RELATIVITY QUESTIONS! (and other common queries)

[PM 2Ring]

IOf course, it's often clear from context, as anything dealing with relativity seems to quickly spawn v/c's all over the place.

Not really, except perhaps in introductory material. Any sane person doing relativity works in units where c = 1. The formulae are a lot simpler that way, and it's only a tiny bit of algebra to convert back to the v/c form if you're so inclined.

[Dopefish]

Fair enough, although in that case you'd probably have bigger cues, such as the course/textbook actually being titled relativity.

I would also think somewhere near the very beginning there would be a mention of 'unless otherwise stated, c has been set to 1' or something to that effect, but maybe it goes without saying when you're far enough along.

[gmalivuk]

All reasonable people should let G=1 as well.

[Sir_Elderberry]

Just set all constants and variables equal to one and sort it out with units when you're done.

[Dopefish]

Susskind actually dicusses what 'classical' refers to starting at 8:26 of this (theres a bit more at 9:00 or so for those who didn't stick around a full 30 seconds). I wasn't looking for it, but when he mentioned it, I immediately thought of this thread. The actual lecture material is potentially relevant too, although I haven't watched it yet.

[phlip]

Just set all constants and variables equal to one and sort it out with units when you're done.

The ideal units set both h and hbar to 1. Also the fine structure constant.

[collegestudent22]

1 = 1. The ultimate equation. Now figure out what the hell it means. (Hint: just change the units and it means whatever you want it to.)

[doogly]

Unit conversions are not the ultimate mathematical statements. They're mostly just boring and parochial.

[mazer.rackham]

I think it is commonly known that if a body approaches a larger body, it's velocity increases, And if it does not collide with that body, it shoots away at that velocity (I don't think I explained that well; hopefully someone understood me). Furthermore, Einstein talks about light curving when passing by a massive object with a strong gravitational field.

So my question is this: If a photon approaches an extremely massive body, at velocity c, and the photon does not hit the massive body, but rather shoots off, will it be travelling at a greater velocity than c? If not, why does light differ than other object? A picture is here: http://i56.tinypic.com/2lxugxi.jpg

And please do not just say that the fastest anything can go is c...

[gmalivuk]

And please do not just say that the fastest anything can go is c...

But it is. And also, photons always travel at c. What happens when they approach or retreat from a massive body is therefore gravitational blueshift or redshift.

[eSOANEM]

If not, why does light differ than other object?

Because it has no invariant mass and so must travel at c in all reference frames.

[collegestudent22]

And please do not just say that the fastest anything can go is c...

But it is.

Unless it has imaginary mass, like a tachyon. Which may not really exist. But it is great for FTL technobabble.

[doogly]

Yes, we now have a lovely fictional science subforum for those.

[sansavis]

Assumptions:

A black hole's event horizon prevents interior light cones from intersecting world lines of external objects.
Gravity propagates as an event.

So the internal gravity of a black hole never affects external objects? What? Where did I misstep?

[nehpest]

Assumptions:

A black hole's event horizon prevents interior light cones from intersecting world lines of external objects.
Gravity propagates as an event.

So the internal gravity of a black hole never affects external objects? What? Where did I misstep?

Yes. A black hole can be uniquely defined by three parameters: mass, charge, and angular momentum. Its gravity is not affected by the internal configuration of mass, if any can be said to exist.

It's actually not that peculiar an idea: in the case of a point charge inside of a conducting sphere, the electric field doesn't depend on the position of the charge.

[dumbzebra]

Light acts as both a particle and a wave. Deal with it.

[gmalivuk]

Is there some query you were making or replying to with that?

[agelessdrifter]

A quick question from my Modern Physics homework:

If you move toward an emitter of yellow light (wavelength 580nm) at half the speed of light, what wavelength do you observe? What would be the answer if the light emitter moved toward you? -- that's verbatim from the text.


I came up with ~502nm for the first part -- I hope that's right, but I don't expect you guys to crunch numbers to verify that for me. My real question is this:
That last part "if the light emitter moved toward you" -- am I correct in thinking I should get the same answer? Isn't "the light emitter moving toward you" physically equivalent to "you moving toward the light emitter"?

[doogly]

Yup, exactly right! The contrast is with the normal doppler effect, with sound. There there is a difference, because sound has a medium. Not so for light!

