Doppler and Searchlight (Headlight)

[Albertgauss]

TL;DR Summary

Where would Doppler red shifted objects get perceived within the searchlight (headlight) effect of relativity, in front or behind the relativistic frame?

The images below were taken from the website

https://www.bing.com/videos/rivervi...E198C7F4F9D4EE240C52E198C7F4F9D4EE2&FORM=VIRE

Starting at about min 11:30.

Question: I know that objects that are physically behind me are going to redshift if I look behind me as I travel near light speed. But the searchlight effect is going to compress all of the universe in front of me. So where do I perceive the redshifted objects appear, in front of me or behind me?

My try at an answer: I think I would see the redshifted objects in front of me. I think the searchlight effect would concentrate all the universe in front of me, but within that central cone of vision, some objects really in front would continue to get bluer (since I approach them) but right besides those objects, objects really behind me will get redder, because they are actually receding behind me. That is, within the central cone ahead of me, I will see both bluer and reddening objects right besides each other. Agree or Disagree? My two slides below try to illustrate.

 

[Orodruin]

The perceived angle at which you go from blue-shift to red-shift is easily calculable and depends on your speed.

That said, objects that pass you will eventually end up behind you (assuming velocities are constant) with a redshift approaching the maximal redshift (given the relative speed).

 

[Albertgauss]

Yes, but when the object physically ends up behind me and starts to redshift, do I perceive the object getting redder behind me or does it get redder in front of me in the searchlight cone?

 

[Ibix]

It's not one or the other - unless it's dead infront or dead behind you it'll always be getting redder.

Stuff right in front or behind you will be Doppler shifted according to $\sqrt{\frac{c\pm v}{c\mp v}}$. Everything else will be somewhere inbetween. The easiest way to work it out is to write the energy-momentum four vector of a light pulse moving at some arbitrary angle to the x axis, then transform to the ship frame. You can read off the redshift as a function of angle and compute the red/blue shift changeover angle easily at that point.

 

[Orodruin]

The easiest way to work it out is to write the energy-momentum four vector of a light pulse moving at some arbitrary angle to the x axis, then transform to the ship frame.

The easiest way is to write down the 4-momentum of the light pulse (or equivalently and perhaps more to the point, the 4-frequency) and take the Minkowski inner product with the observer’s 4-velocity. No transformation necessary.

 

[Ibix]

The easiest way is to write down the 4-momentum of the light pulse (or equivalently and perhaps more to the point, the 4-frequency) and take the Minkowski inner product with the observer’s 4-velocity. No transformation necessary.

Can you get the angle at which the ship perceives zero Doppler from that? I think you need the angular aberration formula at least, which I always have to look up, so I might as well do the transformation.

 

[Orodruin]

Can you get the angle at which the ship perceives zero Doppler from that? I think you need the angular aberration formula at least, which I always have to look up, so I might as well do the transformation.

Sure you can. You just do it the other way around (write down everything in the observer frame - take inner product with emitter 4-velocity to get emitted frequency).

 

[Albertgauss]

It's not one or the other - unless it's dead infront or dead behind you it'll always be getting redder.

I've not seen this statement before. Just to clarify: look at the blue dots I put in my image, let them be blue stars that are millions of light years in front of me but not "dead" in front of me but off the to the side a little. Certainly they are coming closer to me the more I travel towards them. I feel like these would blue-shift up to purple or beyond; it doesn't seem correct to me they would get red until they were physically behind me --> I don't know if my statement here is correct or not.

That said, objects that pass you will eventually end up behind you (assuming velocities are constant) with a redshift approaching the maximal redshift (given the relative speed).

Yes, but for now, I want to keep the blue objects in front of me. They haven't passed me behind yet and they are a long way from doing so. I am considering a scenario where there are still lots of objects ahead of me, getting closer to me on approach, even if they are approaching from way far away.

Of note: the video referenced here has an explanation of the searchlight effect at minutes 2:30 to about 4 minutes.

