# How does curved space-time produce acceleration?

1. Jul 11, 2013

### xcourrier

I apologize for my ignorance but I've looked around a decent bit and still dont quite understand. I understand its no longer a force as Newton described it, but then the bowling ball on a trampoline analogy is useless without a force such as gravity to accelerate objects in the metaphorical "down" on the trampoline.

I have also herd the term geodesics thrown around a bit an I can understand that traveling along one of these lines may simulate a force, but why is it assumed that we are traveling? What's causing the traveling?

I guess the simplest way I can think of describing my confusion is envisioning a dart. If you throw a dart through curved space-time, I can understand how its path can look curved in any number of directions. But what happens if you simply hold on to the dart? Isn't it then no longer traveling through space and only time? What makes it want to travel through space when I release it?

2. Jul 11, 2013

### WannabeNewton

If you hold the dart stationary in your hand, then it is actually accelerating because you are preventing its worldline from being a geodesic of the space-time (in general relativity an object in free fall is not accelerating in an absolute sense and an object not in free fall is accelerating in an absolute sense); worldlines of freely falling particle will naturally be geodesic paths until you apply an external force and end the free fall trajectory (in this case you are applying a constraint force in order to hold the dart in your hand so that it doesn't free fall to the ground); this is one of the tenets of general relativity. When you release the dart and it goes into free fall, its worldline will become a geodesic of the curved space-time background; consequently it will no longer be traveling in just the "time direction" (the geodesics will contain both "temporal" and "spatial" parts because of the space-time geometry associated with the system at hand).

Why geodesic worldlines for freely falling particles? Well consider a man in an elevator whose space-time scale of variations is much smaller than that of the space-time curvature so as to make any tidal gravity negligible (so that we are in effect locally in a uniform gravitational field). Say the man and the elevator are in free fall and the man has a ball in his hand and he lets go of the ball. Because of the equivalence principle, the ball and the elevator and the man will be freely falling at the same rate within this local uniform gravitational field so from the man's perspective he is just floating in place along with the ball and the elevator; assuming the elevator to be enclosed so that the man can only make local experiments regarding the physics inside the elevator, such as the ball drop, he cannot tell if he is in free fall or floating in free space (i.e. inertial). Thus we conclude that freely falling particles are locally inertial (inertial because he can conclude he is just floating in free space and locally because we are taking a space-time region small enough so that the gravitational field is uniform in that region). It turns out that mathematically, the curves in space-time which describe locally inertial particles are exactly the geodesics of the space-time.

Last edited: Jul 11, 2013
3. Jul 11, 2013

### xcourrier

Thanks for the reply, I read it over a few times but it still went a little over my head. The equivalence principle is pretty straight forward, and I get that you are applying a force to hold it up.

But since you are applying a force to hold it up, what is the dart accelerating through? It doesn't seem to be accelerating through time, and it seems stationary in space. Since there is no counter force does that mean space is coming down on it? That would mean that space is constantly "falling" (following a geodesic) towards the earth or any other mass in space, wouldn't it? So mass would be constantly consuming space at an accelerated rate? That doesn't seem to be the case or am I mistaken?

4. Jul 11, 2013

### WannabeNewton

The dart is accelerating precisely just to stay in place in the gravitational field :)! Consider a variation of the elevator thought experiment above: say we have a man in an elevator that is at rest in a uniform downwards gravitational field of magnitude $g$ (e.g. to a good approximation the surface of the Earth). He drops a ball and sees that it falls to the floor of the elevator at a rate given by $g$. Now consider the same man in an elevator that is accelerating upwards in free space with magnitude $g$. He again drops the ball and sees that it falls to the floor of the elevator at a rate given by $g$. He cannot distinguish between these two scenarios so being at rest in a uniform downwards gravitational field of magnitude $g$ is equivalent to accelerating upwards with magnitude $g$. In this sense, if you are holding a dart in your hand then it is at rest in the uniform gravitational field at the surface of the Earth so this is equivalent to it accelerating upwards; this second interpretation is what we use in general relativity after combing gravity with the curved background geometry of space-time.

