In general, let us have a light emitting point A and two light absorbing points B and C, such that the three points fall on a straight line and B is somewhere between A and C. For simplicity, let us consider all photons emitting from A to travel along this same line in the direction of the points B and C. Also, let the distance from A to B be AB and the distance from A to C be AC.   For the photon, special relativity states that there should be maximum space contraction along it's direction of motion. That means that, for the photon, A, B and C are all at the same point in space. This results in the photon effectively "reaching" B and C at the same time. How then does the photon "decide" that it must be absorbed by point B and not C, so as to be consistent with what an outside observer would see (point B "casting a shadow" on point C)?
124025
Disclaimer, not a physicist. When I look at the sky I can see the star Rho Cassiopeiae. In my frame of reference, the photon hitting my retina has traveled for 4000 years and 3.78x10^16 km. In the photon's frame of reference, there's no time elapsed and my retina and the surface of the star are at the same place (no distance traveled). Is that a correct understanding of time dilatation and space compression at relativistic speed?
113155
This is related to  and  The idea is to shine a beam of light from a clock towards a heavy mass, like a black hole, in some manner such that the curvature of spacetime around the mass is sufficient to &quot;bend&quot; the light directly back to the clock. That is, to send the light along a self-intersecting geodesic using a sufficiently curved spacetime. Note that in this case the source and the detector are the same object, so no clock synchronization is required. Hypothetically, this should measure the one-way speed of light. None of the given refutations from the similar posts on this site are convincing. Here are some of the main refutations, as I understand them, and my counter to them: (1) In being curved by spacetime, the light is actually changing direction, hence this is not the one-way speed. To my understanding of General Relativity, the light travels along a geodesic; that is, a curve such that, ignoring technicalities of Differential Geometry, its tangent direction is constant. There is no reflection or acceleration. So in this sense, how can the light with a constant tangent direction be said to change direction? One might say that the path deviates from a straight line in Euclidean space, but given that spacetime is not Euclidean, has no , and has no global notion of direction, this is the only definition of straight or &quot;one direction&quot; that makes sense. Thus, either the light does not change direction and is &quot;one-way,&quot; or we need to be clearer about what exactly we mean by &quot;one-way&quot; in the context of General Relativity. (2) Such an experiment would be impossible or impractical. I think some posters were saying this because the original questions implied that the clock would have to be at the event horizon of a black hole, which is of course impossible or impractical; however, I don't think such an extreme closeness to the black hole is required. Given that the  of a black hole, where light orbits in a circle, is outside of the event horizon, shining the light at some distance outside of the photon sphere should be sufficient to curve the light back to the clock, regardless of the distance of the clock from the black hole (see my rough illustration). At the extreme, we could place the clock directly on the photon sphere, which should be theoretically possible. Regardless, the question is not about practicality, but whether the one-way speed is theoretically measurable.  Update: Essentially what this argument boils down to, I think, is that the average speed of light along a single self-intersecting geodesic may differ from the average speed of light along a non-geodesic self-intersecting path, which is always c. Put that way, however, it does seem a little ridiculous. It is hard to see how the anisotropy of the speed could be defined in such a way as to be consistent with this pathological case. Presumably, we could approximate a single self-intersecting geodesic with a non-geodesic path arbitrarily closely, so, logically, the average speeds over each path should be approximately the same, i.e., c. Furthermore, how would the &quot;preferred direction&quot; of the speed of light behave around a black hole, for instance? Regardless, I still think this experiment does actually solve the problem for the given notion of &quot;direction&quot;. Though the problem is usually posed in the flat spacetime of Special Relativity, with a global notion of direction, so the apparent unanswerability of this question should be seen as a limitation of Special Relativity only, which is then solved by General Relativity.
54513
As explained in the answers to  post, photons apparently exert a gravitational pull on other objects. It has also been explained on this site, that gravity propagates at the speed of light.  I'm wondering, though, how do you reconcile these two facts? I'm trying to imagine the gravitational field made by a photon and it seems like there are some paradoxes. For example, how can gravity propagate ahead of the photon at the speed of light, if the photon is also travelling at the speed of light? My guess is the solution is probably found in relativity, but I certainly can't figure it out.   Right now, the best I can do is to think about gravity as sound, and a photon as an object travelling at the speed of sound. Can anyone help me out?
