Let us assume that a ship is traveling relative to earth at the speed close to C. My understanding is that according to special relativity it should be impossible to tell whether a ship is moving and earth is stationary or vice verse (all laws of physics should behave exactly the same in both cases). However, if we take the frame of reference of earth the atomic clock on the ship would be ticking slower (due to time dilation) -&gt; clocks on earth would appear to run faster for observer on the ship. Shouldn't it be the reverse if the ship was stationary and earth was moving? Why does it not contradict the principle of relativity?
66800
There is a problem with my logic and I cannot seem to point out where. There's a rocket ship travelling at close-to-c speed v without any acceleration (hypothetically), and there is an observer AA with a clock A on Earth, and there's another observer on the rocket BB with a clock B and these two clocks were initially in sync when the rocket was at rest with a FoR (frame of reference) attached to Earth. Now, this rocket's moving and AA tells us that B is running slower than A by a factor of$$ \gamma = 1 /(1-v^2/c^2)^{1/2} $$ $$ t_A/t_B = \gamma $$ where v is the relative 1d- velocity between the two i.e the earth and the rocket! That would mean the time elapsed on A is greater than that on B but this will happen only in the FoR of AA? So $t_B$ in this equation must be the time on B as observed by AA? Is this correct? What do the terms mean in the equations? If the symmetry holds and BB doesn't accelerate, then BB could say that $$ t_B/t_A = \gamma $$ right? where $t_B$ and $t_A$ are the times on B and A with respect to FoR of BB? But I was solving this problem and I took the earth FoR of A, but the prof took the rocket FoR of B? Like how will I know which FoR to solve the problem from? It'd greatly help if the terms in all the above equations were laid down neatly! DO we even need these FoRs?? Because in all the solved problems the prof is't specifying any and is using random ones! Please help!!!! This is the question where I messed up. The first rocket bound for Alpha Centauri leaves Earth at a velocity (3/5)c. To commemorate the ten year anniversary of the launch, the nations of Earth hold a grand celebration in which they shoot a powerful laser, shaped like a peace sign, toward the ship.  According to Earth clocks, how long after the launch (of the rocket) does the rocket crew first see the celebratory laser light?  This must be 25 years. My reasoning is: If v = 3c/5 10v + vt = ct where t is time taken by the light to reach the rocket from Earth as calculated from Earth.. and I solved that for t. and added 10 years to that because the time starts at the launch of the rocket!  According to clocks on the rocket, how long after the launch does the rocket crew first see the celebratory laser light?  This is 20 years. Here, I say: If it takes 25 years as observed by clocks on earth for the laser to reach the rocket, what should be the corresponding time as seen on a clock on the rocket? Using the formula: 25 = $\gamma$t where $\gamma$= 5/4 solved for t!  According to the rocket crew, how many years had elapsed on the rocket's clocks when the nations of Earth held the celebration? That is, based on the rocket crews' post-processing to determine when the events responsible for their observations took place, how many years have passed on the rocket's clocks when the nations of Earth hold the celebration?  For this, I did the following: 10 years on earth = T years on rocket ship where T must be lesser than 10 as observed from Earth FoR! Therefore, T= 4(10)/5 years = 8 years! But, prof says, 10 years on Earth = T years in rocket ship where T must be GREATER than 10 as observed from the rocket FoR??? Therefore, T = 10(5/4) years = 12.5 years!! What does this question actually want?
58607
Relativity has always been explained to me (in books I've read, etc) as a superset of newton's laws - that is; it encapsulates all of Newton's mechanics in addition to other effects (observer effect, time dilation, space-time geometry, etc).   I can kind of imagine 2 of Newton's 3 laws of motion to be incapsulated within Einstein's Special Relativity, but the one about "for every action there is an equal and opposite reaction", I'm struggling to find a place for that within Special Relativity.   Does this sit outside Special Relativity's explanatory power or is it inferred within somehow?
3678
Let's say I fire a bus through space at (almost) the speed of light in vacuum. If I'm inside the bus (sitting on the back seat) and I run up the aisle of the bus toward the front, does that mean I'm traveling faster than the speed of light? (Relative to Earth that I just took off from.)
