### Beetlejuiced

 I’m not pompous; I’m pedantic. There’s a difference. —The Calligraphic Button Catalogue

Just about a year ago, a team of scientists reported that Betelgeuse—the bright red star in Orion’s shoulder—appears to have shrunk by about 15% since 1993. That would mean the diameter’s been shrinking at about 1200 miles an hour for all that time.

Such shrinkage—if it’s really happening (it’s hard to be sure)—could be the precursor to a supernova explosion, which would be way cool. The mathematician John Baez computes that a supernova Betelgeuse might be roughly as bright as the full moon, or maybe up to three times as bright.

Surprisingly, it took almost a year for this information to be widely reported on the Internet, but in the past few weeks, a number of sites have cropped up touting the upcoming supernova, and, as you might expect, a few prophecying doom. You can ignore the doomsayers; at a distance of 600 light years, Betelgeuse is too far away to hurt us.

Browsing the various science forums (such as Discover‘s), I’m struck by how often the following simple question comes up: Given that Betelgeuse is 600 light years away, is it or is it not true that it would it would take 600 years for us to notice any explosion? Or to put this another way: If the sky lights up with a new moonlike object tomorrow night, does that mean the explosion took place 600 years ago?

A pretty good answer—and the one that’s being given on all those science forums—is “yes”. But that can’t be exactly right, at least not for all of us, because at any given moment some of us are sitting in our living rooms while others are driving on the Interstates. Relativity tells us that if we’re moving relative to each other, then we must disagree about the times of distant events.

How much do we have to disagree? By about a half an hour, if I’ve done my arithmetic right. If you, standing on a street corner, say the explosion took place at noon on June 9th of the year 1410, then the driver of the car who has just run over your toes at 70 miles per hour (heading in the general direction of Betelgeuse) must say the explosion took place at 11:30AM on the same day. A half hour is not a lot (though it’s more than the few fractions of a second most of my friends have guessed), but here at the Big Questions, we aim to be precise in every detail.

Edited to add: It’s clear from the comments that I should have been clearer about what I’m assuming. My assumption (which I’d thought I’d stated but now see that I didn’t) is that you (and your friend) see the explosion at the exact moment when he’s running over your toe. You then each calculate how long ago the explosion must have taken place. Your conclusions will differ by half an hour.

#### 66 Responses to “Beetlejuiced”

1. 1 1 rathi

fascinating. could you briefly or even qualitatively explain the math. I would have thought (much like your friends ) that the difference would be roughly the order of how long light takes to reach you once it has “left” the taxi driver.

2. 2 2 Dirk

I for one would like to see the math worked out. I haven’t ever worked with Einstein’s equations but I was under the impression that to see a large time shift effect the relative speeds of the two viewers had to be a large fraction of the speed of light, say 10% or more.

3. 3 3 Stchoty

I got the same result: in general, if the angle between Betelgeuse and the velocity v of the street corner (as measured by the driver) is phi, then

tau = t gamma( 1 + v cos(phi) )

tau ~ t ( 1 + v cos(phi) )

(with tau: proper time of the driver, t: proper time of the person at rest.) according to my calculation.

But, note, that this answer is wrong unless the drivers velocity is constant for the last 600 years or, in other words, his distance to the street corner at 11:30 Jun 9th 1410 of his proper time is approximately 0.27 lightyears. Otherwise, if he is at rest most of the time, both time periods converge.

4. 4 4 Alex

What was the arithmetic used? My gut reaction is that a half hour seems off by several orders of magnitude based on the slow speed involved.

5. 5 5 Ken B

I don’t believe your math. You and the driver both carry a photon detector, and as he runs over your foot the first photons arrive. This is local enough to be well defined; the photons incident on the detectors can be said to be simultaneous. There will be a red shift on one detector, but the car clock and your watch won’t be 30 minutes out.

6. 6 6 Harold

Fascinating. Can you (or anyone else) explain in why there is a difference between the timings of the Beetlejuice event, whereas there is only a miniscule difference in the perceived timings of an even occuring, say, on the sun. What about or alpha centauri or a distant galaxy? 30 minutes is way more than I would have thought, and I don’t understand why it is so much.

7. 7 7 Steve Landsburg

Stchoty: I do not agree with your final paragraph. All that matters is the driver’s inertial frame at the time he makes the measurement, not his past history.

