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I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward.

What troubles me here, is the fact that it is traveling through space-time at the speed of light. Why is it at the speed of light?

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6 Answers 6

First, the fact that an object at rest has energy $mc^2$ is a simple matter of dimensional analysis. If you accept that energy and mass are related, and you know that nature has a natural velocity $c$, then $E=mc^2$ is the simplest thing you can write that describes this. The only complication could have been some numerical factor in front of $m$.

Now, the statement about traveling through time 'at the speed of light' needs to be qualified. You can easily see that it does not make sense if you use ordinary definitions: the speed of light is measured in 'length per time', while a 'speed through time' would be measured by 'time per time', which is just a number.

However, we can make sense of this statement. We think of an observer as tracing a path through spacetime. To denote a point on this path we use a single coordinate that we call $\tau$. The path is defined by the functions $t(\tau)$ and $x(\tau)$: for each value of $\tau$ the observer is at a specific place $x$ at a specific time $t$.

Note that so far $\tau$ is not time: it just just a fictitious coordinate that we use to denote points on the path. I did not even have to specify what units we measure it in.

Let's say we measure $\tau$ in seconds. We can now define a 'velocity vector' $u$ through spacetime, which is the rate at which $t$ and $x$ change when we change $\tau$: $$u=\left(c \frac{dt}{d\tau},\frac{dx}{d\tau}\right).$$ Notice that I snuck a $c$ in there, to make sure $u$ has units of velocity.

The first component of $u$ is a nice definition of our velocity through time. The second component is some way of measuring velocity through space, but it is not the same as the velocity we usually think about, which is $\frac{dx}{dt}$.

Now comes a very nice mathematical theorem, which says that we can always assign values of $\tau$ to points on the path such that $|u(\tau)|=c$ at each point. With this choice of $\tau$, our 'velocity' $u$ is a constant, equal to the speed of light: no matter if we still or move very fast, our velocity through spacetime is the same. If we are sitting still, it just means that we are 'moving faster' in the time direction. If we are moving very fast (by the usual definition of moving...), then our velocity in the time direction will be smaller to compensate. Our only choice is where to point our velocity: a bit more in the time direction, or a bit more in the space direction.

I think this is a very beautiful picture, but it is also a bit misleading. Because $\tau$ is a fictitious coordinate, there are many choices for it and they are all equally good. We could just as well have chosen $|u(\tau)|=2c$, or $|u(\tau)|$ not even constant. So bear in mind that this is all a matter of convention.

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Though "arc length traced out along the curve" depends only on the curve and the underlying geometry, and has the nice physical interpretation of "time, as measured by someone moving along the curve, if that curve is timelike" (or length along the curve, if the curve is spacelike). –  Jerry Schirmer Oct 2 '12 at 2:09
    
Now comes a very nice mathematical theorem, which says that we can always assign values of τ to points on the path such that |u(τ)|=c at each point. Not true. This fails for a massless particle. –  Ben Crowell Sep 3 at 23:39

An object's total velocity through spacetime, its four-velocity, has a component in each dimension of spacetime (4). So, the four coordinates an object has in spacetime are $$x= \left( \begin{array}{ccc} ct \\ x^{1}(t) \\ x^{2}(t) \\ x^{3}(t)\end{array} \right) $$ For the three dimensions of space, we see that these are just the positions as functions of the time as measured by whatever observer whose reference frame we are operating in. At the top, we see ct rather than t. Why? Well, t is measured in units of time, but we need it in units of length to make any sense out of the concept of spacetime. Alternatively, we can convert our positions into units of time by dividing by c. This is done often in astronomy, because the distances are so large (e.g. light-years, light-seconds, etc.).

Now, in order to find the four-velocity, we differentiate each term with respect to proper time. This is similar to the way you find velocity is classical physics - if the position of an object is given by 2t, then it has a velocity of 2. However, note that we are using proper time, $\tau$. The normal t that has been apearing is the time measured by which ever observer who is recording these coordinates, whereas $\tau$ is the time measured on a clock held by our moving observer. So, to find the four velocity, we use this $$\mathbf U = \frac {dx} {d\tau}$$ Where x is the above vector that contains all of the coordinates. Using the fact that time as measured by our observer who is recording the coordinates is related to the proper time of the moving observer by the Lorentz factor, we can write $ct = \gamma c\tau$ Differentiating this with respect to $\tau$, we find that the velocity in the time direction is $c\gamma$. If our 'moving' observer is at rest with respect to the one taking the measurements, then $\gamma = 1$. So, the velocity in the time direction equals c, the speed of light.

We see that for a moving observer, $\gamma$ is greater than one, resulting in a larger velocity through time. This may seem counter-intuitive, since moving observers elapse less time. However, an intuitive way to think about it is this - velocity through space is how muc distance you can cover in some amount of time. So, velocity through time is the amount of time you can elapse in someone's reference frame in some amount of proper time. An observer moving at an enormous speed will record very little proper time, but an observer who he is moving with respect to will observe that it takes an enormous time for him to finish his journey. So, in a small amount of proper time our observer 'travelled' a very large distance through the coordinate time of the observer taking the measurements, hence a higher velocity through time.

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Thanks to Alfred Centauri for pointing out an error. –  Mark M Aug 10 '12 at 1:08

In SR, there is an invariant speed, $c$. This means that if an object's ordinary speed is $c$ in one inertial frame, it is $c$ in all inertial frames.

