Hello, again, lately i've been going over special relativity again and have decided to finally look into the light clock thought experiment to see how he derived time dilation. Of course, it was going well until I checked wikipedia, I know not the best source, for how it was supposed to go and one part got me confused. Please Register or Log in to view the hidden image! How did it get from step two to step three without cancelling out the time signatures. Did I miss some elementary math trick or what, it bewilders me. Also, what is the benefit of the +--- and -+++ sign conventions in the Minkowski metric? The later makes time imaginary and I happen to find it unnecessary as well as unusable but maybe i'm just a bit clueless on how it is applied. More posts and questions will follow.

You move the first term on the right side of the equation to the left side and then factor the delta t' squared.

Hi Anti-stupidity, What do you mean by "cancelling out"? The time \(\Delta t'\) is still there in step 3. No. The metric means we can deal with real time. The spacetime interval is given by something like \(s^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2\) depending on which sign convention you want to adopt. One way to get that minus sign in front of the \((c\Delta t)^2 \) is to write your time component as \(ict\), where \(i=\sqrt{-1}\). Then, when you square it you end up with \(-(ct)^2\). This can be useful, but it isn't the most general or the most elegant way to go about things, and it doesn't translate very readily when you go from special relativity to general relativity. The other solution is to use a metric tensor to get the right signs. This is easily generalised. The underlying mathematical formalism is more complicated. It can't really be avoided once you start talking about general relativity, though. To give you a very rough idea, the metric works with 4-vectors such as (ct,x,y,z). To "square" such a vector, you do something like \(x \eta x^T\), where \(x\) is the 4-vector and \(\eta\) is the metric tensor (\(x^T\) is the transposed 4 vector). It is the operation of the metric tensor that inserts the required minus sign in special relativistic flat space.

Anti-stupidity: With algebra, it's best if you keep in mind what you're aiming for as you work. It's something you get better at with practice. In this case, you start with Please Register or Log in to view the hidden image! That should be \(\Delta t' = \frac{1}{c}\sqrt{(2l_0)^2 + (v\Delta t')^2}.\) The prime on the t in the square root is important there. You know that \(\Delta t = 2l_0/c\). You're looking for a expression for \(\Delta t'\) in terms of \(\Delta t\). Here's what I'd do. Start by squaring both sides: \((\Delta t')^2 = (\frac{2l_0}{c})^2 + (\frac{v}{c}\Delta t')^2.\) Now substitute for \(\Delta t\): \((\Delta t')^2 = (\Delta t)^2 + (\frac{v}{c}\Delta t')^2.\) We want \(\Delta t'\) as the subject, so move all the stuff that involves that over to the left side of the equation: \((\Delta t')^2 - (\frac{v}{c}\Delta t')^2 = (\Delta t)^2.\) Now take out the common factor of \((\Delta t')^2\): \((\Delta t')^2\left(1 - (\frac{v}{c})^2\right) = (\Delta t)^2.\) Divide through by the stuff in the brackets and take the square root of both sides: \(\Delta t' = \frac{\Delta t}{\sqrt{1-(v/c)^2}}\) And we're done. Let's look at what you did. Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Steps 2 and 3 are fine. Please Register or Log in to view the hidden image! Why take that \(\Delta t'\) term back over to the right-hand side? You want to make that the subject. Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! This is correct, but it's doing things the hard way, if you ask me. Please Register or Log in to view the hidden image! Here we're back to where we would have been if you'd just factored out \((\Delta t')^2\) in step 3. That would have made steps 4, 5 and 6 unnecessary. Also, you end up with the \(\Delta t'\) in the denominator in step 7, and then have to move it back to the numerator again, which you do in 8: Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! This is fine. Please Register or Log in to view the hidden image! Why square both sides? We just want \(\Delta t'\) as the subject. It's obvious we need to take the square root of both sides of 9. Please Register or Log in to view the hidden image! ... which is what you did in 11. Please Register or Log in to view the hidden image! And we're done.

In step ten, It was supposed to be a square root but I apparently did not catch it until later. I'm a bit lazy. Thanks for taking the time to look it over.

How would I go about finding the time dilation due to higher derivatives of displacement, like acceleration, jerk, and so on.

I started to think closely about the thought experiment and i'm wondering if time runs differently depending on the perspective that you are looking at the light clock from. Does time dilation really happen or is it really just a visual effect? I've been battling with this for a while and haven't turned up with an answer. I'll see if I can add some more equations expressing what I mean, just give me a bit.

If you concentrate on the spacetime interval you find time dilation drops out of it naturally. I've always found starting by assuming the speed of light is constant is a rather poor start - I like a reason for doing things like that.

I wasted 30 years on that question - it is real - you won't get much sympathy for trying to be cleverer an Einstein - and failing.

Where can I go to get evidence for relativistic affects, not just the classic experiments but more recent ones at least in the past ten years? Especially relativistic kinetic energy (relativistic mass), I hear people saying that this happens all the time in accelerators but where can I get the papers themselves? I do not want reassurance, I want the raw information, that way when I come across anti-relativity cranks or i'm snooping around the thunderbolts forum; i'll be ready. I've been doubting special and general relativity for a while but do not want to turn into a physics crank, please help!

Because this voice play's around in my head all the time saying, "it's proven, it's proven", but I don't have the resources to check such an inference. Am I supposed to go to Jstor and pay for some of the physics papers their or arxiv.org? Or is there some where else that I can get these papers, preferable without having to pay.

The difference between a book and something interactive (like this thread) is that you can get a response tailored to your input. I should confess to a point of principle that it should be possible for anyone (regardless of income or any other factor) to understand not only special relativity but also general relativity. Under difficult circumstances gaining the background maths is likely to be more of a challenge than the actual physics. I happen to have fairly good high school maths - even with that I didn't find it easy to grasp Special Relativity - I had a lot of help from people on the Internet. I can say that when I finally got 'something' it was one of the best moments of my life. I can't let the point pass without a "Thank you" to all involved. The 'something' is to be able to put events, spacetime and high school maths together to predict a result. Not read the paper, ya ya ya, but predict and (sometimes) get it right. For you, Anti-stupidity, I would prescribe the Muon Experiment, do not pass go until you can put the events, spacetime and high school maths together to predict the result. http://www.hyperphysics.de/hyperphysics/hbase/relativ/muon.html - the analysis he gives is (IMHO) pretty crappy - part of the fun is doing a better one (aka your own).