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Invariant spacetime velocity

  1. May 26, 2008 #1
    invariant "spacetime velocity"

    This is related to this thread here. To make responding easier, I have marked questions in red. I have tried to address some concerns pre-emptively. These I have marked in silver.

    If you would like to respond, please respond primarily to the core questions, or the explanation in standard font. If you are responding to a silver clarification, please note that, so that I understand that you are aware that you are not addressing something central. Thanks.

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    Is there any validity in considering an invariant "spacetime velocity"?

    Let me try to explain. According to me, in spatial terms, I am stationary, so I traverse a spacetime "distance" of ct' over a period of t' (this is my proper time, the time we expect to be dilated when compared to someone travelling relative to me). My buddy helps me out in a little experiment by not remaining stationary relative to me but rather having a velocity of v.

    During the period of t', according to me, my buddy travels a distance of vt' to reach an event E, which is simultaneous (in my frame) with the event I reach after a period of t'.

    According to my buddy, he travels a timespace distance of ct before reaching the event E.

    This step may upset some people, but watch closely:

    My total spacetime distance travelled is ct'.

    The magnitude of my buddy's total spacetime distance travelled is sqrt((ct)^2 + (vt')^2) - this is just the hypoteneuse of the triangle with ct (temporal component) and vt' (spatial component).

    Equating these:

    ct' = sqrt((ct)^2 + (vt')^2)

    or

    (ct')^2 = (ct)^2 + (vt')^2

    then rearranging:

    (ct)^2 = (ct')^2 - (vt')^2

    and solving:

    t' = t / sqrt(1 - v^2 / c^2)

    I acknowledge that it is not the simplest way to arrive at the equation for time dilation, but is there a problem other than that?

    Note that this is based on the assumption that relative to me, everything travels at an invariant "spacetime velocity" of c, including myself.

    I am not assuming that I am at rest, but that I have a purely "temporal velocity" of c and anyone who does not have a purely "temporal velocity", but rather has a spatial velocity as well, will have a reduced "temporal velocity" as a result. I am assuming that spatial velocity and temporal velocity are orthogonal.

    Note further that I am talking about "spacetime traversed" by my buddy not the spacetime interval, relative to me, between two events - both of which are inhabited by my buddy. Conceptually, my buddy needed to cut a corner to arrive at a future event (ie event E) after traversing less time than I needed to arrive at an event which was simultaneous with that event in my "rest" frame. So my buddy travelled through less time due to the need to travel through some space.

    (I am aware that, from my buddy's point of view, it is I who cuts a corner to arrive at a future event which is simultaneous with event E in his "rest" frame. I am aware that we will disagree on which events are simultaneous with event E, but that I can work out which events he will perceive as simultaneous.)


    Is it a standard understanding that, relative to me (or any given observer), any inertial thing travels with an invariant "spacetime velocity" of c?

    thanks,

    neopolitan
     
  2. jcsd
  3. May 26, 2008 #2

    Ich

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    It depends on what you mean with "spacetime velocity" etc.
    Velocity in relativity is indeed a four vector with magnitude c, so that is mainstream.
    The "So my buddy travelled through less time due to the need to travel through some space."-thing is often used, but it mixes up coordinates with invariant time, which is why it doesn't help very much to get a deeper understanding of relativity.
     
  4. May 26, 2008 #3

    Fredrik

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    What you're describing is your four-velocity. It's (c,0,0,0) in your rest frame, but has different components in other frames, so it's not invariant.

    The derivation is fine (and totally standard). I think it is the simplest way to arrive at the time dilation equation.

    You just need to realize that what you used to derive it isn't "space-time velocity". You just used the Minkowski metric.

    This is where you used the Minkowski metric.

    It's totally standard to define the four-velocity of an object as the four-vector that has components (c,0,0,0) in an inertial frame where the object is at rest. Since it's a four-vector, its components aren't invariant, but its Minkowski space "square" is (and it's =c2 or =-c2, depending on whether we're using the -+++ or +--- convention in the definition of the Minkowski metric).
     
  5. May 26, 2008 #4
    Actually, I was deliberately trying to get away from the rather bizarre -c2 terminology, since it seems to imply imaginary numbers.

    Perhaps, as an alternative, you could explain why it must be -c2 with a -+++ convention. Or is my step sufficient, with a more (to me) meaningful +--- convention? Here it is again:

    alternatively:

    (ct)2 = (ct')2 - x2 - y2 - z2

    -------------------

    Ich mentioned "invariant time". I launched a search with the string '"invariant time" relativity' and came up with very little, the closest to something that seemed like it applied was a quantum relativity paper. Is the concept standard?