5. Jul 11, 2013

### xcourrier

Right, the equivalence principle. But in that principle the person in the accelerating elevator is experiencing a force g because he is traveling at an accelerated rate through space. So, because of that principle, does that mean the person on the surface of the earth not moving is also traveling through space? Implying that space is "falling" towards earth?

6. Jul 11, 2013

### WannabeNewton

The point is that being at rest in a uniform gravitational field is equivalent to accelerating uniformly in free space. If we then translate this over to general relativity (where, in general, there is no Newtonian notion of a gravitational field since gravity is now a manifestation of the space-time geometry), then we conclude that if we wish to remain stationary in space we must accelerate to do so. So if we are stationary in space then no we are not moving through space (by definition of being stationary) but we must accelerate to stay in place, as explained above. I don't know what you mean by the statement "space is 'falling' towards Earth" because particles are what fall in space, not the other way around. Cheers.

Last edited: Jul 11, 2013
7. Jul 11, 2013

### xcourrier

Thanks for the reply, but this topic may just be beyond my present understanding.

To remain stationary in space (and, I suppose, maintain velocity in time) we must accelerate through space-time. I think I've hit the "I need to sleep on this a bit till it slowly filters in" point :(

The space falling towards earth thing was the only way I could visualize space "pushing" us onto the surface of the earth.

8. Jul 11, 2013

### WannabeNewton

Yessir! I think the argument involving the elevator is the most intuitive way of coming to terms with this, in my opinion. Put another way, we naturally want to be in free fall (our worldlines naturally want to be geodesics and geodesic worldlines are non-accelerating by definition) so if we want to obstruct this (obstruct free fall) then we must somehow accelerate, or be accelerated. For you and me managing to remain stationary on the ground, the normal force from the ground provides the acceleration needed in order to obstruct this natural free fall motion (and so we remain stationary on the ground). So in this sense, the ground pushing up on us is what keeps us stationary and obstructs free fall; in the framework of general relativity the ground pushing up on us and keeping us stationary is a form of acceleration (think again of the elevator).

9. Jul 11, 2013

### pervect

Staff Emeritus
What causes you to age?

Can you "stand still" and stop aging?

Whatever it is that makes you age, prohibits you from "standing still" (for lack of a better word) in time, and hence prevents you from "standing still" in space-time.

Note that when we say space-time, it's not just another word for space. It is actually space and time.

As far as the dart goes, when you hold it, it's following a non-geodesic path. It's when it's in free fall that it follows its natural, geodesic path.

10. Jul 11, 2013

### Staff: Mentor

Let's simplify things a bit. Consider a flat piece of paper with a standard Cartesian grid on it one direction is labeled "t" and the other direction is labeled "x". Now, draw two lines on the paper, one parallel to the t axis and the other which starts parallel and curves gently but noticeably.

Do you see how the straight line can represent an object at rest, and how it is still "moving" through time even if it is at rest in space? Do you see how the curved line can represent an object which is accelerating and how it moves through both time and space? Note that an accelerometer attached to the object at rest would read 0 and an accelerometer attached to the accelerating object would read non-zero.

Now, erase the coordinate grid and draw a new grid slightly rotated to the first and label one direction "t'" and the other direction "x'". Now, the straight line no longer represents an object at rest, but rather one that is moving in a straight line at a constant speed. I.e. it is inertial and an attached accelerometer still reads 0. Note further that the curved line still represents the same acceleration, but a different initial speed, and since the acceleration is the same the accelerometer reads the same non-zero value. Do you follow so far?

Now comes an interesting part. Erase the coordinate grid and do not replace it with another. Without the grid you can no longer say things like "at rest" or "at the same time", but there are still things that you can say. You can measure the length of each curve, you can draw tangent lines to the curves at different points and measure the angle between the tangent lines, you can measure the radius of curvature at each point, etc. Geometrically, in a coordinate-independent sense, you can state that the straight line is straight and the curved line is not, the straight line is the shortest path between any two points on the line, and if you take the tangent vectors at two nearby points on the straight line you find that they are parallel whereas they are not parallel for the curved line. Physically, the angle between tangent lines is the relative velocity, and the deviation from "straightness" is the acceleration measured by an accelerometer. Those are coordinate independent physical statements, because they remain even without the coordinates.