Let me try and recast the question a different way. I'm sitting in my ship. I'm going near lightspeed (I'll use the near lightspeed reference of 99.95% the speed of light in the video above). Physically, millions of light years far ahead of me there are physical blue objects; millions of light years behind me are physically other objects. I know the objects behind me are reddening because I move away from them. If I turn my head to look behind me at these reddening objects (my eyesight is ant-parallel with the ship's motion), will I see them there? Or, if I keep my eyes ahead of me, parallel with the direction of motion of the ship, will the objects physically behind me appear in front of me now but continually getting redder?

No math. If its math that hasn't been done or no one knows/wants to do the calculation, I will simply assume it is an unanswered question of relativity. The video above, in its opening scene, said that all the math and calculations had been taken care of in the video. Plus, for what I need to do, I need to explain this concept without the math, as it is for educational instruction and no one in my audience will understand the math anyway.

 

[Orodruin]

If its math that hasn't been done or no one knows/wants to do the calculation, I will simply assume it is an unanswered question of relativity.

Obviously the math has been done. This is a basic problem that I would expect students to be able to do in my introductory course. The problem is that you are framing your question in a way that makes the answer a resounding "it depends". Some reddening objects will seem to be in front of you, some behind. Also note that time dilation also plays a role, not simply moving towards or away from.

Objects that are straight ahead and objects that are straight behind will maintain their positions regardless of your relative speed. If you want to know about the actual effect you are talking about (which is actually called aberration of light) you cannot just look at objects that are straight ahead or straight behind as the effect only affects objects that are not.

I'm going near lightspeed (I'll use the near lightspeed reference of 99.95% the speed of light in the video above).

I'm also going to stop you right here. An extremely important thing to understand in relativity is that there is no such thing as "I'm going near lightspeed" without proper reference to what you are moving at that speed relative to. If you are moving at 0.9995c relative to the stars in the galaxy, you can just as well describe things as you being stationary and the stars moving at 0.9995c relative to you.

 

[PeroK]

No math.

Whoever then has the effrontery to study physics while neglecting mathematics should know from the start that he will never make his entry through the portals of wisdom.

Roger Bacon (1220-1292)

 

[Ibix]

I've not seen this statement before. Just to clarify: look at the blue dots I put in my image, let them be blue stars that are millions of light years in front of me but not "dead" in front of me but off the to the side a little. Certainly they are coming closer to me the more I travel towards them. I feel like these would blue-shift up to purple or beyond; it doesn't seem correct to me they would get red until they were physically behind me --> I don't know if my statement here is correct or not.

Well, eventually they'll be behind you and be redshifted. And depending how fast you're going and which frame(s) measurements you're using when you say it, "millions of light years ahead" may not take all that long to be behind you.

Yes, but for now, I want to keep the blue objects in front of me. They haven't passed me behind yet and they are a long way from doing so. I am considering a scenario where there are still lots of objects ahead of me, getting closer to me on approach, even if they are approaching from way far away.

In the frame where the stars are at rest, say that approximately half of them are behind the ship. The same is true in the frame where the ship is at rest the same is true. However, because the stars are moving in this frame you do not see them where they are, but rather where they were some time ago, and many more of them will be in your forwards hemisphere than your rear. Ones straight ahead or behind will still be there; everything else will be shifted forwards.

So some things that are behind you will appear in front of you and blueshifted. As time goes on they will become more red as they drift towards the rear of your ship.

If its math that hasn't been done or no one knows/wants to do the calculation, I will simply assume it is an unanswered question of relativity.

No, the maths is fairly trivial. Nothing more complex than a matrix multiplication is needed, and even less than that if you take Orodruin's approach.

Plus, for what I need to do, I need to explain this concept without the math, as it is for educational instruction and no one in my audience will understand the math anyway.

But if you don't understand the maths, how will you know you're teaching correctly?

 

[Dale]

TL;DR Summary: Where would Doppler red shifted objects get perceived within the searchlight (headlight) effect of relativity, in front or behind the relativistic frame?

The images below were taken from the website

https://www.bing.com/videos/rivervi...E198C7F4F9D4EE240C52E198C7F4F9D4EE2&FORM=VIRE

Starting at about min 11:30.

