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Are you as smart as Einstein and Pythagoras combined?

Here is the serious of relativistic events, as seen from some stationary starbase :
The ship flies with constant velocity along red trajectory in space-time.
At point A the ship ejects a pod, which continues to move with its own constant velocity, aa long green trajectory.
After time r=1year at point O the ship ejects another pod, which continues to move with its own constant velocity, along the other green trajectory.
After more time r=1year at point B the ship fires a tachyon, which continues to move with its own constant velocity, along a yellow trajectory.
Later at point H the ship fires another tachyon, which continues to move with its own constant velocity equal to infinity wrt the ship, along the other yellow trajectory. I repeat : the second tachyon moves infinitely fast with respect to the ship.
All four (both pods and both tachyons) collide at point C. When this event occurs the clock of the first pod shows time CA =b=3 years from the ejection, and the clock of the second pod shows time OC = r = 1 year from the ejection.
1) Locate two similar triangles on the diagram.
2) What is time BH?
3)Locate the third similar triangle.
4)Express AB = c in terms of a and b.
The meaning of “similar”:
Geometric object is a sub-set of points of entire plane.
Congruency of two geometric objetcs means:
For any two points A and B distance |AB| is deined.
it is possible to establish a bijection between the elements (points) of the sub-sets(objects like triangles or circles, etc…) which bi-jection also preserves “distance”.
similar vs congruent:
the objects are called similar when It is possible to establish a bijection between the elements (points) of the sub-sets which scales distance by a constant factor.
Triangles in eucleadian geometry are “similar” whenever two angles are “congruent”. But only becuse it is possible to prove that “congruence” of two angles guarantees the existance of bijecton between the points of the triangles, which bi-jection scales the distance between the points by constant factor. Mainly due to the Fifth postulate
*** Is time and distance measured
*** with respect to the ship, or an
*** observer on Point C ?
Good question, I mean it.
1) point C is not observer and cannot be. points are events. time-like lines are legitimate orbservers, such observers can carry clocks. straight lines are inertial observers.
2) I am not specifically interestrested in “time t” and “spacial distance x” measured by a specific observer. I do not want to give such privilige to arbitrary chosen observer. I am intersted in absolute distances independent of choice of observers, called “invaraint intervals”. That’s what geomtries are all about.
The green thingy which looks like a hyberbola is of cource not a hyberbola, it is a CIRCLE. See all distances to the green hyperbolic-looking thingy are equal to r.
Note that triangles AOC and BOC are isoceles. Just like here:
there are some more images to refresh your memories from the 6th grade:
The BLUE thingy which looks like a hyperbola is of cource not a hyperbola, it is a CIRCLE. See all distances to the BLUE hyperbolic-looking thingy are equal to r. That oughta be a cicrcle, AB being the diameter.
Dr. Octavian :
xt space-time plane is absoultely flat, not curved. It is almost eucleadian. Note there is always one parallel straight line.
The space of realtivistic velocities is indeed uniformly curved, but the xt space-time of events is flat.


  1. I doubt anyone on Y!A is as smart as these people…
    Don’t tachyons move faster than light and like move into the past?
    Would that explain yellow trajectory (H to C) seem to be moving backwards in time?
    1) Locate 2 similar triangles? OBC and HBC look similar
    2) What is time at BH? Ha! 1 year!
    Hey! Whom gave me a TU? xx
    3) Locate third similar triangle? BCA
    4) Express AB=c in terms of a and b?
    Where is small c?
    In Anartica?
    There are no cliffs in Anartica!!
    Oops: Small c^2=b^2 – a^2
    **This is a trick question! Since b is hypothenuse, you sneak!!
    The tachyon from H to C, which was realeased from the future arrived at C at the same time the other ones did.
    Hey! Whom gave me another TU! xx
    One last thing Mr Smirnoff, call me thick, but I do not understand how it is that the tachyon from B to C (although faster than the first 2 pods) is not going back in time.
    Can not wait to see other answers.
    Pod 2 is faster than pod 1 since point O is farther away in space wrt C, so is the one shot at point B and at point H, since the ship moves constantly away from C.
    If the yellow line from point B to C was horizontal it would be easy to see that the tachyon would have hit C at the same moment it was ejected.
    I shall risk more TD’s and dare to say, (despite not grasping scary Minkowski space nor relativity) that when tachyon at point B was ejected and hitting point C at the same moment.
    (If green line OC= r= 1 year then yellow line HC= minus1 year and yellow line BC=0)
    Everyone have a nice weekend =)

