Understanding SR: The Relationship Between Speed, Time, and Space Explained

[Fletcher]

Long ago I accepted that the speed you move through time is personal, but I've always wanted to really understand it intuitively. I think I'm almost there.

Suppose me and my ghost are in the same place initially. Now I take off running in a direction, but my ghost remains behind. Since I'm moving through space faster, my motion through time is slower because the space-time interval is the same for both of us (I hope this much is right).

Now my question: Why are the two space-time intervals considered equivalent?

 

[JesseM]

Long ago I accepted that the speed you move through time is personal, but I've always wanted to really understand it intuitively. I think I'm almost there.

Suppose me and my ghost are in the same place initially. Now I take off running in a direction, but my ghost remains behind. Since I'm moving through space faster, my motion through time is slower because the space-time interval is the same for both of us (I hope this much is right).

Now my question: Why are the two space-time intervals considered equivalent?

The whole "speed of movement through spacetime = speed through space + speed through time" is something Brian Greene often uses to explain relativity, but it isn't part of the standard textbook explanation and personally I think it's kind of misleading (for example, hopefully you understand that 'moving through space faster' cannot be meaningful in any absolute space, only relative to a particular frame of reference). The mathematical trick Greene uses to justify this was discussed on this thread, see my first post where I quote from The Elegant Universe to give the technical background.

Anyway, I'm not sure I understand your specific question, in relativity "spacetime interval" refers only to the "distance" in spacetime between two specific events (points in spacetime), you seem to be using it differently. What Greene actually says is the same for different travelers is the magnitude of the 4-vector representing their velocity--if x = (ct, x_1, x_2, x_3) represents an object's "position" in spacetime at a particular moment, 4-velocity is the derivative of this with respect to the proper time \tau (the time as measured by a clock moving along with the object), or dx/d\tau. As Greene points out, the magnitude of this vector is always equal to c, and it's also possible to rewrite it so the square of the magnitude is equal to the speed through space squared plus c^2 times d\tau/dt squared, which Greene identifies as the "speed through time" (really just the speed a moving clock is ticking relative to a clock which is stationary in the frame you've chosen0. So, speed through space squared + c^2 times "speed through time" squared is equal to the constant c in whatever frame you're using, so an increase in the speed through space in that frame means a decrease in the speed the clock is ticking in that frame, and vice versa.

 

[daniel_i_l]

Long ago I accepted that the speed you move through time is personal, but I've always wanted to really understand it intuitively. I think I'm almost there.

Suppose me and my ghost are in the same place initially. Now I take off running in a direction, but my ghost remains behind. Since I'm moving through space faster, my motion through time is slower because the space-time interval is the same for both of us (I hope this much is right).

Now my question: Why are the two space-time intervals considered equivalent?

First of all, when you say "my motion through time is slower" that assumes an absolute measurment of time, it's only true that your ghost will see your clock moving slower than his, but you will also see his clock running slower than yours.
Now let's say that you take off from Earth and go to another planet (X). Event #1 is you leaving Earth and event #2 is you arriving at X.
Lets say that the ghost measures the distance from Earth to X as d_g, and the time that it takes you to get there as t_g. So the interval is
I^2 = (t_g)^2 - (x_g)^2
But in your perspective the spatial distance between the two events is x_0, and since the interval between two events has to be the same in every frame (just like the distance between two points on a grid isn't dependent on the coord system) we know that (t_y)^2 - 0^2 = (t_g)^2 - (x_g)^2
And so the time interval that you measure between the two events is less than what the ghost measures. On this everyone agrees.
This can be further understood if we look from your perspective and measure the time it takes X to reach you after leaving earth. Since X is moving towards you, the spatial distance you measure between you and X is less than what the ghost measured between Earth and X since in his frame both bodies were at rest, so naturally according to you it should take less time to get to X (v/t) than the ghost would think.

And one last thing, it can be a little confusing to talk about "speed through time" and it isn't really needed, the fact that you see a clock moving relative to to you running slow can be explained by the fact that your "lines on simultanity" are different, ie: in your frame, 1:30 on your clock and 1:00 on his clock happen at the same time.

 

[Ich]

Every now and then I tell people that time does not flow slower or faster, but in different directions. A metaphor for different four velocities, which should help to resolve the "paradox" that appears when you use mutually "slower clock rates".
But I never had success.

 

[birulami]

The whole "speed of movement through spacetime = speed through space + speed through time" is something Brian Greene often uses to explain relativity, but it isn't part of the standard textbook explanation and personally I think it's kind of misleading (for example, hopefully you understand that 'moving through space faster' cannot be meaningful in any absolute space, only relative to a particular frame of reference). The mathematical trick Greene uses to justify this was discussed on this thread, see my first post where I quote from The Elegant Universe to give the technical background.

