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4

There is an additional condition coming from the third term on the left hand side of your transformed equation, where you have used what seemed to be the chain rule $$ \frac{\partial x^\lambda}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^\mu}=\delta^\lambda_\mu $$ In reality however, the nontrivial positioning of the covariant and contra variant ...


3

Let us first replace the Minkowski metric tensor $$\eta~=~\eta_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$ with a more general constant metric tensor $$g~=~g_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}.$$ Note that the raised EM tensor $$F^{\mu\nu}~:=~g^{\mu\lambda} F_{\lambda\kappa}g^{\kappa\nu}$$ depends on the (inverse) metric. The Maxwell ...


3

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?" No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see ...


3

There are several possible approaches to this question, but I've always been a fan of the one taken by Edwin Jaynes in his 1965 paper Gibbs vs Boltzmann Entropies. (See sections V and VI for the discussion, which I think can be read in isolation from the rest of the paper.) Here he derives the second law from the empirical fact that we as scientists and ...


3

The problem with your question is that velocity is relative so there is no absolute way to say whether something is travelling through space or not. Any observer can set up some time and space coordinate system $(t, x, y, z)$ to measure positions of spacetime points. The observer can then use these coordinates to measure changes in position with time i.e. ...


2

Mass-energy equivalence (Rest energy = rest mass * c^2) can be applied to a moving body: Total energy = relativistic mass * c^2 From the perspective of a stationary observer, the total energy in a moving body is greater than the rest energy in a stationary body. As velocity approaches c, it is much greater. The ensuing explosion would be much bigger than ...


2

Because: $$ \tau \ne \frac{t}{\gamma} $$ It's true that: $$ \frac{d\tau}{dt} = \frac{1}{\gamma} $$ But in general this does not simply integrate to $\tau = t/\gamma$. This is true only in the special case of non-accelerated motion.


2

As I recall, covariant refers to how an object transforms when you boost to another inertial frame. An example would be the relativistic 4-momentum $P^{\mu}$. Invariant refers to quantities which are unchanged under boosts to different frames. For example the product $P^{\mu}P_{\mu}=m$ has the same numerical value in any frame. Sometimes a relativistic ...


2

This is really a comment on Hypnosifl's answer but it got a bit long to fit in the comment box. Be cautious about interpreting the word time in proper time too literally. We use the phrase proper time because it is the time measured by an unaccelerated observer travelling between the two points. However no such observer can exist in this case because that ...


1

I'm not sure I have fully grasped what you are asking. The equation is for calculating the momentum of the proton in the inertial frame of the observer i.e. the frame velocity is zero by definition. The only thing moving is the proton, at a speed $v$. If, as in your question 2, you have a different frame then the speed of the proton in that frame is given ...


1

vL/c^2 is the difference (for a stationary observer) between two clocks at the two ends of a rod of rest (or proper) length L traveling at velocity v. The clock at the forward-end (in the direction of v) of the rod trails by this amount (as perceived by the stationary observer). The two clocks are synchronized in the rest frame of the rod, but for the ...


1

It depends on the direction of movement of your rod. If it travels perpendicular (its main axis is perpendicular) to its velocity, there is no length contraction. On the contrary, if it travels parallel to its velocity, you can apply the standard length-contraction formulae of relativity (cv http://en.wikipedia.org/wiki/Length_contraction)


1

The expression for time dilation which we come across states the fact that the "time interval between two event is minimum when observed from a frame where two events occure at same place".. In your case the events did not occur at the same place.. so you can instead use the lorentz transformation equation to determine the time interval between event A and B ...


1

You start with $$ E' = \frac{E-up}{\sqrt{1-u^2/c^2}} = \\ \frac{E-up}{\sqrt{(1-u/c)(1+u/c)}}. $$ You know that $p = \frac{E}{c}$, so $$ \frac{E-uE/c}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E(1-u/c)}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E\sqrt{1-u/c}}{\sqrt{1+u/c}} = \frac{E\sqrt{c-u}}{\sqrt{c+u}}, $$ now using $E = hf$, $$ f' = f\sqrt{\frac{c-u}{c+u}}. ...


1

You might be thinking of the Unruh Effect. It says that observers undergoing high acceleration see an increased vacuum temperature. Mandatory Wikipedia link: http://en.wikipedia.org/wiki/Unruh_effect Remember that according to special and general relativity, observers traveling at a constant velocity (speed and direction) all have to observe the same ...


1

I would like to bring the ladder paradox here to explain simultaneity of events.A ladder (an inertial frame) is moving horizontally with a relatively high constant speed with respect to a garage (another inertial frame). The garage has an open door where the ladder can not actually enter if the ladder was at rest in the garage's frame but that is not ...


1

Regarding your assertions: Events $\varepsilon_{AJ}$ and $\varepsilon_{BK}$ were simultaneous in the inertial frame of participants $A$, $B$, $M$. This is a perfectly reasonable statement and it is the sort of language used in everyday physics. Participant $M$ was the middle between $J$ and $K$, in the inertial frame of participants $A$, $B$, $M$. ...


1

The Pauli matrices anticommute, so the product of two of them has to be antisymmetric in its indices. The only antisymmetric 2-tensor in two dimensions is the Levi-Civita symbol $\epsilon$. Hence, you can "guess" the structure of the product in question up to a constant without calculating anything explicitly.


1

IMHO it's important to look hard at the ontology of what's actually there and take care to distinguish between reality and abstraction. For example: I was reading about the light cone in relativity... Relativity is just about the best-tested theory we've got. I "root for relativity". But I will say this: a light cone is an abstract thing. You cannot point ...


1

Well, a rocket traveling at close to the speed of light would be very hard to see at all cause it would go from a moon's distance in one direction to a moon's distance in the other direction in a little over 1 second, and seeing a rocket as far away as the moon would be difficult - but I'm thinking that's not what your asking, so lets pretend that we have a ...


1

Is it possible to travel ONLY through Time and not Space? Well, in a sense, we can't help but travel through time, but in the way that I think you mean, one way to do it is to hover just above the event horizon of a black hole - preferably a super-massive one so the tidal effects wouldn't be a problem. Park a spaceship at 1/10th of 1% greater ...



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