Today I want to answer a question that was sent to me by Ed Catmull who writes:
“Twice, I have read books on relativity by PhDs who said that we travel through time at the speed of light, but I can’t find those books, and I haven’t seen it written anywhere else. Can you let me know if this is right or if this is utter nonsense.”
I really like this question because it’s one of those things that blow your mind when you learn about them first, but by the time you have your PhD you’ve all but forgotten about them. So, the brief answer is: It’s right, we do travel through time at the speed of light. But, as always, there is some fine-print to what exactly this means.
At first, it does not seem to make much sense to even talk about a speed in time. A speed is distance per time. So, if you travel in time, a speed would be time per time, and you would end up with the arguably correct but rather lame insight that we travel through time at one second per second.
This, however, is not where the statement that we travel through time at the speed of light comes from. It comes from good, old Albert Einstein. Yes, that guy again. Einstein based his theory of special relativity on an idea from Hermann Minkowski, which is that space and time belong together to a common entity called space-time. In space-time, you do not only have the usual three directions of space, you have a fourth direction, which is time. In the following, I want to show you a few equations, and for that I will, as usual, call the three directions of space, x, y, and z, and t stands for time.
Now, here’s the problem. You can add directions like North and West to get something like North-West. But you cannot add space and time because that’s like adding apples and oranges. Space and time have different units, so if you want to add them, you have to put a constant in front of one of them. It does not matter where you put that constant, but by convention we put it in front of the time-coordinate. The constant you have to put here so that you can add these directions must have units of space over time, so that’s a speed. Let’s call it “c”.
You all know that c is the speed of light, but, and this is really important, you do not need to know this if you formulate special relativity. You can put a dummy parameter there that could be any speed, and you will later find that it is the speed of massless particles. And since we experimentally know that the particles of light are to very good precision massless, that constant is then also the speed of light.
Now, of course there is a difference between time and space, so that can’t be all there is to space-time. You can move around in space either which way, but you cannot move around in time as you please. So what makes time different from space in Einstein’s space-time? What makes time different from space is the way you add them.
If you want to calculate a distance in space, you use Euclid’s formula. A distance, in three dimension, is the square-root of the of the sum of the squared distances in each direction of space. Here the Δx is a difference between two points in direction x, and Δy and Δz are likewise differences between two points in directions y and z.
But in space-time this works differently. A distance between two points in in space-time is usually called Δs, so that’s what we will call it too. A distance in space-time is now the square-root of minus the squares of the distances in each of the dimensions of space, plus c square times the squared distance in time.
Maybe let me mention that some old books on Special Relativity use a different notation, in which, instead of just putting a minus in the space-time distance, one uses a prefactor for the time-coordinate that is i times c. This has the exact same effect because the i square will give you a minus. The I turns out to be useless otherwise though, so this notation is not used today any more.
But why would you define a space-time distance like this, why not just all plusses? Well, for one, if you do it differently it doesn’t work. It would not correctly describe observation. That’s an answer, but not a very insightful one, so here is a better answer.
Einstein based special relativity on the idea that the speed of light is the same for all observers. You cannot do this in a Euclidean space where all the signs are plusses. But you can do it if one of the signs is different relative to the others.
That’s because a space-time distance that is zero for one observer is zero for all observers. This is also the case in Euclidean space, but in Euclidean space, this just means zero in each of the directions of space.
But what does a zero distance mean in space-time? Well, let’s find out. For simplicity, let us look at only one dimension of space. So if the distance in space-time is zero, this means that the distance in space divided by the distance in time equals plus or minus c. And that’s the same for all observers. So this speed, c, is an invariant speed.
But, well, we are not light, so we do not travel with the speed of light through space, and we do actually cover a distance in space-time. So let us look at this equation for the space-time distance again. Now let us divide this by the time difference. Now what you have on the left side is the space-time distance per time. And under the square root you have roughly something like the squares of the velocities in each of the directions of space. Plus c2.
And there you have it. Relative to yourself, you do not move through space, so these velocities are zero. You then only move into the time-like direction, and in this direction, you move with the speed of light. So, we indeed all travel through time with the speed of light.
I always try to show you equations because physics is all about equations. But to really understand what these equations mean, you have to use them yourself. A great place to do this is Brilliant, who have been sponsoring this video. Brilliant offers a large variety of interactive courses on topics in science and mathematics. They do for example have a course on Special Relativity, that will teach you all you need to know about space-time diagrams, Lorentz-transformations, and 4-vectors.
To support this channel and learn more about Brilliant, go to brilliant.org/Sabine, and sign up for free. The first two-hundred people who go to that link will get twenty percent off the annual Premium subscription.
But, well, we are not light, so we do not travel with the speed of light through space, and we do actually cover a distance in space-time. So let us look at this equation for the space-time distance again. Now let us divide this by the time difference. Now what you have on the left side is the space-time distance per time. And under the square root you have roughly something like the squares of the velocities in each of the directions of space. Plus c2.
And there you have it. Relative to yourself, you do not move through space, so these velocities are zero. You then only move into the time-like direction, and in this direction, you move with the speed of light. So, we indeed all travel through time with the speed of light.
I always try to show you equations because physics is all about equations. But to really understand what these equations mean, you have to use them yourself. A great place to do this is Brilliant, who have been sponsoring this video. Brilliant offers a large variety of interactive courses on topics in science and mathematics. They do for example have a course on Special Relativity, that will teach you all you need to know about space-time diagrams, Lorentz-transformations, and 4-vectors.
To support this channel and learn more about Brilliant, go to brilliant.org/Sabine, and sign up for free. The first two-hundred people who go to that link will get twenty percent off the annual Premium subscription.
