Linked Questions

-2 votes
3 answers
1k views

Why did Einstein put a negative sign in the Pythagorean theorem? [duplicate]

In 4-dimensional spacetime, when we study the spacetime interval, why did Einstein put a negative sign in it? $$x_1=x$$ $$x_2=y$$ $$x_3=z$$ $$x_4=ct$$ $$ds^2=dx^2+dy^2+dz^2-(cdt)^2$$
Bassil Kashour's user avatar
0 votes
1 answer
135 views

Lorentz transformations: new actual notation for a $4$-vector [duplicate]

For the Lorentz trasformations I use this notation \begin{equation*} \left\{\begin{aligned} x&=\gamma (x'+\beta ct')\\ y&=y'\\ z&=z'\\ ct&=\gamma (ct'+\beta x')\\ \end{aligned}\right. ...
Sebastiano's user avatar
  • 2,547
1 vote
1 answer
162 views

Why the imaginary unit in time axis? [duplicate]

Why time is not like other dimensions is a real amount? In relativity time axis is $i*c*t$, where $i$ is the imaginary unit and $c$ is light speed in free space. Did science or philosophy reached to ...
Ahmed Kamal Kassem's user avatar
0 votes
0 answers
182 views

The imaginary time [duplicate]

Some people work with the interpretation that the time basis vector has magnitude sqrt(-1) to justify the negative sign in a -+++ Minkowski metric signature. I came across a Youtube comment that ...
Ken Wang's user avatar
  • 431
0 votes
0 answers
91 views

What is the advantage of using imaginary units for time in the Minkowski Space rather than regular euclidian space as Lorentz used? [duplicate]

I do understand that Lorentz transformations became as a rotation of coordinates as of a hyperbolic rotation. But what is its advantage over real vector? What is the new thing that it introduces and ...
Csopso's user avatar
  • 1
12 votes
4 answers
4k views

Minkowski spacetime: Is there a signature (+,+,+,+)?

In history there was an attempt to reach (+, +, +, +) by replacing "ct" with "ict", still employed today in form of the "Wick rotation". Wick rotation supposes that time is imaginary. I wonder if ...
Moonraker's user avatar
  • 3,155
11 votes
3 answers
3k views

Special relativity and imaginary coefficient of the time coordinate

I read somewhere that part of Minkowski's inspiration for his formulation of Minkowski space was Poincare's observation that time could be understood as a fourth spatial dimension with an imaginary ...
Simon Lyons's user avatar
9 votes
4 answers
1k views

Is there a geometric interpretation of the spacetime interval?

In Euclidean space, the invariant $s^2 = x^2+ y^2+ z^2$ is equal to the length square of the position vector $r$. This is easily understood and can be represented geometrically in a graph. On the ...
Remi's user avatar
  • 93
3 votes
6 answers
777 views

What is the intuition behind the spacetime interval?

In an article that I am currently reading (under the Lorentz Invariants sub-heading), it explains that, just as the distance between two points on a Cartesian plane are obviously invariant of the ...
P0W8J6's user avatar
  • 411
2 votes
2 answers
1k views

Minkowski's complex Euclidean space vs. the real pseudo-Euclidean version

The SR invariance formula makes space-like intervals imaginary (e.g., the distance $x$ in a given frame has interval $ix$). Yet modern physicists consider it bad form to define the distance itself as $...
murray denofsky's user avatar
5 votes
3 answers
608 views

Where does the negative signature case come from in the Pythagorean derivation of distances in spacetime?

I am reading Why does $E=mc^2$ (and why should we care?) by Brian Cox and Jeff Forshaw. I want to understand these three sentences (from page 76/77): Once we follow Occam and make these two ...
sleep's user avatar
  • 175
1 vote
1 answer
660 views

Four-vectors and metric tensor

I think it's safe to say that if $x^\mu=(x^0,x^1,x^2,x^3)$, then $x_\mu=(x^0,-x^1,-x^2,-x^3)$. But I don't really understand why one follows from the other. Could someone explain? Also, I've been ...
nightmarish's user avatar
  • 3,203
0 votes
1 answer
1k views

Is imaginary time a fifth dimension? [duplicate]

I've read that by introducing the concept of imaginary time, the dimension of time can be treated like a spatial dimension mathematically. Assuming, without imaginary time, one considers the universe ...
jbatez's user avatar
  • 111
0 votes
1 answer
660 views

Minkowski metric: Why does it look like it does? [duplicate]

I have been searching for why would we even start with Minkowski spacetime metric as being written as: $$ds^2=-dt^2+dx^2+dy^2+dz^2.$$ No really, so why would we have a negative sign for temporal ...
Beyond-formulas's user avatar
1 vote
2 answers
220 views

Why the infinitesimal distance in space-time contains a minus sign before the dt squared?

Maybe my question seems simple, I am reading about general gravity but the first obstacle faced me is the definition of the distance in a four dimension space. Why does the infinitesimal distance in ...
Malek_Physics's user avatar

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