Question: I know that objects that are physically behind me are going to redshift if I look behind me as I travel near light speed. But the searchlight effect is going to compress all of the universe in front of me. So where do I perceive the redshifted objects appear, in front of me or behind me?

My try at an answer: I think I would see the redshifted objects in front of me. I think the searchlight effect would concentrate all the universe in front of me, but within that central cone of vision, some objects really in front would continue to get bluer (since I approach them) but right besides those objects, objects really behind me will get redder, because they are actually receding behind me. That is, within the central cone ahead of me, I will see both bluer and reddening objects right besides each other. Agree or Disagree? My two slides below try to illustrate.


View attachment 341983View attachment 341984

Directly in front is blueshifted. Directly behind is redshifted. At some intermediate angle there is no Doppler shift. Is your question what that angle is? If not then I am not sure what the question is.

 

[PAllen]

Sure you can. You just do it the other way around (write down everything in the observer frame - take inner product with emitter 4-velocity to get emitted frequency).

Well, if you are given a description in terms of frame where local stars are 'nearly at rest', with a ship traveling on some velocity vector, you would have to Lorentz transform the description. You could write a straightforward vector expression for the angle observed by the ship in terms of light angle in the description frame, but that would really be just as much work as Lorentz transform (IMO).

 

[PAllen]

Just to show what seems required for an all vector approach, we have in frame of stars (suppressing z, giving vectors in component form as (t,x,y)):
a light 4 momentum of $\vec p = E(1,\cos \theta,\sin \theta)$
a ship 4-velocity of $\hat u = \gamma(1,v,0)$, which is also $\hat t'$ of the ship.
You also have, for the ship $\hat x' = \gamma (v,1,0)$, and $\hat y' = (0,0,1)$. This is just a vectorized equivalent of the Lorentz transform. Then, we know that the following must be true:
$E' = \vec p \cdot \hat t'$ and $\cos \theta' = -(\vec p \cdot \hat x') / (\vec p \cdot \hat t')$
These give: $$E' = \gamma E (1-v \cos \theta)$$ and $$ \cos \theta' = \frac {\cos \theta - v} {1 - v \cos \theta} $$
These give you full generality of Doppler and aberration, explaining the 'searchlight' effect.

 

[Orodruin]

Well, if you are given a description in terms of frame where local stars are 'nearly at rest', with a ship traveling on some velocity vector, you would have to Lorentz transform the description. You could write a straightforward vector expression for the angle observed by the ship in terms of light angle in the description frame, but that would really be just as much work as Lorentz transform (IMO).

The point is that you don’t have to if all you want is the frequency as a function of observed angle, which is what @Ibix asked.

 

[Albertgauss]

I thought about the responses here a little bit more. Here's what I came up with.

without proper reference to what you are moving at that speed relative to. If you are moving at 0.9995c relative to the stars in the galaxy, you can just as well describe things as you being stationary and the stars moving at 0.9995c relative to you.

Yes, the above is fine. The stars physically behind me and in front of me are rest with respect to each other. Me in my spaceship (frame) move such that those stars move with respect to me. The stars physically in front of me move towards me at constant velocity of 0.9995c and the stars physically behind me recede from me at that same speed. Also, I would consider the frame assumptions of the video mentioned above.

"millions of light years ahead" may not take all that long to be behind you.

I went ahead and worked out a length contraction for this. If my ship was rest to the first star ahead of me let's say that star is 1 million light years away. If I end up vel=0.9995c towards the same star, that distance to the star is length contracted to be 32,000 light year, so that star stays mostly ahead of me the whole time ( which I calculate remains that way for 31,364 years from t = d/v = (32,000 c yr)/ (0.9995c). You're right that the objects in front of me will end up behind me eventually, but at this point in the trip, they have not done so, yet.

Some reddening objects will seem to be in front of you, some behind

Would that look something like this?

