1. Apr 11, 2007

### Pavel

Hi! Just as I thought I was getting a pretty good grip on relativity, I got thrown back to square one. Please straighten out something for me by taking a role of a passer-by in my thought experiment.

First, I read that according to Special Relativity (SR), observers in two moving inertial frames will perceive time dilation with respect to each other, regardless of the direction of the motion. That is, if I moving at a constant speed towards you, head-on, I will perceive your clock moving slower than mine, and you will perceive my clock to be slower than yours. This is according to SR alone, all other things aside! Correct? Very well. Then I can construct the following thought experiment, which is obviously wrong, but I'd like somebody to explain to me what other relativistic effects are in place and how they solve the "paradox".

I, here on Earth, happen to find out that there's a bomb on Alpha Centaury (AC) that will go off in 3 years, which will kill their civilization. Both Earth and AC are in the same reference frame and it takes 4 years for light to reach AC. Luckily, there's a ship, a passer-by, that happens to fly by Earth tomorrow. This ship flies at a constant speed, almost the speed of light and will fly by both Earth and AC. Being an amateur physicist, I reason: "Hmm, because the passer-by in the ship will literally perceive AC to be much closer and the clock on the bomb on AC will run slower relative to the passer-by (according to SR), if I somehow let the passer-by know about the doomsday of the Alpha Centaurians, the SR's time dilation will allow the passer-by to save them". So, I put a giant billboard on the passer-by's way instructing him to flash the light at Alpha Centaurians to warn them, as he goes by AC. I obviously presume the passer-by always looks in the telescope straight ahead, which will allow him to see the billboard.

Now, please notice, I eliminate any acceleration from the thought experiment: The passer-by flies at a constant speed, both Earth and AC are in the same frame, so I'm talking about inertial frames here moving at a constant speed relative to each other - AC/Earth and the passer-by. You may object by noting that both Earth and AC are in gravitational wells, thus producing acceleration, but I think it's irrelevant - I can always modify the experiment to say that Earth and AC are actually names of the hovering ships, not planets. The "paradox" will still hold.

I know that the passer-by will NOT save the Centaurians, but I don't understand what other effects will make the time on AC to speed up relative to the passer-by. I'm aware of the Doppler effect, but I can't see how that will override time dilation, it has to be some relativistic effect. Is it the mass of the passer-by's ship creating huge mass and therefore gravitational well producing time dilation? But then wouldn't you expect the time dilation of SR be proportional to the GR's dilation from the ship's gravitational well?? So, I'm confused. If you know the answer, please assume the role of the passer-by in the ship and describe what you will perceive starting with looking at the telescope and seeing my warning billboard. In advance thanks, and sorry for being so wordy, just trying to be complete.

Pavel.

2. Apr 11, 2007

### pervect

Staff Emeritus
Hint: have you considered "relativity of simultaneity"? See for instance the wikipedia article http://en.wikipedia.org/wiki/Relativity_of_simultaneity

3. Apr 11, 2007

### JesseM

By "perceive", do you mean what the ship will actually see as it looks through its telescope as it travels, or how it will reconstruct the timing of events in its own frame once it corrects for the delay between when an event happens and when the light from that event reaches the telescope? In the first case, the answer would lie with the doppler effect, which in relativity says that when you are travelling towards an object, you will see its clock sped up, not slowed down; in the second case, Alpha Centauri's clock will be reconstructed to have been ticking slower than the ship's clock throughout the journey, but as pervect said, the answer has to do with the relativity of simultaneity. If two clocks are at rest with respect to each other and synchronized in their rest frame, and they are a distance x apart in their rest frame, then if in my frame the clocks are moving at speed v along the axis joining them, the relativity of simultaneity implies that in my frame the back clock's time will be ahead of the front clock's time by vx/c^2. So, if Alpha Centauri is 4 light years from Earth in its own frame, and the ship sees the Earth and Alpha Centauri moving at 0.8c, then in the ship's rest frame Alpha Centauri's clock is 3.2 years ahead of Earth's clock at the moment the ship passes Earth (but at that moment the ship will see Alpha Centauri's clock as 4 years behind, just as observers on Earth will at that moment, since the light takes 4 years to cross the distance in the frame where Earth and Alpha Centauri's clock are synchronized).

Last edited: Apr 11, 2007
4. Apr 11, 2007

### country boy

The spaceship traveler moving relative to your frame (the inertial frame of Earth and Alpha Centauri) will indeed measure the Earth-AC distance as shorter and the clocks at Earth and AC as going slower. However, he will not agree that the clocks at the two places are sychronized. When he passes AC he will see that the clock there reads more than 4 years later than your clock did when he passed Earth, even if he measures his own transit time as less than 3 years. The different synchronizations of clocks in different inertial frames always saves you in this type of problem.

