UKC

/ Which exploding star do you see first?

Please Register as a New User in order to reply to this topic.
ericinbristol - on 14 Jun 2018

By Michael Merrifield: You are flying a rocket at 0.9c (c being the speed of light) from Star A to Star B. The distance between the stars, in the stars' reference frame is 1.01 light year. Both stars explode simultaneously in your reference frame at the instant you are exactly half way between them. Which flash do you see first?

A Star A first

B Star B first

C Both stars together

Post edited at 22:44
Dave the Rave on 14 Jun 2018
In reply to ericinbristol:

The flash from the speed camera?

graeme jackson - on 14 Jun 2018
In reply to ericinbristol:

>  Which flash do you see first?

Gordon

 

marsbar - on 14 Jun 2018
In reply to ericinbristol:

Yes, but is the rocket on a treadmill?  

DancingOnRock - on 14 Jun 2018
In reply to ericinbristol:

Star B. Because as the light from the stars moves at the same time you will be closer to star B by the time the light reaches you. 

Michael Hood - on 15 Jun 2018
In reply to ericinbristol:

Both stars together. Speed of light is always c in any frame of reference (I think, maybe, not sure )

wintertree - on 15 Jun 2018
In reply to marsbar:

> Yes, but is the rocket on a treadmill?  

Rockets don’t have wheels or ambulatory legs.

Philip on 15 Jun 2018
In reply to ericinbristol:

> Both stars explode simultaneously in your reference frame at the instant you are exactly half way between them. Which flash do you see first?

You have answered your own question in the wording. If both stars explode simultaneously in the rockets reference frame then by definition you see both at the same time.

I suspect you made a mistake and you meant to state that in a non-inertial reference frame two star 1 ly apart explode, what is observed from a rocket travelling at 0.9c midway between them.

Which is again a trivial question.

ericinbristol - on 15 Jun 2018
In reply to Philip:

I took the question from a Tweet by @AstroMikeMerri

https://twitter.com/AstroMikeMerri/status/1007367854981828608

I only partly understood the question but transcribed it accurately in its fundamentals.

wintertree - on 15 Jun 2018
In reply to ericinbristol:

> ...Tweet

The real challenge is unambiguously posing such a question in a tweet.  

For example the relative velocity of the stars is not specified.  It’s also not clear if the “you” is on the rocket.  You don’t have to be on something to pilot it.  Speaking of piloting, as this is apparently an SR not a GR question, the rocket is inertial so it’s not actually being “piloted” so much as ridden.  

Some of this may sound like nit picking, but if there’s one area of physics where absolutely clarity helps it’s “simple” SR questions.

Post edited at 08:12
Spartacus on 15 Jun 2018
In reply to ericinbristol:

Is it like a fireworks rocket, the big ones that cost £9.99?

DancingOnRock - on 15 Jun 2018
In reply to ericinbristol:

To expand on my earlier reply, don’t get the light from the stars and the light from the rocket mixed up. The light from the rocket will still be travelling at the speed of light but that’s a different question. 

The rocket is travelling at approximately the speed of light and will be 6 light months away from both stars. In approx three months time it will meet the light from star B and now be approx 9 light months from Star A so will have to wait approx 6 months for the light from Star A to reach it. IF the rocket was travelling at the speed of light and continued to do so, the light from star A would never reach the rocket. It’s not, but someone else can do the maths of you need a more accurate answer for part 2 of your question. “When does the rocket see Star A explode?”

Post edited at 08:46
DancingOnRock - on 15 Jun 2018
In reply to wintertree:

It says ‘you are flying a rocket’ and ‘at the point you are halfway between them’.

I don’t think this is a relativity question at all. It’s a basic maths one. 

It only becomes a relativity question when the other objects (stars) are moving as well and you start looking at what they can see regarding the rocket and the other other star. 

Coel Hellier - on 15 Jun 2018
In reply to Philip:

> You have answered your own question in the wording. If both stars explode simultaneously in the rockets reference frame then by definition you see both at the same time.

Agreed.  That's what "simultaneous" means.

DancingOnRock - on 15 Jun 2018
In reply to Coel Hellier:

Isn’t the reference frame attached to the rocket only describing positions and (momentums) of the other objects relative to the rocket. Time doesn’t alter. So when the stars explode the rocket won’t ‘see’ the stars explode until the light from them reaches it. 