8. 8 8 Mike

I’ve always wondered if it would be possible to create some sort of telescope that wasn’t a telescope in the sense of magnifying light that was hitting the earth, but rather being able to truly see further and see light before it hit the earth and thus look into the future in the sense that we could have seen abovementioned supernova say 200 years ago (400 years after it happened) instead of in the relatively near future.

I hope that makes sense, but I’m fully aware it looks like I was smoking pot while typing.

9. 9 9 wheninrome15

Whoa. I find that very hard to believe. Could you say a little more on why I should believe it?

What’s wrong with this: Suppose the driver ran over the toes at 11:30, saw the nova, immediately stopped the car and got out. Then he would say to the pedestrian, “It’s 11:31 and I just saw a supernova.” and the pedestrian would say, “I agree it’s 11:31 but the supernova hasn’t happened for me yet.” (Their watches should pretty much exactly agree after getting out of the car). You’re saying the driver could then stand there with the pedestrian and wait half an hour and see the supernova a second time?

10. 10 10 Steve Landsburg

Ken B: Yes, we will agree that the photons arrived simultaneously. It does not follow that we will agree that they started out simultaneously.

11. 11 11 Steve Landsburg

wheninrome15: When the driver stops driving, he revises his ideas about time and space and comes to agree with the pedestrian. The question here is what he calculates while driving, not what he calculates after he’s stopped.

12. 12 12 Brian

There’s a mistake here. The time dilation equation relating tau and t (as quoted by Stchoty) has the following meaning: if I observe you to have been moving at velocity v for t seconds, then you will observe that same time interval to have only been tau seconds long. So if you start driving at 70 mph at 6:00pm and the light hits the earth at 6:01pm, then you will think the light hit the earth 10^-13 seconds prior to 6:01. The 600 years it took the light to reach the earth is irrelevant.

Taken to an extreme limit, imagine light traveling to us from one hundred million light years away. If time dilation occurred the way you say, then you could make a significant shift in the apparent time of the light arriving to earth simply by walking in one direction or another. There’s just no way that the apparent difference in time (between two nearly identical observers) can depend on how far away the light was originally emitted.

13. 13 13 Scott F

So you’re saying the effect of time dilation is not as prevalent when the photons are detected. But rather, the effect is noticed after the measurement is taken, when factoring in the speed of the car and when the photons are detected in trying to figure out when they were created?

14. 14 14 wheninrome15

aha gotcha. that’s what I was looking for.

thanks.

15. 15 15 Steve Landsburg

Brian: If time dilation occurred the way you say, then you could make a significant shift in the apparent time of the light arriving to earth simply by walking in one direction or another.

I maintain that time dilation occurs the way I say, and that yes, you can make a significant shift in the apparent departure time of light arriving to earth simply by walking in one direction or another.

I’ve put the relevant spacetime diagram here. The diagram differs ever so slightly from the blog post, in that the blog post starts from the assumption that you say the explosion happened at noon and concludes that your friend says it happened at 11:30, whereas the diagram starts from the assumption that your friend says it happened at 11:30 and concludes that you say it happened at noon. I hope it’s clear that the two assertions are equivalent.

I am a great fan of your physics blog, so I am hesitant to disagree with you on this. But I do think the diagram speaks for itself. I’ll be glad to know where (if anywhere) you think it goes wrong.

16. 16 16 Harold

This sure does take some thinking about. I suppose 30 minutes in 600 years is only about 0.0006% difference, which does not sound so big. If there were a star roughly in line with Beteljuice, but only 60 light years away, would the calculated “nova time” for the driver and pedestrian be different by 3 minutes?

17. 17 17 Steve Landsburg

Harold: Yes, in that case the time would differ by three minutes. And if Betegeuse were 6 million light years away instead of 600, the difference in calculated times would be about 5000 hours, or roughly 7 months.

See my reply to Brian and the link from there for the calculations.

18. 18 18 Stchoty

“Stchoty: I do not agree with your final paragraph. All that matters is the driver’s inertial frame at the time he makes the measurement, not his past history.”