Back there in that thread you say that Greene plays a mathematical trick. Well, what I find striking is that d\tau/dt very much looks like a velocity and I would just call it the speed of aging.

Harald.

 

[pervect]

Back there in that thread you say that Greene plays a mathematical trick. Well, what I find striking is that d\tau/dt very much looks like a velocity and I would just call it the speed of aging.

Harald.

While you can call it that if you like, don't be surprised if you get a lot of blank looks, criticism, and/or downright incomprehension of what you are talking about when you do.

Long-timers on PF have been around enough to know what someone PROBABLY means when they say "speed through spacetime". Someone who only knows physics might very well be seriously confused by this rather vague statement. The language here is vague enough that one can't really rule out the possibility that someone might mean something other than a 4-velocity when they talk about "the speed through spacetime".

Since a lot of confusion occurs because of the result of such misunderstandings of definitions, one of the first things to do is to ask someone who uses such a term "by that, do you mean a 4-velocity?".

Unfortunately, all too often we get the anwer "Huh, what's a 4-velocity?" which usually leads to a lot of wrangling. After a while, this gets to be rather boring, so we try to avoid it and encourage people to use 4-velocity if they mean 4-velocity, and if they don't know what 4-velocity means, to try and learn what that means so that we can all talk in the same language.

Unfortunately we can't always get everyone to use standard terminology, we can only try to encourage people.

So consider yourself encouraged to use standard terminology whenever possible.

 

[Fletcher]

First, I apologize for using confusing terminology. What I perhaps have trouble understanding is why this "4-velocity" must equal c.

 

[pervect]

Consider a path through space-time, parameterized by the proper time tau

You will have such a path as

t(tau), x(tau), y(tau), z(tau).

The four velocity is just the derivative of this with respect to tau, i.e.

(dt/dtau, dx/dtau, dy/dtau, dz/dtau)

We can then ask a couple of questions:

1) How does tau transform for different observers? This is hopefully simple - it's the same for everyone, it's independent of the observer.

2) How do t,x,y, and z transform for different observers? This is hopefully simple too - via the Lorentz transforms

i.e. http://en.wikipedia.org/w/index.php?title=Lorentz_transformation&oldid=128952772

It is a characteristic of the Lorentz transform that the magnitude of any 4-vector is constant for all observers, i.e.

x^2 + y^2 + z^2 - c^2 t^2 = x'^2 + y'^2 + z'^2 - c^2 t'^2

Because of the above two two facts, we can say that

(dx/dtau)^2 + (dy/dtau)^2 + (dz/dtau)^2 - c^2 (dt/dtau)^2 =
(dx'/dtau)^2 + (dy'/dtau)^2 + (dz'/dtau)^2 - c^2 (dt'/dtau)^2

i.e. it's an invariant, the same in any inertial frame (t,x,y,z) or (t',x',y',z').

For any observer (strictly speaking, someone following a time-like path, i.e. someone who experiences time) in their own reference frame we know that

(dt/dtau) = 1
dx/dtau = dy/dtau = dz/tau = 0

Therefore we know that the [square norm of the] 4-velocity for such an observer is -c^2. In geometric units (which I usually use), it's just -1. The sign depends on the sign convention of the metric - some textbooks might use

c^2 dt^2 - dx^2 - dy^2 - dz^2

in which case the [square-norm would be c^2 not -c^2].

Note that we cannot apply this reasoning to light - there is no frame in which light is stationary, and it does not have a "proper time". There is more to be said about parameterizing the path of light, but I think it might be best left out for now.

 

[Hurkyl]

First, I apologize for using confusing terminology. What I perhaps have trouble understanding is why this "4-velocity" must equal c.

Formally, "the norm of 4-velocity is c" and "length along a string increases at a rate of one meter per meter" are exactly the same thing. The only difference are the physical interpretations we give to the geometric objects.

 

[robphy]

minor typos corrected below

It is a characteristic of the Lorentz transform that the square-norm of any 4-vector is constant for all observers, i.e.

x^2 + y^2 + z^2 - c^2 t^2 = x'^2 + y'^2 + z'^2 - c^2 t'^2

Thereore we know that the square-norm of the 4-velocity for such an observer is -c^2. In geometric units (which I usually use), it's just -1. The sign of the square-norm of the 4-velocity depends on the sign convention of the metric - some textbooks might use

c^2 dt^2 - dx^2 - dy^2 - dz^2

in which case the square-norm of a 4-velocity would be +c^2, not -c^2.

 

[pervect]

minor typos corrected below

OK, I've edited the originals to eliminate the errors to minimize confusion.