[Eebster the Great]

Yes. A black hole can be uniquely defined by three parameters: mass, charge, and angular momentum.

Well, also position and linear momentum of course . . .

Its gravity is not affected by the internal configuration of mass, if any can be said to exist.

Wouldn't there necessarily be a large number of possible internal configurations, given black holes' high entropy?

[doogly]

In their own rest frame the net linear momentum and position are trivial. Usually the black hole is the biggest party in town and things are done in their frame. Of course this isn't necessary, and orbiting frames are sometimes great ideas, but this is why people will often just say Q, M and L are it.

The quantum gravitational microstates corresponding to single macrostate described by Q, M and L is very much a current research question. People cannot currently account for them except in a few extreme cases (extremal kerr might be it, if I recall)

[Eebster the Great]

In their own rest frame the net linear momentum and position are trivial. Usually the black hole is the biggest party in town and things are done in their frame. Of course this isn't necessary, and orbiting frames are sometimes great ideas, but this is why people will often just say Q, M and L are it.

Sure, but if you want to say a black hole is "uniquely defined by three parameters," that of course isn't accurate, because in any single reference frame there are five parameters. Otherwise two congruent black holes in different places would be the same object.

[PM 2Ring]

Wouldn't there necessarily be a large number of possible internal configurations, given black holes' high entropy?

Good point, Eebster. However, if the mass of the black hole is truly concentrated into a zero volume singularity, then there really aren't a lot of different states available. Since all the particles have the exact same location they can't be fermions, because only a pair of fermions of a given type (one of each spin orientation) can be located there by Pauli exclusion, so if particles as we know them are capable of existing there, they have to be bosons. And they can't be composite bosons, because that would require them to have non-zero size, so they have to be elementary bosons, like photons.

OTOH, spacetime inside a BH is radically different to the spacetime we're familiar with in the flatter regions of the cosmos. For one thing, it's transposed compared to "normal" spacetime: space becomes like the time we're familiar with and time becomes spacelike. Any object that falls into the BH will reach the centre in a rather short proper time, but from the frame of the centre that time interval gets dilated tremendously: in the limit it takes infinite time to actually reach the centre. Another way to say the same thing is that the temporal location of the centre is in the infinitely distant future. So the black hole may still manage to retain some internal structure after all.

Or we can adopt Hawking's strategy for eliminating the BH information loss paradox: from the frame of a distant observer an object falling into a BH gets time dilated as it approaches the EH, appearing to take an infinite amount of time to actually cross the EH. This also applies to the mass that originally collapsed to form the BH in the first place, or at least it applies to the last bits of matter that pushed the thing over the Schwarzchild limit. So, for the observer at infinity, it takes an infinite amount of time for a BH to actually form. If that seems to contradict the usual description of black holes, remember that the basic Schwarzchild black hole is an artificial, mathematical construct: it's the only thing that exists in its universe, and it's eternal, so it never underwent a process of creation.

[somebody already took it]

I was wondering is someone could explain to me why faster-than-light communication is equivalent to time travel. Specifically, there are two things I read from Wikipedia I have questions about:
1. Zero-mass particles traveling in a vacuum measurements of time and velocity in different frames are related by Poincaré transformations. So is there an intuitive way to understand Poincaré transformations? The Wikipedia article on them is pretty high on jargon.
2.This:

Some observers with sub-light relative motion will disagree about which occurs first of any two events that are separated by a space-like interval. In other words, any travel that is faster-than-light will be seen as traveling backwards in time in some other, equally valid, frames of reference, or need to assume the speculative hypothesis of possible Lorentz violations at a presently unobserved scale (for instance the Planck scale). Therefore any theory which permits "true" FTL also has to cope with time travel and all its associated paradoxes, or else to assume the Lorentz invariance to be a symmetry of thermodynamical statistical nature (hence a symmetry broken at some presently unobserved scale).

How does one go about proving that "some observers with sub-light relative motion will disagree about which occurs first of any two events that are separated by a space-like interval"? And how does that imply there *must* be a frame of reference where FTL travel is seen as traveling backwards in time?

(my source: http://en.wikipedia.org/wiki/Faster_tha ... ossibility)

[doogly]

Alright, lemme get out my MS Paint.

[Yakk]

FTL in one frame of reference isn't sufficient for time travel.