I calculated the proper time elapsed between the event on each observers world line that is simultaneous to the explosion and the event where both observers meet. (The difference arises because “the event simultaneous to the explosion” is different for both observers.) The proper time is in general a functional of the world line between both events and does indeed depend on the entire history of the respective observer. (Which is quite easy to calculate, if the world line is straight, i.e. the observer is not accelerated, to which my result applies.) If the driver is at rest for most, but not all of the time then his world line fails to be straight but his proper time integral differs from the other observer only on a small fraction of his world line.

How did you calculate it?

19. 19 19 Ken B

@Steven
Ah, you mean this is the train-in-the-barn example. How does a train I measure at 100 yards fit into the barn you measure 80. In this case your barn ends in BG and my train is 600 light years long (plus or minus). Also the passing trains example.
@Brian
No, I think the point is that distnace is also affected, and the longer the distance the greater the absolute size of the effect. 30 minutes may be wrong but the effect will be greater for greater distances.

20. 20 20 Steve Landsburg

Stchoty: Okay, I see what you did; you and I are answering different questions. My calculation is here. (This corresponds to your calculation in the case where the driver’s speed never changes.)

I’ll say more about in the next comment….

21. 21 21 Steve Landsburg

Stchoty: The magnetic force on a moving electron depends only on the electron’s current velocity, not on its past history. The electron could of course go back and calculate proper times along its own past worldline, but those proper times would have no effect on its current behavior. In exactly that spirit, it seems to me that the natural interpretation of the moving observer’s opinion about the time of the explosion is: time as measured in his current inertial frame. That’s the only calculation that affects the way he currently sees and responds to the world around him.

22. 22 22 Al V.

Steve – I don’t follow your diagram. Since me, my home, and my friend did not exist in 1410, how can the origin be in 1410? And thus, wouldn’t the run be less than 600 years?

23. 23 23 Stchoty

OK, the difference, as far as I see, is that in your scenario both observers meet prior to the explosion, whereas my observers meet together with the light ray arriving from betelgeuze. Strange how our results match anyway.

But don’t you agree that the quite counterintuitive result is caused by the fact, that both observers are in constant motion relative to each other, rendering the situation a little unrealistic. Imagine the world line labeled “time (measured by your friend)” bends after a short time period to become parallel to the “time (measured by you)”. Then both proper times would be nearly equal. At least I guess. It’s a little difficult to calculate this case in your scenario, because the notion of “simultaneous to the explosion” becomes more intricate if not ambiguous for the accelerated driver.

24. 24 24 Steve Landsburg

Al V.: something was at the same latitude and longitude as your current home back in 1410. that’s what i’m calling “home”.

25. 25 25 neil wilson

You all think I am insane anyway but….

The driver here on earth is going to be staying here on earth. The explosion is going to arrive here on earth at basically the same time from everyone. If NASA sends someone to the moon when it is between the earth and the star he will be going far faster than 70 MPH.

Are you saying that people at different ends of the earth, where i am rotating towards the start and you are rotating away from the star have different views of the explosion?

26. 26 26 Steve Landsburg

Neil wilson: Are you saying that people at different ends of the earth, where i am rotating towards the start and you are rotating away from the star have different views of the explosion?

Yes.

27. 27 27 Steve Landsburg

Stchoty: My scenario does not assume the observers meet prior to the explosion. They meet just at the moment when the explosion becomes visible. (The steep red line is not meant to be a world line; it is a line of constant space-location for the driver, which is to say that it’s only assumed to be *parallel* to his world line).

28. 28 28 Stchoty

Of course the electromagnetic force on an electron is a local phenomenon depending only on the present event on the electrons worldline, and therefore is independend on any definition of simultaneity.

But the problem we discuss depends essentially on such a definition. Now the problem is: if the observer is accelerated then there is as far as I know, no “good” universally valid definition of simultaneity. The definition “simultaneous in the current inertial reference frame” becomes ambiguous for large distances and even reverses the order of some events that are timelike to each other.

29. 29 29 Stchoty

P.S. If you are in possession of Misner, Thorne, Wheeler, Gravitation, they elaborate on this problem in chapter 6, as far as I remember.

30. 30 30 Brian

Let me make sure I understand your claim.
Say that up until 11:29am, both my friend and I are stationary, and our watches are completely synchronized. Are you saying that if my friend starts moving at 70 mph at 11:29, then he will see the explosion at 11:30? Meanwhile I, who remained stationary, don’t see it until noon?

