New answers tagged

1

There is no frame in which "space" is at rest. Using the length contraction formula $$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$ for the distance to the star is slightly misleading. The distance $L$ in your frame corresponds to two points, $x_0 = 0$ for you and $x_1 = L$ for the star, that you are observing simultaneously, say at time $t_0$. In the star's frame ...


1

Aside from simply looking at the Lorentz transformation, seeing a divergence and concluding "meh, it doesn't work", another way to gain insight into the divergence is through the statement: no finite sequence of finite boosts will get you to a speed $c$ relative to your beginning inertial frame. Imagine yourself in a spaceship with orientation controls ...


3

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.) The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/...


-1

Lorentz transformations apply to objects with nonzero mass. For an object with mass, it would require an infinite amount of energy to reach light speed.


3

This is an excellent question which is much more subtle than it first appears. On a first (or even a second) reading, you seem to answer your own question and then ignore your own answer! What I mean is, you clearly understand that while in an accelerating vehicle you experience a 'jerk' which does not happen with uniform motion, and you also acknowledge ...


2

The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ...


4

The flaw is that you've failed to do an experiment which will tell you whether the frame is inertial or not. If you do such an experiment -- for instance take a test mass, initially at rest with respect to the frame, release it, and see if it remains at rest -- you will immediately discover that the frame is not inertial.


-4

This link provides an answer to this and other questions on the experimental basis of special relativity: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Tests_of_Einsteins_two_postulates There are several experiments using terrestrial sources, where the light source is moving, for example the speed of gamma rays from the decay of ...


1

You are quite correct that we don't have to bring the twins together at the same point in space. It's enough to make their relative speed zero i.e. bring them into the same inertial frame. The twin paradox generally involved bringing the twins together just for simplicity. When two observers are at the same point in space they can directly compare their ...


4

A boost is "the only other" plausible choice for transformations that mix spacetime co-ordinates, aside from rotations. The two notions in this sense are complementary. See for example: Palash B. Pal, "Nothing but Relativity", Eur. J. Phys. 24, pp315-319,2003 Given certain "reasonable" assumptions about our Universe; roughly The first relativity ...


2

My understanding (which is based somewhat on Jackson's chapter on SR in Classical Elecrodynamics) is that the invariance of the interval is not enough to derive the Lorentz transformations - you also need the second postulate (that the speed of light is constant in all frames). The invariance of the interval follows from the fact that spherical light waves ...


1

I think you're overlooking the fact that the ordering of distant events is ambiguous in special relativity. In general, observers will disagree on whether distant events are simultaneous. To recap the events you are considering: Event AB: A and B pass one another. Clocks A and B both read noon. Event BC: B passes C. B's clock reads 1 o'clock. Anyone (...


0

The symmetry argument is invalid. Clock B shows 1:00 in all frames of reference when it reaches space station C, so there is a mistake in the question.


4

It depends what you mean by their clocks both being at noon when they're at positions A and B, and their speeds differing by 161,000 miles per second. By the way the question was stated, I'll assume you meant their positions were measured in an outside observer's frame, i.e., at noon in the outside observer's frame ship A is at position A and ship B is at ...


0

No, they can't, not as you've formulated the problem. Spaceship B and clock B are at the same point in spacetime, so all frames will agree whether or not two events at the spaceship and the clock are simultaneous. Both A and B will observe B's computer monitor changing from "no" to "yes" when clock B reads 1:00. I don't understand your symmetry argument ...


0

They sure can. Every moving frame has its own time perception, and if different people are in differently moving frames they perceive different time(s?). That might even result in exterior event sequences being switched - for example, star A becomes a supernova before star B for one frame, and the other way around for another frame. Don't try to apply ...


1

Consider converting the time coordinate to have units of length; then the expression is $$ ct' = \gamma c t - \gamma \beta x $$ where $\beta = v/c$. A look at a spacetime diagram should convince you that the term $-\gamma\beta x$ accounts for the fact that observers in the two coordinate systems will disagree about which events are simultaneous: the $x'$-...


0

We express equation by differences \begin{equation} \Delta t' = \gamma \left(\Delta t-\dfrac{v\Delta x}{c^{2}}\right) \tag{01} \end{equation} Now, let in the unprimed system two simultaneous events $\:E_{1}\left(x_{1},t_{0}\right), E_{2}\left(x_{2},t_{0}\right)\:$ at a distance $\:\Delta x= x_{2}-x_{1}\:$ apart as in Figure. Also, let on the middle point $\:...


3

Regarding total momentum conservation, the point is that in non-inertial reference frames inertial forces are present acting on every physical object. Momentum conservation is valid in the absence of external forces. However, if these forces are directed along a fixed axis, say $e_x$, or are always linear combinations of a pair of orthogonal unit vectors, ...


0

There are 2 points: as you see from formulas, Lorentz transformations are linear in coordinates the actual, interesting terms involved in the Lorentz transformations are the spatial coordinates and $ct$, not just $t$ If you consider that, you see how the time transformation from one frame to the other is now simpler; it is even simpler to see that the ...


3

Not through our normal sight. First of all, doppler shift will totally change the color of everything, objects ahead of us shifting towards blue/ultraviolet, and these left behind becoming more red/infrared. Then, Lorenz Contraction combined with changed time and distance it takes light reflected off objects (and separately, emitted by objects) to reach our ...


2

Another, less messy way to do this is as follows. Link the said matrix to the identity matrix by a path defined by: $$\Lambda:\mathbb{R}\to \mathscr{M}_{4\times4};\;\Lambda(\zeta) = \left(\begin{array}{c|c}\cosh\zeta & -\hat{B}^T \,\sinh\zeta\\\hline -\hat{B}\,\sinh\zeta &\mathrm{id} +\hat{B}\,\hat{B}^T\, (\cosh\zeta-1) \end{array}\right)\tag{1}$$ ...


4

Hints : $$ \text{your matrix :}\:\: \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \text{is this matrix :}\:\: \eta= \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & \boldsymbol{0}^{\mathsf{T}} \\ \boldsymbol{0} & \mathrm{I} ...


0

"The stationary person would see him accelerating from 0.5c to 1.499c, faster than light" The statement itself is a large mistake... The person in the space ship knows that he had accelerated from 0.5 to 0.99 time the speed of light, but the person outside would only see him accelerating from 0.5c to 0.99665551839c. It is because relative velocity is given ...


0

I would work out your barycentric co-ord first, then compare them with a matrix subtraction and normalisation (Most efficient in code). This would give you a vector multiplier to from 0-1 (percentage) per component. Next work out the maximum velocity in each dimension given the maximum offset. The simply times your normalised components by this velocity.


2

To answer this, You can't add velocities as in Newtonian mechanics. In relativity, If frame B is moving at velocity u and C is moving at velocity v both with respect to an inertial frame A, then relative velocity between B and C is $\frac{u+v}{1+ uv/c^2}$. So substituting your arguments here would now leave us with relative speed less than c. Just because ...


0

The following might help your cause (I don't have time to look at all your question), since you allege its lack in the literature. Your allegation implies a wish to dig into this kind of question seriously, so here are a couple of things you will need to learn to do so. To work out the matrix for a general boost, simply take the matrix for a simply written ...


1

What is meant is that physical laws are the same between (inertial) reference frames so that if you observe two bodies undergoing an elastic collision then you will experimentally determine that the momentum before the collision is the same as the momentum after the collision. An observer in another frame will also note that in his reference frame, the ...



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