21099
I know there have been similar questions but I'm still unclear what the overall consensus is. (1) I assumed and have read that photons travel in straight lines unless deflected by gravity but there are conflicting theories. (2) I have heard that single photons takes every possible path but that makes no sense. Why would a single photon traveling from here to the moon go to every other place in the universe along the way.  (3) I can understand a single photon traveling as a particle or packet of energy but I have a hard time understanding a single photon traveling as a wave. I have never understood any attempts to answer this.  (4) Does a single photon have a frequency and what causes the frequency? If not then how does it have energy? I have many questions about single photons and their frequency.
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!
56627
I am having a serious trouble finding an agreement between the two facts in Special Relativity:  (1) given two events $A$ and $B$ in spacetime separeted by a spacelike interval $s$, the distance in space between the two events will be the shortest in the reference frame which the two events occur simultaneously:   The invariance of the interval says $s^2=(\Delta x)^2-(c\Delta t)^2$ and $\Delta x$ is minimum when $\Delta t=0$  (2) the length contraction says $L=\dfrac{L_0}{\gamma}$ where $\gamma=(1-(v/c)^2)^{-1/2}$ , $L_0$ is the proper length of an object in a reference frame which the object is at rest and $L$ is the length of the object in a reference frame which the object moves with speed $v$ in the $x$ direction.  This says that the lenght of the object in the $x$ direction is the largest in the comoving reference frame of this object.  The question is: According to (1), the reference frame in which two events $A$ and $B$ occur simultaneously is the reference frame that gives the shortest distance in space between the two events. Let $A$ and $B$ be the two end points of a meter stick in a reference frame in which the stick is at rest and at time $t=0$. This two events occur simultaneously, so the distance in space between the two is the shortest acording to (1). But according to (2) shouldn't this distance be the largest?  What is wrong?
3247
From what I understand, the  states that there is a fundamental natural limit to how accurately we can measure velocity and momentum at the same time. It's not a limit on equipment but just a natural phenomenon.  However, isn't this just an observational limit? There is a definite velocity and momentum, we just don't know it. As in, we can only know so much about the universe, but the universe still has definite characteristics.  Considering this, how do a wide range of quantum mechanical phenomena work? For example, quantum tunneling - its based on the fact that the position of the object is indefinite. But the position is definite, we just don't know it definitely. Or the famous light slot experiment? The creation of more light slots due to uncertainty of the photon's positions?  What I am basically asking is why is a limit on the observer, affecting the phenomenon he is observing? Isn't that equivalent to saying  because we haven't seen Star X, it doesn't exist? It's limiting the definition of the universe to the limits of our observation!
107594
In his 1905 , Einstein derives the Lorentz transformation using the two postulates of SR;constancy of $c$ for all inertial frames and the Invariance of the laws of physics for all inertial frames.   I'll summarize his mathematical derivation and then ask one specific question about it.  So we consider two frames $(x,y,z,t)$ and $(\xi,η, ζ,\tau)$ in relative motion along the x-axis with velocity $v$, and we're interested in finding a spacetime transformation that relates thier coordinates.   We consider some arbitrary point $x'=x-vt$. This point is at rest in  $(\xi,η, ζ,\tau)$ since it's moving with $v$, Therefore this point has $x',y,z$ coordinates that is independent of time, in other words the distance between that point and the origin of $(\xi,η, ζ,\tau)$ is constant.    We consider the following scenario: hit a beam of light from the origin of $(\xi,η, ζ,\tau)$ at $\tau_0$ arriving at the point $x'$ at $\tau_1$ and then being reflected and arrive at the origin of $(\xi,η, ζ,\tau)$ at $\tau_2$.  So that we have: $1/2(\tau_0+\tau_2)=\tau_1$. Since $\tau$ is a function of $(x,y,z,t)$ we have:  $\dfrac{1}{2}[\tau(0,0,0,t)+\tau(0,0,0,t+\dfrac{x'}{c-v}+\dfrac{x'}{c+v})]=\tau(x',0,0,t+\dfrac{x'}{c-v})$  Assuming that $x'$ is infinitely small then taylor expanding this equation and approximating it to first order we get:  $\dfrac{\partial \tau}{\partial x'}+\dfrac{v}{c^2-v^2}\dfrac{\partial\tau}{\partial t}=0$  Solving it then we have:   $\tau=a(t-\dfrac{v}{c^2-v^2}x')$  where $a$ is some unkown function of $v$ (in fact $a=1$).  Finally consider a beam of light emitted from $(\xi,η, ζ,\tau)$ at the origin, it's $\xi$ coordinate is given by  $\xi=c\tau=ca(t-\dfrac{v}{c^2-v^2}x')$  It's given by $\dfrac{x'}{c-v}=t$ in $(x,y,z,t)$, plugging in for $t$ we get:  $\xi=a\dfrac{c^2}{c^2-v^2}x'$  He then states:      Substituting for $x'$ its value, we obtain   $\xi=a\dfrac{1}{\sqrt{1-v^2/c^2}}(x-vt)$   ...   My question is:  1) the three equations $\tau=a(t-\dfrac{v}{c^2-v^2}x')$ and $\xi=c\tau$ and $x'=x-vt$ when combined together gives :  $\xi=c\tau=ca(t-\dfrac{v}{c^2-v^2}x')=a\dfrac{c^2}{c^2-v^2}x'$, since $x'=x-vt$ by plugging in we get:  $\xi=a\dfrac{c^2}{c^2-v^2}(x-vt)=a\dfrac{1}{1-v^2/c^2}(x-vt)$ not $a\dfrac{1}{\sqrt{1-v^2/c^2}}(x-vt)$ .  But He says     Substituting for $x'$ its value, we obtain   $\xi=a\dfrac{1}{\sqrt{1-v^2/c^2}}(x-vt)$   ...   All these equations are copied from Einstein's original paper, So what is wrong with my calculations that does not make it match up with that of Einstein?