Last edited: Jul 11, 2013
11. Jul 11, 2013

### A.T.

It's simply postulated, based on the fact that nothing we know can stop moving and stop aging. So everything advances in space-time.
The lack of forces, that could force it on a non-geodesic path. See video below at 0:40.

The proper acceleration has a spatial component (upwards), but that doesn't imply actual movement in space. It must have proper acceleration in order to stay stationary in space.

Note that acceleration is not the same a velocity. In uniform circular motion you have proper radial acceleration inwards, but no radial velocity. When hovering in a gravitational field you have proper radial acceleration outwards, but no radial velocity

12. Jul 11, 2013

### A.T.

Xcourrier, If you haven't seen it already, check out this post by DrGreg:

https://www.physicsforums.com/showpost.php?p=4281670&postcount=20

Note that all worldlines in the diagram are geodesics (locally straight lines). So they have no proper acceleration (what an accelerometer measures, a frame independent value).

But they change their direction relative to some of the coordiante charts. This indicates coordinate acceleration (change of velocity in some reference frames).

So when you think about acceleration, it is important to keep these two concepts apart.

13. Jul 11, 2013

### OCR

A really good explanation, one of the best I've read...

OCR

14. Jul 11, 2013

### xcourrier

It seems this is really the part that abstracts the concept. Whenever dealing with things in higher dimensions, you can normally break it appart into components of the lower dimensions (like x and y coordinates for a 2d plane). Yet somehow uniform motion in time and uniform motion in space (or being stationary) produce an acceleration in space-time. I understand this is a product of the "warping," but it seems like this would be much easier if I could visualize the warping of the 3d space and the warping of time separately. Is that possible at all?

The General Relativity: Einstein vs Newton video was helpful, I had seen it before but I still had issues translating it to the 3d spacial dimension. Drawing a perpendicular coming off the page (or screen I guess) doesn't quite work lol. I know I'm trying to take things too litteral, but the further I can get before the wall of abstraction hits, the better :)

15. Jul 11, 2013

### WannabeNewton

The kinematics of general relativity are described at the local scale by decomposing point-wise mathematical objects of interest into components purely along the time direction and components purely along the spatial directions (the "temporal" and "spatial" parts of the object) with respect to a frame setup by some observer-this is the so called kinematical decomposition; this can also be done on a global scale for field-like mathematical objects in special cases by using a family of observers. However I don't know your level of mathematics so I can't say if the technical details will be of any help.

However don't confuse this with an actual picture of space-time analogous to an x-y Cartesian grid; you cannot visualize 4d space-time in its entirety (what you can do is suppress extra dimensions and draw the so called embedding diagrams).
See here: http://www.einstein2005.obspm.fr/parallel/Cosmology/CIPKO_vised.pdf

16. Jul 11, 2013

### xcourrier

Basic calculous but go ahead and throw some equations out if you think they could help at all. I could use some direction to start looking things up

Edit: those links look awesome. spending the next few hours looking over them. thanks a ton!

17. Jul 11, 2013

### WannabeNewton

If you have multivariable calculus, basic linear algebra, 2nd year mechanics, and some 2nd year EM under your belt then you can try using the following textbook: https://www.amazon.com/Gravity-Introduction-Einsteins-General-Relativity/dp/0805386629 it's an undergraduate GR text that's rather popular at the undergrad level (my university has been using it for a while now for the undergrad GR class). An alternative text is: https://www.amazon.com/A-First-Course-General-Relativity/dp/0521887054/ref=pd_bxgy_b_img_y (the one I used when first learning GR).

Last edited by a moderator: May 6, 2017
18. Jul 11, 2013

### WannabeNewton

Thanks!

19. Jun 1, 2015

### Steinein

I haven't finished reading all the posts here, but I have a couple of questions if you good folks don't mind. I've run into this thread while searching for explanations of ideas that underlie the theory of general relativity. I think I now understand Einstein's reasoning. The reasoning is that gravity is not a force, but rather a property of the phenomenon known as mass. Mass distorts space (and time).