Philip on 15 Jun 2018
In reply to ericinbristol:

It's badly worded in the original. It's hard to see how you make this a puzzle.

If the 2 stars and an observer equal distance from both sees the even as simultaneous, then you could question what a rocket sees at 0.5 ly from 1 star at 0.9c travelling towards the other star. But that is too simple.

More interesting would be how far from Star A would rocket at 0.9c need to be such that it sees both simultaneously. But that is only using SR and classical mechanics.

What about a rocket accelerating to 0.9c from rest at Star A such that it reaches 0.9c at 0.5ly. At what point in space and what time T after T=0 does the simultaneous explosion in inertial frame occur such that the rocket observes it simultaneously. Assuming the clocks were synchronised at t=0 what is the time on the rocket.

I'll leave the latter as an exercise for the reader.

 

GrahamD - on 15 Jun 2018
In reply to Philip:

> It's badly worded in the original. It's hard to see how you make this a puzzle.

I agree with the posts above - its a trick question and the key is understanding "Both stars explode simultaneously in your reference frame".  The rest of the the information is chaff.

 

Philip on 15 Jun 2018
In reply to Coel Hellier:

If you want some fun, the original author is being overly defensive on Twitter.

Philip on 15 Jun 2018
In reply to GrahamD:

Try telling the author on Twitter. He's adamant it's valid, and he's been to Harvard!!

tom_in_edinburgh - on 15 Jun 2018
In reply to Philip:

> I'll leave the latter as an exercise for the reader.

The reader respectfully declines.

 

SuperLee1985 - on 15 Jun 2018
In reply to DancingOnRock:

This was my initial thought to, but it seemed a bit too simple/obvious.

I then started wondering if Time Dilation Effect or Length Contraction would alter the expected outcome but the Rocket is the only object that is moving so from it's frame of reference it would not notice these effects.

I then started thinking about dopler shifting, which will cause the star you are moving toward to be blue shifted and the star you are moving away from to be red-shifted. Because starts generally produce more light in the IR spectrum than the UV spectrum, the star you are moving towards will appear significantly brighter than the one you are moving away from which will become quite dim. So even taking this into accent you will still see Star B first and you may not see Star A at all.

DerwentDiluted - on 15 Jun 2018
In reply to ericinbristol:

Dissapointed. I thought this was a teaser for a new TV showing combining the much loved 'I'm a celebrity get me out of here' format with uncleared minefields littered with UXO.

Something I would make time to watch.

Post edited at 12:36
Philip on 15 Jun 2018
In reply to ericinbristol:

I was wrong, it's not badly worded. It's so simple. It's Star B.

You're in a rocket going from A to B, A is behind you, you don't ever see it. :-P

krikoman - on 15 Jun 2018
In reply to wintertree:

> Rockets don’t have wheels or ambulatory legs.


"Most" rockets don't, I think you'll find.

krikoman - on 15 Jun 2018
In reply to ericinbristol:

>  Which flash do you see first?

None because you're just entering the event horizon of a super massive black hole in between to two stars.

 

DancingOnRock - on 15 Jun 2018
In reply to SuperLee1985:

It’s suspicioulsy similar to one my 14 year old bought home from school which involved a spaceship travelling at some fraction of c and communicating with a base station on earth. It transmits after 3 months, when does it recieve the reply?

wintertree - on 15 Jun 2018
In reply to DancingOnRock:

> It says ‘you are flying a rocket’ and ‘at the point you are halfway between them’.

Flying then - still not a useful term.  I could be sat stationary half way between the stars flying a remote control rocket.

”Two stars A and B are stationary with respect to each other.  You are on a rocket travelling between A and B at relativistic speed.  At the half-way point you observe ...”

> I don’t think this is a relativity question at all. It’s a basic maths one. 

> It only becomes a relativity question when the other objects (stars) are moving as well

But they are moving from the reference frame of the rocket.  No inertial reference frame is “special” or more stationary than another.

You are right in a sense that the answer to this question doesn’t depend on if the speed of the rocket is reletavistic or not but that’s the point - the question is written so as to make people think it does matter.

If the event had been simultaneous in the reference frame of the stars, then there would as I think you say be more of an SR aspect to the question.

wintertree - on 15 Jun 2018
In reply to krikoman:

> "Most" rockets don't, I think you'll find.

