Hey guys, studying for a physics II quiz on special relativity (Einstein for the win). Basically. T = Tproper * Root( 1-(v^2/c^2)) Right? And: L = Lproper * Root( 1-(v^2/c^2)) Right? Thanks in advance
Looks right, given the usual assumptions, but I'm just a guy on the 'Net. Check this out: [thread=61223]how to tex[/thread] \(T = T_{proper} \sqrt{1-(v^2/c^2)}\) \(L = L_{proper} \sqrt{1-\frac{v^2}{c^2}}\) Code: [ tex]T = T_{proper} \sqrt{ 1 - (v^2/c^2) }[/tex] [ tex]L = L_{proper} \sqrt {1 - \frac{v^2}{c^2} }[/tex]
okay. And proper time is time measured from a stationary clock that observes the events happening right?
You got this one backwards, the correct one is: \(T_{proper} = T\sqrt{1-(v^2/c^2)}\) The derivation is simple if you start from the invariance of the Lorentz interval: \((c d T_{proper})^2=(cdT)^2-(dx)^2\) Yes.
Doesn't that sort of disagree with the idea that a clock in motion runs slow? Since the proper time would be less then the time. So the equation "L" is the opposite of the one for "T", what is the derivation of that?
No, it doesn't disagree, you can also derive it from: \(x'=\gamma (x-vt)\) \(t'=\gamma(t-vx/c^2)\) Differentiate: \(dx'=\gamma (dx-vdt)\) \(dt'=\gamma(dt-vdx/c^2)\) Proper time means dx=0. This gives you: \(dt'=\gamma dt\) Using the notation \(dt'=dT\) and \(dt=dT_{proper}\) you get : \(dT=\gamma dT_{proper}\) i.e.: \(dT_{proper}=dT \sqrt{1-(v/c)^2}\) Start with \(dx'=\gamma (dx-vdt)\) \(dt'=\gamma(dt-vdx/c^2)\) and make dt'=0 (you are marking the endpoints of the rod you are trying to measure simultaneously). Remember that \(dx=dL_{proper}\) and \(dx'=dL\)
Right, my bad. Yes, the proper time is less than the frame time. If a moving clock takes T=20 to elapse Tproper=10, then it's running slow.
The proper time is always the shortest time between any two events. Since proper time is the time measured by a clock carried along with the thing that you're interested in, the shorter proper time means "moving clocks run slow". i.e. they record less time than other clocks between the same events.
The most reliable way is to define your events and work from the Lorentz transforms. Tach's done it through differentiation (it *looks* right to me Please Register or Log in to view the hidden image!), but you can also just consider the events at each of of a length, and at each end of a time interval. A time interval is easy, because the hypothetical moving clock is a pointlike object in space, so its spacetime worldline is one dimensional, so the start and end events are well defined. For example, consider a time interval on moving clock that starts as it passes x=0: Start event: \(t=0, x=0\) End event: \(t=\Delta t, x=v \Delta t\) Transforming those events to the clock's rest frame to get the proper time: Start event (easy): \(t'=0, x'=0\) End event (after a little algebra): \(t'=\Delta t \sqrt{1-(v^2/c^2)} \ \ , \ \ x'=0\) Now, a length is a little more involved, because the hypothetical moving ruler is an extended object (not pointlike), so its worldline is not one dimensional, so the start and end events are not well defined. So, you first need to think about what spacetime events really define the length of a moving object. It's not so hard, after a little thought about spacetime and events: A length is the spatial difference between two events with no time difference. So, to get the length of the moving rod we need two events that happen at the same time, one at each end. I'll leave that as an exercise for you to work through!
Pete, I think your expression for t' is is wrong in post #9. Remember that \(\gamma = 1/\sqrt{1-(v/c)^2}\) is always greater than (or equal to) one.
Whee! Please Register or Log in to view the hidden image! Special Relativity Op Art by PhysForum Photos, on Flickr
No, it isn't. Setting dx=0 is basically saying the object's position in that frame doesn't change. Such things are done all the time in Riemannian geometry, ie the length of a curve through a manifold is \(s = \int ds = \int \sqrt{|ds^{2}|} = \int \sqrt{|g_{ab}dx^{a}dx^{b}|}\) Putting in the Lorentzian metric and you get \(s = \int \sqrt{|-dt^{2}+d\mathbf{x}\cdot d\mathbf{x}|}\) where \(d\mathbf{x} \cdot d \mathbf{x} = \sum_{i=1}^{N}dx^{i}dx^{i}\). Suppose the path is parameterised by \(\lambda\) then you have \(s = \int_{\lambda_{0}}^{\lambda_{1}} \sqrt{|-\left(\frac{dt}{d\lambda}\right)^{2} + \sum_{i} \left(\frac{dx^{i}}{d\lambda}\right)|}\). Setting \(dx^{j}=0\) is then seen to be equivalent to setting \(\frac{dx^{j}}{d\lambda} = 0\). In other words the path doesn't vary in the \(x^{j}\) direction. If you were doing motion confined to a plane within space then that's the sort of condition you'd get. Ching's general attitude, enormous level of ignorance and exception amount of self delusion about his own understanding reminds me of Jack_.