Moving at constant speed is equivalent to standing still. When an object is moving through space without any forces acting on it, it is following a straight line and will continue to do so indefinitely, unless a force interferes with it. These imaginary "straight lines", including the one that describes the path of our object, are curved by mass. The object will continue on its path, now moving on a curved path. Now imagine that I'm standing on the ground of a planet whose mass is the same as our Earth's, and which has no atmosphere - i.e. there is no air, there are no particles that could slow down the movement of an object that's falling to the ground (there are no external forces either). Is it then true that when I'm observing an object falling to the ground (after being pushed from a pillar of, say, 100 meters), I'm actually observing it from a non-inertial frame of reference, and the object's motion is in fact inertial motion? Put another way - I am the one who is technically accelerating, as the situation is equivalent to me accelerating in a spaceship at 9.81 m/s/s, and observing that same object float freely in space from my cabin. I would perceive it as "falling down" at the same rate (9.81 m/s/s). The object falling to the surface of the planet is simply following its path in space, while moving at a constant speed.

If that is true, however, why then do objects behave as if they are indeed accelerating when dropped from a height? Since the Earth's surface is not moving towards them, and they are in free fall (which is inertial motion according to Einstein, as we've established), why then does an object dropped from a greater height hit the ground harder than an object of same mass dropped from a lower height? Why are objects falling at all? What is the initial impulse that causes the object to move in a constant way along its geodesic? If I'm holding a pen in the air, I understand the motion of both myself and the pen is equivalent to the motion of that spaceship I mentioned earlier. From this it follows that if I were to release the pen from my grip, I would stop applying the force that caused it to accelerate upwards, and it would now find itself in inertial motion. Shouldn't that imply that the pen would continue to move in the same direction at a constant speed? Why is the pen moving downward, and hitting the Earth with force proportional to the height from which it was dropped, if the Earth is not moving toward it and it itself is not accelerating?

Edit: I get it's the deceleration that causes the forcefulness of the impact, but it's just so unintuitive, since the surface is not moving, and I don't get why the object is moving towards the center of mass either. I guess my question really is as follows: Why does an object have to accelerate to stand still in curved spacetime?

20. Aug 4, 2015

### wingingit

Steinein -

Like you, I followed a link here on my search for a better explanation of the concepts behind GR. I've been scouring the internet for a couple weeks now, linking from one page to the next, and I swear if I watch one more video of a guy dropping a ball in a rocket ship or falling elevator, I'm going to try my own gravity experiment by leaping off the nearest bridge. And don't get me started about bowling balls on rubber sheets - there must be an awful lot of bedwetting bowlers in the physics community.

The problem is that there are lots of grade-school explanations of GR, and a fair number of graduate-level descriptions, but there's not much in between. As far as I can tell from my search, Einstein's thought process for GR was "Some guy in a rocket ship... curved spacetime." The same goes for SR: "Two guys with clocks in trains... E=mc2." It's like the underpants gnomes' business plan. (Actually, the offerings for SR aren't quite so bad - there are at least a couple places that make the connection fairly well. I suspect its greater popularity just means more videos overall, so we're in a monkeys+typewriters=Shakespeare situation.) I'm not sure why this is, but I've seen it with other difficult concepts, as well.

Actually, the explanations here are some of the better ones I've seen, but even here, we seem to jump from rocket balls to geodesics with very little in between. I get that the physics for the guy in the accelerating ship works the same as for the guy standing on the street in Cleveland. And I know what a geodesic is, and why you would use one to describe movement on a "straight" line through curved spacetime. What's missing is the clear, CONCEPTUAL explanation of how we go from one to the other (i.e. no references to Riemannian manifolds!)

However, I think I've kind of pieced together a rough understanding, and so I'd like to post it here, both as an answer to your question, and in hopes that greater minds will chime in and (nicely) straighten out some of my more bizarre logical twists. Here goes...