FTL in multiple different arbitrary frames of reference is sufficient for time travel.

Basically you start with two frames of reference (A and B) that are moving relative to each other. The slower the FTL, the faster they have to be moving relative to each other. You send a message from A to B via FTL. They then echo it back from B to A. The second message arrives before the first one left.

With sufficiently fast FTL, this is doable at relatively slow velocities. With FTL just barely above c, you need velocities really close to c to pull this off.

Other than doing the math, you can intuitively grasp why by thinking about time/length contraction. The "clock" of the fast moving other is "slow" enough that its rebroadcast ends up arriving before the first message was sent, in a sense.

You can also do the math, or do the geometrical solution (which I'm guessing doogly is about to pull out and show you).

[collegestudent22]

You can also do the math, or do the geometrical solution (which I'm guessing doogly is about to pull out and show you).

Will he? Because from my geometric solution, I am only seeing a time travel effect when the angle out of the light cone exceeds 90 degrees from vertical - an impossibility, as the horizontal line between "past" and "future" corresponds to truly instantaneous travel, so this would require actual time travel. (Perhaps with a blue 1950s London police box.) Otherwise, A will send it, B will receive it some time later (but before it sees the photons from A), and then send it back. It will arrive in A's future. The only thing changing is at what angle the distinction between "future events" and "causal future events" (and a distinction between 'causal' events and 'visible' events) actually is.

The only "time travel" is in what is visible, not what actually occurs. And seeing the photons from the source after the effect is not really the same thing as time travel, any more than not having a line of sight with the photons from the source. Or is it "time travel" to get a signal from a copper wire before the light gets through the wire (hint: photons cannot travel through a copper wire)? Sure, in B's perspective, the light from the source is seen after the signal is received, but from A's perspective the response is still received after the original signal is sent.

Specifically, the (rather simple) math I am working off of goes like this:

A \rightarrow B: FTL_2 = t_c - t_1 = t_{AB}

B \rightarrow A: FTL_2 = t_c - t_2 = t_{BA}

A \rightarrow B \rightarrow A: FTL_T = t_{AB} + t_{BA} = \tau

And unless the FTL time is actually negative, you end up with some passage of time (positive \tau), no matter what. The common conception of time travel is getting into one's own past. That would be impossible here - A cannot get into A's past. All that is possible here, as far as I understand it, is that A can effect what is "B's past", but even then, only from the perspective of B, and only if they determine their past solely as events not in the light cone.

[Robert'); DROP TABLE *;]

...only if they determine their past solely as events not in the light cone.

Past events are in their light cone; that's basically what the backwards-extending light cone defines.

[Yakk]

You forgot to rotate the "present line" in your geometry. That horizontal line is made up in a sense.

Yes, A will get the message in the past. Because the "present" line you draw rotates when you move at speed.

There is no universal present.

When I say "do the math", I mean do the right math. Not some arbitrary math. The math you did is wrong, because it presumes one universal clock. It doesn't use time dilation at all. Doing the wrong math won't tell you what is happening. The math you did is (at best) Newtonian, which (not surprisingly, because in a Newtonian universe the speed of light isn't special) doesn't have time travel via FTL.

Here is a classic (stationary observer) light cone:

Code: Select all

     .                   .
      .                 .
       .               .
        .             .
         .           .
          .         .
           .       .
            .     .
             .   .
              . .
   ____________.___________
              . .
             .   .
            .     .
           .       .
          .         .
         .           .
        .             .
       .               .
      .                 .
     .                   .


Here is a light cone when the observer is under acceleration:

Code: Select all

   .               .
    .             .
      .           .
       .         .       
         .       .          __
          .     .        ___
            .   .     ___
             . .   ___
               .___
            __. .
         ___  .   .
      ___    .     .
   ___       .       .
            .         .
            .           .
           .             .
           .               .
          .                 .
          .                   .


Stick two of these next to each other.

Now, send a message from the "normal" light cone, above that horizontal line, to the tilted one.

The tilted one then sends a message back, above its "horizontal" line (which is rotated by its velocity -- yes, that is actually how it works, your "present" line rotates when you go fast), back.

Can you see how it could send a message into the past lightcone of the first sender?

All of this relies on time and space warping via relativity.

