Special Relativity - Two Postulates/ Fundamental Effects Flashcards
What is a Van de Graf generator?
Generates very high electric potential. Transfers charge to build a static source. Can be used to generate a large voltage in a controlled way.
Its relation to particle acceleration. 1 eV is natural units for energy. Energy gained from accelerating one electron over one volt.
When the charge is released as a stream of electrons into a vacuum tube, we can measure energy of incident electrons, due to increase in temperature. Instead of a linear relationship, the speed went horizontally asymptotic as it approached c. T = (1/2)mv^2 won’t do!
The Two Postulates of SR?
Relativity Postulate (1): All inertial frames are equivalent, and the law of physics are the same in all of them. There is no preferred reference frame, all motion is relative.
Speed of Light Postulate (2): The speed of light in vacuum is the same in all inertial reference frames. No matter the frame, you are always stationary w.r.t a frame moving at v = c.
The second is defined due to the radioactive decay of Caesium, and the metre by the distance light travels in 1/c th of a second.
Biggest consequences here, the loss of simultaneity, time dilation, and length contraction.
Traverse rod thought experiment?
Two 1 m long parallel rods. One stationary, the other moving at close to the speed of light… what is the result?
If we imagine that these lengths have contracted in the stationary frame, and therefore expanded in the moving frame, then this cannot be. The marks of the pens idea doesn’t work! Proof by contradiction, both cannot be true, transverse length does not change. If this wasn’t the case, postulate 1 would be useless.
Proper length?
The stationary length (at least viewed as so) of an object is known as it’s proper length
Characteristics of a frame of reference in Special Relativity?
Yet again, massless interpenetrable array of rods over space, but also array of synchronised clocks at the grid points.
The importance of time, place, and events. And proper time.
When using SR, it is vital you define for sure the whereabouts and timings of all occurring events. If the difference in time at two synchronous events is 0, or two simultaneous events meet at the same place (event) after the same time interval, then it is a proper (characteristic) time.
The time interval between two events with the same special coordinates is a proper time interval.
Light bouncing between mirrors thought experiment?
In frame S’ where both mirrors are stationary in their reference frame. Light is emitted (0,0,0), reflected (0,lo,lo/c), and comes back to the original mirror (0,0,2lo/c), and due to the two events being in the same spacial coordinates, it is a proper time (2lo/c).
In frame S where both mirrors are moving w.r.t the frame. Light is emitted (0,0,0), reflected
(vt/2,lo,L/c) and comes back (vt,0,2L/c). Knowing that L = ct/2 when ‘t’ is the time between events 1 & 3, we can plug this all into Pythagoras to find improper t. Using the proper ‘t’ equation and taking out the Lorentz factor, we then derive time dilation.
Characteristics and implications of the Lorentz factor?
Gamma must always be greater than or equal to 1.
If v goes to 0, gamma goes to 1 and no time dilation occurs.
If v goes to c, then gamma goes to infinity, infinite time dilation occurs.
Lorentz FACTOR, is what it says on the tin, it means S will see S’ clock running slower by a factor of gamma.
Paradox, S’ will also see S’s clock running slower due to it moving w.r.t S’ observer.
Moving clocks run slower.
Explain the Muon lifetime experimental proof for time dilation.
Muons produced from cosmic rays in the atmosphere. Muons decay. Muon lifetime = 2.2 micro seconds. Take a distance of 20 km from surface, and speed = 0.995c. Using d=vt or decay law naively wouldn’t work due to the speed of the muons.
Using t=yto, gamma = 10, so in the lab frame, muons have a lifetime of 22 microseconds!
Using probability equation, after calculating time using speed and distance yields an incorrect probability compared with experiment. Using mean decay time found using time dilation equation though yields the correct result.
If you wanted to relate this to length contraction, one could say that from the muons perspective, the mountain appears ten times shorter than what it is in the laboratory frame. So it still reaches.
A bit on decay curves.
Remember the integral that gets you to N=Noe^(-ht)
If t= 1/h then you get the characteristic decay time where 1/e of the sample remains.
The mean decay time T = 1/h where h is the decay constant, h = ln(2)/t(1,2).
Some particles live very long, so their mean lifetime is greater than the time it takes for half to decay.
Proper length and length contraction?
If the time interval between two events happening such as the synchronising of clocks or the front of a train passing two points have a time that are equal or are measured in the same spatial coordinates, it is a proper length.
Train passing example of length contraction?
With a train passing a platform and seen by a stationary observer, the platform itself is a proper length and the length measured when the train passes each end is improper, as they’re seen at different times.
lo is length of the platform seen from frame S. Such that when the train has a velocity ‘v’, lo=vt, where t is an improper time. From the S’ frame, x1’ and x2’ leads to t being a proper time, of an improper length. Hence, this leads to l=vto.
Now! The ratio of these two lengths by the observers l/lo=vto/vt, goes to l=lo/y.
The reciprocity of time and length.
Just as a decaying muon can be seen by Earth to ‘exist’ for a lot longer than it will in its frame in order to reach the ground. The muon, which see’s the Earth coming toward itself from it’s supposed reference frame, will see the Earth contract in a way, which means by using its characteristic time of decay, it will still reach the ground due to how much the length has contracted, and how far the Earth will have travelled.
The solution to the twins paradox?
So, from both twin A & B’s perspective, each other should ‘travelling’ and therefore time should contract both ways, this cannot be. The answer? B (moving) does not remain in one inertial frame, they go, then go back, time dilation result requires a single frame of reference.
B accelerates when turning around, hence, non-inertial.
The rod-shed paradox
Rod and shed paradox shows that from a stationary shed frame, the rod is seen to contract by a factor that allows it to fit in the shed and shut both doors at the same time.
In rod frame, the shed is seen to contract and the leading door lags and reaches the back of the rod after the back door hits the front of the rod. What’s wrong here? What is being specified, the idea of the doors closing ‘at the same time has no bearing here, as simultaneity is lost and things are seen at different times by different frames.
What is seen as simultaneous in one frame Is not in another.