Let's start with the accelerometer, first mentioned on this page by DaleSpam. Also, again following DS's lead, let's get rid of some extraneous dimensions, so my poor brain doesn't explode. The principles will translate just fine to higher orders, I think. We'll keep the z-axis - the traditional up-down axis that we most commonly associate with gravity. And we'll keep the t-axis, of course. Finally, we are also going to eliminate the Earth, and its tidal effects, n-body diagrams, etc. Let's put our accelerometer on a really huge planet far out between galaxies. Its surface is so big we can consider it a plane, its radius so big that its g is effectively constant over our distances, and it is so massive that our own mass is negligible.

So our simple 1D accelerometer is just a metal frame, with a lead ball suspended between two springs - one at the top and one at the bottom. As we know from a gazillion rocket-ball videos, the accelerometer in free fall matches the accelerometer in free space - no acceleration. The one on the planet surface matches the one in the accelerating ship - yes acceleration. And if you can't believe a ball on springs, who can you believe, right? Seriously, though, we've kind of leap-frogged Einstein here, but he has something to say about this. Specifically, as I understand it, he was answering the question, "Why does the ball have to represent acceleration? Couldn't the ball just be pulled down by the gravitational 'force', the invisible string of classical physics?"

The answer to this is no, and the reason is simply that forces don't work like that. We all know that F=ma, so if I tie one invisible string to M1, and another to M2, and tug on them with the same force, they will accelerate at different rates, dependent on their masses. And I don't know from the strong or weak forces, but I can tell you that two protons, each with the charge of 1 proton, will rocket away from each other; but two bowling balls, each with the charge of 1 proton - yeah, not so much. EM force is also independent of mass.

But gravity doesn't follow F=ma - it pulls on each mass with exactly the amount of force necessary to attain a given acceleration. There's a name for forces that are proportional to mass: we call them accelerations. And so we come back to Einstein. Einstein's specific genius, from what I can see, was the ability to call a spade a spade. Physicists were trying to figure out why light seemed to have the same velocity regardless of the velocity of the observer. Einstein comes along and says, "Well, hey, maybe that's just the fastest anything can go." Obviously a complete load of tosh... except SR has worked, over and over. Okay, then, how about this: gravity doesn't behave like other classical forces, acting instead like an acceleration. Einstein says, "Huh, well maybe that's because it is." Once again, obvious BS, except that it works, both then and now - the more precisely we measure, the better his predictions look (go check out Gravity Probe B).

So Einstein agrees with the accelerometer, and we will too, at least until the LHC spits out its first graviton. But so far, I don't think I've said anything you couldn't pick up elsewhere. Pay close attention, though, because here's where the magic happens.

Imagine that you pick up our accelerometer and take it way, way high up, though still in the approximately constant-g zone of our giant planet, and drop it. There's no air resistance and no starting velocity, so v(t)=gt, at least from the perspective of an observer on the planet surface. As a graph, with t on the horizontal and z on the vertical axis, we are looking at a hyperbolic, with the slope of z(t) ever increasing (or decreasing, whatever - signs don't matter here.)

But according to our accelerometer (and Albert E.) that's not true - there is no acceleration; as far as the accelerometer is concerned, it's floating blissfully through empty space at a constant inertial speed (or no speed, inertial reference frames being relative, after all.) That means that, from the accelerometer's perspective, which according to E. is the real perspective, its z(t) line, called its "world line", represents a constant velocity, and so it should be arrow-freakin'-straight. So we grab hold of that hyperbolic z(t) line, and like a bent metal bar, we yank on it until it's straight, like it should be. But hey, that world line is still tied to the t-z coordinate system we originally drew it on, so when we yank the world line straight, we're putting a bend in one or both axes.

I'll get to what this means in real life, but for now, just take a second and contemplate what this means visually. In order for the world line to be straight, as the accelerometer knows it is, the coordinate system has to be bent. This, at least as I understand it, is the origin of curved spacetime.

Okay, but what does it actually mean? Well, now we run up against my poor math background, but here's what I see. According to the accelerometer, it is moving at a constant velocity, so dz/dt is constant throughout its journey. Picture your basic sloped mx+b line from 8th grade. But the outside observer sees the value of dz/dt increasing at each measurement interval.