What happens if my friend stops his car 20 minutes later? Does he return to my reference frame and get to see the explosion again at noon? What happens if I start moving at 70 mph at 11:59? Do I move into the frame where the explosion happened at 11:30 and therefore miss the explosion entirely?

The biggest problem is that at 11:30, in any reference frame, the light still hasn’t reached the Earth. It’s 300 million miles away, so there’s no way that any observer on the Earth will see it.

I’ll draw you a space time diagram, as I see it, and upload it in a little bit.

31. 31 31 Seth

AI V: Actually, what makes up you and your home existed in 1410. It’s just organized differently now. But, to Professor Landsburg’s point, I think the spatial orientation is the anchor.

Very interesting discussion. Does the 70 mph observer calculate a different distance to the explosion as well or just time?

32. 32 32 Steve Landsburg

Seth: Yes, he calculates a different distance as well, but here the discrepancy is far less impressive. If you (on the street corner) say the distance is 600 light years, then your friend (in the car) will say that the difference is 599.999999999996999999999999992 light years (again, assuming I’ve done my arithmetic right). Pretty hard to tell that apart from 600. (The difference, I think, is maybe 20 miles or so.)

33. 33 33 Harold

Brian: I don’t pretend to understand this, but as to your question, I think both pedestrian and observer see the light at 12:00, i.e. the same time. It is only the calculated time the nova actualy happened (either 600 years ago, or 600 years and 30 minutes ago), that changes.

It reminds me of the museum guide saying “that T Rex skeleton is 65,000,015 years old” how does he know so exactly? Because it was 65 million years old when he started at the museum, 15 years ago.

34. 34 34 Steve Landsburg

Harold: Your most recent comment is exactly correct. Nice analogy with the museum guide.

35. 35 35 Brian

Okay, here is the space-time diagram as I see it: http://s1012.photobucket.com/albums/af249/gravityandlevity/spacetime.png (sorry for the low image quality)
I assumed that you and your friend have been in the same reference frame for all the time since the nova actually exploded (600 years ago), and then at 11:30 he starts moving at 70 mph.
I think that your friend will observe the explosion to be only 10^-11 seconds earlier than you do.

36. 36 36 Seth

Thank you Steve. Actually, that seems to be an interesting framing bias. 20 miles and 30 minutes seem to be about the same magnitude to me (especially since I can travel 20 miles in 30 minutes at 40 mph). Expressing the distance in light years makes it look small, just as expressing the time in years.

Interesting post. It reminds me that the universe isn’t what it seem. It doesn’t quite follow our Newtonian instincts. At least if we assume the speed of light is constant. I’m not so sure it has been.

37. 37 37 Steve Landsburg

Brian: If I understand your diagram, it assumes that your friend has driven almost all the way to Betelgeuse when he notices the explosion. But I was assuming your friend is still here on earth, just running over your toe with his car, at the moment when you *both* notice the explosion.

38. 38 38 Brian

Sorry Steve, I did something a little confusing. I sort of reversed the picture, and imagined that the explosion is a stationary event and that you are traveling toward it at the speed of light (rather than vice-versa). It just makes the picture easier to draw.

39. 39 39 Stchoty

“Stchoty: My scenario does not assume the observers meet prior to the explosion. They meet just at the moment when the explosion
becomes visible.”

Hm, ok, but then I don’t understand in what respect we are answering different questions.

From your calculation I’m unable to tell how much time elapsed for the driver between the meeting of both observers and the event
that is (for him) simultaneous to the explosion. Don’t you have to define this? If you don’t, then I’m not sure what the question is you are answering.

“(The steep red line is not meant to be a world line; it is a line of constant space-location for the driver, which is to say that it’s only assumed to be *parallel* to his world line).”

Ok, but as far as I understand the problem, there are two events, that should be defined on each observers worldline (or any line parallel to it): 1) the event that happens at the same (proper) time as the explosion, 2) the event where the light ray crosses the world line. Then I can calculate the time that passed on this worldline, between both events.

I understand that you somehow cut short these calculations, but unless I can fill in this information I’m unable to understand what your answer means.