Apologies I couldn't get 0.9995c as we were talking about before but this should be fine for a picture. Actually, this one comment of Orodruin may have led me to ask my question even more simple than what I initially asked. When the redshifted objects appear in the forward light cone (and these objects are not directly behind me but maybe in the rear view and off to the side), do they appear red-shifted in the light cone now? Where the green arrow points and it says "What Color now?" --> that is what I am asking. I don't know if:

1) The photons from those used-to-be-red-shifting objects would appear to be approaching me from the front and therefore would be blue-shifted now

or

2) maybe those objects appear to be in front of me but keep getting redder because they are actually behind me.

you cannot just look at objects that are straight ahead or straight behind as the effect only affects objects that are not.

Yes, I agree. I am trying to consider the objects like you said. By analogy, if I was on a highway taking a road trip, some objects down the road are approaching me but are not directly in the highway itself. The gas stations, road signs, and turnoffs are all off to the side of the highway. I already passed some objects on the highway already, but, they, too are not on the highway and so also are not directly behind me. In this analogy, the stars I approach and recede would be "off to the side" in the same way.

and many more of them will be in your forwards hemisphere than your rear. Ones ... behind will still be there; everything else will be shifted forwards.

Yes, I agree, they are there. But they will retain their red-shift of the pure Doppler or do they now switch to blue-shift because the photons that came from these objects now approach me from the forward direction? I agree with you here, but the color coming in now is what I am asking from the image of the object I perceive.

Directly in front is blueshifted. Directly behind is redshifted. At some intermediate angle there is no Doppler shift. Is your question what that angle is? If not then I am not sure what the question is.

Sorry, I hit reply too soon and I can't fix this post. I'm in "Edit" mode now where I can't get the quote feature to work and I did want to add about Ibix comment about the changeover angle which would support choice number 2. That quote was:

"You can read off the redshift as a function of angle and compute the red/blue shift changeover angle easily at that point."

which I can't reconcile to Dale's quote above.

Also, I retract the math comment. I know what I was trying to say, but it didn't come out the right way here. Yes, math is important here

 

[PAllen]

Aberration does not change energy. Thus if a star is redshifted per the rocket, that is true weather you consider the angle to the rocket in the star frame or in the rocket frame. So in the star frame, all red shifted stars are behind the rocket. In the rocket frame some of these are in front. The boundary is the particular angle where no shift occurs. In the star frame, this is behind the rocket. In the rocket frame it is in front of the rocket. The two angles are symmetric about the perpendicular, related by $\theta’ = \pi - \theta$.

 

[Albertgauss]

And just to make sure, PAllen's Equations on Monday night at 10:36 pm are the same (or nearly so) as the equations under discussion between Ibix and Odonriun Monday afternoon before? I don't remember these transformations for awhile and want to read up on them again. A website with a description of this math would be greatly appreciated.

 

[PAllen]

And just to make sure, PAllen's Equations on Monday night at 10:36 pm are the same (or nearly so) as the equations under discussion between Ibix and Odonriun Monday afternoon before? I don't remember these transformations for awhile and want to read up on them again. A website with a description of this math would be greatly appreciated.

I gave what I thought was a complete derivation. If you have questions, just ask. It might take me a day to respond, depending on time of day, but then someone else would probably take up the slack. I find many textbook discussions of aberration to needlessly longwinded. Also, be aware that there are different formulas for aberration and general Doppler that look different, but are actually mathematically equivalent.

 

[Ibix]

And just to make sure, PAllen's Equations on Monday night at 10:36 pm are the same (or nearly so) as the equations under discussion between Ibix and Odonriun Monday afternoon before?

Let the star rest frame be S and the ship S'. Using the primed frame, a light pulse approaching the ship has $p'^a=(E',-E'\cos\theta',-E'\sin\theta',0)$, where $\theta'$ is the angle away from the ship's straight ahead direction. An observer at rest in S has $u'^a=(\gamma,-\beta\gamma,0,0)$, where $\beta=v/c$ is the ship's speed as a fraction of $c$ and $\gamma=1/\sqrt{1-\beta^2}$, so measures the light pulse to have energy $E=\eta_{ab}p'^au'^b=\gamma E'(1-\beta\cos\theta')$. Thus the redshift seen by the ship as a function of angle away from the nose is $$\frac{E'}{E}=\frac{f'}{f}=\frac{\sqrt{1-\beta^2}}{1-\beta\cos\theta'}$$You can check that this reduces to the textbook longitudinal and transverse Doppler shift formulae when $\theta'$ is 0, 90° and 180°. Note that it is 1 when $\cos\theta'=\frac 1\beta(1-\sqrt{1-\beta^2})$. Everything nearer the nose than this is blueshifted and more so the closer to the nose you get, everything outside that angle is redshifted and more so the closer to the tail you get.

