*}$$

So we then find that $\frac{dx}{dt}=g \sqrt{X_0(2 T + X_0)}$ which goes to infinity as $T$ increases

 

[JCMateri]

I think I understand it better now. The world-lines in the Minkowski diagram consist of a foliation of hyperbolae indexed by $x$. The radial lines represent time. Where an hyperbola intersects a radial line corresponding to a given time value ($t$ in the Rindler frame) tells us its position in the $X,T$ frame. It also tells us when ($t$ in the Rindler frame and $T$ in the stationary frame) the accelerated object catches up (from its starting position at $(X.T)=(X,0)$) to an object whose world-line is the radial line with starting position at $(X,T)=(0,0)$. Thanks for your help.

 

[JCMateri]

I no longer think the last part is correct where I said:

It also tells us when ( $t$ in the Rindler frame and $T$ in the stationary frame) the accelerated object catches up (from its starting position at $(X.T)=(X,0)$) to an object whose world-line is the radial line with starting position at $(X,T)=(0,0)$.

 

[Ibix]

I no longer think the last part is correct where I said:

I agree. I think you edited your post - it was shorter when I "like"d it. There are no objects following the lines of constant Rindler time, since those are spacelike. In fact, they are not lines except on the 1+1 dimensional drawing - they are 3d planes in a full 3+1 dimensional case, and are a Rindler observer's notion of "all of space, now" (which goes wrong where they cross at the Rindler horizon).