    I feel that what Fredrik said is true, that strictly speaking individual components of the 4-vector are not invariant, but the 'Minkowski space "square"' is. (The individual components are Lorentz invariant, but that is really just saying the same thing.) "Invariant time" would be dependant on being at rest spatially and being at rest spatially is frame dependent so ... no invariant time. Or is "invariant time" another one of those concepts which sound like one thing but are really another - like "proper" time (where proper time is defined here - it is not proper as in being "right", just that it is distinguished from co-ordinate time).

    cheers,

    neopolitan
     
  6. May 26, 2008 #5

    Dale

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    Yes, there is. The wikipedia link given above is decent. Basically, if you want to establish relativistic analogs for all of your classical quantities the natural approach is to take your position four-vector and differentiate it wrt time to get a four-velocity. But, the "differentiate wrt time" part requires a little thought in relativity since coordinate time is frame variant, so if you differentiate a four-vector with a frame-variant scalar you will not wind up with a four-vector. So instead of differentiating wrt coordinate time you differentiate wrt proper time which is invariant. By differentiating a four-vector wrt an invariant scalar you get something which is manifestly a four-vector.

    It just happens that this four-vector has a norm of c.

    Not just inertial objects. Even an accelerating object has a four-velocity with a norm of c at all times. The four-acceleration (second derivative of the position four-vector wrt proper time) is always perpendicular to the four-velocity.

    Btw, the four-velocity is less useful than the four-momentum. There are two primary reasons for that. First, the four-velocity is undefined for a photon since the proper time is zero. Second, the sum of two four-velocities is not a four-velocity since it does not have magnitude c. In contrast, the four-momentum of a photon is well-defined and the sum of two four-momenta is a four-momentum.
     
  7. May 26, 2008 #6

    Ich

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    The concept is standard, but as you suspected correctly, the standard terminology is "proper time". I called it "invariant" (it is in fact invariant) to show what I mean with "mixing up":
    c²tau²=c²t²-x², where we have an invariant on the left side, and two coordinates (well, coordinate differences to be exact) on the right.
    The rearranged formula
    c²tau²+x²=c²t² has the (rather meaningless) combination of proper time and spatial coordinate distance on the left and coordinate time on the right side. The only merit of this arrangement is that you can plot it in an euclidean plane and use the ordinary pythagorean theorem, but it obscures the main point, that coordinates may vary but intervals do not.
    (take the derivative wrt tau before squaring to retrieve your formulas, with the extra caveat that v is commonly meant to be dx/dt, not dx/dtau. Another reason not to use the rearranged version).
     
  8. May 26, 2008 #7

    Fredrik

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    You have to define the Lorentz interval either through [itex]ds^2=-dt^2+dx^2+dy^2+dz^2[/itex], or [itex]ds^2=dt^2-dx^2-dy^2-dz^2[/itex]. (I'm using units such that c=1). No matter which one you choose, some "distances" in space-time will be imaginary. There's no way around that, except not talking about distances. That's actually what I do.

    I prefer -+++ myself, but I wouldn't say that the distance you have moved through space-time is an imaginary number. I define proper time along a time-like curve as the sum of a bunch of contributions of the form [itex]\sqrt{-ds^2}[/itex] (it would be [itex]\sqrt{ds^2}[/itex] with the +--- convention), so instead of saying that the Minkowski space squared "distance" of your journey through space-time is -t'2 (or -c2t'2 if we want to mention c explicitly), or that this "distance" is =it'. I would just say that the proper time of your world line is t'. (Note that proper time is the same as what you call "spacetime distance" when we're using units such that c=1).

    Then I would say that your buddy has moved a spatial distance vt' in your rest frame, and use Pythagoras: t'2=v2t'2+t2. Solve for t', and you have the time dilation equation.

    I don't think that term is standard. I would assume that someone who uses it means "proper time".

    I think the correct term here is Lorentz covariant.
     
    Last edited: May 26, 2008
  9. May 26, 2008 #8
    It is "standard understanding" for Euclidean relativity. This is non-mainstream (although various endorsers have in the meantime been published in peer-reviewed journals), therefor it is in this thread in the Independent Research section.
     
  10. May 27, 2008 #9
    http://en.wikipedia.org/wiki/Lorentz_covariance

    Yes. The scalar is invariant, while the individual physical quantities (such as spatial and temporal intervals) are covariant such that the scalar (and the spacetime interval) is invariant.

    cheers,

    neopolitan
     
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