Category: Relativity

Train station diagrams are spacetime diagrams!!

zimbles2 Comments

I found something really cool in a book that I saw in a library, namely that there exist something like a train schedule diagram. Which is exactly a spacetime diagram but then for a simple real world application. It has the distance on y-axis and the time on the x-axis like so:

https://i.stack.imgur.com/wLvfo.jpg

So actually I just found out after posting this blog that Numberphile just did a video about this, that’s a coincidence:

Velocity is rotation of your axis

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Objects:

Imagine if two object lie along the y – axis with a distance of say 5 meters. Now create a new x’ and y’- axis by rotating it at a certain angle. The distance between the objects is still 10 meters, but in the y’ axis it will now be say 4 meters and in the x- 3 meters.

Events:

Imagine if two events occur at the same spot with a pause of 4 seconds. Now create a new x’ and t’ axis by rotating it at a certain angle (how?). The duration of the new events in the new axis-system is now 5 seconds, and the difference in distance between the events is 3 meters.

See the analogy? Now how do you rotate the axis system in spacetime? And what kind of rotation is this? Answer: velocity and hyperbolic.

 

Afbeeldingsresultaat voor rotation

Gravity

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  1. Time moves faster for your head than for your toes.
  2. The law of Cosmic Laziness states that objects want to move slowest possible through time

Thus you get pushed towards earth, i.e. you feel gravity

Maybe a possible nice analogy to explain gravity as spacetime curvature.

zimblesLeave a comment

So:

First watch this great video by MinuteEarth:

A short explanation of gravity in General Relativity is usually given as follows:

  • Mass tells spacetime how to curve, curved spacetime tells mass how to move.

This is actually wrong, since Mass is not the (only) thing that dictates spacetime curvature, but actually 4-momentum-flow! Mass is just one of the 4-momentum-flows! There are actually 16 flows, or actually 10 because of the symmetry of the Stress-Tensor.

For rivers it goes like this:

  • The waterflow in Rivers follow the spatial curvature of the landscape, and the the wateflow carves out the landscape changing its spatial curvature.

Does 4-momentum flow ‘carve out’ spacetime just like the River flow ‘carves out’ the landscape?

The underlying driving force of rivers is obviously gravity and going downhill. The underlying driving force of 4-momentum flow is actually …. entropy!

 

Who actually moves slower through time? 

zimbles1 Comment

One of the hardest thing for me to comprohend of special relativity was the fact that an object moving fast through space moves slow through time, but that object sees us also as moving slower through time… this doesnt make sense!

But just as two rockets moving away from each other in outer space, where the question cannot be answered who actually is moving. The question who is actually going slower through time, is also meaningless. It is the exact same question. 

Imagine just two straight lines crossing each other in a plane, which line is actually moving away from the other? This question doesnt make sense, and is exactly the same question as above.. The angle between the lines is what can be measured.

Deriving the Unruh effect in a simple manner.

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unruh

 

This derivation is based on this paper.

Equations used:

Special relativity:

dt = d\tau \gamma

\frac{dv}{dt} = a (1-\frac{v^2}{c^2})^{3/2}

Integrating over dt we get:

v(\tau) = c \tanh (a \tau / c ) (1)

Equation(s) derived:

T = \frac {\hbar a}{2\pi k c}

Derivation:

First find the hyperbolic orbit of the accelerated (Rindler) observer in the z direction, which are:

t(\tau) = \frac{c}{a} \sinh (\frac{a\tau}{c})

z(\tau) = \frac{c^2}{a} \cosh(\frac{a\tau}{c})

Then consider a plane wave with angular frequency omega_k and wave vector K parallel to the direction z. The lorentz transformation of the angular frequency omega_k looks like:

\omega_k'(\tau) = \frac{\omega_k - Kv(\tau)}{\sqrt{ 1-v^2(\tau)/c^2)}}

If we substitute equation (4) in the last equation we get:

\omega_k'(\tau) = \frac{\omega_k - K c\tanh (a\tau /c) }{\sqrt{ 1-\tanh^2 (a\tau /c) )}}

With K = \omega_k/c we get:

\omega_k'(\tau) = \frac{\omega_k (1 - \tanh (a\tau /c) ) }{\sqrt{ 1-\tanh^2 (a\tau /c) )}}