79363
I'm in trouble with time dilation: Suppose that there's two people on the Earth (A,B), they are twins and each other has a clock. (So they are at the same reference frame). B travels in a spaceship and is orbiting around the Earth, as B's speed has increased there's some time dilation. So it's supposed that if A "looks" to B, B's clock will run slower than A's, and viceversa. And if one day B decides come back to the Earth, he will be younger than his twin and the hour of his clock would be different (earlier than A's clock) because of time dilation.  My first question is, why each person see the other's clock as the slow one? (In fact, I found some information that may solved it but I'm not sure if it's the right answer, and if it is, I don't understand it. -> It's because everyone thinks about himself as the reference frame. In other words, that for B, he is the stationary and who moves is A, so he sees A's clock run slower as B is "stopped" and the same for A. B is who is moving and A the stationary.)  The other question I have is that if it's true that each one sees the other as the slow one there's something that i missed out. We all agree that B (after going to the space and came back to the Earth) he is younger than his brother. So, why B don't see A moving and aging very fast if A see his brother and his things going very slow? Let's imagine they were looking to each other all the time, if both of them see the other going slowly once they meet and "discover" B is the young one... how is that possible?  [I think that in one episode of Cosmos of Carl Sagan (though may be I'm mistaken) he said that a neutrino "borned" in the Big Bang could have seen the creation of the Universe until today in few seconds due to its high speed, here starts my doubts: or i misunderstood something or there's contradictory information]  (I don't know anything about physics, only what is taught at high school as I'm a teenager so it'd be better to answer with no kind of calculation as I'd not understand it.)
91613
I'm not familiar with any complicated physics equation, however I do understand some basics. Suppose there is two objects, both of them are moving away from each other in a 3-dimensional space, which they both have the speed of half the speed of light ($c/2$). Relative to the reference frame of object A, object B would be moving away from itself equal to the speed of light.     A          B &lt;-----      -----&gt;   -c/2       c/2   to a 3rd observer    0          c    to A   -c          0    to B   This seems to break the law of "nothing can move faster than the speed of light". I believe that there is some other explanation for this case. I have done some research, and there are quite a few questions on this topic. Take  and  for example. However, they are about adding velocities together, which isn't quite the same as in the case I was describing.   is also similar to my case, however the two objects described are both moving toward the same direction, and I'm not shooting any beam or laser toward the other object.  Most probably in reality there are some extremely complex laws and equations which makes this question more complicated. However, the two objects are not interacting with each other and I'm not trying to "detect" the speed of the other object in the reference frame of A or B. Also none of the objects are moving at a significant fraction of the speed of light relative to the third observer, so special relativity doesn't apply(?) I just could not think of any reason why wouldn't B be moving away from A in the speed faster than light.  A simple explanation would be great.
84572
Alright, so David Griffiths in his "Introduction to Electrodynamics" states that the Twin Paradox is not a paradox at all since the traveling twin returns to Earth. By returning to Earth, the twin had to reverse direction, thus undergoes acceleration, and therefore cannot claim to be a stationary observer.   However, what if the traveling twin simply Skypes the twin that is on Earth. The twin on earth will still appear older, which would make no sense since in that case the rocket can be seen as the stationary frame of reference while the Earth "travels" at a speed close to the speed of light. No acceleration is undergone, yet the paradox remains.   Is Griffiths just completely glossing over important nuance again?
43477
In the first paragraph of Wikipedia's article on , it states one of the assumptions of special relativity is     the laws of physics are invariant (i.e., identical) in all inertial systems (non-accelerating frames of reference)   What does this mean? I have seen this phrase several times, but it seems very vague. Unlike saying the speed of light is constant, this phrase doesn't specify what laws are invariant or even what it means to be invariant/identical.   My Question   Can someone clarify the meaning  of this statement?   (I obviously know what an inertial frame is)
26814
I would like to know what happens with time dilation (relative to surface) at earth's center .  There is a way to calculate it?  Is time going faster at center of earth?   I've made other questions about this matter and the answers refers to:  $\Delta\Phi$ (difference in Newtonian gravitational potential between the locations) as directly related, but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it.  Any clues? Thanks
62157
The usual simple statement of geodesics is that when moving from point $(x_o,y_o,z_o,t_o)$ to point $(x_1,y_1,z_1,t_1)$ the object will follow the path of greatest proper time experience, i.e., a clock traveling the geodesic will record more seconds than one going another path.  I believe that is true even if you allow rocket thrusters or other energy addition devices. But what if you consider a circular orbit which, after 1 year, results in going from $(0,0,0,0)$ to $(0,0,0,1)$?  Wouldn't the proper time be maximized by simply sitting still, i.e., traveling only through time and not space?  The orbiting clock would presumably run slower just as the orbiting clock in the international space station runs slower than one on Earth.  In fact, if you want a physical model, assume a clock on a 400 km tall flag pole synchronized with the ISS during a pass-by (the ISS orbits at 400 km above the planet surface).  One orbit later, wouldn't the ISS clock record less time than the flag pole clock? Is that a violation of the concept &quot;orbits are geodesics, for which the shortest distance between two 4D points is the path of greatest time&quot;?
64486
Consider the scenario where you measure the time it takes for light to travel to the left 10 meters and to the right 10 meters. Both measurements will take the same time, even though we are moving through space at crazy speeds. This must mean that light is not moving relative to "space" as a whole. What does it move relative to? The light emitter? If so, try shooting two beams of light 10 meters from the wall. The first time the emitter is stationary; the second time it is moving at 100 m/s. Am I mistaken in thinking that it would hit the wall slightly faster? Wouldn't this light be moving faster than the light emitted from the stationary source?
