Spacetime intervals again - still figuring out the formula

[PeteSF]

I find this hardly relevant because the Minkowski "distance" between events is not a real distance in the sense as we use it in space anyway.

...but it is for events with spacelike separation, yes?

 

[Mortimer]

The point is that with a spacelike separation the invariant ds becomes complex (because ds^2<0). If you don't like that you can multiply the whole interval with i and get a real number again. That's really just a matter of convention. But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ds) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).

 

[PeteSF]

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.

What?
Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes?
FTL speeds are required for information transfer between such events, but is that relevant?

 

[Mortimer]

Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").

but is that relevant

Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come across the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

 

[Ich]

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").

Spacetime has to be measured with RODS and CLOCKS, according to Einstein.
Either a length, measured (at least symbolically) with rods, or a time difference, measured with clocks, are physically meaningful and well defined. There exists a prescription how to measure proper length, and it is invariant. What more do you want?

But JesseM is right, as world lines are necessarily timelike, ds=dtau will be used more often in calculations and is therefore more useful in a way. But it´s not the whole truth about the spacetime interval. You need both interbpretations posslible to be consistent.

 

[robphy]

Measuring displacements in Minkowski spacetime

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).

Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").


Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come across the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

This may be helpful.


Here's a procedure to measure displacements in [Minkowski] spacetime using a clock, a light source, and detector. (This is the "radar method".)
Given two events, P and Q, what is the square-interval?

Consider an inertial observer meeting P.
One his worldline, there is
an event S when at a light ray can be sent to Q, and
an event R when that light ray's reflection is received from Q.

Let t_P, t_S, t_R be the times read off that observer's clock. (Obviously, t_R \geq t_S.)

According to this observer, the spatial-displacement from P to Q is
\Delta x_{Q\mbox{ from }P}=c(t_R-t_S)/2 ,
that is, half of the measured round-trip time multiplied by the speed of light.

According to this observer, the time-coordinate of the distant-event Q
is DEFINED by t_Q= (t_R+t_S)/2,
that is, the average of the send and receive clock-readings.

So,
according to this observer, the time-displacement from P to Q
is \Delta t_{Q\mbox{ from }P}=t_Q-t_P,
or
\Delta t_{Q\mbox{ from }P}=(t_R+t_S )/2 -t_P.
that is, the average of the send and receive clock-readings minus the clock-reading t_P.

The following quantity can tell us about the causal relationship of Q from P.
( t_R - t_P)(t_S - t_P).

If the events on this worldline happen in the sequence S-then-P-then-R,
then Q is spacelike-related to P. This quantity is negative.

If the events on this worldline happen in the sequence P-then-S-then-R,
then Q is in the timelike-future of P. This quantity is positive (since it's the product of two positive numbers).

If the events on this worldline happen in the sequence S-then-R-then-P,
then Q is in the timelike-past of P. This quantity is positive (since it's the product of two negative numbers).

If any two events coincide, then that quantity is zero.
If it's S and P that coincide, then Q is in the lightlike-future of P.
If it's R and P that coincide, then Q is in the lightlike-past of P.
If S, R, and P coincide, then Q coincides with P.


Now, consider any two inertial observers that meet P and
perform this procedure to make measurements of Q.
So, each observer will have a different set of send and receive events and
thus a different set of clock-readings for send and for receive.

According to special relativity, the quantity
c^2( t_{receive} - t_P)( t_{send} - t_P) is invariant.
Indeed,
<br /> \begin{align*}<br /> (c\Delta t_{Q\mbox{ from }P})^2-(\Delta x_{Q\mbox{ from }P})^2<br /> &amp;=c^2\left(\frac{t_R+t_S}{2} -t_P\right)^2-\left(c\frac{t_R-t_S}{2} \right)^2\\<br /> &amp;=c^2\left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(c\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\\<br /> &amp;=c^2\left( \left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\right) \\<br /> &amp;=c^2\left( \frac{2(t_R-t_P)(t_S-t_P)}{4} - \frac{-2(t_R-t_P)(t_S-t_P)}{4} \right) \\<br /> &amp;=c^2 (t_R-t_P)(t_S-t_P) \\<br /> \end{align*}<br />
(This calculation is simpler if we assume t_P=0.
To keep to the spirit of emphasizing the time-measurements, one should start at the bottom with (t_R-t_P)(t_S-t_P) and obtain the expression (c\Delta t)^2- \Delta x^2.)


So, to comment on the sections I quoted above,
there is a way to measure (with a physical setup) the separation of two spacelike-related events.

 

[JesseM]

JesseM, no, I meant those four lines you used to get to the final formula.

OK, if you're comfortable with these formulas:

Time dilation:
d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2} (1)

Velocity = distance/time:
v = \sqrt{dx^2 + dy^2 + dz^2}/dt (2)

Then those last four lines are just a matter of algebra:

First, start with equation (1):
d\tau = dt \sqrt{1 - v^2/c^2}

Then square both sides of equation (2) and substitute in for v^2:
d\tau = dt \sqrt{1 - (dx^2 + dy^2 + dz^2)/(c^2dt^2)}

Then factor out 1/dt^2 from the expression inside the square root:
d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

Then do \sqrt{(1/dt^2)*stuff} = (1/dt)\sqrt{stuff}, and the 1/dt cancels out with the dt that was already outside the square root:
d\tau = \sqrt{dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)}

If there's still any of this algebra you're not clear on, let me know which step is giving you trouble.