The motion through this fourth dimension is parameterixed as a spatial distance parallel to proper time. There is this fourth dimension and every observer measures it with time and a vector tangent to this fourth dimension has length equal to the speed of light.
ReplyDeleteThis is fun to talk about to people not very familiar with the lore of special relativity. Everything in the universe moves at the speed of light. Then as they ask, “how can this be?” you can then introduce special relativity. Everything moves at the speed of light. It is just that massless things move along spatial distances at the speed of light, and massive things move in space plus time at the speed of light. For massive things they just move in different spacetime directions.
In fact, in physical reality time is not so special. When you move in direction x and then in direction -x it depends on our choice. The space is just new always in your travel direction. I can interprete that I move always forward - only at different speeds, in some chosen coordinates. Okay, those coordinates move in opposite direction in relation to my current velocity, but we can define. To ensure that a particle moves always only forward we must postulate that coordinate-speed to be c.
ReplyDeleteVice versa, we can conclude that all elementary actions are in speed c forward. Do they define space? Maybe well. We measurers then define time when taken a group of actions as an object. We are a part of the fabric we measure. The spacetime is emergent via measuring, interacting, with time.
Is there only one action created all measurable? The litte god at speed c? Thought it was the case we would not benefit from that info.
Could we say that we are falling into the black hole of future at the speed of light?
ReplyDeleteYou could say that but it wouldn't make much sense.
DeleteWas this asked by Ed Catmull of Pixar?
ReplyDeleteThat's who he claimed to be, yes. I have no reason to doubt it but also didn't bother checking as I don't really see how it matters.
DeleteNicely explained, as usual. But I must object to the history component. S.R. was not based on the spacetime ideas of Minkowski (who was one of E.’s math teachers). In fact, E. resisted the spacetime formulation for many years, calling it “pointless erudition”. Eventually he changed his mind, after publishing his first general relativity paper. It’s natural to project contemporary understanding backwards in time, but that leads us to ahistorical notions of how ideas developed.
ReplyDeleteHi Lee,
DeleteSorry about this. When I said "based on" I actually didn't mean a temporal but a conceptual order. I see that I should have been more careful here.
Yes, I was going to make the same comment. :-|
DeleteThe views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth, space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.
Address to the 80th Assembly of German Natural Scientists and Physicians, (Sep 21, 1908)
Einstein was by his own admission not very good at maths (compared to other physicists, of course), and often worked intuitively, only later formulating things mathematically. With time, he came to appreciate mathematics more and more, especially after he discovered that differential geometry could help hilm formulate general relativity.
I actually didn't mean a temporal but a conceptual order
DeleteWhich event occurred first depends on the frame of reference. :-D
Phillip Helbig4:21 AM, August 31, 2020
DeleteHow's the paper on fine-tuning coming along? It's already the end of August. You better hurry up.
I see the CUP are bringing out a book next year "Fine-Tuning in the Physical Universe" in their "If my Aunty had just 2 more b*lls she'd be my Uncle" series. Dozens of academics and the Cambridge University Press writing and publishing absolute unscientific sh*te. What is going on in the world?
"Einstein based his theory of special relativity on an idea from Hermann Minkowski"
DeleteSabine, you really ought to edit/correct this. Although Einstein was Minkowski's student, I am not aware that any of Minkowski's ideas contributed to Einstein's 1905 papers.
It's not possible to change a video on YouTube. You'll have to live with it.
DeleteOh, I see. I didn't realize the text was a transcript of the youtube vid. Thanks for the reply!
DeleteHello Sabine,
ReplyDeleteIf i understand correctly, ds is a measure of spacetime. ds/dt implies change in this measure, so change in spacetime. For change to happen time needs to pass. What time passes for spacetime? (i thought time is part of spacetime. Does spacetime have its own time, a different time to our time?)
Time is one a projection of velocity space just like a perspective picture on paper have different depth distance projections dependent on the view angle... And a paper is a part of real view and you can tilt it again... An analogue for proper time and coordinate time (proper picture on paper and re-projected picture tilted by an axis).
DeleteThe statement "we travel through time at the speed of light" has no physical meaning. The path of a particle through spacetime can be described by a function p(r) = (x(r),y(r),z(r),t(r)) that gives the position (x,y,z,t) as a function of a parameter r. This way the point p(r) traces out some path in spacetime.
ReplyDeleteNow, the same path is described by many different parameterizations. For example, if we pick q(r) = p(2r) that's the same path. The point varies twice as fast with r, but that has no physical meaning. The total path that is traced out in spacetime is still the same path. That is, the set {p(r) | r in [-∞,∞]} is the same as the set {q(r) | r in [-∞,∞]}, and it is only this set that has physical meaning.
In fact, we can also pick q(r) = p(r^3 + r) or some other parameter transformation.
We might look at the vector p'(r) tangent to the path. The length of this vector depends on the parameterization, but its direction doesn't. So p'(r)/|p'(r)| is a true physical property of the path, but the length |p'(r)| is just an artefact of the parameterization that we chose to describe the path.
In order to make the parameterization somewhat more canonical, we usually pick a parameterization where |p'(r)| = c. As it happens, the four velocity is defined to be p'(r) given this choice of parameterization.
So the length of the four velocity is c simply by definition. This statement has no physical meaning. The statement "we travel through time with the speed of light" is uninteresting. It tells you nothing about physics. It only tells you something about the conventions we use when we describe paths through spacetime. We could very well have picked the convention |p'(r)| = c/2, and then we'd have the statement "we travel through time with half the speed of light", or even |p'(r)| = 2c, and then we'd "travel through time with double the speed of light".
The proper length of a curve isn't any parameterization of a a curve, it's a very specific one. Spend a little more time with your Special Relativity book and maybe you'll find that physicists are not quite as stupid as you think.
