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If I use Mathematica to solve ##a^b=b^a## for ##b## as a function of ##a## then I get $$b=-\frac{a \ W_0\left(-\frac{\ln(a)}{a}...

Mathematica has a function called "RegionPlot" that makes them very easy. Without the "pretty" options for the labels and the gridlines...

Here is a plot of the region where ##a^b>b^a## I think your rule covers the upper right and lower left quadrants. (the quadrant lines...

But only ##g## can be used to raise and lower indices, and ##g## is symmetric.

They are the same because g is a symmetric tensor.

I noticed the following question generated quite a lot of interest on the Internet. Which is larger ##e^{\pi}## or ##\pi^e##? Most (if...

Have you considered simply trying to "run it up the flagpole"? In other words: write a clear paper, send it to the appropriate journal...

Awesome, congratulations!

I finished - and printed - my collection of short stories! There Be Monsters A collection of fantastical maritime-themed short stories...

After a very brief discussion this thread will remain closed. To actually measure the one way speed of light would require an...

So you are using the standard time dilation formula that directly assumes an isotropic speed of light. Thus you will measure an...

You are wrong. Do the maths, don't wave your hands and think you're sure you know how it will work out. Or, equivalently, draw a...

How much it falls behind depends on your synchronisation convention, because the "time dilation" formula aquires an anisotropic term...

You're simply imposing it, without measuring anything. The problem is not imposing that the one-way speed of light is c, the problem is...

you can, but you’re using the time dilation formula in the calculation…. and the derivation of that formula includes the assumption that...