We know satellite is moving , so there should be time dilation due to motion, also we know due to Earth's Gravitational field there should be time dilation, what should I do, just calculate both of them and add?
22562
In the Wikipedia article on  I see two formulae for the gravitational time dilation. One contains $\sqrt{1 - r_0/r}$ but the other $\sqrt{1 - 1.5 r_0/r}$. What formula should I use to calculate the time dilation due to gravity inside a satellite? The speed induced time dilation is of course just special relativity.  With the first of these two formulas I can perfectly reproduce the graphs on time dilation as a function of $r$,  I find that the two contributions to time dilation cancel at 1.5 Earth radii, all numbers are OK... But why not the second formula?
123195
I'm reading about the twin paradox in special relativity - if there are two identical twins, one of whom who sets off in a high speed rocket to a planet, and then heads back, will find the twin who remained on Earth to have aged comparatively more due to time dilation. However, from the perspective of the travelling twin, it was the Earth that was moving and so the Earth twin should be the younger one. The 'paradox' can be resolved by realising that the travelling twin does not remain in an inertial frame for all time, since he has to turn around at the planet and hence accelerate.   Now, there are three possible curvatures the universe could have - positive, negative, or flat. As far as I'm aware, we're quite sure it's flat. However, if it was positive, does this mean that the travelling twin, if he travelled long enough, would loop back to Earth eventually? And therefore he would be able to return to Earth without having changed inertial reference frame, and then we would actually have a paradox?  I'm thinking perhaps that travelling through curved space itself constitutes acceleration, since in a way you're changing direction? In which case the travelling twin wouldn't remain in an inertial reference frame and we could solve the 'paradox' as before?
89736
Consider this scenario. I enter a space ship at 10:00 AM, it starts to move at a speed 100 times faster than  light, I travel for 55 mins. And then slow down to very much less than speed of light and land on another planet at 11:00 AM. Now I look at Earth with a telescope. What period of Time am I going to see?
16681
Time tends to slow down near objects with large amounts of positive mass relative to observers in micro gravity.  Considering that negative mass is the opposite of normal mass and would time tend to speed up near a body of negative mass relative to observers in micro gravity?
3206
If I'm standing at the equator, jump, and land 1 second later, the  Earth does NOT move 1000mph (or .28 miles per second) relative to me,  since my velocity while jumping is also 1000mph.   However, the Earth is moving in a circle (albeit a very large one),  while I, while jumping, am moving in a straight line.   How much do I move relative to my starting point because of this? I  realize it will be a miniscule amount, and not noticeable in practise,  but I'd be interested in the theoretical answer.
105645
I have seen the classical twin paradox before. It uses a twin stationary on Earth and the other traveling away and back. I have seen many contradictory solutions for it, some use general relativity, others use special relativity, either way, I am still troubled by it. They always try to break the symmetry through the traveling twin's acceleration and deceleration, but never quite succeed.  So, let's do away with the classical twin paradox and let's explain a much simpler, perfectly symmetrical version of it where both twins are moving towards each other.  So imagine we have Twin A in a spaceship, and Twin B in another, a light year apart from each other at the beginning of the experiment. They both start traveling at the same speed towards each other when the first light of one twin since the beginning of the experiment reaches the other, so they effectively start moving one year into the experiment.  If I understand relativity properly:   From Twin A's frame of reference, he's stationary and Twin B is moving at a constant speed towards him, therefore, because of time dilation, Twin B's clock is ticking slower. From Twin B's frame of reference, he's stationary and Twin A is moving at a constant speed towards him, therefore, because of time dilation, Twin A's clock is ticking slower.   Regardless of what their observations might be because of the Kepler effect and what not, time dilation dictates that a moving clock will absolutely tick slower than a stationary one. So, because Twin B is moving relative to Twin A, Twin B's clock is absolutely ticking slower than Twin A's. The same is supposed to be true the other way around. This is obviously a contradiction.
80619
I've looked up  and on Griffths' didatic Introduction to Electrodynamics, but, still, am struggling to understand why time has the opposite sign of the space coordinates. Any insights are welcome.
97250
My question has a few parts concerning the speed of light in general relativity.  Firstly, time changes in response to gravity and speed. Therefore, as gravity effects time in an area of space, should the speed of light "change?" When I say "change" I understand that light's speed is finite but wouldn't the length of second change and then change light's speed in meters per second?  Secondly, does the speed of light actually change with gravity? Time stops at the speed of light but does that mean light cannot experience time? Can light experience slower time while travelling through warped space time?
