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I am reading Goldstein and trying to understand the special theory of relativity. I am not sure how did he make this following argument.

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The books explain that $ds^2$ is invariant in spacetime. He didn't discuss Lorentz's transformation yet. How can the author says for $ds^2$ to be the same in both frame transformation should include relative velocity in both space and time? I understand after looking at the Lorentz transformation. But just looking at $ds^2$, you can make this argument?

Qmechanic
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ran singh
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2 Answers2

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This has to do with the homogeneity of space and time and with the isotropy of space. If we accept these characteristics for spacetime, and the existence of invariant quantities, i.e., quantities that are the same in any reference frame, then the transformation of coordinates cannot depend on a particular point (homogeneity) nor on the direction of velocity (isotropy). Therefore, it can only depend on the relative velocity modulus o relative velocity squared.

geofisue
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Starting with two events, in a primed an unprimed frame:

$$ E_1 = (t_1, x_1) = (t'_1, x'_1) $$ $$ E_2 = (t_2, x_2) = (t'_2, x'_2) $$

The invariant intervals are ($c=1$):

$$ (ds)^2 = (t_2 - t_1)^2 - (x_2-x_1)^2 = (dt)^2 - (dx)^2 $$ $$ (ds')^2 = (t'_2 - t'_1)^2 - (x'_2-x'_1)^2 = (dt')^2 - (dx')^2 $$

Since they are equal:

$$ (dt)^2 - (dx)^2 = (dt')^2 - (dx')^2 $$

so

$$ (dt)^2 - (dt')^2 = (dx')^2 - (dx)^2$$

Obviously, the spatial separation of the two events are not equal in each frame..they are moving. Thus, the time differences aren't equal either.

For example, if the $S$ frame is stationary, and the two events occur at the same space point in $S$, then:

$$ (dt)^2 - (dt')^2 = (dx')^2 $$

JEB
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