Beginner's Minkowski diagram / space-time interval question

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Good morning,

We are trying to understand Special Relativity through internet, and we would like to ask a couple of questions about the subject. We plan to post more questions as more doubts arise.
Both doubts concern the Minkowski space-time diagram.
First doubt: in the temporal dimension, why is it necessary to choose ct instead of t? Most pages say it's for the time coordinate to be in the same units of the space coordinates (since metre/second*second = metre). If the only problem was that the time coordinate should be at the same units as x, y and z, then why c, and not any other velocity?
Second doubt: consider, for instance, a diagram with 3 spatial dimensions and 1 temporal dimension. The space-time interval between the origin and a point (x, y, z, ct) would be:
[tex]s^{2}=x^{2}+y^{2}+z^{2}-c^{2}t^{2}[/tex]
What is the reason for the minus sign, instead of a plus sign, in c²t²?

Thank you in advance for your patience.
 

A.T.

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First doubt: in the temporal dimension, why is it necessary to choose ct instead of t? Most pages say it's for the time coordinate to be in the same units of the space coordinates (since metre/second*second = metre). If the only problem was that the time coordinate should be at the same units as x, y and z, then why c, and not any other velocity?
It is just convenient to have light's world line at 45° in the diagram.
Second doubt: consider, for instance, a diagram with 3 spatial dimensions and 1 temporal dimension. The space-time interval between the origin and a point (x, y, z, ct) would be:
[tex]s^{2}=x^{2}+y^{2}+z^{2}-c^{2}t^{2}[/tex]
What is the reason for the minus sign, instead of a plus sign, in c²t²?
If coordinate time t is a dimension, then proper-time is the pseudo-Euclidean interval of timelike world lines. If you don't like it you can rearrange the formula so that proper-time is a dimension and the coordinate time the normal Euclidean interval:
http://www.adamtoons.de/physics/relativity.swf
But this is a different type of diagram. You can compare them here:
http://www.adamtoons.de/physics/twins.swf
 

dx

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First doubt: in the temporal dimension, why is it necessary to choose ct instead of t? Most pages say it's for the time coordinate to be in the same units of the space coordinates (since metre/second*second = metre). If the only problem was that the time coordinate should be at the same units as x, y and z, then why c, and not any other velocity?
Say you choose t to be in seconds and x to be in meters. What would the slope of a light signal world-line be? Answer this question and you will see immediately why this is not a convenient choice.

Second doubt: consider, for instance, a diagram with 3 spatial dimensions and 1 temporal dimension. The space-time interval between the origin and a point (x, y, z, ct) would be:
[tex] s^{2}=x^{2}+y^{2}+z^{2}-c^{2}t^{2} [/tex]
What is the reason for the minus sign, instead of a plus sign, in c²t²?
That's just the geometry of flat spacetime. You can take it as the basic law.
 
Thank you for your answers.
dx: If the y axis is time in seconds and the x axis is space in meters, then the slope would be [tex]\frac{1}{3} \times 10^{-8}[/tex]. Is it correct or is there any misconception?
What do you precisely mean by "flat spacetime"? Is it opposed to a spacetime with gravity?
A.T.: What do you mean by "pseudo-Euclidean interval"?
Third doubt: From what we understand, by using coordinates (x, y, z, ct), the space-time interval is defined as what we wrote above, with the term -c²t². If we used only t as the time axis, would the distance function be written differently? If so, how would it be?
Fourth doubt: From some reading, it appears that the negative sign of -c²t² comes from ict, since i² = -1. Is this true? What is the reason for the i?
You don't need to answer all the doubts at once.
Thank you in advance for your patience.
NOTE: http://en.wikipedia.org/wiki/Introduction_to_special_relativity is very a interesting read.
 
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A.T.

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A.T.: What do you mean by "pseudo-Euclidean interval"?
In Euclidean space you just add up all the squares to compute the squared distance. This is called Pseudo-Euclidean because of that minus sign.
 

Fredrik

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dx: If the y axis is time in seconds and the x axis is space in meters, then the slope would be [tex]\frac{1}{3} \times 10^{-8}[/tex]. Is it correct or is there any misconception?
That's right. It would be indistinguishable from a horizontal line in a diagram that isn't ridiculously huge.

What do you precisely mean by "flat spacetime"? Is it opposed to a spacetime with gravity?
Technically a "flat spacetime" is one in which the Riemann curvature tensor vanishes. Minkowski space (i.e. the spacetime of SR) is flat. And yes, that means that gravity isn't included.

Third doubt: From what we understand, by using coordinates (x, y, z, ct), the space-time interval is defined as what we wrote above, with the term -c²t². If we used only t as the time axis, would the distance function be written differently? If so, how would it be?
It would be the same. I prefer the term "line element" over "spacetime interval", and I think that if you want to call something "spacetime interval", it should be [tex]\sqrt{s^2}[/tex] or [tex]\sqrt{-s^2}[/tex].

I also prefer to let the axis drawn vertically be the "t" axis, not the "ct" axis, and to use units such that c=1 (e.g. light-seconds per second). This is essentially the same thing as letting the vertical axis be "ct", but you never have to write any c's.

Fourth doubt: From some reading, it appears that the negative sign of -c²t² comes from ict, since i² = -1. Is this true? What is the reason for the i?
Some authors of older books defined the time-position four-vector as (ict,x,y,z) so that you can use the definition of a scalar product that you're already familiar with to get (ict,x,y,z)²=-c²t²+x²+y²+z². No one does it that way anymore (I think). I prefer to use matrices myself (and units such that c=1). I write the time-position four-vector as

[tex]x=\begin{pmatrix}t\\ x\\ y\\ z\end{pmatrix}[/tex]

and I define the "scalar product" g by

[tex]g(x,y)=x^T\eta y[/tex]

where

[tex]\eta=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/tex]

This g isn't really an inner product, since g(x,x) can be negative. It's a "symmetric bilinear form". (An inner product is a positive definite symmetric bilinear form). That would have been the only appropriate term for it if the vectors had been members of some arbitrary vector space, but these vectors are members of a tangent space of a manifold (don't worry about how or why), and because of that, g is also the "metric" or "metric tensor" of that manifold. (Don't confuse that term with the "metric" of a "metric space").
 
Thank you for your answers.
Thank you for the explanations. We would need more mathematical knowledge to understand Fredrik's explanation for the fourth doubt.
We still have some doubts that are disturbing us.
Consider the formula for the distance between the origin and a point (x, y, z) in a 3D space:
[tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex]
In a 3D space, the distance traveled by a beam of light in an amount of time t would have to be equal to ct:
[tex]\sqrt{x^{2}+y^{2}+z^{2}}=ct[/tex]
Squaring both sides:
[tex]x^{2}+y^{2}+z^{2}=c^{2}t^{2}[/tex]
Passing c²t² to the first member of the equation:
[tex]x^{2}+y^{2}+z^{2}-c^{2}t^{2}=0[/tex]
Here, one derives a particular case of the Minkowski metric: the one where equals zero.
But it is a very simple and intuitive reasoning; why can we derive a particular case of the Minkowski metric using this simple reasoning? It would be very good to understand it in terms of this simple fact, because it would explain both the reason of the minus sign and the appearance of c together with t.
How/why, using this reasoning, we can generalize to the Minkowski metric (s² = x² + y² + z² - c²t²)? If, in this reasoning, x² + y² + z² is the square of the distance traveled by the light, why can we generalize and say it is the line element of a frame moving with velocity v with respect to another frame (that is, x² + y² + z² would equal v²t² in the formula)?
Thank you in advance.
 
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