Also, in SR, there is the notion of a 4-vector which is a spacetime vector rather than a spatial vector. The Lorentz transformation that takes us from one inertial coordinate system to another is a kind of spacetime rotation that leaves the "length" of a any 4-vector constant.

Just as there is the notion of a spatial velocity vector, there is the notion of a spacetime velocity 4-vector; the 4-velocity.

But the "length" of the 4-velocity is a speed and, as mentioned above, it is invariant under Lorentz transformations.

Putting this all together in a completely informal way, it's straightforward to see that the speed of any 4-velocity vector is $c$, the invariant speed. Our speed in spacetime is $c$.

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But the "length" of the 4-velocity is a speed[...] Not true. |v| is not interpreted as a speed. A counterexample is that the |v| of a massless particle is always zero, but this is clearly not its speed. Even for a massive particle, |v| is not interpreted as a speed. the speed of any 4-velocity vector is c, the invariant speed. Our speed in spacetime is c. This is also wrong. Given that a vector has an invariant magnitude, you can't infer anything about what its magnitude is. Every four-vector has an invariant magnitude, and four-vectors can have any magnitude you like. –  Ben Crowell Sep 3 at 23:47
    
@BenCrowell, 4-velocity isn't defined for light-like particles so that takes care of your counter example. The magnitude of the 4-velocity for time-like particles is $c$ which is a speed. This is all elementary so it isn't at all obvious to me why you would claim it is wrong. This is an old post written and if there were anything blatantly wrong about it, I suspect it would have garnered a comment before now. –  Alfred Centauri Sep 4 at 12:32

The answers above are all correct as far as they go, but the next step is even more stunning.

If I am A, the object at rest, then I am moving through spacetime at the speed of light. My friend B gets on a spaceship and rockets away at a high constant velocity and to me his time appears to slow down because of his velocity through space.

But if my fried B takes his position to be at rest, then I A am moving away and he sees my time as slowing down.

How can time slow down for both of us? This is the paradox of relativity. There is no absolute time.

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None of this material about the twin paradox has anything to do with the erroneous notion that observers "move through spacetime" at c. –  Ben Crowell Sep 3 at 23:38

I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward.

This is wrong.

What troubles me here, is the fact that it is traveling through space-time at the speed of light. Why is it at the speed of light?

It isn't. This idea seems to be something that the popularizer Brian Greene has perpetrated on the world. Objects don't move through spacetime. Objects move through space. If you depict an object in spacetime, you have a world-line. The world-line doesn't move through spacetime, it simply extends across spacetime.

Greene's portrayal of this seems to come from his feeling that because the magnitude of a massive particle's velocity four-vector is traditionally normalized to have magnitude $c$, it makes sense to describe the particle, to a nonmathematical audience, as "moving through spacetime" at $c$. This is simply inaccurate. A good way to see that it's inaccurate is to note that a ray of light doesn't even have a four-vector that can be normalized in this way. Any tangent vector to the world-line of a ray of light has a magnitude of zero, so you can't scale it up or down to make it have a magnitude of $c$. For consistency, Greene would presumably have to say that a ray of light "moves through spacetime" at a speed of zero, which is obviously pretty silly.

The reason we normalize velocity four-vectors for massive particles is that the length of a tangent vector has no compelling physical interpretation. Any two tangent vectors that are parallel represent a particle moving through space with the same velocity. Since the length doesn't matter, we might as well arbitrarily set it to some value. We might was well set it to 1, which is of course the value of $c$ in relativistic units. But this normalization is optional in all cases, and impossible for massless particles.

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A logical analysis of absolute motion ongoing within an absolute 4 dimensional Space-Time continuum, leads you to the awareness of the constant motion of all objects. That constant motion turned out to be the speed of light. If one analyzes the concept of constant motion of all objects located within Space-Time, and that this ongoing motion is the equivalent to the magnitude of the motion known as the speed of light across space, this leads you to understanding Special Relativity and all of its equations, and does so in the simplest manner. (Shown further below)

Here we stack motion vectors c and v along with an object and its length. The length of the object is set to being the same length as the motion vector c to simplify the creation of the equations. This is done in one geometric representation. Letter "c" represents our constant motion across Space-Time, and "v" represents motion across space only. As shown via the few examples below, all of the SR equations can be derived using this simple geometric representation of constant motion.

. . . . .enter image description here

$$To \enspace begin,\qquad L^2=L'^2+b^2\qquad \qquad and\qquad \qquad c^2=a^2+v^2 \qquad $$ $$Thus\qquad L'^2/L^2+b^2/L'^2=1\qquad and\qquad a^2/c^2+v^2/c^2=1$$ $$ \enspace \qquad \qquad And\enspace \qquad \qquad x^o=y^o, \qquad \qquad Thus\enspace \quad L'^2/L^2+v^2/c^2=1 \qquad \qquad $$ $$\rightarrow \quad L'^2/L^2=1-v^2/c^2\quad \rightarrow \quad L'^2=L^2(1-v^2/c^2)$$ $$\rightarrow \quad L'=L\sqrt{1-v^2/c^2}$$ $$To \enspace begin,\qquad a^2/c^2=1-v^2/c^2\qquad and\qquad t'^2/t^2=a^2/c^2 \qquad $$ $$Thus\qquad t'^2/t^2=1-v^2/c^2\quad \rightarrow \quad t'^2=t^2(1-v^2/c^2)$$ $$\rightarrow \quad t'=t\sqrt{1-v^2/c^2}$$

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This has nothing to do with the question. –  Ben Crowell 6 hours ago

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