Delete"The proper length of a curve isn't any parameterization of a a curve, it's a very specific one."
DeleteCertainly...but that does not help you.
"Spend a little more time with your Special Relativity book and maybe you'll find that physicists are not quite as stupid as you think."
I'm not saying that physicists are stupid. I'm saying that the argument in your video is meaningless. Put a 2 in front of the square root of the space-time distance and your video becomes an argument that we travel through time at twice the speed of light (note that twice that square root is still zero for light).
In order to properly argue that we travel through time at the speed of light, you must define what the speed at which we travel through time means, by an experiment that can determine this speed. Then the fact that this experiment produces the numerical value of c would be a physically meaningful fact.
In your video you have simply defined it to be sqrt((cdt)^2 - dx^2 - dy^2 - dz^2)/dt and taken dx=dy=dz=0, so this comes out to be c. So one could argue that we have the following experiment: we measure dt, then compute dt/dt and multiply the resulting quantity by c. This experiment produces the numerical value of c, indeed. However, in order to even determine the result of the "experiment", we already had to know the numerical value of c. In other words, if we did not already know c, then this experiment would not be able to measure it.
No doubt there is a way to define the speed at which we travel through time in such a way that it's both defined by an experiment for which you don't need to know c already, and the numerical value that comes out of the experiment is c. After all, there is the physically meaningful fact about clocks that if a clock travels along a trajectory p(r), then between p(r1) and p(r2) the time elapsed on the clock is the integral of |p'(r)|dr from r1 to r2. However, I would be quite surprised if you were able to come up with a definition for "the speed at which we travel through time" that makes "we travel through time at the speed of light" is a clear way to describe this fact about clocks to anybody who does not already know it.
After a literature search, I am unable to locate a reference for the claim: "we do travel through time at the speed of light." My copy of Poisson and Will reads: "In spacetime we promote ' t ' to one of the coordinates and resist the temptation of using it also as a parameter. We wish the parameter to be Lorentz-invariant and for this we select the 'proper time'. The trajectory of the particle in spacetime will be described by parametric equations with proper-time assuming the role of the parameter." Reading on: "the proper-time of a photon is not defined and the photon's velocity-vector is also not defined." (2014, pages 194-198, Gravity). The phrase in the above essay: "Relative to yourself, you do not move through space, so these velocities are zero. You then only move into the time-like direction, and in this direction, you move with the speed of light. So, we indeed all travel through time with the speed of light"
Delete... does not make any physical sense to me.
Gary,
DeleteIf you want that in a more common textbook terminology that "you do not move relative to yourself" merely means "in your restframe". By definition, in your restframe, v=0. I was merely trying to avoid having to explain what a restframe is.
James,
Delete"note that twice that square root is still zero for light"
Yes, but it cancels out if you compute the invariant speed.
Maybe Jules is writing about what is in MTW about reparameterizing the proper interval, After all a second on Earth is different than on some other planet, so observers from different planets will have different counting rates for time. I suppose a variable parameterization is possible, for the Earth's rotation is slowing down over hundreds of millions of years.
DeleteYes, of course you can reparameterize a curve, this is why I said above that proper length isn't any parameterization but one specific parameterization. And since people constantly come here to "teach" me undergrad stuff let me mention that I know you can't parameterize a light-like curve by proper time and I certainly never said you can.
DeleteSabine said… “Now, of course there is a difference between time and space”
ReplyDeleteThis is a foundational assessment that I’m reasonably certain, although not intuitive, is arguably contradicted by objective analysis of observations; at least if limited to what is directly visible as changing per time. You wrote a piece on the function of atomic clocks and like every time keeping method, they assign a unit of time measurement to a change of position in space (distance), and that unit represents the proportion (“for lack of a better term”) of “motion energy” relative to distance (i.e. the motion energy at the equator is greater than the arctic circle however both are proportional to their respective distance’s representing 1 day of time per rotation).
Everything we can observe at least is in constant motion, and physically speaking distance has no meaning without motion, we couldn’t even observe or be aware of distance without motion (another agreement that space-time is inseparable). Any direct observable would indicate time is not different from space; it is a measurement for a change of position in space being “motion energy” per distance.
I’ve come across arguments that are filled with bias and intuition in contradiction to observable facts; “the universe doesn’t have to be anything” we need to keep trying to see it for what it is, not what we expect or want.
I was taught that it was just a useful way of thinking about the relativistic tradeoff between moving in space and in time. The more you move in space, the less you move in time. Photons move at the speed of light, so they aren't affected by the passage of time. As you move faster, your time scale gets slower.
ReplyDeleteEven better, you can sort of use the same intuitions you developed thinking about space alone. If you move at a constant velocity on a plane, for example, the faster you move parallel to the X axis, the more slowly you move along the Y axis and vice versa.
The equations have enough similarity so that one's intuitive sense can be extended to deal with time if one considers that everything is moving at the speed of light.
So, we are moving through space at the reciprocal (multiplicative inverse) speed of light, aren't we?
ReplyDeletec = 299792458 m/s,
1/c = 0.0000000033356409519815204957557671447492 s/m
So, for every meter we've walked, we've swallowed (spent) 0.0000000033356409519815204957557671447492 seconds, even at rest.
But, if we are at rest, how come we can walk even 1 meter and still be still?
Sorry, I don't know what you mean. A velocity in space is space per time, not time per space.
DeleteHello Sabine,
ReplyDeleteLet me try again. Cannot write delta so, d=delta. ds is partly made from time. So ds/dt has a different meaning than dr/dt, because dr (space distance) is not made with time. I believe you kind of say this by saying it is "the distance in spacetime per time" and not call it speed. But then you say it is the movement through time and say it is a speed, why? For me, logically ds/dt is something that does not have a meaning, how am i wrong?