Either the dz of the observer is getting bigger for each dz of the accelerometer (picture the vertical z-axis bowing away from the straight mx+b accelerometer world line, so that each time you run a horizontal shadow from the world line to the z-axis, it is ever further along the observer's z-axis than what it would be on the "true" vertical z-axis of the accelerometer's world line. Savvy?) OR the dt of the observer is getting smaller for each dt of the world line (picture the horizontal t-axis bowing up toward the mx+b accelerometer world line, so each vertical shadow from the world line to the t-axis is ever less far than the "true" horizontal dt of the world line) ... OR some combination of both.

Makes your brain hurt, I know, and sounds like total BS, but 1) if you accept the accelerometer's zero-reading in free fall, as Einstein did, this is the natural outcome, and 2) like the time dilation of SR, it WORKS. The effects predicted here have been tested using satellites in various orbits, and the results are consistent with curved spacetime.

This can be extended to explain the accelerating-while-standing-still dart mentioned above, though I'm going to skip the dart and just use my guy on the ground. First, let's look at the simple explanation - the observer on the ground sees the accelerometer accelerating toward the ground. The accelerometer sees the observer, and the ground, accelerating toward it (whether it thinks "Oh, no, not again!" like the bowl of petunias, I will leave up to you.) SOMEBODY is accelerating, the accelerometer knows it's not itself, and Einstein agrees, so who's left?

Looking from a slightly more complex perspective, it helps to remember that GR specifies that gravity curves spaceTIME, not necessarily just space. When we were looking at the zt graph before, I said that either the z-axis, the t-axis, or both, could be curving. So when I tell you that you are accelerating just by standing still on the planet's surface, one way to look at it is that you are experiencing reduced dt's, rather than increased dz's. (Mind you, this may not matter all that much - my understanding of relativistic spacetime, versus Galilean spacetime, is that any of the axes are pretty much interchangeable, but hey, if it makes you feel better...)

Incidentally, this is where geodesics come in - a geodesic is a "straight" line on a curved surface, like me taking an airplane from here to Podunk, Iowa. It's a straight shot... for a given value of "straight". If spacetime is curved, the "straight" line is also curved, just as the Earth's surface between here and Podunk is curved. It feels like straight, because that surface is the only reality I know, but from an additional dimension, it's obviously curved. Where I would disagree with other posters, though, is that I think the free falling accelerometer is the "true" straight line. It is the elevator going at a constant velocity upward that follows a geodesic (paralleling the outward-bowed z-axis).

So as I said, you are accelerating because of reduced dt's, which make your velocity into an acceleration. What's that? You say you don't have velocity because you aren’t' moving? Oh please. Relativity, remember? Moving is a frame of mind, or at least a frame of reference. Our observer and accelerometer may disagree on who is accelerating, but there is no disagreement that their velocities are completely relative. Without the acceleration issue, there would have been no disagreement. So there is a reference frame in which you are moving; continually reduce the length of dt, and your v turns into a. Once again, sounds like BS, but is confirmed by experimentation. Clocks on satellites further away from Earth experience different time than those closer in, but you need a hydrogen maser clock to note the difference.

So there you have it, from rocket-balls, to curved spacetime, to accelerating-while-standing-still. I hope you appreciate it, because I'm about to get pilloried for this.

Best regards,
Doug

21. Aug 4, 2015

### soothsayer

Steinein,

If a more mathematical derivation of acceleration from the Schwarzschild metric would help you understand how acceleration comes from geodesics in GR, you can find that here: http://mathpages.com/rr/s6-07/6-07.htm

It is as simple as setting up the metric (with no angular momentum) and solving for dr/dt, using a geodesic relation or two and a lot of algebra. You will see that coordinate acceleration is a natural consequence of such metrics. In essence, it's not so different from taking an x(t) equation and deriving acceleration, except the calculus is pretty much already done for you. Of course, if you really want to start from scratch and convince yourself of the validity of the Schwarzschild metric, that's going to take a little more work...