92097
In the book The Elegant Universe, Greene describes a situation in which there are two space travelers, George and Gracie, moving in relationship towards each other at a constant velocity with no other available vantage points. Both George and Gracie have an equal right to declare that they are stationary and that the other is moving.From George's perspective, he is stationary while Gracie is moving and he observes (consequently? not sure this is direct consequence) her clock to tick slower than his. However, doesn't George also have the right to claim/perceive himself as the one who is in motion, and if so wouldn't he perceive Gracie's clock to tick faster than his? This seems to contradict the first perception. How would contradicting perceptions be available to the same observer? I assume I'm off somewhere.
56665
My question is not about why speed of light has a particular constant value which has been sufficiently addressed in other questions on SE already. I want to know whether the fact that speed of light has to be constant is a consequence of other fundamental laws of Universe, or is it just a fundamental law itself.  My motivation for asking this question is thinking about Special Theory of Relativity. One way of understanding why time dilation and length contraction happen is to look at it from the perspective of constancy of speed of light. Alice who is on the moving source of light and Bob, a fixed observer see the speed of light as constant. In order for Alice to see the speed of light as a constant value, time slows down and length expands (in forward direction) thus keeping the speed constant.   In the  I'm doing, the Professor tells that this constancy of speed of light could be more generally referred to as, "Principle of wave constancy". That is, the speed of any wave is a property of the medium (and thus it remains constant). So, the speed of a wave in water, for example, remains constant even if the source of the waves (say a paddle keeps moving). Frequency and wavelength may change but the speed remains constant. In a similar fashion, light is an electromagnetic wave and thus its speed is constant. And the constant speed is defined by the medium (which happens to be vaccum i.e., no medium for light).   My questions are twofold now -    Is the above reasoning correct to explain why speed of light is constant? (i.e., thinking of it as constancy of any wave in general) Theoretically, could there be other waves that could travel through vaccum at a different speed? If yes, time dilation and length contraction will have to be different.   I'd appreciate it if the answer is kept non-technical (essential math is okay). More importantly, I request you to try and explain without using General Theory of Relativity. Because I'm thinking how the scientific understanding by 1905 had already knew that speed of light has to be constant.
50359
This is probably an obvious question, but I don't see it answered anywhere.  Imagine we have an object in the Universe that is traveling, relative to the Milkyway at about .99999999999999999999999999c and I realize that's not possible, but imagine.  Speed is relative, so to the object, the Milkyway is flying past it and to us, the object is flying past us.  That's all good and relative.  But the photons that our superfast object encounters are gonzo blue-shifted, and relativity says that's OK too.   But lets say the object is traveling so fast, the photons it encounters are energetic enough to spontaneously undergo pair production.  That seems difficult to explain.  A relatively low energy visible light photon to our perception won't split into an Electron and a Positron, but a very energetic photon can, but it's the same photon just observed by 2 different reference frames?    I understand that observers won't agree on when things happen or on energy but the two observers should agree on events that take place and pair production for one but not the other seems impossible.     Is there a simple layman's way that this can be explained?
49355
I've very recently started to try to understand special relativity. I'm want to get a decent understanding of the twin paradox. I'll post what I've done so far and highlight what's gone wrong for me.  The situation is that Alice and Bob both begin at point $x_1$ in Alice's coordinate system $(x,t)$ (we have orientated the axis so that y and z will not matter). Bob then instantaneously moves off with velocity $v$ in the positive $x$ direction. Bob's coordinate system is now $(x',t')=(\gamma (x-vt), \gamma (t-v\frac{x}{c^2}))$. In Alice's coordinates, Bob reaches point $x_2$ after having moved distance $d$, then instantly turns around and travels back to Alice at $-v$. I want to show that Alice's space time interval $\Delta s_A^2$ is greater than Bob's $\Delta s_B^2$, as each path's space time interval is proportional to the proper time passed along that path.    According to Alice: $\Delta s_A^2=-c^2\Delta t^2 + \Delta x_A^2$ where $\Delta t=\dfrac{2d}{v}$ and $\Delta x_A=0$ as she didn't move, but the time elapsed is the time it took bob to travel the distance and then come back. Thus $\Delta s_A^2=-\dfrac{4c^2d^2}{v^2}$  Still according to Alice, Bob's $\Delta s_B^2=-c^2\Delta t^2+\Delta x_B^2$, where the change in time is the same, but now bob has moved distance $d$ twice, thus $\Delta x_B=2d$ ($\Delta x$ refers to total distance traveled rather than displacement which is 0 in this case). Now we have $\Delta s_B^2=-\dfrac{4c^2d^2}{v^2}+4d^2$.     The |size| of Bob's spacetime interval now definitely smaller than Alice's, and this would be all OK, except that when I do the calculations in Bob's frame, they don't agree. This is contradictory to the fact that $\Delta s^2$ is conserved under Lorentz transformations.    According to Bob: $\Delta {s'}_A^2=-c^2\Delta t'^2 + \Delta {x'}_A^2$ where $\Delta t'=\dfrac{2d'}{v}$. I'm not sure if I'm right in saying that $v$, the relative velocity of the two frames, is the only velocity upon which both Alice and Bob will agree on, other than the speed of light. Anyway, $d=\gamma d'$ as lengths contract by $\gamma$ i.e. $d'$ is smaller than $d$ by factor $\gamma$. Also, $\Delta {x'}_A=2d'$, so we have $\Delta {s'}_A^2=-\dfrac{4c^2d'^2}{v^2}+4d'^2=-\dfrac{4c^2d^2}{\gamma ^2 v^2}+\dfrac{4d^2}{\gamma ^2}=4d^2\dfrac{v^2-c^2}{\gamma^2 v^2}\neq \Delta s_A^2$  Although I haven't included them here, my calculations for $\Delta {s'}_B^2$ and $\Delta s_B^2$ agree. Sorry for the ultra-long post, but any help would we well appreciated!
