29

Nope, the "$c$" in the Lorentz Transformations doesn't just apply to the speed of propagation of information in a medium, though I understand why you might think of it that way. Instead, the Lorentz Transformations are fundamental to the structure of space and time itself. In particular, they imply that the speed of light $c$ is frame independent: ...


17

You are right that the traveler need not age much when voyaging from one star to another many lightyears away. If the distance plotted from Earth is $d$ in the measurements appropriate to the rest from of Earth or Sun, then the proper time accumulated by the traveler is $$ \tau = \frac{1}{\gamma} \left(\frac{d}{v}\right) $$ if they journey mostly at speed $v$...


11

Does special relativity imply that I can reach a star 100 light years away faster than in 100 years? The time that you experience in your spaceship on your way to the star could be more than or less than (or equal to) 100 years depending on how fast your spaceship travels. The time that an earthbound observer sees you take to reach the star is necessarily ...


10

No. Lorentz transformation is a consequence of the postulate of special relativity which says that $c$ is the same in all inertial reference frames. Analogous statement is not true for sound. If a bat moving at $1/3$ of the speed of sound along the $x$ axis measures in its frame of reference the speed of sound propagating away from it along the $x$ axis, it ...


7

Your statement about aging is incorrect. Instead, time dilates relative to stationary, so they each see the other moving more slowly through time, and therefore aging more slowly. Also, you say several times "with no frames of reference". There are frames of reference, it's just that there's no absolute frame of reference.


4

Reference frames are just coordinate systems, nothing more. Spacetime is not very much different from Euclidean space. The main difference is that there is a flipped sign in the spacetime version of the Pythagorean theorem. Just as you can put Cartesian coordinates on Euclidean space, you can put inertial coordinates on spacetime. The Lorentz transformation ...


4

Yes, this is predicted by special relativity. Due to time dilatation, in your frame of reference, time will proceed slower than in the reference frame on earth. The closer your speed is to light speed, the larger the effect. As seen from earth, your trip will last more than a hundred years. In your own reference frame, the trip can approach instantaneous ...


4

Don't worry about Lorentz transformations, just realise that the proper time interval $\Delta \tau$ for something to travel between two events is the same judged in all inertial frames (an invariant). $$(\Delta \tau)^2 = (\Delta t)^ - (\Delta x)^2 = (\Delta t')^2 - (\Delta x')^2\ ,$$ where $x, t$ refer to spacetime coordinates in the Earth's frame of ...


4

"Simultaneously" is a comparison which requires two events and one reference frame. If event $A$ is simultaneous with event $B$ according to reference frame $S$ then that means that the time coordinate of $A$ is the same as the time coordinate of $B$ both according to $S$. In principle, you could set $B=A$ and then have one event and one frame, but ...


3

Without a frame of reference it would be difficult to say which is moving and which is not. Not merely difficult, it would be impossible. “Moving” means that the velocity is non-zero, and all velocities are relative to some reference frame. So without a reference frame the word “moving” is meaningless. So which one would be aging faster and which slower? ...


2

The two questions are related. First, proper time is defined as the spacetime interval between timelike separated events. $$d\tau = \sqrt{-ds^2}$$ If the interval is timelike, then $ds^2<0$. This means $-ds^2$ is positive and the proper time is a real number. The spacetime interval is $$ds^2 = d\vec{s}\cdot d\vec{s} = -dt^2 + dx^2 + dy^2 +dz^2$$ If we ...


2

Light always travels at c in every reference frame - this is one of the core tenets of special relativity. If the lamp is turned on at halfway between A and B, they will both receive the light at the same time as seen by a static observer. There is no way once the light is turned on to travel by vehicle and deliver light faster than the leading edge of the ...


2

As all motion is relative, I assume you mean the astronaut is not moving relative to the Earth. Gravity and relative motion can affect time dilation. So time will run a little slower on the surface of Earth, due to its gravity, than it will for the astronaut further from the surface.


