- #1

Ragnar

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How do we prove that the spacetime interval is invariant? Also why is it so important?

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- Thread starter Ragnar
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- #1

Ragnar

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How do we prove that the spacetime interval is invariant? Also why is it so important?

- #2

bernhard.rothenstein

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Have a look please atHow do we prove that the spacetime interval is invariant? Also why is it so important?

Thomas Moore, 'A Traveler's Guide to Spacetime, Mc.Graw Hill,Inc. 1955\ Starting with Chapter 4

It is important among others because it is the starting point for the derivation by Einstein of the Lorentz-Einstein transformations.

I hope I gave you a good and accessible refence.

- #3

nakurusil

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How do we prove that the spacetime interval is invariant? Also why is it so important?

Use the Lorentz transforms for x',t' in the expression of the spacetime interval [tex]ds'^2=c^2t'^2-x'^2[/tex]

The invariant(s) (there are quite a few more, like , for example [tex]E^2-(pc)^2[/tex]) are very important because they aid in solving problems where relative motion is involved.

- #4

masudr

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The Lorentz transformation is defined so as to keep the spacetime interval invariant. More precisely, any [itex]\Lambda[/itex] such that

[tex]\Lambda \eta \Lambda = \eta[/tex]

where [itex]\eta = \mbox{diag}(1,-1,-1,-1)[/itex] is a transformation which keeps the spacetime interval invariant.

EDIT: in component form, using Einstein summation

[tex]\eta_{a'b'} = \eta_{ab}\Lambda^a\mbox{}_{a'}\Lambda^b\mbox{}_{b'}[/tex]

[tex]\Lambda \eta \Lambda = \eta[/tex]

where [itex]\eta = \mbox{diag}(1,-1,-1,-1)[/itex] is a transformation which keeps the spacetime interval invariant.

EDIT: in component form, using Einstein summation

[tex]\eta_{a'b'} = \eta_{ab}\Lambda^a\mbox{}_{a'}\Lambda^b\mbox{}_{b'}[/tex]

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- #5

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It's important because it is a quantity that all observers will agree upon, in spite of their general disagreement in the component-displacements.

Its analogue in Euclidean geometry is the [square-]distance between two points.

- #6

Hurkyl

Staff Emeritus

Science Advisor

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The space-time interval serves the same role in (the geometry of) Special Relativity as the distance formula serves in Euclidean geometry.Also why is it so important?

- #7

bernhard.rothenstein

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Do you teach or only use special relativity. If you teach I would send you a story..The Lorentz transformation is defined so as to keep the spacetime interval invariant. More precisely, any [itex]\Lambda[/itex] such that

[tex]\Lambda \eta \Lambda = \eta[/tex]

where [itex]\eta = \mbox{diag}(1,-1,-1,-1)[/itex] is a transformation which keeps the spacetime interval invariant.

EDIT: in component form, using Einstein summation

[tex]\eta_{a'b'} = \eta_{ab}\Lambda^a\mbox{}_{a'}\Lambda^b\mbox{}_{b'}[/tex]

- #8

neutrino

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- #9

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Spacetime Physicsdeals with invariant interval; the exposition is enlightening. You can download the first chapter of the first edition from Edwin Taylor's website: http://www.eftaylor.com/download.html#special_relativity

Great find. Ed's site must have been recently updated.

See the famous "Parable of the Surveyors". (I've been working on a variation and extension of this parable.)

The last sections of Chapter 1 include the rapidity discussions that have removed from the second edition.

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- #10

masudr

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Do you teach or only use special relativity. If you teach I would send you a story..

I learn & use only, I'm afraid.

- #11

yogi

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How do we prove that the spacetime interval is invariant? Also why is it so important?

Although the transforms upon which the invariance of the interval is based were developed by lorentz and Einstein - it was Minkowski that first pointed out the physics - the fact that in our universe, space and time can be unified and the unification is easy to visualize - any two events in spacetime are separated by an interval which has the same spacetime magnitude in every possible uniformly moving frame which can be imagined. Almost all problems in SR can be quickly solved by using this fundamental concept.

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