40. 40 40 Steve Landsburg

Brian: At any given time (from your point of view) you measure distance along a horizontal line. At any given time (from your friend’s point of view) he measures distance along a line parallel to the one you’ve labeled “30 minutes”. Because we are both seeing the explosion from earth, you’ve got to run one line from each family all the way back to earth, 600 light years from the explosion. I don’t understand the significance of the particular lines from each family that you’ve chosen, or of their intersection point.

Does my picture make sense to you? If not, does it come close enough to making sense that you can pinpoint what you think is wrong?

41. 41 41 Steve Landsburg

Brian: Sorry, I responded to your diagram having overlooked your earlier “Let me make sure I understand your claim” comment.

Say that up until 11:29am, both my friend and I are stationary, and our watches are completely synchronized. Are you saying that if my friend starts moving at 70 mph at 11:29, then he will see the explosion at 11:30? Meanwhile I, who remained stationary, don’t see it until noon?

I am assuming that you and your friend both *see* the explosion at the same moment—at noon, June 10, 2010, just as your friend is running over your toe with his car at 70 mph.

You each then calculate, in your own inertial frames, when the explosion must have happened. Your past history is irrelevant to these calculations.

I claim that your calculated times will differ by half an hour.

42. 42 42 Steve Landsburg

Stchoty: I am assuming that the light from the explosion hits both you and your friend at 12 noon today, just as he is running over your toe with his car (so everyone agrees that the explosion first became visible at the exact moment your friend passed by you).

You then ask yourself when the explosion must have happened, and calculate that it happened 600 years ago, using your meter sticks to determine how far away Betelgeuse is, etc. Your friend does exactly the same thing.

My assertion is that your calculations of the time of the explosion will differ by half an hour. This assertion has nothing to do with your friend’s path yesterday or tomorrow; it only assumes he’s still traveling 70mph when he does his calculation.

43. 43 43 Stchoty

“Stchoty: I am assuming that the light from the explosion hits both you and your friend at 12 noon today, just as he is running over your toe with his car (so everyone agrees that the explosion first became visible at the exact moment your friend passed by you). You then ask yourself when the explosion must have happened, and calculate that it happened 600 years ago, using your meter sticks to determine how far away Betelgeuse is, etc.
Your friend does exactly the same thing.”

I think you should be more careful on this point. What exactly does it mean to you, that “your friend calculates the time the explosion happened”? To me it means he calculates his proper time that has elapsed since the event on his world line happening simultaneously to the explosion at Betelgeuse (and subtracts it from ‘noon, June 10, 2010′).

The answer 1) depends on the notion of simultaneity used to define this event, and 2) it depends on the whole path of the observer between this event and the light rays arriving (because of the definition of ‘proper time’). And I don’t see any other meaningful way of calculating it.

Regarding 1) there is another problem of how to define simultaneity for accelerated observers or to deal with an observer that is in uniform motion all the time for 600 years.

It’s still unclear to me how you evade to adress points 1) and 2) in solving this problem.

44. 44 44 Ron

I believe you’ve got the slope wrong in your spacetime diagram.
The formula is not V/C, as your diagram states. The formula for
the Lorentz contraction for observed length (distance) is
L * sqrt(1-v**2/c**2)
Note that this is a function of v squared over c squared, not
v over c.

Lets use the above equation to calculate the measured length seen
by the 70 mph driver. I get that the driver’s measure of the
distance to the star is 22 light minutes shorter than the
non-driver (or 22.332987 to be more exact). Therefore, the
calculated time of the explosion would differ between the driver
and the stationary observer by 22 minutes.

Calculation source (details of SQRT routine not included):

/* rexx */
numeric digits 40
L2= 600 * sqrt( 1 – (70*70/(186000*186000)) )
Diff= 600 – L2 /* answer is in units of light years */
DiffMin = Diff * 365 * 24 * 60 /* convert light years to light minutes */
say DiffMin
exit 0

45. 45 45 Steve Landsburg

Ron: If you look at my diagram, you’ll see that it shows exactly the distance contraction that you’re quoting. So we are doing basically the same calculation, presumably with different rounding.

Edited to add: Actually, I now think your calculation is wrong because it fails to account for the driver’s belief that Betelgeuse is moving toward him, so the light from the explosion originated not from Betelgeuse’s current location but from its location at the time of the explosion. So the difference in our answers is more than just rounding error. I stand by my original analysis.