Then do some complicated algebra shown with \tanh (a\tau/c) = b

First step: \frac{1-b}{\sqrt{1-b^2}} = \frac{\sqrt{ (1-b)(1-b) }}{\sqrt{(1-b)(1+b)}} = \frac{\sqrt{ (1-b) }}{\sqrt{(1+b)}}

Use this hyperbolic identity

Second step

Conclusion:

\omega_k'(\tau) = \frac{\omega_k (1 - \tanh (a\tau /c) ) }{\sqrt{ 1-\tanh^2 (a\tau /c) )}} = \omega_k e^{-a\tau/c} (2)

Now consider the time-dependent phase, which is determined by:

\phi (\tau) = \int_{\tau} \omega'_k(\tau') d\tau'

Plugging in equation (2) and integrating we get for the phase:

\phi (\tau) =\frac{c}{a} \omega_k e^{-a\tau/c}

Where the frequency spectrum S is determined like:

S(\Omega) = |\int_{\infty}^{\infty} d\tau e^{i \Omega \tau } e^{i \phi }|^2

Substituting our formula we get a hard looking integral:

S(\Omega) = |\int_{\infty}^{\infty} d\tau e^{i \Omega \tau } e^{i (\omega_k c/a) e^{a \tau/c} }|^2

This integral looks hard and I can’t solve it, if we look at page 4 of the paper we can see it equals an equation that looks like a Planck Law.

S(\Omega) = \frac{2 \pi c}{\Omega a}\frac{1}{e^{2\pi \Omega c/a} - 1}

Which looks like the planck law

E= \hbar \Omega \frac {1}{e^{\hbar \Omega / kT} -1}

If we compare the following factors:

(e^{2\pi \Omega c/a}-1)^{-1}

(e^{\hbar \Omega / kT} -1)^{-1}

Now we equate them for reasons I don’t get:

(e^{\hbar \Omega/ kT} -1)^{-1} =  (e^{2\pi \Omega c/a}-1)^{-1}

And we get:

\frac {\hbar \Omega}{kT} =  2 \pi \Omega \frac{c}{a}

Rearranging:

\hbar / kT =  2 \pi c/a

Solving for T, temperature:

T =  \frac {\hbar a}{2 \pi kc}

We get the unruh temperature.

figure1

 

I never found a correct derivation of E=mc2

zimblesLeave a comment

So there are lots of derivations of this equation floating around on the web. But so often they are wrong! They often derive E=delta mc^2, which is actually not that hard.

A really ugly derivation uses the Taylor expansion of the relativistic kinetic energy equation and then just says that the first term is e=mc^2. But you cant get it like that!

Examples of wrong derivations:

An example is:

Click to access 0308039.pdf

This derives the delta equation.

Another example: this is even a whole webpage devoted to ‘explaining’ E=mc2:

http://www.emc2-explained.info/Emc2/Derive.htm#.WHjnu1wpm7A

WRONG!!! Arg so frustrating!! Even de advanced page is wrong: http://www.emc2-explained.info/Emc2/Deriving.htm#.WHjn91wpm7A

This is derivation of E=delta mc^2 instead of E=mc^2. In other words, the equation derived tells us that an object with mass M and kinetic energy E has extra inertia equal to E/c^2. It doesn’t say why ‘rest mass’ is equal to energy. I want to know!

http://www.askamathematician.com/2011/03/q-why-does-emc2/

This guy seems to have a lot of knowledge, but still derives it wrong! This is with the infamous Taylor expansion.

Now one of the most sad things I know is that even in the great Feynman lectures it is derived wrongly with the Taylor expansion…

http://www.feynmanlectures.caltech.edu/I_15.html

Better derivations:

http://www.adamauton.com/warp/emc2.html

Although this derivation doesn’t really show how spacetime geometry shows the equivalence of mass and energy. It just shows that photons have some apparant inertia.

Another awkard derivation is just postulating the existence of 4-vectors, then defining relativistic momentum as mass times the four vector, and then calling one of the components (the ‘time’) component as energy. Then saying, see: E=mc^2. It is correct but really unsatisfactory and artificial.

 

I think the best derivaiton is with the Lorentz/Poinceare group and showing a symmetry in spacetime. But I don’t know that one yet. I believe that one can be even pretty simple if you understand the notation.