2

An observer is, by definition, at rest in their own frame. So what you are asking for is a test that would detect if an observer were moving relative to an external “absolute” frame, usually called the aether frame. The direct experimental signature of that is anisotropy in the speed of light. See this list of experimental evidence for SR: http://www.edu-...


2

In the light diagram there is only one point such that an event that happens in that point happens in the "present" moment for both the frames (in the sense that I interpreted from your question), and it's the origin. And this means that if we define as the event $A$ the one in which "one hand of one observer from the blue frame touches one ...


1

First: ditch A and B. Their names are now Unprimed and Primed. So Unprime is on the platform in frame $S$, and Prime is on the in-bound train in frame $S'$, moving at $\beta=\sqrt{3}/2$ ($\gamma = \frac 1 2$) in the $x$ direction. "Unprime" has two events: starting a stop watch at: $$ E_0 = (t_0=0, x_0=0) $$ and stopping it $T=32$ seconds latter at:...


1

If $A$ and $B$ (say, two boxes) are moving relative to each other at constant velocity, the time dilation that $A$ perceives for $B$ is the same as the one that $B$ perceives for $A$ : the situation is symmetric. I don't think this has anything to do with what's inside $A$ or $B$.


1

It's most important to focus on question 2. I'll work with $c=1$. Time dilation is computed as $\sqrt{ds^2/dt^2}$ on suitable assumptions of $dx^i/dt$. Special relativity in Minkowski space gives time dilation due to motion, say with Cartesian coordinates $x^i$ satisfying $dx^i=\beta^i dt$; using this in $ds^2=dt^2-dx^2$, $\sqrt{ds/dt}=1/\gamma$. In general ...


1

Your reasoning is true. If an observer on, for example, earth were to measure the time passing on the spaceship, they would find that the time that passed for the astronaut is $$t'=t\sqrt{1-\frac{v^2}{c^2}}=\frac{t}{\gamma}$$ where $t'$ is the time elapsed on the spaceship moving with relative velocity $v$ and $t$ is the time that has passed on earth. One ...


1

To simply answear your question. Yes. (Nice name btw) For more details it seems that you want to find some absolute reference. Which is precisely what special relativity is breaking. Everything is relative. In your scenario for instance the first sentence is wrong "Clock A is stationary and clock B is moving fast" Nothing is stationary nor moving. ...


1

Call one rod 1 and the other 2. The rods are identical so in their respective rest frames they have the same length: $L_1=L_2=L$. Let's forget about $V$ for a second and define the speed of rod 1 in the observers frame $u_1$ and the speed of rod 2 as $u_2$. In this frame the rods appear contracted: $$L_i=\frac{L}{\gamma_{u_i}}=L\sqrt{1-\frac{u_i^2}{c^2}}$$ ...


1

It is one of the postulates of special relativity that the speed of light $c$ is the same for all non-inertial observers. Being a postulate, this means there is no known cause for it - we know that $c$ is constant, but we don't know why (For some mor informtion on this, you may read Why and how is the speed of light in vacuum constant, i.e., independent of ...


1

I don't think the answer to your question is any different from the answer to the normal ladder paradox. Your question says: "The beam is projected in such a way that all colours of the spectrum reach the path of the rod at the same time from the perspective of a stationary observer in the room." But that doesn't mean that all colours will reach ...


1

I think this answer will disappoint you since it doesn't provide much insight into the other question, but with respect to the rest frame of the rod, the light arrives at an oblique angle due to aberration. The cross-sectional width of the rod along the path of the light is narrower than the light beam, so the red and violet light will make it past. It will ...


1

Spacetime diagram showing how faster than light travel would allow traveling back in time Since that would very directly break causality, this is a great argument as to why you can't do it. So I tried to make a diagram to clarify to myself more precisely how that would go like: SVG source. Legend: points: A: Spaceship departs from Earth moving towards C ...


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