Edited to add further: Brian’s comment here shows that a correct calculation based just on the Lorentz contraction (as you’ve attempted) would result in a discrepancy of just a tiny fraction of a second, not 22 minutes. But the bulk of the effect does not come from the Lorentz contraction.

46. 46 46 Steve Landsburg

Stchoty: It seems to me that the paragraph you quoted responds to both your points 1) and 2). The whole setup assumes that it makes sense to ask “how long ago did the explosion happen?” in the first place. Any given observer answers this question based on calcuations made in his own inertial frame.

Sure, an observer who has accelerated is going to say “gee, yesterday, prior to the acceleration, I would have calculated differently”. But he still gets a unique answer within his current reference frame, just as an electron knows how to behave based solely on calculations made in *its* inertial reference frame, without regard to past accelerations.

47. 47 47 Brian

Thanks for the clarification, Steve. It’s clear that I misunderstood the problem.

When the first light from the explosion reaches you, you will say “look, a ray of light hit me from a source that is 600 light years away. Therefore the explosion happened 600 years ago.”

You friend, who is traveling at 70 mph, will see that same distance length-contracted by about 10^14 (as you noted yourself above). He will say “look, a ray of light hit me from a source that is (1 – 10^-14)*600 light years away. Therefore the explosion happened (1 – 10^-14)*600 years ago”. Since light appears to travel at the same speed relative to all observers, this is the end of the problem.

So your friend will calculate the time of the explosion to be only about 100 microseconds earlier than you will.

I think I see the mistake in your calculation as well, and it’s a fairly simple one. In your space-time diagram, you claim that “the slope of that line” (the amount by which your friend’s space-time axis is skewed relative to yours) is equal to (70 mph)/(the speed of light). In fact, the slope is given by the Lorentz factor (1 – (v/c)^2)^(1/2) which is approximately equal to 1/2 * v^2/c^2 which is about 5 * 10^(-15). If you correct this slope then your calculation will give about 100 microseconds.

48. 48 48 Brian

Oops, I meant that the slope is given by (1 – Lorentz factor). Obviously, if the Lorentz factor is 1 (which means v = 0), then your friend has the same space-time axis as you.

49. 49 49 Steve Landsburg

Brian: We now agree about the problem, but I do not agree with your calculation, or with your criticism of my diagram.

A driver traveling at speed v, passing through location 0 at time 0, will pass through location v at time 1. Therefore, with time on the vertical axis, his worldline passes through (0,0) and (v,1),
and therefore has slope 1/v. All of his various “fixed location” lines are parallel to this and therefore have slope 1/v.

His “fixed time” lines must be Lorentz-perpendicular to his “fixed location” lines; that is, the Lorentz inner product of a point on the fixed-time line through the origin with a point on the fixed-distance line through the origin must be 0. This requires the “fixed location” line to have slope v, as I have drawn it.

The distance from the origin to the explosion, along that fixed time line, is given by the Lorentz metric as sqrt( horizontal^2 – vertical^2), which is 600 times the Lorentz contraction factor that you are claiming should be the slope of the line. But the Lorentz contraction shows up as that distance, not as a slope.

Your 100 microseconds is, I think, an arithmetic mistake.

50. 50 50 Stchoty

“But he still gets a unique answer within his current reference frame, ”

Steven, every observer calculates a unique answer *independent* of any coordinate system. Like the length of a curve in euclidean space does not depend on the cartesian coordinate system you choose, so is every statement about the time measured by any given observer (between to well defined events) independent of the local inertial reference frame he (or anyone else) uses to describe these events. “Time” as measured by an observer is just the length of a segment of his world line calculated in the Minkowski-metric. Neither this length nor the calculation relies on any reference frame.

If, on the other hand, you just calculate the differences between two time coordinates in one given inertial frame (as you seem to do), then this does not have any obvious geometric (not to say “physical”) meaning for an observer that is not permanently at rest in this system. If the observer is not in uniform motion how do you know the ‘correct’ inertial frame to calculate the answer in? Why didn’t you choose another one?

“just as an electron knows how to behave based solely on calculations made in *its* inertial reference frame, without regard to past accelerations.”