Re: winginit's explanation, it should be added that the accelerometer indeed measures no acceleration in free fall because there IS no proper acceleration, only coordinate acceleration, relative to the observer. The observer is the one that is undergoing proper acceleration, this is what we call "weight", and it is caused by the fact that the ground is preventing the observer from moving along the local spacetime geodesics.

Also, I don't really know of any explanations for GR between the "grade school" and "graduate school" versions, but I would argue that the latter is more of an "undergrad" explanation, as that's where I learned all this.

22. Aug 8, 2015

### Steinein

http://aether.lbl.gov/www/classes/p139/exp/experiment3.html

At first I thought this was quite straightforward and went along with it, derived the formula and continued with the next experiment. But then I went back and got stuck. First of all, how does Mrs. Einstein observe the photon at all if it's not travelling at her? Also, doesn't Mr. Einstein also see Mrs. Einstein's clock as running slow from his perspective? But if he is the one travelling at, for example, 0.99 c, Mrs. Einstein will be the one aging more rapidly in comparison, yet he sees her clock as running slow?

Also, now I can't even understand why this horizontal distance at all? Doesn't that imply that the photon travelled that distance at a speed that's less than c? I can't in my mind understand why it would take it more time to reach the mirror? How come an object fired horizontally from height h hits the ground at the same time as an identical object that's simply dropped from the same height? I know this is completely different, but I'm baffled and can't think straight anymore. I can't detect what's wrong with my reasoning and it's killing me.

Also, http://www.pitt.edu/~jdnorton/Goodies/rel_of_sim/

^^ number 4 - the rotation of bodies in transverse motion

If the observer standing on the body sees both ends crossing the line at the same time, and the guy moving relative to it doesn't, isn't it still possible to determine whether both ends actually crossed or didn't cross the line simultaneously?

I always end up just thinking about a photon from A travelling towards the guy moving to the left, and having less of a distance to traverse than the photon from B. I'm sorry about the retarded questions.

23. Aug 9, 2015

### soothsayer

Steinein, these are all excellent questions that pretty much anyone who has ever studied SR has asked. They are not at all "retarded". Hopefully I can clarify for you:

This is what we call a "thought experiment", obviously because it's not an experiment that could really be performed in real life. You could not have a train moving at 0.99c, for example, just as Mrs. Einstein could not actually see the beam of light inside of the train. Some things get lost in translation if you try to view them from a mind of an experimental physicist. Actually, I imagine she might be able to see the photon if the train was transparent and filled with dust or fog, but that might obfuscate the physics a little bit. Rest assured, if Mrs. Einstein COULD see the photon, this is what she would witness.

As to your second question, about Mr. Einstein also seeing Mrs. Einstein's clock as moving slower than his, this is what's known as the "Twin Paradox", which I imagine is a term you've heard before. Both Mr. and Mrs. Einstein witness the other one's clock as moving slower than their own. Who is actually aging slower than the other? This paradox can actually be resolved if you consider that both Mr. and Mrs. Einstein are correct, but that a paradox only arises if and when they return to one another and compare notes and find that they disagree. At this point, Mr. Einstein has undergone a gigantic initial acceleration to reach 0.99c at the start of the experiment, and then another even larger acceleration to stop and return back to where our observer Mrs. Einstein is. The thing about acceleration is that it is susceptible to time dilation as well, but whereas velocity is relative, acceleration is NOT. That means that when Mr and Mrs Einstein are reunited and compare notes, they can both agree that MR. Einstein was the one accelerating, and as the math works out, they will both agree that he aged a slower than she did during the trip. This is the same reason as why when analyzing gravitation time dilation, there is no Twin Paradox effect: if Mr. Einstein was lowered into the gravitational well of a black hole, he would see Mrs. Einstein's clock speeding up, and she would see his slowing down--spacetime curvature, like acceleration, is invariant, not relative.

An object fired horizontally from height h hits the ground the same time as an object simply dropped at height h at the same time, because they undergo the same acceleration due to gravity, and thus travel the same distance parallel to the gravitational field lines at the same time. The ball fired horizontally has a component of velocity perpendicular to the gravitational field. While it moves the same speed in the y direction, it does NOT in the x direction--its velocity is greater than that of the other object, so it can travel farther in the same amount of time.