“In the following, I want to show you a few equations, and for that I will, as usual, call the three directions of space, x, y, and z, and t stands for time.”
ReplyDeleteWe may do another interesting calculation about the speed of time in the following way.
We calculate the 4-dimensional speed of an object as a vectorial sum:
v(sum)^2 = (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 + (dc(tau)/dt)^2
where (tau) is the proper time of the object.
If we now insert for the proper time (tau) the according Lorentz transformation:
(tau) = gamma * (t – vx/c^2) (with v and x here the vector sum for all 3 directions) then the result is
v(sum)^2 = c^2 .
That means that the 4-dimensional speed for any object is c at any time.
Is this just a curiosity? Or the other way: If we understand that the proper time of an object reflects the internal motion in this object, then it means that the inside of an object (take an elementary particle) moves permanently at c. And this already explains half of special relativity, i.e. dilation.
And we can now make a surprising statement: c is not the maximum speed in this world but the only speed existing in reality. All other cases of speed which we observe are configuration speeds. Which means we can have a configuration of several constituents of an object moving exclusively at c whereas the object as a whole is at rest (or at any other speed).
This here is at the end Lorentzian relativity.
That's mostly correct, except that c is of course not the only speed existing in reality because by "speed" we usually refer to the spacial part only.
Deleteantooneo said... "That means that the 4-dimensional speed for any object is c at any time."
DeleteI may be wrong, but I’d interpret that a little differently as c being the space-time scale to which the 4-dimensional coordinates are relative.
In the interpretation which I have proposed here, this summed up c is the 3-dimensional speed. Because it is the superposition of the external speed of the object and the internal speed inside the object, which is reduced in relation to the motion of the whole object. This internal speed represents dilation in the interpretation of Lorentz.
DeleteThe Lorentzian relativity does not work with Einstein’s 4-dimensional space-time but space and time are used conventionally. The advantage is much simpler mathematics describing relativistic phenomena (particularly for GR), but despite of that it yields similar results as the complicated way of Einstein, in some cases even better ones.
As philosopher Hans Reichenbach (once cooperating with Einstein) has indicated, the change from Einstein towards Lorentz is in an analogy a Copernican step.
The distance or interval in spacetime is
Deletes^2 = (ct)^2 - x^2 - y^2 - z^2
Now suppose we have a body moving with x = vt and we set y = z = 0. We then have
s^2 = (ct)^2 - (vt)^2 = (c^2 - v^2)t^2.
For s = cτ, where τ is the propertime as measured by a clock on the body. I leave it as an exercise to show this gives a Lorentz gamma function for time dilation. Now suppose v = 0 then we are at rest with this body and we have s = (ct)^2. One can then see this interpretation of the interval as a distance as also about moving along this distance at the speed of light.
Lawrence,
DeleteLawrence said… “One can then see this interpretation of the interval as a distance as also about moving along this distance at the speed of light.”
So I could be sure one way or another, doesn’t that imply, distance relative to c is the local scale to which the other coordinates are relative?
As you know, atomic clocks, which use light to keep time, proved time is different locally due to relative velocity. For c to be a constant for each local observer it must also scale (distance relative to time) to the proper time of each clock, otherwise each local observer’s clocks would not only disagree on time they would disagree on the value of c since speed is a measure of distance per time.
If you have one clock you always know what time it is. If you have two clocks you are never sure.
DeleteCoordinates and time transform into each other, a sort of pseudo-Euclidean rotation, but the speed of light is the absolute invariant. Each clock measures a proper time, but my proper time might not be what I see of yours.
LC
Lawrence Crowell:
Delete“Each clock measures a proper time, but my proper time might not be what I see of yours.”
The proper time of each clock in motion is reflected by the internal motion of the particles which build the material of the clock. This is the Lorentzian view. This internal motion, which goes formally on at c, is reduced for an object (and so any clock) at motion following the equation:
c(object)^2 = c(nominal)^2 – v(motion)^2
which I have deduced above.
I am not sure of what you mean by c(object)^2 = c(nominal)^2 - v(motion)^2. The speed of light in flat spacetime is constant. Above I do write something similar with coordinate and proper time.
DeleteLawrence,
DeleteThose atomic clocks in the relativity test showed, when brought back together into same inertial frame that, they recorded time differently. While I know, and I am sure c is invariant for each observer it must also be invariant to each clock’s record of time otherwise the observer’s would not see c as being invariant, c is distance per time. You can’t have the distance value of c be the same for each observer per different rates of t unless that distance also scales proportionately to the clocks rate of time. I’m not disagreeing c is invariant for each observer, just refining what it really means. For example two observer’s at relative velocity to each other but in close proximity might greatly disagree on the distance to a galaxy near the edge of the universe as measured in light years if one is traveling near the speed of light, because while c is locally invariant it is not mutually invariant.
Sigh, confusion seems to have taken over. Clocks on other frames slow down, but there is also Lorentz contraction of lengths. So while you will observe a clock on some frame moving with a velocity v there is also the contraction of lengths that removes this problem of apparent speeds of light.
DeleteLawrence,
DeleteLawrence said… “Sigh, confusion seems to have taken over. Clocks on other frames slow down, but there is also Lorentz contraction of lengths. So while you will observe a clock on some frame moving with a velocity v there is also the contraction of lengths that removes this problem of apparent speeds of light.”
That’s exactly what I was saying, c is invariant to how each observer views their own space-time, however due to Lorentz contraction and time dilation a light year would be a different measure between them (i.e. they would not agree on a measurement of how many light years to a distant object). Therefore c is not mutually invariant (it has a different value for each) it’s only locally invariant (has the same relative value to their view of space-time, i.e. 299 792 458 m / s.