Sorry I should have said “driven wheels”.  I’m not aware of any rocket powered vehicles with driven wheels, anyone else?

JoshOvki on 15 Jun 2018
In reply to DancingOnRock:

> when does it recieve the reply?

If it goes into most of my work colleagues emails, then the answer is never, they will never receive a reply.

Rob Parsons on 15 Jun 2018
In reply to Philip:

> If you want some fun, the original author is being overly defensive on Twitter.


Bloody hell: "... But what do I know with my degrees from Oxford and Harvard an decades of teaching relativity to thousands of undergraduates?"

What an advert for intellectual debate at Nottingham University. Not.

Philip on 15 Jun 2018
In reply to Rob Parsons:

> Bloody hell: "... But what do I know with my degrees from Oxford and Harvard an decades of teaching relativity to thousands of undergraduates?"

> What an advert for intellectual debate at Nottingham University. Not.

Maybe that's how they do it at Harvard.

Eric9Points - on 15 Jun 2018
In reply to ericinbristol:

So are we agreed that the observer in the rocket sees a blue star and a red star?

Ron Rees Davies - on 16 Jun 2018
In reply to wintertree:

>>I’m not aware of any rocket powered vehicles with driven wheels, anyone else?

Stephenson's Rocket?

 

Post edited at 07:17
DancingOnRock - on 16 Jun 2018
In reply to wintertree:

It’s nonsense. If the stars explode simultaneously in time, you won’t see them explode until 6months later. If they appear to explode simultaneously then they’ll appear to explode simultaneously, when they actually exploded 6months ago. 

He seems to be just trying to explain the meaning of ‘simultaneous’ when it’s applied to reference frames. 

So can we assume that we ignore time when it comes to events? 

In which case the question is pointless. 

Post edited at 08:12
Pero - on 16 Jun 2018
In reply to ericinbristol:

The key is to analyse the problem in the reference frame of the rocket.  Note that "simultaneous" in the rocket's reference frame is not the same as "simultaneous" in the stars' reference frame, in which the rocket is moving.  This is called the "relativity of simultaneity" and is, perhaps, the most difficult concept in SR to digest.  To an observer at rest relative to the two stars (and half way between them to keep things simple), the stars do not explode simultaneously.

Anyway, in the reference frame of the rocket:

Star A and star B are equidistant from you when they simultaneously explode.  Although Star A was moving away from you and Star B towards you (both at 0.9c) when they exploded, this does not affect the speed of the light from these stars, so the light from both explosions reaches you simultaneously.

If the stars exploded simultaenously in their reference frame (assuming they are relatively at rest), when the rocket was half way between them, then that would be a different problem with a different answer: the rocket would see Star B explode first.

Post edited at 12:20
Mike505 on 16 Jun 2018
In reply to ericinbristol:

Obviously star B unless your rocket has rear view mirrors?

skog on 16 Jun 2018
In reply to DancingOnRock:

> It’s nonsense. If the stars explode simultaneously in time, you won’t see them explode until 6months later. If they appear to explode simultaneously then they’ll appear to explode simultaneously, when they actually exploded 6months ago.

I think - and I'm certainly no expert - that rather than posing a trick question, he's using the question to get people to think about this: there is no such thing as a universal 'simultaneous' which applies everywhere, 'simultaneous' actually means different things depending on what's being compared and how it's behaving relative to the other objects, and even something which happens simultaneously from your perspective won't be exactly simultaneous from mine.

Ramblin dave - on 16 Jun 2018
In reply to skog:

Yes. Pero's explanation a couple of posts ago is bob on, and the guy also gives a good, readable breakdown in the twitter thread.

Ramblin dave - on 16 Jun 2018
In reply to Rob Parsons:

> Bloody hell: "... But what do I know with my degrees from Oxford and Harvard an decades of teaching relativity to thousands of undergraduates?"

To be fair, he was replying to someone who was telling him fairly rudely that he didn't know the basic definitions of what he was talking about. I know that we've had enough of experts these days, but sometimes it is worth thinking about who's more likely to have misunderstood something fundamental...

Post edited at 14:58
DancingOnRock - on 16 Jun 2018
In reply to skog:

But surely appearing to be simultaneous and actually being simultaneous are quite different things. 

If two things happen simultaneously, then they happen at the same universal time.

skog on 16 Jun 2018
In reply to DancingOnRock:

> If two things happen simultaneously, then they happen at the same universal time.