1) An electron knows how to behave without regard of its history, because the force acting upon it does only depend on the event the electrons world line passes (and it can solve its equation of motion ;-))

2) This has IMHO nothing to do with the problem at hand. Here we explicitly ask about specific proper time intervals for two different observers. By the very definition of “proper time”, the answer depends on all the velocities (i.e tangent vectors to his world line) the observer had in space time during this interval (though, not on his acceleration, as I hope I have not asserted.)

51. 51 51 Steve Landsburg

Stchoty:

Here we explicitly ask about specific proper time intervals for two different observers.

I agree that this is a perfectly good question. But it’s not the question I was asking.

The two inertial observers must agree about the speed of light. They must disagree about the distance to Betelgeuse, by a factor of sqrt[1-v^2], independent of their past activity. Because distance=rate * time, and because they disagree about distance and agree about the speed of light, they must disagree about how long the light took to travel.

All I am saying, then is that a) two inertial observers must agree about the speed of light, b) those two observers, if their relative velocity is not zero, must disagree at the moment they pass each other about the distance to Betelgeuse, and c) if they calculate the time it took for light from Betelgeuse to reach them, using the formula distance = rate x time, then they must get different answers.

I agree that if one of the observers has accelerated then his watch will continue to show the affects of that acceleration. But we’re not talking (or at least I wasn’t talking) about what’s on their watches; I was talking about how they currently calculate the time of the explosion.

Edited to add: What I said here was not correct; the driver must correct not only for the Lorentz contraction but for
the fact that Betelgeuse appears to be moving toward him. Brian has this exactly right in the final paragraph of the comment after this one. I stand by my original spacetime diagram as an accurate picture of what both observers report. I do not claim that the driver is reporting the time on his wristwatch, which, as you note, depends on his entire worldline and not just his current motion.

52. 52 52 Brian

Steve, I finally understand what’s going on here. I apologize for how unnecessarily long-winded and confusing this discussion has been; I feel I have been particularly unhelpful.

Both of us are essentially looking for an answer to the following question: “Two observers look off into the distance and see the star Betelgeuse; one is at rest relative to the Earth (and, presumably, Betelgeuse) and the other is moving forward at 70 mph. They ask themselves the question ‘If I sent a ray of light from here to Betelgeuse, how long would it take to get there?’ ” Where we differ is in the existence of one further assumption made by the observer.

In my approach, I assumed that the observer would understand that, while he may be moving now, he will always remain on the Earth. Therefore, any apparent motion of Betelgeuse relative to himself is “fake”, in the sense that it isn’t going to last very long. So the observer can just measure the distance from himself to Betelgeuse (which is very slightly Lorentz-contracted) and divide by the speed of light. This calculation will suggest that the light will arrive in 600 years – 100 microseconds.

In your approach, you imagine that the observer sees two things: the ray of light moving away from himself (at the speed of light) and the star Betelgeuse moving toward himself (at 70 mph). If he assumes that these two facts will remain constant, then he will estimate that the light will hit Betelgeuse in about 600 years – 32 minutes (and, indeed, length contraction plays almost no role).

Of course, this answer can be turned around to match the original problem statement, where both observers saw the supernova exploding in 2010 and asked themselves when the explosion happened. Such an observer would see the light moving toward him and Betelgeuse also moving toward him, and might calculate the time of the explosion by time-reversing both trajectories and calculating their point of intersection. That observer would arrive at the conclusion that the supernova occurred (600 years + 32 minutes) ago, as you said originally.

53. 53 53 Steve Landsburg

Brian: yes, this is right. Your last paragraph in particular is highly clarifying.

54. 54 54 Neil

This seems to be closely related to a problem I know of as the “Andromeda paradox”. The A paradox has to do with the difference, perceived between two earthly observers in relative motion, in the timing of events occuring in the Andromeda galaxy in each observer’s “now”. In contrast, this problem has to do with the timing of events occuring in the past.

55. 55 55 Steve Landsburg

Neil: If I understand you correctly, this is not a different phenomenon (nor does it deserve to be called a paradox). The events happening in a distant galaxy “now” are, perforce, unknown to us. We can’t start arguing about their timing until sometime in the future when we learn about them. So at the time when the argument occurs, the events are in the past—just as would be the case in my Betelgeuse example. Which makes this exactly the same phenomenon.