I didn't go into this, but there are also optical effects called Terrell rotations. There is, or was, a YouTube video that illustrates this. The voice is a robo-female with a sort of Bengali and British accent. It shows what happens if the speed of light is some ordinary speed and one drives down a road. Then it gets really weird by flying through solid figures.
DeleteLawrence Crowell says:
Delete“I am not sure of what you mean by c(object)^2 = c(nominal)^2 - v(motion)^2.”
Sorry that this equation is difficult to read. I do not know how to use indices in this text. The summands mean the following:
c(object) is the internal oscillation at c in the object. This oscillation is reduced by dilation when the whole object is in motion.
c(nominal) is the nominal speed of light as we all know it. This value is always the same in a measurement.
v(motion) is the actual macroscopic speed of the object under consideration.
The equation
c(object)^2 = c(nominal)^2 - v(motion)^2
describes the process of dilation in a handy way. For an object at rest (so v(motion) = 0) the internal oscillation, which represents the proper time of the object, is the coordinate time. If now the object is put into motion (with v = v(motion)), then the proper time is reduced as given by this equation.
It is by good reason that this equation reflects Pythagoras. Because it is the result of the superposition of the external speed (v(object)) and the internal speed (c(object)).
To remind you again: This follows the Lorentzian interpretation of SR. It says that the *measurement* of the speed of light yields always the same result. But the cause of this result is the fact that any measurement device is subject to dilation and contraction if in motion, so this result does not reflect the physical reality.
Lawrence,
DeleteI'm aware of some optical effects but it's more than that. If one of those observers was traveling near light speed they would not only view that distance object being closer, if they traveled to it, it would take much less time to reach also. Again, all I've been saying is a simple fact that while they both will measure light speed as 299,792,458 m/s, the meaning of that distance/time relationship is different for each in not only their respective views of the Universe but also if moving through the Universe.
Sabine, you forgot to link in your measurement video which has the measurement of "experimental time (sec)" using caesium-133 atom
ReplyDeleteNOW, blend in those two (relativistic time and exper time (sec)) in one video and you've got my attention!!
In the restframe s=t. So your final formula is ds/ds = c, which makes no sense.
ReplyDeletes has the dimension of a length. t has the dimension of a time. My formula is fine. Yours not.
DeleteBut that is their only difference. In the restframe t and s/c are both the proper time.
DeleteBut you are right: it should be ds/ds = 1. In the restframe this formula has the same meaning as your formula ds/dt = c. In both cases we divide proper time by proper time. In mine with the same units, in yours with different units.
DeleteAnd of course nothing can be learned by dividing a physical quantity by itself. Your answer to Ed is wrong.
DeleteFranzi,
DeleteYou are the one who divided a quantity by itself. I did not. It is correct that one does not learn anything from what you said.
Also, you constantly come here with very basic mistakes and then expect me to sort out your problems. I don't have time for this. Good bye.
I saw this on the front page of Hacker News. Well done Sabine!
ReplyDeleteOn another note:
https://reddit.com/r/physics won't allow this blog post on the forum. They banned me for even trying!
Instead of quality content like Sabine's, they have posts like, "Kite Aerodynamics: How Did a Kite Lift a 3-Year-Old?"
You must have made them angry somehow! Keep up the good work!
"https://reddit.com/r/physics won't allow this blog post on the forum. They banned me for even trying!"
DeleteHuh, wtf? Don't remember having had anything to do with reddit. I used to have an account (maybe still have one) but barely ever used it. (Only so many accounts you can take care of.)
Hello Sabine,
ReplyDeleteI do not understand your explanation, or rather it creates a logical inconsistency for me. I was hoping to learn something from this by presenting the way i see it and hopefully you pointing out where i take a wrong turn.
*I'll use the view that spacetime exists, meaning there is a four dimensional structure where each point in it holds information.
*The information that each point holds does not change, so spacetime does not experience change.
*If we take two specific points in this structure and connect them with a straight line (no gravity, no warping) the length of this straight line is a given (and is called ds).
*Let's assume these two specific point are on my world-line.
What do you mean by moving from one point to the other? What is that is moving? Do we place a figurative me on-top of spacetime and that is what is moving? Movement is defined in a certain way. A change of position in space, so space changes, and that leads to space experiences time, so we can say how long it takes for this position to change, and that leads to speed. Spacetime does not experience change/time.
I noticed you do respond to comments, and as said i wish to learn something, so if no respond to the above, maybe why?
1) you are talking nonsense
2) I cannot understand something meaningful from your writing.
3) you are so far behind in understanding, it will take to much
time to explain where you are wrong.
4) I do not answer this type of questions. I'm not a free teacher.
(it falls under rule 5 in your comment rules)
5) none of the above
Thank you, and have a wonderful day.
Ophir,
DeleteI think yours is a terminology problem. What we mean by "moving" on a curve is merely that you parameterize the curve and each value of the parameter corresponds to a point on the curve. As I have explained previously here (sorry for the link rot), it isn't possible to talk about something like a "passage of time" without having some system that can form a memory, so if the only thing you are talking about are curves in space-time, then the passage of time is a rather meaningless statement indeed. I am guessing that's what you mean to say. That's correct. So when we are talking about curves in space-time we implicitly (but usually not explicitly) assume that we are talking about a system that has a way of forming a memory.
Sabine,
DeleteFirst, i feel the need to say thank you. By that i mean i acknowledge the fact you are doing this for a time, (seen by the link to an older post with a link to an even older post. By the way really liked both posts, even though the cartoon does not display) and it can probably get annoying answering the same questions one person at a time, or one post at a time.