 

As I understand it, there is no such thing as universal time, and that's the point which was being made.

Time distorts, it isn't flat - for example, the reason you can't travel at the speed of light is that no matter how fast you're going away from something, the light travelling away from it is going at 300 million metres per second away from you AND from it; speeding up relative to another object does not cause you to speed up relative to light, which will always be going 300 million metres per second faster than you.

 

wintertree - on 16 Jun 2018
In reply to DancingOnRock:t

> If two things happen simultaneously, then they happen at the same universal time.

But there is no “universal time” in SR.  Events simultaneous to inertial frame may not be simultaneous to a different inerial frame.

Pero - on 16 Jun 2018
In reply to DancingOnRock:

> But surely appearing to be simultaneous and actually being simultaneous are quite different things. 

> If two things happen simultaneously, then they happen at the same universal time.

Universal time was/is one of the postulates of Newtonian mechanics.  However, it is not compatible with an invariant speed of light (for all inertial observers).  This caused a headache at the end of the 19th Century when experimental evidence for the invariance of the speed of light was mounting.  Eventually, as we all know, Einstein cracked it: "until at last it came to me that time itself was suspect".

Einstein took instead the invariant speed of light as a postulate and that was the end of universal time and universal simultaneity.

 

nufkin - on 16 Jun 2018
In reply to skog:

> always be going 300 million metres per second faster than you. 

Doesn't light go at 300 million metres per second regardless of the speed of the source? 

Robert Durran - on 16 Jun 2018
In reply to nufkin:

> Doesn't light go at 300 million metres per second regardless of the speed of the source? 


Yes, in the sense that if you run towards me at half the speed of light while shining a torch at me, you would measure the light going away from you at the speed of light. But the point is that I would also measure the light arriving at me at this same speed of light. Unlike if you run towards me at 5m/s and throw a ball at me which travels away from you (relatively) at 10m/s, I would measure it arriving at me at 15m/s.

skog on 16 Jun 2018
In reply to nufkin:

> Doesn't light go at 300 million metres per second regardless of the speed of the source? 

Yes. And regardless of the speed and direction of the destination, it also arrives at 300 million metres per second.

It just doesn't work the way we're wired to think about things!

Bulls Crack - on 16 Jun 2018
In reply to ericinbristol:

The one out of the front window..obv

Pero - on 16 Jun 2018
In reply to skog:

> Yes. And regardless of the speed and direction of the destination, it also arrives at 300 million metres per second.

You have to be careful.  The postulates of SR (including the invariance of the speed of light) apply to inertial reference frames.  These postulates - like Newton's laws - do not necessarily hold for an accelerating reference frame.

skog on 16 Jun 2018
In reply to Pero:

Ok thanks. You're already beyond what I think I know, but could you give an example of what that could mean?

Pero - on 16 Jun 2018
In reply to skog:

a) Suppose you have a star 4 light years from Earth.  The Earth rotates in 24 hours.  In the Earth's reference frame, the star travels about 24 light years (in a circle) in 24 hours.  Considerably more than the speed of light.

b) Suppose a rocket accelerates from Earth towards that star, reaching 0.8c in a few days or weeks, say.  Owing to length contraction, the star is now only 2.4 light years away in the rocket's reference frame, having moved 1.6 light year in a few weeks.

Post edited at 20:43
skog on 16 Jun 2018
In reply to Pero:

Ok, I see thanks - those make sense (kind of)!

But anyone on any of those locations would still see light itself arrive and depart at 300 million metres per second, wouldn't they?

Pero - on 16 Jun 2018
In reply to skog:

> Ok, I see thanks - those make sense (kind of)!

> But anyone on any of those locations would still see light itself arrive and depart at 300 million metres per second, wouldn't they?

Strictly speaking the postulates of SR apply to inertial reference frames.  If you go back to the original problem and consider two rockets.  R1 is travelling at a constant speed of 0.9c relative to stars A and B.  R2 is also moving from A to B but accelerating.  If we assume that the positions and speeds of R1 and R2 coincide at the half way point, then their instantaneous reference frames coincide at the point.  But, as R2 is accelerating, it pulls ahead of R1 and the light from star B must reach it before the light from star A.

Trying to take a constant speed of light from both stars in this case (over the duration of the problem) for R2 would not work.

