A Why is Tau-AB^2 not t^2 + x^2?

BLevine1985
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question about proof page 21
why is Tau-AB^2 equal to t^2 - x^2 ?It seems it should be t^2 + x^2 according to the geometry of the diagram...
 

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BLevine1985 said:
why is Tau-AB^2 equal to t^2 - x^2 ?
Because that's how the metric is defined in an inertial frame in special relativity.

BLevine1985 said:
It seems it should be t^2 + x^2 according to the geometry of the diagram...
No, because the diagram is not depicting a Euclidean geometry. It's depicting a Minkowskian geoemetry. That means you cannot interpret lines and angles in the diagram as if they were Euclidean; they aren't.
 
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Thank you. That was very helpful!
 
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BLevine1985 said:
Thank you. That was very helpful!
You're welcome!

Also, welcome to PF!
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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