Second, i still do not understand.
You do tread carefully, but bottom line there is a sentence "So, we indeed all travel through time with the speed of light.". This sentence has a meaning in English language.
I do not think you mean the memory is forming at the speed of light. There is no mention of the other T (using your post idea and convention). So "we" is not our memory, to my understanding.
If i do understand your answer then the sentence should read: So, we indeed all travel through ("travel through" not meaning travel through) time with the speed of light.
What am i missing?
Ophir,
DeleteI can't follow, sorry. The purpose of my video was exactly to explain what this sentence means because if you merely make this statement without any explanation it is rather meaningless. Then you come here to say that if you only read the last sentence it makes no sense. I can only shrug my shoulders to that. Maybe try and read it again.
Sabine,
DeleteI know you know the following i'm not trying to teach you, but there is a need to create a common example.
1)A person holds a stopwatch. It shows zero. He activates the stopwatch until it shows one second. What is the distance, between these two events in spacetime in meters? ~300 million.
Can the person measure the distance between these two events in meters? No. The way the person measures the distance between these events is one second.
2)There are two things that are a constant distance in a person's space from each other. The distance is 300 million meters on a ruler. What is the distance between the two world-lines of the two things in spacetime in seconds? 1 second.
Can the person measure the distance in seconds. No. The way the person measures the distance between these things is 300 million meters.
3)The person sees one thing explode when he starts his stopwatch and when the stopwatch reaches one he sees the second thing explode. What is the distance in spacetime between the explosions? In what units? We measure spacetime in two different ways, and in this case the person uses both of them. Well, for the distance to be a clear thing you have to measure it in one way. Well we know what one second is worth in meters, so convert one way to the other and we have the distance as a clear thing. This much meters, or this much seconds.
This is all said in your video.
Now lets go to the first situation and take one way to measure distance of spacetime and divide it by the other way to measure distance of spacetime. Why?
We already know the conversion factor from the ds=0 example.
In spacetime a second is a scale to measure spacetime-distance and a meter is a scale to measure spacetime-distance.
In space and time a meter is a scale to measure position and a second is a scale to measure change.
In the first example again. the distance in spacetime between the two events is the same distance no matter how you measure it. That is why, as you said, if you divide the same way to measure the distance by the same way to measure the distance
you get that one second is one second, and one meter is one meter.
What does "traveling through" have to do with this?
There is an equivalence between spacial and temporal distances. An apple isn't an orange, but they have equivalent values of calories, fibre and vitamin C for example. You can convert some of your 'time-speed' into 'space-speed' at a certain date.
DeleteLet me try again. (Part 1)
DeleteTo measure distance between two points you need:
1) a line of numbers
2) calibrate the line of numbers and give the specific calibration a name (meter, feet, second, david... Whatever name you want. It is an identification)
3) place the line of numbers along what you want to measure.
4) read the numbers off the line at the two points, and the distance is the difference between the two points
With spacetime we have a problem. We cannot do step 3. We do not have access to do such an operation.
However we can measure distance in space and time.
Time is a line of numbers in spacetime. If we measure time in seconds for example, then time is a calibrated line of numbers in spacetime.
Time and space are perpendicular in spacetime.
So we can measure two distances in spacetime. One in our time direction and one in our space direction. This is what we can do.
(Part 2)
DeleteWe have two objects in our space. One object explodes, then the second object explodes.
What is the distance between the two explosions in spacetime?
We cannot do part 3 of the procedure, so we cannot measure this. But we can measure the distance in our space between the two objects, and let's say we do that with a ruler (a line of numbers) with a calibration called meter, and we get 5 meters. We can also measure the time between explosions, and let's say we do that in seconds, and we get 2 seconds. Time in seconds is a calibrated line of numbers in spacetime, so the time difference we measure is a distance in spacetime in the direction of our time line. This means we can say the answer to the question is: the distance between the two explosions in spacetime is 2 seconds in our time direction "plus" 5 meters in our space direction (which is perpendicular to our time direction). Not very "pretty" but an answer.
Now, i do not like the name of my time calibration. I want to call it feet. Same calibration but i call it different. So the distance in spacetime is 2 feet in my time line direction "plus" 5 meters in my space direction.
Now if everybody had the same time and space and would use the same calibration, then this would even be useful. We could compare spacetime distances. If you got 2 feet and 5 meters, then it would have been the same distance in spacetime.
But, relativity tells us: We do use the same calibration, because we all calibrate space and time with c (and this is a very very important thing to notice). On the other hand, we do not all have the same space and time. So our previous answer is not useful. We cannot compare with others. I measure 2 feet in my time line direction and you measure 2 feet in your time line direction, and direction is part of the answer
But, we can come up with a way to make one distance out of our two measurements, and in a meaningful fashion. If everybody uses this same method, we could compare this one distance, so things will be useful again.
To use this method of changing our two measurements to one measurement, we need to have the two spacetime distances that we measure to be calibrated the same. This means we have to see what is the ratio between our two different calibrations. The ratio is c. It is c because c is used to calibrate both time and space in one shot. But if you do not understand that refer to the video ds=0 example. So 1 feet = 300 million meter.
Now everybody converts feet to meter, or meter to feet (whatever we decide, but all do the same thing) and all use the same specific meaningful method to do the "plus" part, and we get a distance in spacetime that is useful. We can compare the distance in spacetime of different people.
(Part 3)
DeleteOk. In the last part of the video it is suggested that we do the following:
Measure our time in seconds, let' say we measure 1 second. Calculate what is the distance in spacetime in meters between the two points of 'start time measure' and 'stop time measure'. We get 1 second * 300 million = 300 million meters, or with our new name 1 feet* 300 million = 300 million meters. Now take the answer and divide it by our time measurement. We get 300 million meter divided by 1 second = c. No surprise over here. c is the ratio between our two calibrations of spacetime distances (second and meter). Now the video suggests this is a speed. It is not a speed it is a conversion ratio between two calibrations. I hope it is clear from the long explanation above, but...