5. Apr 11, 2007

### Eli Botkin

Pavel:
There is no paradox. AC cannot be saved by anyone who depends on info from Earth to do this.

As you point out, a signal from Earth will take at least 4 years to arrive at AC, at least 1 year after AC’s demise.

If a signal is sent from the passer-by ship immediately after it passes Earth, even if the ship is moving at very high speed toward AC, that signal will arrive at AC AFTER Earth’s signal would have arrived at AC, regardless of the choice of frame. You can see this if you draw an SR Minkowski diagram.

Focus on EVENTS and their perceived temporal order, rather than on clock dilation which deals with the transformation of an event’s time-coordinate from one inertial frame to another. The temporal order of 2 events is the same in all frames if the events are timelike-connected, that is if the line joining the events is an allowable worldline (the worldline's absolute speed < c).

6. Apr 11, 2007

### windscar

As the object traveled close to the speed of light, it would see the distance light had traveled from the billboard contracted to almost zero for one secound. So it would take a long amount of time for the ship to recieve the message as it passed by depending on how close it came to earth. Hopefully not too close, because it would appear to be a black hole unless it was using some sort of "warp drive". They themselves would then need to shoot a signal into a gravity well that they observed Earth and AC being in. For them, Earth and AC wouldn't be very far away but would be more of a matter of the time the signals take for the "rest" frame to travel to the frame traveling close to C.

7. Apr 11, 2007

### Pavel

Thank you for the insightful replies guys, they were very helpful. Wikipedia seems to have a few good articles on resolving this as well. Looks like the Wiki book on SR might be a very helpful read. But as you indicated, there appears nothing special - I simply need to focus on temporal order of the events from both frames.

I'd like to dive deeper into the passer-by's ship becoming a black hole (due to high speed) story, but I'll review GR and acceleration first. I'm sure it'll change the whole picture.

One more thought occurred to me while reading SR and I'll appreciate any insight. It says there's no preferred (absolute) frame and it's meaningless to talk about who is faster than whom, it's all relative. Well, in pure theory. But practically speaking, at least in this Universe, can't we use CMBR as the kind of ether to measure our direction and speed against? And then use those numbers to compare them with those of some alien civilization, if one exists? Granted, I can't say that the Earth's speed (corresponding to reading 2.7K of blackbody temperature from CMBR) is not absolute, but if I have sensitive enough measuring tools, I should be able to read different temperatures in different directions of the Earth's movement (Doppler effect). On top of that, if an alien civilization reads say 2.2K, then can't I conclude that I'm, in fact, moving faster than the aliens?? Then if we had enough civilizations scattered around, couldn't we come up with some mean temperature that would allow us to agree on the absolute frame, corresponding to some temperature?? What's wrong with this logic?

Pavel.

8. Apr 11, 2007

### JesseM

That doesn't happen--see If you go too fast do you become a black hole? from the Usenet Physics FAQ.
I believe 2.7K is the temperature an observer at rest with respect to the CMBR will measure, any observer in motion relative to it would see a higher temperature, not a lower one. In any case, observations of temperature can tell you something about your speed vs. the aliens' speed relative to the CMBR, but why would you think the CMBR is "at rest" in any absolute sense? The notion that there is "no preferred reference frame" is just saying there is no difference in the laws of physics from one frame to another, there may be differences in observations of particular physical landmarks like the sun, the galaxy or the CMBR. But if you perform a certain experiment in an isolated, windowless room on Earth, and the aliens perform the same experiment in an isolated, windowless room on their planet, you'll both get the same results, because the laws of physics are the same in both frames.

9. Apr 12, 2007

### Mentz114

Yes. There are clear signs of movement in the CMBR measurements. The frame of the CMBR is the frame of the matter in the universe at the time it became transparent. But it's not an absolute frame.

There's a good pic of the dipole anisotropy of the CMBR here

http://antwrp.gsfc.nasa.gov/apod/ap990627.html

You can be sure of one thing, you are moving faster than the aliens WRT the CMBR. Any further deductions at your own risk.

10. Apr 14, 2007

### windscar

I read this post and it is a giant load of crap. E=mc^2 has nothing to do with this idea, and E=mc^2 isn't used to determine relativistic mass.