116695
I'm trying to wrap my head around how geodesics describe trajectories at the moment.  I get that for events to be causally connected, they must be connected by a timelike curve, so free objects must move along a timelike geodesic. And a timelike geodesic can be defined as a geodesic that lies within the light cone.  I want to know why exactly null geodesics define the light cone. Or, why null geodesics define the path of light.  Also, if there's a better explanation why matter follows timelike geodesics, that would also be welcome.
26571
I'm a newbie with physics but I'm wondering how a ray of light can essentially be represented. I have always known that a ray of light proceeds in a straight line until it encounters another object (or material) that refracts/reflects it. But a light ray should be part of an electro-magnetic wave, is this correct? If it is so, it should obey to the  and this doesn't seem to me to describe a straight line ray.  I'm having problems visualizing how light is emitted and how it relates with the wave equation. Can someone with a clear understanding of the problem explain it to me in simple terms?
95948
Often this is asked if it was pointed towards a black hole. so I thought, if a scientist decided to sacrifice himself and jumped in a black hole, released some dry-ice mist and pointed a laser pointer away from him, what would he see.   (The scientist is just to help you understand. Maybe this experiment is impossible because he'd get spaghettified first)  And the laser pointed is pointed directly away from a black hole, not slightly to the left or right, where I'd assume it would curve and fall back.
83328
From this ,     Let’s say Alice is holding a light clock, and Bob is watching her run by, while holding it, with speed V.  Alice is standing still (according to Alice), and the time,$ \tau$, between ticks is easy to figure out: it’s just $\tau = \frac{d}{C}$.  From Bob’s perspective the photon in the clock doesn’t just travel up and down, it must also travel sideways, to keep up with Alice. The additional sideways motion means that the photon has to cover a greater distance, and since it travels at a fixed speed (EEP y’all!) it must take more time   My question is, why does photon need to travel sideways to hit the Alice light clock?
118791
If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is  $$ v \oplus w = \frac{v+w}{1+vw} $$  As expected, $v \oplus 1 = 1$,  as "nothing can go faster than light".  Similarly, $v \oplus -1 = -1$. (same thing in the other direction)  But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.  So what would the ground observer see? Is this even a meaningful question?  Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.
24386
This is not a contradiction and I know it is impossible but still consider a thought experiment by me and point out if something is wrong. See the following picture and then the explanation follows.    Rest frames are easy to understand. They are just for clarification. Lets move on to the moving frames. The velocity of the box is $0.1\ c$. Now a photon is emitted (I am not taking a light ray to avoid complication in discussion). After emission the box also moves a certain distance ahead. So the photon takes more than one second to reach the wall. But even the source moves that much distance ahead. When the light is reflected back, the source and the wall again move forward as the box is moving. But after reflection the wall has no role to play. We are concerned about the source. The source moves a certain distance ahead and therefore light takes less than one second to reach back. But the difference of the first case and the second case is not the same. To explain, I will give some equations.  Moving frame 1:  Velocity of the box: $0.1\ c = 30,000km/s$  Time taken by light to reach the wall $=\frac{330,000\ \text{km}}{300,000\ \text{kms}^{-1}}= 1.1s$  Distance moved ahead by the box from original position: $30,000\ \text{km}$  Moving frame 2:  distance moved ahead by the box from original position  $&gt;30,000\ \text{km}$ as the box also moves ahead after reflection.  let $d$ be the distance moved ahead by the box after reflection.  Time taken by light to reach the source: $\frac{270,000\ \text{km}\ -\ d}{300,000\ \text{kms}^{-1}} &lt;0.9s$  Therefore we see that this could possibly determine the absolute motion.  So in total we see that the light will take lesser time to reach the source. Please correct me if I am wrong and give your opinion why is it so.  EDIT 2 for clarification  The box is the compartment in space in which you are moving and I am calculating the time the light takes to reach back the source. I am not even adding velocities. Please correct me and tell me where did I add velocities. For clarification just let me give a simple example. Suppose person A is standing still and B is running towards a light source. Obviously he will see the light before person A even though the speed would be the same. So, what I am doing over here is not increasing the speed of light, but instead decreasing the distance to be covered by it.