But, in fact, as you suggest, if R2 measured the speed of the light just as it reached the rocket, it would find the light from both stars, measured over a short enough time, would be c.  In a general reference frame, therefore, the speed of light is locally (instantaneously) c - whenever measured locally.  But not if measured "globally", as we are free to do in this problem for a non-accelerating rocket. 

Post edited at 21:28
In reply to ericinbristol:

Errrr..... whichever one is in the direction you’re looking? You can’t see what’s behind you.

skog on 16 Jun 2018
In reply to Pero:

Thanks.

I think I can see what you're saying, but I'm not going to pretend to understand it properly!

I'm going to have to do some reading...

 
wintertree - on 16 Jun 2018
In reply to Ron Rees Davies:

> >>I’m not aware of any rocket powered vehicles with driven wheels, anyone else?

> Stephenson's Rocket?

Your post sent me down quite a tangent trying to find out why “rocket” was a common enough word in the early 1800s for a train to use it.  Various etymology sites suggest the word wasn’t used for describing   stored propellant jet engines until some time later.   The word was apparently used to describe cylindrical things at the time of Stephenson’s Rocket, as a derivative of the Italian word for bobbin.

I’d always assumed it was called the rocket because it was - comparatively - fast, but now I wonder if it’s because it looked cylindrical...

Hopefully there’s a UKC expert out there to comment...

DancingOnRock - on 16 Jun 2018
In reply to Pero:

The star appears to travel 24 light years. But it doesn’t actually travel 24 light years. 

In the original problem, one star is travelling -0.9c and the other +0.9c and the rocket is stationary. However , this is only as far as the maths is concerned. As far as the physics is concerned the rocket is not making the stars move and as there are billions of other stars you must consider the rocket to be the object that is moving. 

wintertree - on 17 Jun 2018
In reply to DancingOnRock:

> As far as the physics is concerned the rocket is not making the stars move and as there are billions of other stars you must consider the rocket to be the object that is moving. 

But in general the stars are not stationary with respect to each other.  In this case we can’t telll as the badly worded question didn’t specify the relative motion of the two stars.

There is no concept of stationary in SR, only non accelerating.  

What you wrote sounds similar to Mach’s principle perhaps.

Pero - on 17 Jun 2018
In reply to DancingOnRock:

> As far as the physics is concerned the rocket is not making the stars move and as there are billions of other stars you must consider the rocket to be the object that is moving. 

It was actually Galileo who first realised there is no such thing as absolute motion.  What this means is that you cannot assign a specific universal absolute velocity to anything.  The rocket is moving at 0.9c relative to the stars, but in no sense is the rocket absolutely moving at 0.9c.

One example might be a game of snooker.  Relative to the table the balls start at rest and then the white ball moves.  But, the table is on the surface of the Earth, which is spinning at about 1,000 km/h.  And, the Earth is moving relative to the Sun; the Sun relative to the galactic centre; and the galaxy relative to the Andromeda galaxy, say.

The point is that there is no experiment you can make on the cue ball to determine that it is or is not absolutely at rest.  You can never say: object X is definitely, absolutely at rest or definitely absolutely moving at 17.3 m/s.  All motion is relative to something else.

The other point is that in both Newtonian mechanics and SR the laws of physics apply to ALL inertial reference frames.  There is no special frame in which the laws of physics apply.  You are free, therefore, to consider the snooker table at rest and study the motion of the balls in that reference frame.  As you are free to consider the reference frame of the rocket, which is just as valid as the reference frame of the stars.

If this were not true, then you would have to consider the motion of your snooker table through the Cosmos before you could analyse the game!  In the same way that you want to insist that the rocket is "really" moving, I could insist that a snooker table is "really" moving and at no time would the cue ball actually stop moving.

 

crossdressingrodney - on 17 Jun 2018
In reply to Coel Hellier:

It's not quite that simple though, right? You could have two stars exploding simultaneously (in your reference frame) and yet if they were different distances from you, you wouldn't observe a simultaneous flash. The flash is only simultaneous in this question because the stars are an equal distance from you (in your reference frame) at the point when they explode. So I don't agree the question answers itself.

I think it illustrates quite neatly that calculations that would be tricky in one reference frame (stationary wrt the stars - perhaps the one we naturally think of) become easy in another. It also tests the understanding that the answer only depends on the position of the stars when they explode, not their velocities.


Please Register as a New User in order to reply to this topic.