Spacetime intervals again - still figuring out the formula

[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.