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

This equation is what leads people to beleive that as V approaches very close to C, your mass would approach infinity. And then you would become a black hole. This would be more of a question of how this mass is distributed from an object as it approaches the speed of light. This equation alone tells nothing of the sort. Now if the speed of an object increases the curvature of space-time as it approaches the speed of light, the curvature of space-time around this moveing object would be relative of the original mass and size of this object. Now if you speaking about a small space craft, the point from wich space-time is going to infinity would be rather and small, and I would say yes it could create some sort of singularity like a black hole, given space-time would curve to almost infinite mass around a small object. My experiment to prove this is would come from the expansion of the universe itself. It expands exponentialy the farther out you look. At the edge of the visable universe stars travel close to the speed of light relative to us... and we can not "see" any stars or galaxies from past this point, because light cannot travel from them to us from this relative velocity.

Last edited by a moderator: Apr 14, 2007
11. Apr 14, 2007

### Thrice

As Eli Botkin said, focus on the things that are frame independent. There's probably some frame out there that thinks you're moving close to c. Why aren't you turning into a black hole?

12. Apr 14, 2007

### pervect

Staff Emeritus
I don't know why you think that, as you don't quote any references of your own, but the sci.physics.faq is a reliable resource (it is however written mainly for a lay audience, which is both a strength and a weakness)

If you want more references look at

which is an actual "post" (as opposed to the FAQ).

Information about Steve Carlip, the author of the above post, (which also cites the FAQ entry and expands a bit on it) can be found at

http://www.physics.ucdavis.edu/Text/Carlip.html#Honors

(If you're really paranoid, I imagine you can verify Steve Carlip's credentials from other sources).

13. Apr 14, 2007

### Pavel

I wish I were more informed on the subject to jump into the “black hole” debate, but there’s still a lot of basic stuff to conquer. By the way, I do appreciate your time and effort explaining all this.

From what I understand, the formula Windscar posted $$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$ is the time dilation factor, used for both calculating time delay and mass increase. Now, are these two different ways to look at the same effect or are these two completely separate effects? It looks like any way you look at the definition of mass, there’s time involved in one of the factors. So do I deduce increase in mass simply due the time factor changing or does the mass increase regardless of what time does? If the former, then what about accelerated particles producing some damage on the wall they hit. The amount of kinetic energy, using classical physics, should yield a certain amount of damage to the wall the particle hits. If I have a relativistic mass increase, the particle should do more damage, from an observer’s perspective, the only perspective I care about, right?. When I measure the damage, the time factor is out, is it not? Would that be a way to see if time dilation and mass increase are two separate effects?

Thanks,

Pavel.

14. Apr 14, 2007

### JesseM

Are you familiar with the full version of E=mc^2 used for objects that have a nonzero momentum, namely:

$$E^2 = {m_0}^2 c^4 + p^2 c^2$$

If you substitute in the formula for relativistic mass that you posted above, along with the formula for relativistic momentum:

$$p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

...then you'll find that you end up with the formula $$E = mc^2$$, where m is the relativistic mass rather than the rest mass. So the formula E=mc^2 can in fact be used to determine relativistic mass.

Last edited: Apr 14, 2007
15. Apr 14, 2007

### JesseM

The factor $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ appears in many formulas in relativity, it is denoted with the symbol $$\gamma$$, the greek letter "gamma". Not just in time dilation and relativistic mass, but also in the formula for length contraction, or:

$$L = \frac{L_0}{\gamma} = L_0 * \sqrt{1 - \frac{v^2}{c^2}}$$

(L0 is the length of the object in its own rest frame, L is its length in another frame where the object is moving at speed v along the axis whose length is measured)

The relativistic gamma factor also appears in the "Lorentz transformation" which transforms the coordinates of an event in one frame to the coordinates in another. If an event happens at space coordinate x and time coordinate t in my frame, and I want to figure out its coordinates x' and t' in another frame which is moving at speed v along my x-axis, and whose origin (x'=0) passed by my own frame's origin (x=0) at a time coordinate of 0 in both frames (t=t'=0), then the transformation is:

$$x' = \gamma (x - vt) = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$t' = \gamma (t - \frac{vx}{c^2}) = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}$$

The time dilation equation and the length contraction formula can be derived directly from the Lorentz transformation, while the formula for energy/relativistic mass requires slightly more complicated proofs based on the assumption that the laws of physics should work the same way in each of the frames given by the Lorentz transform.
I don't know of any way to deduce an increase in relativistic mass (which as I mentioned in my last post is equivalent to an increase in energy) from time dilation alone. Like I said, the relativistic gamma factor appears in a lot of different places, not just those two equations.