15302
My understanding of Special and General Relativity comes from books which attempt to explain them to non-experts in these fields.   Let A and B be observers, which at some time t(0), are together at some location P fixed to the earth's surface. A and B are provided with "clocks", which at time t(0), are identical lumps of the same radioactive element-that has a very long half-life. A and A's "clock" remain fixed to the earth's surface. B and B's "clock" leave A and stay together, but eventually return to A after some interval of time T (as measured by A's "clock"). If, during this time T, B's velocity relative to A is sufficiently close to the velocity of light or B remains close to a sufficiently massive body, it seems that time can "pass more slowly" for B than for A. So that when B rejoins A, B's "clock" will contain more of the radioactive element than A's "clock"   My question is: Are there any scenarios that B together with B"s "clock" can undergo (during the time T) which will cause time to "pass more quickly" for B than for A? So that when B rejoins A, B's "clock" will contain less of the radioactive element than A's "clock". If there are no such scenarios, is there any fact or law in Relativity theory which rules them out?
26361
This question is mainly inspired after watching the movie known as Interstellar We knew that for time dilation caused by relativistic motion between A and B. A will measure B's clocks slowing down, and B will measure A's clock slowing down by the same rate, while they both measure their own clocks ticking normally In interstellar, there's a planet that experience severe gravitational time dilation relative to earth's frame due to being in close proximity to the black hole Gargantula, therefore for every hour spent there measured by the astronaut's frame, 7 years will have passed in Earth's frame Thus this means to Earth's frame, they will measure the astronaut's clocks to be ticking slower relative to them  But what about earth's clock as measured by the astronauts? Will they measure Earth's to be ticking faster than theirs (at the same magnitude it is slowed in earth's frame measuring theirs)? Because if they both measured the other's clocks as running slower, then how could the twin paradox like aging difference be possible?
71213
The common explanation/trope for the equivalence principle always has something to do with you being inside an elevator or spaceship, and your supposed inability to differentiate, say, gravity from a rocket, or free-fall from being motionless in empty space. For example, Wikipedia  "being at rest on the surface of the Earth is equivalent to being inside a spaceship (far from any sources of gravity) that is being accelerated by its engines." Many other sources say similar things, often making broad, sweeping generalizations like "there is no measurement you can make to distinguish the two scenarios."  Isn't that just plainly wrong? Gravitational fields aren't homogeneous. Here on the Earth, a clock on the floor runs more slowly than a clock on the table, and we have clocks precise enough to measure such small differences due to the gravitational gradient. But doesn't a clock in an accelerating spaceship run at the same rate no matter where in the ship you put it?  I searched the top 50 or so questions about the equivalence principle and found  that talks about tidal effects in gravitational fields, but the explanation is very confusing. As far as I can tell, it seems to be saying the "attraction" between the particles that arises as a result of slightly different gravity vectors is somehow equivalent to the actual mutual gravitational attraction the particles should have absent the external gravitation field. I don't see how that could possibly be the case.  Also in that answer is a link to  of tidal forces and the equivalence principle, which seems to be saying the opposite: that tidal forces are, in fact, a distinguishing characteristic between gravitational acceleration and rocket acceleration, and that the equivalence principle only exactly applies to small enough points in space over a small enough duration, where tidal effects are negligible.  I know what seems most correct to me, but as I'm not an expert in this field, would an actual expert please shed some light on these varying views of the equivalence principle?