36330
I've been reading about different ideas of getting the observer out of quantum theory. One Sean Carroll seems to like is Everett's many worlds concept.  I had a thought that I need someone here to shoot down. It could be something as basic as the double slit experiment that proves me to be wrong. Here's my thought:  As I see it, we never really "measure" the position of something. We sense a photon that is emitted by what we are "looking" at. Why couldn't the wave function of a particle just break down at the moment it emits the photon at a specific point in space-time regardless of whether there is an observer or not. After the emission, the particle location would go back to being a probability function.  What am I missing?
103439
I've often heard that Einstein shattered the notion of absolute motion (i.e. all things move relative to one another) and that he established the speed of light as being absolute. That sounds paradoxical to me; I cannot understand how the two concepts can be reconciled.  Before going further, I'd like to say: 1) Over the years, I've seen many layman's explanations on these topics (including the nice YouTube video by Vsauce, ). I understand everything that's said (or, at least, I think I do). Just nothing I've found seems to address this apparent contradiction. 2) More recently, I've tried to find the answer on my own. That includes searching the posts on this site. Some come close (), but nothing I've been able to find seems address specifically what I'm asking.  Back to the question: Relativity shows us that there is no universal frame of reference by which to judge motion, so object A might be reckoned as moving at 10&nbsp;m/s relative to object B or as stationary relative to object C. This is fine for me. I can grasp that the universe has no intrinsic coordinate system, that we only think that way on Earth because we have the ground to move over.  Then there's the speed of light (in a vacuum). The speed of light is the ultimate "speed limit," it's often said. But if there is no universal frame of reference, how can there be any such speed? The very idea only make sense if there is a universal frame.   If one object is moving (uniformly) at 60%&nbsp;c and another object is also moving at 60%&nbsp;c, but in the exact opposite direct, then from the perspective of either one (if they could still see each other) the other would appear to violate that speed limit. All these spacetime bending consequences used to explain why nothing can move past this speed only seems to enshrine the concept that there is some ultimate speed standard. If there is only relative speed, then the concept of light having a specific speed in the vacuum should be a nonsensical one since it having speed (x m/s) would only make sense when measured against some other body.   Since I was very young, it has always sounded to me like motion is only mostly relative, that until you get close to the speed of light, the effects of an absolute frame of reference are negligible. Perhaps that there is an actual fabric of space which everything moves relative to, which is why there is something to expand between galaxies (faster than light can propagate) in the metric expansion of space. Growing up, I always thought this would just start makes sense with time. Now I'm up to a first year (college) level in physics, I even know basic calculus, yet I'm still hopelessly confused.  EDIT:  Thank you to whoever suggested this may be a duplicate of . It and others are very much related and at least partially answer my question. Unfortunately, explaining that distances become shorter and time becomes slower as a way to stop you from exceeding the speed of light does not explain how that speed is not an absolute.  By my reckoning, if all speed is relative, then no matter how fast you go light should always race away from you at the same apparent speed. I.e. there should be no speed limit. For there to be a speed which you cannot exceed or you would catch up (and make time irrelevant) requires the very concept of some external speed by which light can travel and nothing else can reach - thus my logical paradox continues unabated.
98760
This was explained to me many years ago, by a physics teacher, with the following analogy:  "If someone on the beach wants to reach someone else that is in the water, they will try to travel as much as they can on the beach and as little as possible on the water, because this way they will get there faster."  I'm paraphrasing of course, but this is as accurate as I recall it. This explanation makes no sense to me. Was he telling me the light knows where it is going? It wants to get there faster? It chooses a different direction? (No need to answer these questions, this was just me trying to understand the analogy.)  My attempts to clarify the issue were without success and many years later I still don't know.    Why does light change direction when it travels through glass?