All photons (moving in the same medium) travel the same speed as all other photons, regardless of what reference frame you are viewing them from. In Mr. Einstein's field, the photon is simply "dropped" downward, and "bounces" back up from the mirror and returns to the place it started in a time = 2h/c. Mrs. Einstein sees a photon that is "dropped" but also "fired" in the perpendicular direction, due to what she sees as movement of the train and the mirror in the x direction. That photon travels a greater distance. If photons were like massive objects, there wouldn't be a discrepancy--Mr. Einstein and Mrs. Einstein see the photon as having different velocities. This is fine, because velocity is relative and no one inertial reference frame is more "correct" than any other, but we know photons are not like massive objects; their velocity is invariant and not relative. Mr and Mrs Einstein must both see the photon as having the same velocity in their respective frames. The only way we resolve this paradox is to invoke time dilation/length contraction.

Re: simultaneity, to establish that one observer is "correct" and the other one isn't is to say that there is a preferred reference frame. Consider that this experiment is being carried out on Earth. The object is not only falling, but also rotating along with the Earth. There is nothing special or absolute about this reference frame. An alien observer at a distance from Earth, not rotating along with it, would see a different result than the two observers in the example. Is there a clear way to determine which of the three observers has the most unbiased perspective? You could have special mechanisms at the endpoints of these horizontal lines 2, 1, and 0, that detect and log when the endpoints of the object pass by, but even the reference frame inhabited by any detection equipment is completely arbitrary (at rest with respect to the rotation of the Earth, perhaps? Nothing special about that). We could remove Earth from the scenario in an attempt to simplify the experiment, but if it were done out in deep space away from any massive objects or gravitational fields, the notion of absolute reference frame makes even less sense, and there isn't even clear directionality.

24. Aug 10, 2015

### Steinein

Thank you very much for your effort sir.

My understanding was that time dilation occured because of prolonged signal propagation; on the contrary, on wikipedia it says that it's the fabric of spacetime itself that's changed. When I looked up what spacetime was, I found it was described as a mathematical model. Does that mean that I will not find a satisfactory answer as to the nature and mechanism of these phenomena?

My reasoning was that light simply had a greater distance to travel. However, since slowing of the wave is not witnessed (everyone measures the speed as c), no one actually experiences time as running slower or faster at different velocities. I also thought that length contraction was simply an observational phenomenon witnessed by an outside observer. What is the explanation for actual length shortening?

When I looked up why photons apparently move even though time should stand still from their perspective, the answer was that their paths were maximally contracted? A photon is massless and thus already maximally contracted, and it would follow from this answer that it's not only the speeding object that's shortening but its entire path? I could make no sense of this.

Last edited: Aug 10, 2015
25. Aug 10, 2015

### Staff: Mentor

Steinein, you're mixing together a number of different concepts, and that's going to cause confusion.

First of all, time dilation due to relative motion in SR is something different from gravitational time dilation in GR. You need to understand them separately; they're not the same thing and they don't work the same way.

Second, length contraction is an SR phenomenon (actually, that term can be used to refer to two different SR phenomena); there is no such thing as "gravitational length contraction" in GR, so again, if you're trying to understand how gravity works, length contraction is not involved.

Finally, photons (or any massless objects) are fundamentally different in relativity from ordinary objects with nonzero rest mass, and you have to treat them differently. Many of the intuitions you have about ordinary massive objects in relativity are simply wrong for photons. For example, it's not correct to say that photons are "maximally length contracted" or "experience zero time"; the correct thing to say is that the concepts of "length contraction" and "time dilation" (in the SR sense) do not apply to photons.

I would recommend taking all of these issues one at a time, instead of trying to lump them all together. Each one is probably worth a separate thread on its own. Or, even better, work your way through a good textbook on SR (Taylor & Wheeler's Spacetime Physics, for example--or online, this one here). Trying to look up answers to individual questions on Wikipedia or other sources is probably not going to help you very much with understanding, nor is asking individual questions here; it looks to me like you need a solid grounding in the fundamentals of SR first, which is really beyond the scope of what we can do here just answering questions.