Last edited: Apr 14, 2007
16. Apr 15, 2007

### Pavel

Ok, let me ask you about it this way. I set up an inertial mass between two springs in space, make it oscillate and count ticks per unit of time. The number of ticks obviously depends on the amount of inertial mass. I then speed up the system. Given SR, I should expect the number of ticks to decrease. Does this decrease happen due to time dilation or mass increase? Is it possible to tell? Well, maybe. What if I now calculate the $$\gamma$$ factor and somehow reduce the inertial mass to adjust for the relativistic mass increase? My question then is, will I now observe the same number of ticks when the system was at rest or will the number still be lower? If it’s the same, then should I conclude that time dilation and relativistic mass increase are the same thing being called two different names depending on the context of measurements? And if the number of ticks is still lower, then should I conclude that time dilation is a separate effect in addition to mass increase? I hope I’m making sense here. If my math was up to speed, I’d figure this out on my own, but until then I have to resort to thought experiments and ask you about the resulting observations. Again, thanks for your help.

Pavel.

17. Apr 16, 2007

### windscar

First off, if you assume that you would not observe some kind of black hole effect from observing an object approaching the speed of light, then you would in affect be putting a limit on the amount of curvature of space time as an object approached C. The thing is that Einsteins equation's don't put any such limits on relativity. And if someone discovered a way to prove where these limits take place they would solve a lot of problems faced today in theoretical physics because these equations wouldn't explode into infinities. No one really knows' but based on these equations, it should, because as V approaches C any M or T would approach infinity. Then the real question to answer this would be if as M approached infinity, would the mass be concentrated into a small enough point to create a black hole. Well, the thing is (and think should answer this question and Pavel's question) Special Relativity was not made into a field type of theory. It give no indication of the curvature of space-time near an object that is approaching the speed of light. Time dialation was found by figureing out how much of the hypotenuse of a right traingle should contract to show what an observer at rest should actually observe. Then all these other values were derived from that, when it could very well be the time dialation were sin(theta) =(vt/ct'). As for the matter of using the increase in mass to find further time dialation, I would say no. I don't beleive the mass increase is actual increase in the amount of matter inside the atoms of the object. I see it as the objects speed is curving the space time around it relative to an observer at rest. If space-time is indeed one single thing, then the amount of time dialation should be the same even though it has curved the space around it by the same factor, the Lorentz Factor.

18. Apr 16, 2007

### pervect

Staff Emeritus
I'm not sure why you think that - there is in fact no limit on the curvature.

The details can be found in the Aichelburg-sexyl ultraboost

http://en.wikipedia.org/wiki/Aichelburg-Sexl_ultraboost
http://www.citebase.org/abstract?id=oai:arXiv.org:gr-qc/0110032
(online at)
http://arxiv.org/abs/gr-qc/0110032

Components of the curvature tensor increase ("gravitational field") increase without limit, approaching impulse functions in the limit as the velocity approaches c.

Why doesn't the force increase without limit? Basically, because of gravitomagnetism. The "gravitational field" (the curvature tensor, interpretable as tidal gravity) can be decomposed into parts that are somewhat analogous to the electric and magnetic fields of electromagnetism. The magnetic-like parts happen to oppose the electric-like parts in this case. Thus two parallel moving masses appear to have a stronger "electric-like" attrraction, but they also have a "magnetic-like" repulsion. The total force transforms covariantly, as it must.

But there's a much simpler way of describing the situation.
You're missing the most basic point of relativity - Velocity is relative

As other posters have already observed, relative to some observer, you are moving at a high velocity - arbitrarily close to 'c'. You would know if you were a black hole, because you would implode to a singularity. Because you don't implode to a singularity, we know that your scenario must be wrong, even without a more detailed analysis.

19. Apr 16, 2007

### Pavel

From what I understand, that's not necessarily the case. The larger the black hole is, the less density it needs to have to shield itself from the rest of the Universe. If I stuff enough stars into a galaxy to create sufficient gravitation thereby making it a black hole to the rest of the Universe, I can still can have a Sun inside such galaxy with planets revolving around it. No implosion necessary. I didn't make this up, is it wrong?

I also would appreciate any insight into my question about time dilation effect vs. mass increase. Is it the same thing or two different effects? In advance, thank you very much.

Pavel.

20. Apr 16, 2007

### MeJennifer

Actually an observer falling into a black hole does not necessarily have fall into the singularity, it depends on what kind of black hole.
For instance in the case of a Kerr metric it is not necessarily the case.

Last edited: Apr 16, 2007