88440
I'm trying to derive the relativistic velocity addition equation using the time dilation equations but I get a wrong result. Assume the widely used scenario in which there are 3 clocks: O, A and B. A is moving relative to O at velocity $v$. B is moving relative to A at velocity $w$. Find the velocity of B relative to O (in O's frame of reference). The time dilation equations says:  $$t_A = t_O \sqrt{1-v^2/c^2}$$  $$t_B = t_A \sqrt{1-w^2/c^2}$$  $$t_B = t_O \sqrt{1-u^2/c^2}$$  From these 3 equations we want to derive $u$. Then we get:  $$t_O \sqrt{1-u^2/c^2} = t_O \sqrt{1-v^2/c^2} \sqrt{1-w^2/c^2}$$  $$1-u^2/c^2 = (1-v^2/c^2) (1-w^2/c^2)$$  $$1-u^2/c^2 = 1-v^2/c^2-w^2/c^2+v^2 w^2/c^4$$  $$u^2 = v^2 + w^2 - v^2 w^2/c^2$$  This is different from the equation  $$u = \frac{v + w}{1+v w/c^2}$$  Why doesn't the derivation of velocity using the time dilation equations alone work?
9305
I know this question has been asked in other forms, generally regarding the balance of forces.  This time I want to focus on motion.  I've got a laser accelerometer on my desk.  It tells me that I'm accelerating at $9.8~\rm m/s^2$.  For the first experiment I'm travelling in space.  I pick a nearby star and discover that I move about $490$ meters in $10$ seconds from that star.  For the next experiment I'm on the surface of Earth.  I measure the same acceleration with my laser accelerometer.  I pick a spot (the center of the Earth) and discover I don't move at all in $10$ seconds.  How is acceleration without motion possible?
58833
Can you resolve this paradox: Time and space are supposed to be treated symmetrically in SR, yet time dilates while length contracts. I've been working on this for weeks, and I can't resolve it.  The mirror experiment shows that time and space both expand for the moving observer, in contradiction to the results one derives from the LT, yet a well-known text, Eisberg's Fundamentals of Modern Physics (pp. 20-21) incorrectly concludes the usual asymmetrical results from that, by changing the definition of time dilation! He defines it as meaning that time measured in the moving frame is longer, when tau is actually well-known to be shorter. Time dilation refers to the mover's time UNITS LOOKING longer to the stationary observer because the total number of time units is FEWER, the opposite situation from the mirror experiment.
44177
If I were to stand by a pyramid, which weighs about 20 million tons, I would slow down by a trillion million million million of second. Don't know if that's exactly right, but you get the point. Also, if you went into a black hole, time would stop at the event horizon (as measured by a clock of a distant outside observer). Why is this?
61731
No matter how fast you go, you will aways perceive the speed of light as constant. Taking that as a fact, the special relativity theory was formulated. Now, for what I understand about general relativity, just by standing here on the surface of the planet under the effect of gravity, my frame of reference is a little different from someone in geostationary orbit, and time goes by a little slower for me.  So, considering me and someone in geostationary orbit, does both of us would perceive light at the same speed? If so, does that means that the light is slower down here than up there?
43945
In curved spacetime particles follow timelike geodesics, which should have maximal proper time (at least locally). I thought this path usually corresponds to a global maximum, and there are only strange exceptions. Propably I was wrong, or I really don't understand something, because it doesn't seem to be true for the case of orbitals in Schwarzschild geometry.  The Schwarzschild metric is (with +--- signature)  $$ g_{\mu\nu} = \textrm{diag}\left(1-\frac{2GM}{r},-\left(1-\frac{2GM}{r}\right)^{-1},-r^2,-r^2\sin^2\theta\right) $$  An observer with fixed $r,\theta,\phi$ coordinates will age according to $d\tau_1 = \sqrt{g_{tt}}dt$. An observer, which orbits the center at the same constant radius with $\theta=\pi/2$ will age according to $d\tau_2 = \sqrt{g_{tt}-\omega^2r^2\sin^2\theta}dt$, so he ages slower during the same amount of coordinate time $dt$. If the two observers start from the same point, they will meet again after a full rotation of the orbiting observer, and the orbiting observer will age less. What I don't understand is that the orbit is the geodesic (if the angular velocity is set properly), so why didn't he age more? He should have been on the path with maximal proper time.
91149
What is the time dilation in Mars, compared to earth? Can we accurately calculate it? What information is needed to do these calculations?
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.)
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?