[doogly]

the horizontal line between "past" and "future" corresponds to truly instantaneous travel

Nope. Leaving the cone is what is strange and forbidden and all that good stuff. There absolutely nothing special about the horizontal line; that's a pure coordinate artifact.

[gmalivuk]

Indeed, the horizontal line is only the simultaneous slice from one perspective. Any perspective that is in motion relative to that first one will have a slanted "right now" line relative to the horizontal.

Also, there are at least a couple threads about this already, which contain further information and discussion. I'd link to them, but I'm hungry and going to go get breakfast instead.

[doogly]

[Malconstant]

Oh this thread is delightful.

[IIAOPSW]

howdy,

So I just read the wikipedia page on "the speed of gravity" and there is one part which perplexes me. It is the fact that gravitational field always points towards the "instant" position of the emitter (as opposed to the light delayed position). Why? I had always thought if that if the light hasn't caught up to you yet, than neither has anything else including gravity. What about the dead stars that we are just observing today, what "instant" position of theirs are we being pulled towards. how is this possible?

[Robert'); DROP TABLE *;]

It would be astounding if the wikipedia page suggested anything to do with an "instant" position, since it doesn't exist in Relativity.

[Macbi]

howdy,

So I just read the wikipedia page on "the speed of gravity" and there is one part which perplexes me. It is the fact that gravitational field always points towards the "instant" position of the emitter (as opposed to the light delayed position). Why? I had always thought if that if the light hasn't caught up to you yet, than neither has anything else including gravity. What about the dead stars that we are just observing today, what "instant" position of theirs are we being pulled towards. how is this possible?

The formula for the force has a term that depends on the velocity of the body doing the pulling. By coincidence this is such that if that body is moving at constant speed then the force points toward its "instant" position. If the body suddenly changes direction then this will no longer be true. The force points toward where you would expect the body to be, if it travelled at a constant velocity.

[BlackSails]

I imagine what you want is the gravitational analogue of the http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential

[eSOANEM]

This is more a QM question than a relativity one, but does involve SR so I'm going to post it here.

When I was first introduced to the Schroedinger equation, I was told that the reason he found the form for the kinetic energy operator was that he was guided by the laws of quantum radiation.

Subsequently, I decided to follow this through by starting with the de Broglie relation, substituting \sqrt{2mE} for the momentum and using the resulting wavelength to get an expression for the angular wavenumber. Unsurprisingly, assuming the wavefunction is a sin wave, and the kinetic energy operator involves a second derivitive, the usual operator becomes apparent as the way to extract the energy from the wavefunction.

My question concerns my substitution for the momentum. I recently decided to follow this same of reasoning through with a relativistic substitution for the momentum in terms of the kinetic energy. When looking for this substitution, I remembered from a discussion here that E^2 = p^2 c^2 + m^2 c^4. Rearranging, we get that p = \pm\frac{\sqrt{E^2 - m^2 c^4}}{c}

Out of interest, I then decided to see where these two relationships between momentum energy were the same expecting it to be for low energies but, to my surprise I found them to be the same when the energy was equal to (1\pm\sqrt{2})mc^2 which is a very high energy indeed.

This confused me somewhat, but I decided to continue and, of course, reached an operator quite unlike the usual one which gave completely different values even for relatively simple cases (such as the ground state of an infinite potential well).

Obviously, I've made a mistake somewhere, I was just wondering what it was. I think it's probably that I used the wrong relativistic substitution and that, by including the rest energy I am messing things up somewhat, but in that case, I don't know the substitution I should be using (is it the same, but replacing E (the energy including rest energy) with E'+mc² where E' would be the kinetic energy not including the rest energy?).

Edit: actually, I'm pretty sure that that is what I should have done, it gives me an answer which approximates the normal operator at low energies.

Edit: following through with this substitution, I get that the usual operator T = \frac{\hbar^2}{2m} \nabla^2 returns values \frac{E^2}{2mc^2} greater than it should. This is all fine, but I'm struggling to get a nice operator from this.

[doogly]

Yeah, schro isn't relativistic. You are on your way (perhaps, if you're doing this correctly) to the Klein Gordon equation. Which was actually discovered before the Schrodinger, interestingly enough.

[eSOANEM]

Thanks. I've worked it all the way through and I have arrived at the Klein-Gordon equation albeit through a much more long-winded and less elegant method than that used on wikipedia. Thanks for telling me what I was aiming for so I could check.