Aberration is easier to do the other way round. In the star rest frame, the ship has four velocity $(\gamma,\beta\gamma,0,0)$ so its $x'$ direction is $x'^a=(\beta\gamma,\gamma,0,0)$. A light ray arriving at the ship from angle $\theta$ has four momentum $p^a=(E,-E\cos\theta,-E\sin\theta)$, which has component $-\eta_{ab}x'^ap^b=-E\gamma (\beta+\cos\theta)$ in the $x'$ direction. Since this must be equal to $-E'\cos\theta'$ we can derive $E'/E=\gamma (\beta+\cos\theta)/\cos\theta'$. Then we can use our previous expression to get rid of the $E'/E$ and get $$\cos\theta'=\gamma^2\frac{\beta+\cos\theta}{1-\beta\cos\theta}$$
Hopefully that agrees with @PAllen.

do they appear red-shifted in the light cone now?

I know what you are trying to say, but don't call it a light cone. That's something else entirely in relativity.

 

[Orodruin]

Ibix and Odonriun

Who is this person? They sound wise.
(My emphasis)

 

[PAllen]

Let the star rest frame be S and the ship S'. Using the primed frame, a light pulse approaching the ship has $p'^a=(E',-E'\cos\theta',-E'\sin\theta',0)$, where $\theta'$ is the angle away from the ship's straight ahead direction. An observer at rest in S has $u'^a=(\gamma,-\beta\gamma,0,0)$, where $\beta=v/c$ is the ship's speed as a fraction of $c$ and $\gamma=1/\sqrt{1-\beta^2}$, so measures the light pulse to have energy $E=\eta_{ab}p'^au'^b=\gamma E'(1-\beta\cos\theta')$.

Here we agree except for two formalities - sign convention and which frame's angle to use in formulas. For @Albertgauss, this is a skill you have to learn comparing different texts/sources - conventions will be different and superficially different formulas are most often consistent (unless one source has made an error). So far, @Ibix and my results are identical except for differences in these choices of convention.

Thus the redshift seen by the ship as a function of angle away from the nose is $$\frac{E'}{E}=\frac{f'}{f}=\frac{\sqrt{1-\beta^2}}{1-\beta\cos\theta'}$$

I chose to write this in terms of $\theta$ rather than $\theta'$. The formulas are equivalent, though it is not obvious algebraically. It can be seen by noting that if $E=\eta_{ab}p'^au'^b=\gamma E'(1-\beta\cos\theta')$ as derived above, by computing in S, you must have:
$E'=\gamma E(1-\beta\cos\theta)$. This then gives my form of the Doppler formula (noting that my angle sign convention is the opposite of yours).

...

Aberration is easier to do the other way round. In the star rest frame, the ship has four velocity $(\gamma,\beta\gamma,0,0)$ so its $x'$ direction is $x'^a=(\beta\gamma,\gamma,0,0)$. A light ray arriving at the ship from angle $\theta$ has four momentum $p^a=(E,-E\cos\theta,-E\sin\theta)$, which has component $-\eta_{ab}x'^ap^b=-E\gamma (\beta+\cos\theta)$ in the $x'$ direction. Since this must be equal to $-E'\cos\theta'$ we can derive $E'/E=\gamma (\beta+\cos\theta)/\cos\theta'$. Then we can use our previous expression to get rid of the $E'/E$ and get $$\cos\theta'=\gamma^2\frac{\beta+\cos\theta}{1-\beta\cos\theta}$$
Hopefully that agrees with @PAllen.

Here you have goofed in your elimination algebra. If you do it right you get something equivalent to my aberrations formula (delta sign conventions). I did this in a way that needed much less algebra, but I worked out that your approach would come out the same with correct algebra. The $\gamma^2$ is wrong.