 

[NanakiXIII]

I don't understand how you got to the third formula

d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

 

[jtbell]

It may be a little bit more obvious if we first rewrite the preceding equation in a slightly different format:

d\tau = dt \sqrt{1 - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

Now let's proceed:

d\tau = dt \sqrt{\frac {dt^2}{dt^2} - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} dt^2 - \frac {1}{dt^2} \frac {dx^2 + dy^2 + dz^2}{c^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} \left( dt^2 - \frac {dx^2 + dy^2 + dz^2}{c^2}\right)}

 

[JesseM]

I don't understand how you got to the third formula

d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

Well, do you understand how to factor variables out of equations in algebra? For example, do you understand why if you factor x^2 out of (x^3 + 2x^5 + 5) you'd get (x^2)*(x + 2x^3 + 5/x^2)? If so, just factor (1/dt^2) out of the equation (1 - (1/dt^2*c^2)(dx^2 + dy^2 + dz^2)) in the same way.

 

[NanakiXIII]

Yes, but if you factor out 1/dt^2, aren't you just multiplying everything by dt^2?

Maybe I'm doing something wrong, I'm not that great at algebra, but if I factor out 1/dt^2...

dt^2-c^2*dt^2(dx^2+dy^2+dz^2)

I know, that looks kind of messy, but I don't know how to use Latex.

 

[JesseM]

Yes, but if you factor out 1/dt^2, aren't you just multiplying everything by dt^2?

Yes, exactly, although you have to include that (1/dt^2) on the outside, just like if you factor x out of (x^2 + 2x), you have to divide the whole thing by x and then include x on the outside, like (x)*(x + 2).

Maybe I'm doing something wrong, I'm not that great at algebra, but if I factor out 1/dt^2...

dt^2-c^2*dt^2(dx^2+dy^2+dz^2)

That shouldn't be a c^2*dt^2 multiplying the space interval, it should be 1/c^2. Remember, the original thing in the square root was:

1 - (dx^2+dy^2+dz^2)/(c^2*dt^2)

So, if you multiply everything by dt^2, it will cancel out with the dt^2 in the denominator of (dx^2+dy^2+dz^2)/(c^2*dt^2), giving:

dt^2 - (dx^2+dy^2+dz^2)/(c^2)

Maybe it'd help if I put the fraction in Latex--do you see why if you multiply \frac{dx^2 + dy^2 + dz^2}{c^2 dt^2} by dt^2, you'll get \frac{dx^2 + dy^2 + dz^2}{c^2}?

 

[NanakiXIII]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

And yeah, the Latex is a lot easier to read.

 

[JesseM]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

As I said earlier, I can't think of any conceptual explanations that don't involve any math--if someone else can think of one hopefully they'll jump in, but it may just be that there is no such purely conceptual explanation.

 

[robphy]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

And yeah, the Latex is a lot easier to read.

So, does the geometric interpretation using a hyperbola that I described in post #21 https://www.physicsforums.com/showpost.php?p=541106&postcount=21
make sense?

How about the physical interpretation that I described in post #56
https://www.physicsforums.com/showpost.php?p=549398&postcount=56 ?

 

[Mortimer]

Ah yeah, I see. Ok. So that explains why there's a minus mathematically. But I'm going to be annoying and ask again for a different explenation so that I'll understand it. So that it makes sense.

In a last attempt to try and make you understand it (and promote Euclidean special relativity) I have put my remarks of posts #33, 36 and 42 in a web page that you find http://www.rfjvanlinden171.freeler.nl/4vectors. Maybe it helps.

 

[NanakiXIII]

robphy, the first post just makes little sense to me. The second isn't much better, I've forgotten what it's about by the time I'm halfway. It just seems to advanced.

Ok, I just thought of something when looking at that site. ds^2=c^2dt^2-dA^2, so if you turn that around, c^2dt^2=ds^2+dA^2. This has been said before, I think. But I just realized, no idea why that happened just now, that that means that c^2dt^2 is the 'long side of the triangle', of which I can never remember the proper name. This makes no sense to me. Why is the temporal dimension bigger or equal to the entire interval, when it is but one of four dimensions used to calculate that interval? I suppose this is exactly the same as what I asked before, only worded differently. But maybe it'll help understand what I don't understand, if anyone didn't.

 

[Mortimer]

But I just realized, no idea why that happened just now, that that means that c^2dt^2 is the 'long side of the triangle'

I think you are getting it now! (it's hypothenuse b.t.w.)
I repeat two pieces of the webpage that are important in this respect:

the components of the Minkowski 4-vector can have no physical meaning. Their function is purely mathematical

and

ds is now no longer the 4D displacement but just the displacement in the time dimension. The factor cdt that plays this role in the Minkowski style of this 4-vector has become the actual 4D displacement

That's it. Simply do not try to seek a physical meaning behind the Minkowski 4-vector. That meaning is only obvious in the Euclidean 4-vector.

 

[robphy]

That's it. Simply do not try to seek a physical meaning behind the Minkowski 4-vector. That meaning is only obvious in the Euclidean 4-vector.

I invite the readers of this thread to read this discussion I had with Mortimer
https://www.physicsforums.com/showthread.php?t=73582 in which I elaborate on the physical interpretation of the velocity 4-vector and its components.

 

[quasar987]

From post #19, in relation to post #9:

But when you calculate the differential ds^2, those intervals become differentials as well. That is x -> dx and t -> dt.

Would you mind writting mathematically what you mean by this statement? Thx.