I have heard many physicists (ex:- Michio Kaku) saying "Warp speed" from Star Trek doesn't violate any known physical laws. But doesn't it violate causality? Say, we make warp drive possible and travel towards Alpha Centauri (4.22 light years away) in warp speed and reach there, say, within a year (+/- few months) and blow it up. Now using Special Relativity, we could devise a frame of reference wherein the observer would see the Star blow up before we ever left planet Earth. Wouldn't that violate causality and make warp speeds unattainable?

  • $\begingroup$ He is asking how it violates causality. I know it violates causality, but what I do not understand is why many Physicists feel it doesn't violate any law of Physics and how Alcubierre drive does not violate causality. $\endgroup$ – Dhruva Patil Jan 25 '14 at 9:31
  • 1
    $\begingroup$ I disagree that this is a duplicate. This question is asking as to why it is claimed that it doesn't violate causalty, even though it seems to. By the way, @DhruvaPatil, I think you should not taken Michio Kaku's popular science talks very seriously, they are often wrong (I bet he knows that himself), not necessarily in this case. $\endgroup$ – Abhimanyu Pallavi Sudhir Jan 25 '14 at 11:58
  • 1
    $\begingroup$ This is not an answer, but you can read this: motls.blogspot.com/2013/07/… $\endgroup$ – Abhimanyu Pallavi Sudhir Jan 25 '14 at 12:02

This is more of a meta answer, since it isn't really Physics, but it got a bit long to put in a comment. You say:

I have heard many physicists (ex:- Michio Kaku) saying "Warp speed" from Star Trek doesn't violate any known physical laws.

You need to think about precisely what this statement means. If we take the Alcubierre drive as an example it is a perfectly legitimate solution to Einstein's equation. So if General Relativity is a Law of Physics then the Alcubierre drive indeed doesn't violate any laws of physics, and it can be used to create closed time-like curves with the loss of causality that this implies.

But there are two problems with saying the Alcubierre drive doesn't violate any laws of physics:

Firstly the Alcubierre drive requires a ring of exotic matter to work, and exotic matter violates a number of energy conditions. If these count as laws of physics (and so far observation suggests they do) then the Alcubierre drive does violate laws of physics.

Secondly most of us believe that general relativity is an approximate theory that breaks down under a number of conditions (specifically in the quantum regime). It's been argued that when you take quantum effects into account a closed timelike curve causes an instability that destroys it. All this is speculative since we have no theory of quantum gravity, but if true it also means that the Alcubierre drive does violate laws of physics.

So when you see a bald statement like xxx doesn't violate laws of Physics, the statement is meaningless unless you specify the assumptions it is based on.

| cite | improve this answer | |
  • $\begingroup$ Didn't know about the CTCs until know. Thanks for pointing it out. Cleared up my doubt. One more thing, is the energy condition theory not a firmly established one? Since many Physicists (and now, it seems NASA too) are ignoring it. $\endgroup$ – Dhruva Patil Jan 25 '14 at 10:11
  • $\begingroup$ @DhruvaPatil: dark energy violates the strong energy condition, but I suspect most physicists consider this as a bit of a special case, and it doesn't imply that what we usually think of as exotic matter exists. Re NASA: I suspect you're thinking of Harold White. If so his views are some way from the mainstream, though it would be great if he was right :-). $\endgroup$ – John Rennie Jan 25 '14 at 11:12
  • $\begingroup$ +1, and thirdly it violates SR, which it shouldn't, motls.blogspot.com/2013/07/… $\endgroup$ – Abhimanyu Pallavi Sudhir Jan 25 '14 at 12:04
  • 1
    $\begingroup$ @Dimensio1n0, Thanks for the link. That is pretty much the kind of answer I was looking for. $\endgroup$ – Dhruva Patil Jan 25 '14 at 13:03
  • $\begingroup$ @JohnRennie Yes, I was talking about Harold White. And if dark energy does violate the energy condition, and dark energy has been proven(?). Then surely, the energy condition(s) must be wrong, right? $\endgroup$ – Dhruva Patil Jan 25 '14 at 13:09

General relativity has several complications in regards to inertial frames, which I'm ignoring here because I think there's a more fundamental misunderstanding about special relativity and FTL going on here.

Now using Special Relativity, we could devise a frame of reference wherein the observer would see the Star blow up before we ever left planet Earth. Wouldn't that violate causality...?

No, it would not. The fact that you could devise such a frame is completely irrelevant to whether or not causality is violated. What's important is that nothing ever sends a signal to its own causal past, i.e. within or along its past light cone.

In general, an FTL signal of a fixed speed does not violate causality. To actually send something into your past light cone, you need FTL signals of at least two speeds in some particular inertial frame. That in some other frame you've sent a signal into the (coordinate) past doesn't matter. If you only ever use one particular FTL speed in one particular frame, no causal violations will occur.

Although physically restricting FTL to a single speed (again, in some particular inertial frame) would preserve causality, it would violate the principle of relativity instead, since there will be a unique, distinguished frame in which the signaling is instantaneous.

As for warp drives, we can take that as indication that a single warp bubble might or might not violate causality (depends on the geometry), but unrestricted ability to construct warp bubbles will surely enable us to violate it. This we can add to the many issues of constructing a warp drive.

| cite | improve this answer | |
  • $\begingroup$ I am not sure I get what you are trying to say. I have not read a great deal on Relativity, but doesn't Simultaneity of relativity say 2 events casually connected must occur in the same order in any frame of reference (moving relative to the event or the events rest frame)? $\endgroup$ – Dhruva Patil Jan 25 '14 at 11:44
  • $\begingroup$ @DhruvaPatil: A locally FTL signal sends info outside the light cone. If you have just one FTL speed in a particular inertial frame, then different frames will disagree as to which order FTL-connected events occurred in, yes. But this is not yet a genuine problem. So different observers disagree as to which is the cause and which is the effect--so what? To actually generate something paradoxical, such as killing your own grandpa or whatnot, you need to send a signal into the past light cone of some event. $\endgroup$ – Stan Liou Jan 25 '14 at 12:03
  • $\begingroup$ @DhruvaPatil: Addendum: in the context of relativity, though, if you can send an signal at some FTL speed, then it's possible to send it at any FTL speed. This is forced by the principle of relativity. That's why relativity identifies being causally connected with being able to communicate non-FTL. However, this doesn't change the point that more than one FTL speed (in one in.frame) is required for a causal paradox--it simply means that restricting to one speed violates something other causality instead, as I said in the post above. $\endgroup$ – Stan Liou Jan 25 '14 at 12:11
  • $\begingroup$ What I am not getting is how having two FTL signals of different speeds could possibly send 1 of the signals into your past light cone. P.S.I hope you don't mind my dumbness as I just don't seem to understand it. $\endgroup$ – Dhruva Patil Jan 25 '14 at 13:16
  • $\begingroup$ In addition to the link John Rennie provided, this website explains why instantaneous signaling in two different inertial frames leads does this. Also, this website has a nice spacetime diagram (I haven't read their explanation, though). Note that every locally FTL signal is instantaneous in some inertial frame, so two different FTL speeds in one inertial frame = instantaneous signaling in some pair of inertial frames. $\endgroup$ – Stan Liou Jan 25 '14 at 13:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.