If it is a speed, then it is a speed in what? The distance is in spacetime. This suggests that it is a speed (again this cannot be treated as speed, but let's go with it) in spacetime. Spacetime itself does not have its own time, so it cannot have speed. Speed is the rate that something changes. If the something does not change it does not have a concept of speed. The other option is that it is a speed in our space, but the distance we calculated is not in our space (it is in spacetime, it is said so in the video too).
So the claim here is that if we look at a ratio in a spacetime setting in the same way that we look at this ratio in a space and time setting, then we can say things that we say in our space and time setting about a spacetime setting. That does not work (it is meaningless). It is a logical inconsistency, and that is why it gives us speed in something that does not have speed. Things have a meaning, we need to see what they mean in the setting they are presented in.
The explanation in this post seems simple and clear to me. Spacetime has four axes, all with the dimension of distance. Distance along the time-axis of spacetime is c multiplied by the elapsed time. The standard scientific method of verifying theoretical equations with observations confirms this. (The choice of c was not arbitrary.) In high school physics one learns that distance equals rate multiplied by elapsed time, so rate equals distance divided by the elapsed time, which in the case of spacetime movement along the time axis is c.
DeleteUnderstanding why c works and other constants don't gets into some more complicated mathematics, as Dr. Crowell and others indicate, but once you accept that, the math is at high school level or below. If you want Relativity explained, that is a bigger issue, not covered in the post.
Hi,
DeleteOne last try.
First to set the record straight. I'm not saying the video is not clear, not saying anything about the presentation at all, not trying to show i"m smart, not trying to say you are not smart, not trying to say i'm smarter than you, not trying to say you are not thinking, not anything in human relationships. If something i wrote gives this impression, i apologize. It is not the intent.
We have point A and point B in our space. We measure the distance between them with a ruler putting the 0 reading on point A. We get that point B is one meter. We use another ruler, and we get point B is 3.3 feet.
We can say the meter got to point B from point A in one step and the feet got from point A to B in 3.3 steps, so they both moved from point A to point B but the meter moved 3.3 faster. That is a way to describe the situation. But the fact of the matter is that nothing moved. 3.3 is a conversion between two calibrations.
When you say we are moving through spacetime, and it does not matter how careful you are trying to explain what that means (that it is is a conversion factor between two calibrations), a person naturally thinks something is actually moving in spacetime. Nothing moves in spacetime. It is a feeling that we have that we are at this point or that point in spacetime. Sabine wrote a post about that. It is a feeling that we move from one point in spacetime to another point in spacetime. It is called a block universe. Nothing moves in a block. Now i believe this fact is hard to comprehend for us humans because we cannot shutdown time. We cannot shut down the feeling that we are moving in spacetime. I would go as far as saying that we cannot really really comprehend it, just like we cannot really comprehend 4 dimensions.
So describing something (that can be explained in a way that is true to what it really means) using things that give us the wrong impression about what we really struggle with is not good practice in my opinion.
A person will have a harder time understanding spacetime because of this.
Or, as someone once said: "Time ist the direction you're moving when you don't move at all...!"
ReplyDeleteDr. Sabine
ReplyDeleteWhich Brilliant courses teach about the geometry of space-time (Minkowski, Hilbert, Riemann)?
How does this fit into GR?
ReplyDeleteI think I'm going to vomit. Bible Boy and Panpsychic Boy have merged their crank theories into
ReplyDelete"A panpsychist account of cosmological fine-tuning, to avoid the deep difficulties facing theistic and multiverse accounts."
(Luke Barnes Retweeted
Philip Goff @Philip_Goff)
This is all the fault of that alcoholic, Brian Schmidt. He needs to get himself along to AA and stop promoting the mentally delusional as scientists.
A better way to put it is that we travel through spacetime at the speed of light, and 𝘰𝘯𝘭𝘺 the speed of light, our speed never varies. In the special case where we are motionless in all space dimensions then that speed is entirely in the direction of time, and conversely this explains why we can never exceed 𝘤 in space, as this is the only speed we ever travel at.
ReplyDeleteThat's correct.
DeleteThanks for this, a great video on a great topic :)
ReplyDeleteAside: one of my personal a-ha moments when trying to get my head round SR as an amateur (reading Susskind’s SR book) was when it was shown that the magnitude of the 4-velocity is always c. So basically (my way of understanding, which is hopefully correct, goes!) everything is always moving at the speed of light in a 4 dimensional space with signature -+++… ergo in a frame in which you are not moving in space, all your velocity is in time and you are moving in time at c. While if you move with nonzero velocity in space, your time component changes accordingly to keep the overall magnitude of your 4-velocity as c, hence time dilation.
Question –
*there is the fact that there is a speed c which is a constant related to the geometry of spacetime, than which one cannot move faster, etc.
*and there is a fact that massless particles like light photons move at this speed
Are these 2 facts the same fact, ie is there a conceptual reason why these 2 speeds must always be the same? Or is this empirical fact something that we know by observation alone?
Rollo,
DeleteYes, that's correct. As to your question. It's a very good question. The brief answer is that the latter statement follows from the former. The empirical component is the observation that photons are (to excellent precision) massless. This does not strictly speaking follow from electromagnetism or special relativity (though it is required by gauge-invariance).