  2. Can I just clarify something… when you say ‘similar triangles’, are we talking about the side lengths in terms of times, distances, or space-time intervals? Because it makes a difference to how I answer the question; obviously two triangles considered in Euclidean space will not necessarily be similar in Minkowski space, and v.v. 🙂
    Okay, I may just be talking rubbish but I’ll put down my thoughts so far; I’ve given up trying to figure out the whole thing, so here’s what I get so far. To be added later probably 🙂
    I’m going to use |AB| to mean the spacetime interval; [AB] to mean the space interval (in light years) and {AB} to mean the time interval (in years) between A and B. We know from special relativity that
    |AB|^2 = [AB]^2 – c^2{AB}^2, if we define our signature as -+++.
    Also, B and C lie of the same ‘temporal’ line (which is in fact a hyperbola, Alex: yes, it is the same -spacetime- distance everywhere from O, but in relativity, we are in Minkowski space, and the privileged curves in Minkowski space which are all the same space-time interval from the origin are definitely hyperbolae. We are dealing with a negatively curved space here. This means we also have to remember that the triangle inequality is reversed: for any triangle PQR, we have that the longest side must be *greater* than the sum of the other two, not less.)
    This means we can say:
    {BH} = {CH} [1i].
    We are given that
    |OA| = |OB| = |OC| [2i]; and that
    {OA} = {OB} = {OC} [2ii] meaning
    [OA] = [OB] = [OC] [2iii].
    Also, {AB} = 2 yrs and {AC} = 3 yrs [2iv].
    We can also say that
    {BC} = 0 [3i], and therefore
    |BC|^2 = [BC]^2 – {BC}^2 = [BC]^2 [3ii]
    i.e. a, the space-time interval between B and C, is the ordinary spatial distance between them.
    Also, since [2i] and [2iv], then
    [AB]^2 – c^2{AB}^2 = [AC]^2 – c^2{AC}^2
    [AB]^2 – 9 lyrs^2 = [AC]^2 – 4 lyrs^2
    [AB]^2 = [AC]^2 + 5 lyrs^2 [4].
    (Am I butchering units here? I can’t tell anymore 🙁
    Now I have to go to an electro night covered in makeup, so I’ll come back and try again in a bit. Bye 🙂
    ADDED: do you know what, I’ve seen where the confusion is – my bad 🙂 it is indeed a circle in Minkowski space, but appears as a hyperbola in Euclidean space due to the bizarre Lorentzian signature. I do apologise; I was thinking about it backwards 🙂
    Here’s one point at least:
    4) we do have c^2 = a^2 + b^2, don’t we? Since if the ‘hyperbola’ is actually a Minkowski circle, then the angle ACB is a right angle in ‘invariant’ terms, making this a Pythagorean triangle, since C is opposite the right angle it is the hypotenuse.
    … also, I have to say, I’m having trouble defining what you men by ‘infinitely fast’ in this case. That can’t mean ‘in zero time’, since the blue hyperbola defines the line of temporal simultaneity; this tachyon h seems to be travelling actually backwards in time, according to the diagram. Is the line h meant to be parallel to the x-axis? We can of course consider it as a tachyon moving forward in time from C to H (still a tachyon – look at the slope) but I still feel I’m missing some vital piece of information…

  3. Point of clairification. Is time and distance measured with respect to the ship, or an observer on Point C ? (C appears to be the appropriate choice)
    (And if time is measured with respect to the Point C, an exception applies in determining the velocity of the second tachyon which you state is wrt the ship)
    No, I am not as smart as Einstein and Pythagorean combined, or separately, so I’ll just answer the question the way I want to.
    First, the ship is hitting a point is space and time. Let’s adopt the ship’s reference frame and presume time is based on it. The interesting thing is that the point, not the ship, is now moving.
    Pod 1 hits the point H in 3 years, its time.
    t.1 = t.s √ (1-(v1)^2)
    t.1 = 3 yrs.
    t.s = 3 yrs/√ (1-(v1)^2)
    Pod 2 hits the point H in 1 year its time. It launches at 1 year ship’s time.
    t.2 = (t.s -1) √ (1-(v2)^2)
    t.s = 1 yr/√ (1-(v2)^2) +1
    Solve for v1 related to v2
    t.s = 1 yr/√ (1-(v2)^2) +1 = 3 yrs/√ (1-(v1)^2)
    ……….. gotta run, I’ll finish this when I get back.


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