Why Are the Terms Squared in the Lorentz Transformation?

Dark_knight90
Messages
14
Reaction score
0
Hello
This is a part of a simple paper about special relativity

[PLAIN]http://img15.imageshack.us/img15/8789/91001769.jpg

I don't understand the assumption in the red box .. why are they all squared ?

thank you
 
Last edited by a moderator:
Physics news on Phys.org
Do you remember the Pythagorean rule? In two-dimensional space, \Delta x and \Delta y are the lengths of two sides of a right triangle, and \Delta r is the length of the hypotenuse:

(\Delta r)^2 = (\Delta x)^2 + (\Delta y)^2

What you have is the three-dimensional version.
 
That's basically an extension of the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two sides. You can then repeat the process, adding in the square of the length of the third side giving you the square of the total length.
 
Got it .. Thank you :)
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
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...
Back
Top