Maybe I'll explain this in more detail in another video. Best,
Sabine
Hi Sabine,
ReplyDeleteI think your statement can be misleading; a better way to state this is that if 4 dimensions exist such that a meter has a meaning along the time axis, then we (like any energy) have the same pace as light along the direction of time. But I see no reason for a factor 1 to 1 between space and time displacement to be mandatory. It could be 1 in time for 2 in space for instance. The only thing mandatory is that the present has 3 dimensions and moves all together in the direction of time. Is that correct?
Best,
J.
I can't make sense of your explanation, sorry.
DeleteSabine,
ReplyDeletein order to discuss a velocity in time, you need that dimension to physically exist "now", including past and future. In this way you may (or may not) be able to define a meter along the axis of time. Assume you can, then my question.
Best,
J.
This could be a big experimental result. Measurements of wave function collapse with gamma radiation sources finds no correction that would be attributed to a role for gravitation. This may throw ideas such as Penrose's in doubt. I will post this on a previous blog entry on measuring quantum gravity.
ReplyDeletehttps://www.sciencemag.org/news/2020/09/one-quantum-physics-greatest-paradoxes-may-have-lost-its-leading-explanation
Sabine,
ReplyDeleteThanks for an excellent and informative video! This belated comment is on how minor algebraic manipulations that convert Minkowski spacetime into Euclidean form can make the relationship between time and distance particularly vivid.
If the Deltas are implicit your spacetime equation was:
s = sqrt( -x^2 -y^2 -z^2 +(ct)^2 )
Squaring both sides gives:
s^2 = -x^2 -y^2 -z^2 +(ct)^2
Shuffling terms doesn’t change the meaning, so here is a positive-only version of the above sum-of-squares (SOS) equation:
(ct)^2 = s^2 + x^2 + y^2 + z^2
A nifty feature of positive-only SOS equations is that both sides of the equation can be interpreted as boxes — distorted cubes. These two boxes share a body diagonal and are embedded in a Euclidean space of one less dimension than the total number of squared terms. Another useful feature is that any group of squares on one side can be replaced by a single squared term that represents a “box diagonal”, more commonly known as a vector. Thus:
r = sqrt( x^2 + y^2 + z^2 )
or
r^2 = x^2 + y^2 + z^2
Substituting back gives:
(ct)^2 =s^2 + r^2
Sabine noted that for no motion, s/t=c, or s=ct. However, for the full case of motion the time will be different from t, so let’s call this more general case s=ct’. This gives:
(ct)^2 =(ct’)^2 + r^2
Dividing by c^2 gives time:
t^2 = t’^2 + (r/c)^2
… which is the Pythagorean equation for a right triangle.
Notice that even in its 4D form this equation uses an all-positive signature Euclidean space, rather than the mixed signature of Minkowski space.
-----
The biggest advantage of Euclidean spacetime is that it transforms the often-obscure relationships of Minkowski SR into simple, easy-to-understand right triangles.
Here’s an example:
Draw a circle, then divide it in half with a vertical line. Label this vertical diameter line t, the time elapsed by a clock in the rest frame.
Next, start at the bottom of t and draw a line to any point on the side of the circle. Label this new line r/c. It represents how far some clock traveled during the rest clock interval t.
Finally, connect the top of t to the top of r/c, and label this last line t’. Geometry requires that this new line will form a 90° angle with r/c, thus completing the right triangle. This last line represents how many ticks the moving clock will manage to get in before the clock-at-rest completes t ticks.
Oddly, that’s it! If your other clock just stays put, as in Sabine’s example, the first side becomes a dot, and the second side t’=t. The faster the clock moves, the longer the distance line becomes, and the fewer ticks the moving clock can get in — a very literal, in-your-face tradeoff between distance and time. Finally, if the object gets very, very close to the speed of light, the distance-traveled side gobbles up all of the available t diameter, and the moving clock has no ticks at all.
This right-triangle relationship doesn’t end with time. Since such triangles amount to a different way of representing the Lorentz factor, all of the major features of Minkowski SR also have right triangle representations in Euclidean SR.
-----
So my point in all of this?
Well, first, I just wanted to show that switching over to Euclidean SR makes the relationship between space and time as point-blank as possible, with the distance vector of the moving clock essentially taking over the time vector of the rest frame, converting it from ticks to travel.
But the other point is to say that, as in all areas of math, choice of representation matters. Minkowski space does a better job of keeping our usual intuitions of space and time as pre-existing, whereas Euclidean spacetime is much more relational, requiring both frames to “talk” about what direction means what. But both represent the same underlying relationships, and thus add complementary value to understanding SR.
Makes sense. If you were to move through space-time at a speed close to the speed of light then time "stops". You can therefore say you are moving at the same speed as time flies when you're not moving.
ReplyDelete"Now, of course there is a difference between time and space, so that can’t be all there is to space-time. You can move around in space either which way, but you cannot move around in time as you please"
ReplyDeleteIMHO this difference doesn't exist.
Since there is no absolute space, there are no immaterial coordinates or stationary "points" in space. There are only moving particles, and space is nothing but the space between particles. So you cannot move "around in space", you can only move around in relation to a particle. And due to the high velocities of the celestial bodies and the expansion of the universe, you can never travel back to the same "point" in space.
For instance the Earth circling the sun moves back and forth in two spatial dimensions relative to the sun.
But it also moves back and forth in time relative to the sun!
Due to the elliptic nature of its orbit, Earth accelerates and decelerates regularly, resulting in an alternately growing and shrinking time dilation. Thus, Earth time regularly advances and falls back relative to sun time.
In fact right that way the case is in the relativity.
DeleteYou can imagine it so that every orthogonal space dimension can be exchanged with time in turns. We only project the changing continuum of 4d to linearity in 3:1. The choice is ours humans - of course well argumented with our emergent rest frame.