56. 56 56 Neil

Yes, Steve, they probably are the same. But the Andromeda paradox, as Roger Penrose relates it, involves another angle you will like–the issue of free will. A decision as to whether to invade earth is being made on some distant planet in Andromeda (I don’t worry too much about that affecting me.) For the observer sitting in an armchair, the decision is yet to be made, but for an observer walking around the room the decision is already made (say to invade earth). How can something be yet to be decided and be decided at the same time?

57. 57 57 Roger Schlafly

Steve, this problem has nothing to do with relativity. You are just formulating a question in an ambiguous way, and then giving an answer that seems surprising only because your interpretation of the question is contrary to common sense.

The essence of your point is that if a car is driving 70 mph and you extrapolate that car backwards 600 years, then the car might be in Jupiter’s orbit somewhere. Then you are pointing out that a car in Jupiter’s orbit might see the supernova 30 minutes later than Earth, because that is how long it takes light to get from Earth to Jupiter. That’s all you’re saying.

But your interpretation of the problem is contrary to common sense because cars do not drive from Jupiter to Earth. That 70 mph is completely dwarfed by the 1000 mph velocity from the rotation of the Earth, and the much larger velocity from the revolution about the Sun. Your same argument would say that people in the USA and China would disagree about the age of the supernova, and that their disagreement would depend on the time of day. No one measures time that way.

58. 58 58 Steve Landsburg

Roger Schlafly:

Your same argument would say that people in the USA and China would disagree about the age of the supernova, and that their disagreement would depend on the time of day.

Absolutely.

No one measures time that way.

How do you know? People in the USA and people in China will disagree about the age of the supernova, but they’ll both round off their estimates to 600 years. So when they both say “600 years”, that doesn’t mean they’re in agreement.

59. 59 59 dave

@neil what if there was no free will? and therefore no ‘decision’.

60. 60 60 Neil

@dave: I’ve got no problem with that because I hold the view that there is no free will, because the conditions that lead the decision makers to their decision are pre-determined. But, according to The Big Questions, Steve does think there is free will, so he might have a problem with it.

61. 61 61 Stchoty

Steven, thanks for your last comment. I think I now better understand your argument.

“I agree that if one of the observers has accelerated then his watch will continue to show the affects of that acceleration.
But we’re not talking (or at least I wasn’t talking) about what’s on their watches; I was talking about how they currently calculate the time of the explosion.”

I think we have some fundamental disagreement about the meaning of relativity, which is both exciting and strange. To me your last two statements contradict each other. There is no other time for an observer than the time on his watch, so “time of the explosion” should mean to any observer “time his watch displayed at the event simultaneous to the explosion.” Questions about time in relativity should IMHO always be answered in terms of proper time, because this is the time that is actually measured by some observer.

You, if I understand you correctly, are using a different notion of “time” in your last statement, meaning something like “time coordinate in the observers current inertial frame.” To me there is no meaning to such a time coordinate other than as “proper time
of some unaccelerated observer”, which is why I concluded that the driver has to be in uniform motion for approximately 600 years, for your statement to make sense.

62. 62 62 Steve Landsburg

Stchoty: I fully agree with your characterizations of the questions you and I were answering. If we continued to discuss this, we might or might not continue to disagree about which question is more natural. But I think we at least fully understand each others’ positions.

63. 63 63 Roger Schlafly

Steve, China and the USA do not have different notions of time that they change every 12 hours based on their orientation towards Betelgeuse. It is not just a matter of roundoff.

64. 64 64 Steve Landsburg

Roger: If my arithmetic is right, the rotation of the earth causes people in China and the U.S. to disagree about the timing of events on the sun by up to .0007 seconds. We disagree about the timing of events on earth, of course, by far far less than that. Only very rarely do we contemplate events farther away than the sun. So it’s not surprising that we don’t bother to make corrections for these discrepancies. I’d say that ignoring time discrepancies of far less than .0007 seconds is a very good example of what is meant by roundoff.

65. 65 65 David

Bad Astronomy – one of the best science blogs let alone for this. Phil Plait is a great writer, I’m glad you linked his post on this.

66. 66 66 Ben

Steve: Does this mean that an observer in a car moving 70 mph decides that the GPS signal he just received was transmitted 5 nanoseconds earlier than the observer who just got his toes run over? It seems like this error would be significant; do GPS receivers actually do this relativistic correction?